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NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes| EduGrown

In This Post we are  providing Chapter 13 Surface area & Volume NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Surface area & Volume Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Surface area & Volume NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 13 Surface, Area & Volume

Exercise 13.1

Unless stated otherwise, take π = 22/7.

1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Answer

 Volume of each cube(a3) = 64 cm3
⇒ a3 = 64 cm3
⇒ a = 4 cm
Side of the cube = 4 cm
Length of the resulting cuboid = 4 cm
Breadth of the resulting cuboid = 4 cm
Height of the resulting cuboid = 8 cm
∴ Surface area of the cuboid = 2(lb + bh + lh)
= 2(8×4 + 4×4 + 4×8) cm2
= 2(32 + 16 + 32) cm2
= (2 × 80) cm2 = 160 cm2

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Answer

Diameter of the hemisphere = 14 cm
Radius of the hemisphere(r) = 7 cm
Height of the cylinder(h) = 13 – 7 = 6 cm
Also, radius of the hollow hemisphere = 7 cm
Inner surface area of the vessel = CSA of the cylindrical part + CSA of hemispherical part
= (2πrh+2πr2) cm2 = 2πr(h+r) cm2
= 2 × 22/7 × 7 (6+7) cm2 = 572 cm2

3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Answer

Radius of the cone and the hemisphere(r) = 3.5 = 7/2 cm
Total height of the toy = 15.5 cm
Height of the cone(h) = 15.5 – 3.5 = 12 cm

Curved Surface Area of cone = πrl = 22/7 × 7/2 × 25/2
= 275/2 cm2

Curved Surface Area of hemisphere = 2πr2
= 2 × 22/7 × (7/2)2
= 77 cm2

Total surface area of the toy = CSA of cone + CSA of hemisphere
= (275/2 + 77) cm2
= (275+154)/2 cm2 
= 429/2 cm2 = 214.5 cm2
The total surface area of the toy is 214.5 cm2

4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Answer

Each side of cube = 7 cm
∴ Radius of the hemisphere = 7/2 cm
Total surface area of solid = Surface area of cubical block + CSA of hemisphere – Area of base of hemisphere

TSA of solid = 6×(side)2 + 2πr2 – πr2
= 6×(side)2 + πr2
= 6×(7)2 + (22/7 × 7/2 × 7/2)
= (6×49) + (77/2)
= 294 + 38.5 = 332.5 cm2
The surface area of the solid is 332.5 cm2

5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Answer
Diameter of hemisphere = Edge of cube = l
Radius of hemisphere = l/2
Total surface area of solid = Surface area of cube + CSA of hemisphere – Area of base of hemisphere
TSA of remaining solid = 6 (edge)2 + 2πr2 – πr2
= 6l2 + πr2
= 6l2 + π(l/2)2
= 6l2 + πl2/4 
= l2/4 (24 + π) sq units

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
 
Answer
 
Two hemisphere and one cylinder are given in the figure. 
Diameter of the capsule = 5 mm
∴ Radius = 5/2 = 2.5 mm
Length of the capsule = 14 mm
∴ Length of the cylinder = 14 – (2.5 + 2.5) = 9mm
Surface area of a hemisphere = 2πr2 = 2 × 22/7 × 2.5 × 2.5
= 275/7 mm2 

Surface area of the cylinder = 2πrh
= 2 × 22/7 × 2.5 x 9
= 22/7 × 45
990/7 mm2

∴ Required surface area of medicine capsule
= 2 × Surface area of hemisphere + Surface area of cylinder
= (2 × 275/7) × 990/7
= 550/7 + 990/7
= 1540/7 = 220 mm2

Page No. 245

7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.)

Answer

Tent is combination of cylinder and cone.
 
Diameter = 4 m
Slant height of the cone (l) = 2.8 m
Radius of the cone (r) = Radius of cylinder = 4/2 = 2 m
Height of the cylinder (h) = 2.1 m

∴ Required surface area of tent = Surface area of cone+Surface area of cylinder
= πrl + 2πrh
= πr(l+2h)
= 22/7 × 2 (2.8 + 2×2.1)
= 44/7 (2.8 + 4.2)
= 44/7 × 7 = 44 m2

Cost of the canvas of the tent at the rate of ₹500 per m2
= Surface area × cost per m2
= 44 × 500 = ₹22000

8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the
same height and same diameter is hollowed out. Find the total surface area of the
remaining solid to the nearest cm2.

Answer

Diameter of cylinder = diameter of conical cavity = 1.4 cm
∴ Radius of cylinder = Radius of conical cavity = 1.4/2 = 0.7
Height of cylinder = Height of conical cavity = 2.4 cm
TSA of remaining solid = Surface area of conical cavity+TSA of cylinder
= πrl + (2πrh + πr2)
= πr (l + 2h + r)
= 22/7 × 0.7 (2.5 + 4.8 + 0.7)
= 2.2 × 8 = 17.6 cm2
 
9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.
 
Answer
 
Given,
Height of the cylinder, h = 10cm and radius of base of cylinder = Radius of hemisphere (r) = 3.5 cm
Now, required total surface area of the article = 2 × Surface area of hemisphere + Lateral surface area of cylinder
Exercise 13.2
 
1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
 
Answer
 
Given, solid is a combination of a cone and a hemisphere.
 
Also, we have radius of the cone (r) = Radius of the hemisphere = 1cm and height of the cone (h) = 1cm
∴ Required volume of the solid = Volume of the cone + Volume of the hemisphere
 
2. Rachel, an engineering, student was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel mode. (Assume the outer and inner dimensions of the model to be nearly the same.)
 
Answer
 
Given, model is a combination of a cylinder and two cones. Also, we have, diameter of the model,BC = ED = 3 cm.
 
3. A gulab jamun, contains sugar syrup upto about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).
 
Answer
Let r be the radius of the hemisphere and cylinder both. h1 be the height of the hemisphere which is equal to its radius and h2 be the height of the cylinder. 
Given, length = 5 cm, diameter = 2.8 cm
 
4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).
 
Answer
Given, length of cuboid (l) = 15 cm
Breadth of cuboid (b) = 10cm
and height of cuboid (h) = 3.5 cm
∴ Volume of cuboid = l×b×h = 15 ×10 × 3.5 = 5253
 
5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water upto the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel,one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
 
Answer
 
6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use π = 3.14)
Answer
 
7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
 
Answer
 
 
8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be
345 cm3. Check whether she is correct, taking the above as the inside measurements and π = 3.14.
 
Answer
 
She is not correct.
Exercise 13.3
 
1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
 
Answer
 
Given, the radius of the sphere (r) = 4.2 cm
Radius of the cylinder (r1) = 6 cm
 
2. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
 
Answer
 
Let r1, r2 and r3 be the radius of given three spheres and R be the radius of a single solid sphere.
Given, r1 = 6 cm , r2 = 8 cm and r3 =10 cm
 
3. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
 
Answer
 
Given, the height of deep well which form a cylinder (h1) = 20 m
 
4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4m to form an embankment. Find the height of the embankment.
 
Answer
 
Given, the height of deep well which form a cylinder (h1) = 14 m
 
5. A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
 
Answer
 
Let the height and radius of ice cream container (cylinder) be h1 and r1.
 
6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 10 3 5 . . cm cm cm  × ?
 
Answer
 
We know that, every coin has a shape of cylinder. Let radius and height of the coin are r1 and h1 respectively.
 
7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
 
Answer
 
Let the radius and slant height of the heap of sand are r and l.
Given, the height of the heap of sand h = 24 cm.
 
8. Water in a canal, 6 m wide and 1.5 m deep is flowing with a speed of 10 kmh−1. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
 
Answer
 
Given, speed of flow of water (l) = 10 kmh-1 = 10 × 1000 mh−1
Area of canal = 6 × 1.5 = 9m2
 
9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 kmh−1, in how much time will the tank be filled?
 
Answer
 
Given, speed of flow of water = 3 kmh−1 = 3×1000 mh−1
∴ Length of water in 1 h = 3000 m
Exercise 13.4
 
1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
 
Answer
 
Let the height of the frustum of a cone be h.
 
2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
 
Answer
 
Let the slant height of the frustum be l and radius of the both ends of the frustum be r1 and r2.
 
3. A fez, the cap used by the turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.
 
Answer
 
Let the slant height of fez be l and the radius of upper end which is closed be r1 and the other end which is open be r2.
 
4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ` 20 per litre. Also find the cost of metal sheet used to make the container. If it costs ₹8 per 100 cm2. (Take π = 3.14)
 
Answer
 
Let h be the height of the container, which is in the form of a frustum of a cone whose lower end is closed and upper end is opened. Also, let the radius of its lower end be r1 and upper end be r2.
 
5. A metallic right circular cone 20 cm high and whose vertical angle is 60º is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.
 
Answer
 
Let r1 and r2 be the radii of the frustum of upper and lower ends cut by a plane. Given, height of the cone = 20 cm.
∴ Height of the frustum = 10 cm
 
 
Exercise 13.5 (Optional)
 
1. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.
 
Answer
 
Since, the diameter of the wire is 3 mm.
When a wire is one round wound about a cylinder, it covers a 3 mm of length of the cylinder.
Given, length of the cylinder = 12 cm = 120 mm
∴ Number of rounds to cover 120 mm = 120/3 = 40
Given, diameter of a cylinder is d = 10 cm
∴ Radius, r = 10/2 = 5 cm
∴ Length of wire required to complete one round = 2πr = 2π(5) = 10π cm
∴ Length of the wire in covering the whole surface
= Length of the wire in completing 40 rounds
= 10π × 40 = 400π cm
= 400 × 3.14 = 1256 cm
Now, radius of copper wire = 3/2 mm = 3/20 cm
 
2. A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate).
 
Answer
 
Here, ABC is a right angled triangle at A and BC is the hypotenuse.
 
3. A cistern, internally measuring 150 cm × 120 cm× 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. how many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
 
Answer
 
Given, internally dimensions of cistern = 150 cm×120 cm×110 cm
∴ Volume of cistern = 150×120×110 
= 1980000 cm3
Volume of water = 129600 cm3
∴ Volume of cistern to be filled = 1980000 – 129600 = 1850400 cm3
Let required number of bricks = n
 
4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
 
Answer
 
Given, area of the valley = 97280 km2
 
5. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.
 
Answer
 
Given, oil funnel is a combination of a cylinder and a frustum of a cone.
Also, given height of cylindrical portion h = 10 cm
 
6. Derive the formula for the curved surface area and total surface area of the frustum of a cone. Using the symbols as explained.
 
Answer
 
Leth be the height, l be the slant height and r1 and r2 be the radii of the bases (r1>r2) of the frustum of a cone. We complete the conical part OCD.
The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let slant height of the cone OAB be l1 and its height be h1 i.e.,OB = OA = l1 and OP = h1
The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let slant height of the cone OAB be l1 and its height be h1 i.e., OB = OA = l1 and OP = h1
Then in ∆ACE,
 
7. Derive the formula for the volume of the frustum of a cone given to you in the section 13.5 using the symbols as explained.
 
Answer
 
Leth be the height, l be the slant height and r1 and r2 be the radii of the bases (r1>r2) of the frustum of a cone. We complete the conical part OCD.
The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let the height of the cone OAB be h1 and its slant height be l1.
i.e., OP = h1 and OA = OB = l1
Then, height of the cone OCD = h1 −1
∆OQD ~ ∆OPB (AA similarity criterion)

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NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles | EduGrown

In This Post we are  providing Chapter 12 Area related to circle NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Area related to circle Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Area related to circle NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 12 Area related to circle

Exercise: 12.1
 
Unless stated otherwise, use π =22/7.
 
1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
 
Answer

Let the radius of the third circle be R.
Circumference of the circle with radius R = 2πR
Circumference of the circle with radius 19 cm = 2π × 19 = 38π cm
Circumference of the circle with radius 9 cm = 2π × 9 = 18π cm
Sum of the circumference of two circles = 38π + 18π = 56π cm
Circumference of the third circle = 2πR = 56π
⇒ 2πR = 56π cm
⇒ R = 28 cm
The radius of the circle which has circumference equal to the sum of the circumferences of the two circles is 28 cm.
 
2.  The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
 
Answer
 
Let the radius of the third circle be R.
Area of the circle with radius R = πR2
Area of the circle with radius 8 cm = π × 82 = 64π cm2
Area of the circle with radius 6 cm = π × 62 = 36π cm2
Sum of the area of two circles = 64π cm2 + 36π cm2 = 100π cm2
Area of the third circle = πR2 = 100π cm2
⇒ πR2 = 100π cm2
⇒ R2 = 100 cm2
⇒ R = 10 cm
Thus, the radius of the new circle is 10 cm.
 
3. Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.
 
 
 
Answer

Diameter of Gold circle (first circle) = 21 cm
Radius of first circle, r1 = 21/2 cm = 10.5 cm
Each of the other bands is 10.5 cm wide,
∴ Radius of second circle, r2 = 10.5 cm + 10.5 cm = 21 cm
∴ Radius of third circle, r3 = 21 cm + 10.5 cm = 31.5 cm
∴ Radius of fourth circle, r4 = 31.5 cm + 10.5 cm = 42 cm
∴ Radius of fifth circle, r5 = 42 cm + 10.5 cm = 52.5 cm
Area of gold region = π r12 = π (10.5)2 = 346.5 cm2
Area of red region = Area of second circle – Area of first circle = π r22 – 346.5 cm2
                              = π(21)2 – 346.5 cm2 = 1386 – 346.5 cm2 = 1039.5 cm2

Area of blue region = Area of third circle – Area of second circle = π r32 – 1386 cm2

                                          =  π(31.5)2 – 1386 cm2 = 3118.5 – 1386 cm2 = 1732.5 cm2
 

Area of black region = Area of fourth circle – Area of third circle = π r32 – 3118.5 cm2

 

 

                                             = π(42)2 – 1386 cm2 = 5544 – 3118.5 cm2 = 2425.5 cm2

Area of white region = Area of fifth circle – Area of fourth circle = π r42 – 5544 cm2

                                             = π(52.5)2 – 5544 cm2 = 8662.5 – 5544 cm2 = 3118.5 cm2
 

4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

 
Answer
 
Diameter of the wheels of a car = 80 cm
Circumference of wheels = 2πr = 2r × π = 80 π cm
Distance travelled by car in 10 minutes = (66 × 1000 × 100 × 10)/60 = 1100000 cm/s
No. of revolutions = Distance travelled by car/Circumference of wheels
                              = 1100000/80 π = (1100000 × 7)/(80×22) = 4375
4375 complete revolutions does each wheel make in 10 minutes.
 
5. Tick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
          (A) 2 units                     (B) π units                  (C) 4 units              (D) 7 units
 
Answer
 
Let the radius of the circle be r.
∴ Perimeter of the circle = Circumference of the circle = 2πr
∴ Area of the circle = π r2

A/q,

2πr = π r2
⇒ 2 = r
Thus, the radius of the circle is 2 units. (A) is correct.
Exercise: 12.2
 
Unless stated otherwise, use π =22/7.
1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
 
Answer
 
Area of the sector making angle θ = (θ/360°)×π r2 
Area of the sector making angle 60° = (60°/360°)×π r2 cm2
                         = (1/6)×62π  = 36/6 π cm2 = 6 × 22/7 cm2 = 132/7 cm2
 

2. Find the area of a quadrant of a circle whose circumference is 22 cm.

 
Answer
 
Quadrant of a circle means sector is making angle 90°.
Circumference of the circle = 2πr = 22 cm
Radius of the circle = r = 22/2π cm = 7/2 cm 
Area of the sector making angle 90° = (90°/360°)×π r2 cm2
                    = (1/4)×(7/2)2π  = (49/16) π cm2 = (49/16) × (22/7) cm2 = 77/8 cm2

3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
 
Answer
 
Here, Minute hand of clock acts as radius of the circle.
∴ Radius of the circle (r) = 14 cm
Angle rotated by minute hand in 1 hour = 360°
∴ Angle rotated by minute hand in 5 minutes = 360° × 5/60 = 30°
Area of the sector making angle 30° = (30°/360°)×π r2 cm2
                         = (1/12) × 142π  = 196/12 π cm2 = (49/3) × (22/7) cm2 = 154/3 cm2
Area swept by the minute hand in 5 minutes = 154/3 cm2

4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14)
 
Answer

Radius of the circle = 10 cm
Major segment is making 360° – 90° = 270°
Area of the sector making angle 270° 
                                 = (270°/360°) × π r2 cm2
                                           = (3/4) × 102π  = 75 π cm2
                                = 75 × 3.14 cm2 = 235.5 cm2
∴ Area of the major segment = 235.5 cm2
Height of ΔAOB = OA = 10 cm
Base of ΔAOB = OB = 10 cm
Area of ΔAOB = 1/2 × OA × OB
                         = 1/2 ×10 × 10 = 50 cm2
Major segment is making  90°
 Area of the sector making angle 90°                                                                
= (90°/360°) × π r2 cm2
= (1/4) × 102π  = 25 π cm2
 = 25 × 3.14 cm2 = 78.5 cm2
Area of the minor segment = Area of the sector making angle 90° – Area of ΔAOB
                                            = 78.5 cm2 –  50 cm2 = 28.5 cm2
 

5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
   (i) the length of the arc
   (ii) area of the sector formed by the arc
   (iii) area of the segment formed by the corresponding chord

 
Answer
 
Radius of the circle = 21 cm
 
(i) Length of the arc AB = θ/360° × 2πr
                                        = 60°/360° × 2 × 22/7 × 21
                                        = 1/6 × 2 × 22/7 × 21 = 22
The length of the arc is  22 cm.
 
(ii) Angle subtend by the arc = 60° 
Area of the sector making angle 60° = (60°/360°) × π r2 cm2
                                          = (1/6) × 212π  = 441/6 π cm2
                               = 441/6 × 22/7 cm2 = 231 cm2
∴ Area of the sector formed by the arc is 231 cm2

(iii) Area of equilateral ΔAOB = √3/4 × (OA)2 = √3/4 × 212 = (441√3)/4 cm2
Area of the segment formed by the corresponding chord 
                     = Area of the sector formed by the arc – Area of equilateral ΔAOB
                     = 231 cm2 – (441√3)/4 cm2 
 

6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)

 
Answer
 
 
Radius of the circle = 15 cm
ΔAOB is isosceles as two sides are equal.
∴ ∠A = ∠B
Sum of all angles of triangle = 180°
∠A + ∠B + ∠C = 180°
⇒ 2 ∠A = 180° – 60°
⇒ ∠A = 120°/2
⇒ ∠A = 60°
Triangle is equilateral as ∠A = ∠B = ∠C = 60°
∴ OA = OB = AB = 15 cm
Area of equilateral ΔAOB = √3/4 × (OA)2 = √3/4 × 152 
                                          = (225√3)/4 cm2 = 97.3 cm2

Angle subtend at the centre by minor segment = 60°
Area of Minor sector making angle 60° = (60°/360°) × π r2 cm2
                                                                                     = (1/6) × 152 π  cm2 =  225/6 π  cm2
                                                  =  (225/6) × 3.14 cm2 = 117.75  cm2
 

Area of the minor segment = Area of Minor sector – Area of equilateral ΔAOB

                                            = 117.75  cm2 – 97.3 cm2 = 20.4 cm2


Angle made by Major sector = 360° – 60° = 300°
Area of the sector making angle 300° = (300°/360°) × π r2 cm2
                                                   = (5/6) × 152 π  cm2 =  1125/6 π  cm2
                                                  =  (1125/6) × 3.14 cm2 = 588.75  cm2
 

Area of major segment = Area of Minor sector + Area of equilateral ΔAOB

                                            = 588.75  cm2 + 97.3 cm2 = 686.05 cm2
 

7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)

 
Answer

 
Radius of the circle, r = 12 cm
Draw a perpendicular OD to chord AB. It will bisect AB.
∠A = 180° – (90° + 60°) = 30°
cos 30° = AD/OA
⇒ √3/2 = AD/12
⇒ AD = 6√3 cm
⇒ AB = 2 × AD = 12√3 cm
sin 30° = OD/OA
⇒ 1/2 = OD/12
⇒ OD = 6 cm
Area of ΔAOB = 1/2 × base × height
                         = 1/2 × 12√3 × 6 = 36√3 cm
                         = 36 × 1.73 = 62.28 cm2

Angle made by Minor sector = 120°
Area of the sector making angle 120° = (120°/360°) × π r2 cm2
                                                   = (1/3) × 122 π  cm2 =  144/3 π  cm2
                                                  =  48 × 3.14 cm2 = 150.72  cm2
∴ Area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB
                                                                        = 150.72  cm2 – 62.28 cm2
                                                                                                = 88.44 cm2

8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m
long rope (see Fig. 12.11). Find
 (i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)
 
Answer
 
Side of square field = 15 m
Length of rope is the radius of the circle, r = 5 m
Since, the horse is tied at one end of square field, it will graze only quarter of the field with radius 5 m.
 
(i) Area of circle = π r2 = 3.14 × 52 = 78.5 m2
Area of that part of the field in which the horse can graze = 1/4 of area of the circle = 78.5/4 = 19.625 m2
 

(ii) Area of circle if the length of rope is increased to 10 m = π r2 = 3.14 × 102 = 314 m2

Area of that part of the field in which the horse can graze = 1/4 of area of the circle 
                                                                                            = 314/4 = 78.5 m2

Increase in grazing area = 78.5 m2 – 19.625 m2 = 58.875 m2

 

 

9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 12.12. Find:

(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
 
Answer
 
Number of diameters = 5
Length of diameter = 35 mm
∴ Radius = 35/2 mm

(i) Total length of silver wire required = Circumference of the circle +  Length of 5 diameter
                                                              = 2π r + (5×35) mm = (2 × 22/7 × 35/2) + 175 mm
                                                              = 110 + 175 mm = 185 mm

(ii) Number of sectors = 10
Area of each sector = Total area/Number of sectors
Total Area = π r2 = 22/7 × (35/2)2 = 1925/2 mm2
∴ Area of each sector = (1925/2) × 1/10 = 385/4 mm2

10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
 
 
Answer
 
Number of ribs in umbrella = 8
Radius of umbrella while flat = 45 cm
Area between the two consecutive ribs of the umbrella =
                                            Total area/Number of ribs
Total Area = π r2 = 22/7 × (45)2 = 6364.29 cm2
∴ Area between the two consecutive ribs = 6364.29/8 cm2
                                                                   = 795.5 cm2


11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
 
Answer
 
Angle of the sector of circle made by wiper = 115°
Radius of wiper = 25 cm
Area of the sector made by wiper = (115°/360°) × π r2 cm2
                                                                        = 23/72 × 22/7 × 252 =  23/72 × 22/7 × 625 cm2
                                                                        = 158125/252 cm2
 
Total area cleaned at each sweep of the blades = 2 ×158125/252 cm2 =  158125/126 = 1254.96 cm2
 

12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.

(Use π = 3.14)
 
Answer
 
Let O bet the position of Lighthouse.
Distance over which light spread i.e. radius, r = 16.5 km
Angle made by the sector = 80°
Area of the sea over which the ships are warned = Area made by the sector.
Area of sector = (80°/360°) × π r2 km2
                               = 2/9 × 3.14 × (16.5)2 km2
                               =  189.97 km2


 
 
 
13. A round table cover has six equal designs as shown in Fig. 12.14. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm2 . (Use √3 = 1.7)

Answer


Number of equal designs = 6
Radius of round table cover = 28 cm
Cost of making design = ₹ 0.35 per cm2
∠O = 360°/6 = 60°

ΔAOB is isosceles as two sides are equal. (Radius of the circle)
∴ ∠A = ∠B
Sum of all angles of triangle = 180°
∠A + ∠B + ∠O = 180°
⇒ 2 ∠A = 180° – 60°
⇒ ∠A = 120°/2
⇒ ∠A = 60°
Triangle is equilateral as ∠A = ∠B = ∠C = 60°
Area of equilateral ΔAOB = √3/4 × (OA)2 = √3/4 × 282 = 333.2 cm2

Area of sector ACB = (60°/360°) × π r2 cm2
                                            = 1/6 × 22/7 × 28 × 28 = 410.66 cm2
Area of design = Area of sector ACB – Area of equilateral ΔAOB
                        = 410.66 cm2 – 333.2 cm2 = 77.46 cm2
 

Area of 6 design = 6 × 77.46 cm2 = 464.76 cm2



Total cost of making design = 464.76 cm2 × ₹ 0.35 per cm2 = ₹ 162.66
 
14. Tick the correct answer in the following :
Area of a sector of angle p (in degrees) of a circle with radius R is
  (A) p/180 × 2πR             (B) p/180 × π R2                   (C) p/360 × 2πR           (D) p/720 × 2πR2 
 
Answer
 
Area of a sector of angle p = p/360 × π R2
                                                         = p/360 × 2/2 × π R2 
                                           =  2p/720 × 2πR2 

Hence, Option (D) is correct.

Exercise: 12.3
 
Unless stated otherwise, use π =22/7
1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
 
Answer 
 
PQ = 24 cm and PR = 7 cm
∠P = 90° (Angle in the semi-circle)
∴ QR is hypotenuse of the circle = Diameter of the circle.
By Pythagoras theorem,
QR2 = PR2  + PQ2 
⇒ QR2 = 72  + 242
⇒ QR2 = 49  + 576
⇒ QR2 = 625
⇒ QR = 25 cm
∴ Radius of the circle = 25/2 cm
Area of the semicircle = (π R2)/2
                                    = (22/7 × 25/2 × 25/2)/2 cm2
                                                = 13750/56 cm2 = 245.54 cm2
Area of the ΔPQR = 1/2 × PR × PQ
                              = 1/2 × 7 × 24 cm2
                                        = 84 cm2
Area of the shaded region = 245.54 cm2 – 84 cm2 = 161.54 cm2
 
2. Find the area of the shaded region in Fig. 12.20, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.
 
Answer
 
Radius inner circle = 7 cm
Radius of outer circle = 14 cm
Angle made by sector = 40°
Area of the sector OAC = (40°/360°) × π r2 cm2
                                                = 1/9 × 22/7 × 142 = 68.44 cm2
Area of the sector OBD = (40°/360°) × π r2 cm2
                                                                       = 1/9 × 22/7 × 72  = 17.11 cm2
 
Area of the shaded region ABDC = Area of the sector OAC – Area of the sector circle OBD
                                                      = 68.44 cm2 – 17.11 cm2 = 51.33 cm2
 
3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.


Answer
 
There are two semicircles in the figure.
Side of the square = 14 cm
Diameter of the semicircle = 14 cm
∴ Radius of the semicircle = 7 cm
Area of the square = 14 × 14 = 196 cm2
Area of the semicircle = (π R2)/2
                                    = (22/7 × 7 × 7)/2 cm2 = 77 cm2 
Area of two semicircles = 2 × 77 cm2  = 154 cm2
 
Area of the shaded region = 196 cm2 – 154 cm2 = 42 cm2
 
4. Find the area of the shaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.
 
Answer
 
OAB is an equilateral triangle with each angle equal to 60°.
Area of the sector is common in both.
Radius of the circle = 6 cm.
Side of the triangle = 12 cm.
Area of the equilateral triangle = √3/4 × (OA)2 = √3/4 × 122 = 36√3 cm2
Area of the circle = π R2 = 22/7 × 62 = 792/7 cm2
Area of the sector making angle 60° = (60°/360°) × π r2 cm2
                                                                              = 1/6 × 22/7 × 62 cm2  = 132/7 cm2
Area of the shaded region = Area of the equilateral triangle + Area of the circle – Area of the sector
                                          = 36√3 cm2 + 792/7 cm2 – 132/7 cm2
                                                        = (36√3 + 660/7) cm2 
 
5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. 12.23. Find the area of the remaining portion of the square.
 
Answers 
 
Side of the square = 4 cm
Radius of the circle = 1 cm
Four quadrant of a circle are cut from corner and one circle of radius are cut from middle.
Area of square = (side)2 = 42 = 16 cm2
Area of the quadrant = (π R2)/4 cm2 = (22/7 × 12)/4 = 11/14 cm2
∴ Total area of the 4 quadrants = 4 × (11/14) cm2 = 22/7 cm2
Area of the circle = π R2 cm2 = (22/7 × 12) = 22/7 cm2
Area of the shaded region = Area of square – (Area of the 4 quadrants + Area of the circle)
                                           = 16 cm2 – (22/7 + 22/7) cm2
                                                         = 68/7 cm2
 
6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.


Answer
 
Radius of the circle = 32 cm
Draw a median AD of the triangle passing through the centre of the circle.
⇒ BD = AB/2
Since, AD is the median of the triangle
∴ AO = Radius of the circle = 2/3 AD
⇒ 2/3 AD = 32 cm
⇒ AD = 48 cm
In ΔADB,
By Pythagoras theorem,
AB2 = AD2  + BD2 
⇒ AB2 = 482  + (AB/2)2
⇒ AB2 = 2304  + AB2/4
⇒ 3/4 (AB2) = 2304
⇒ AB2 = 3072
⇒ AB = 32√3 cm
Area of ΔADB = √3/4 × (32√3)2 cm2 = 768√3 cm2
Area of circle = π R2 = 22/7 × 32 × 32 = 22528/7 cm2

Area of the design = Area of circle – Area of ΔADB
                              = (22528/7 – 768√3) cm2
 
Page No: 236
 
7. In Fig. 12.25, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Answer

Side of square = 14 cm
Four quadrants are included in the four sides of the square.
∴ Radius of the circles = 14/2 cm = 7 cm
Area of the square ABCD = 142 = 196 cm2
Area of the quadrant = (π R2)/4 cm2 = (22/7 × 72)/4 cm2
                                             = 77/2 cm2
Total area of the quadrant = 4 × 77/2 cm2 = 154 cm2 
 
Area of the shaded region = Area of the square ABCD – Area of the quadrant
                                          = 196 cm2 – 154 cm2
                                                        = 42 cm2 
 
8. Fig. 12.26 depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If
the track is 10 m wide, find :
(i) the distance around the track along its inner edge
(ii) the area of the track.
 
Answer
 
Width of track = 10 m
Distance between two parallel lines = 60 m
Length of parallel tracks = 106 m
DE = CF = 60 m
Radius of inner semicircle, r = OD = O’C 
                                               = 60/2 m = 30 m
Radius of outer semicircle, R = OA = O’B
                                               = 30 + 10 m = 40 m
Also, AB = CD = EF = GH = 106 m
 
 
Distance around the track along its inner edge = CD + EF + 2 × (Circumference of inner semicircle)
                                                                          = 106 + 106 + (2 × πr) m = 212 + (2 × 22/7 × 30) m
                                                                          = 212 + 1320/7 m = 2804/7 m
 
Area of the track = Area of ABCD + Area EFGH + 2 × (area of  outer semicircle) – 2 × (area of  inner  semicircle)
                           = (AB × CD) + (EF × GH) + 2 × (πr2/2) – 2 × (πR2/2) m2
                           = (106 × 10) + (106 × 10) + 2 × π/2 (r2 -R2) m2
                           = 2120 + 22/7 × 70 × 10 m2
                                 = 4320 m2 
 
9. In Fig. 12.27, AB and CD are two diameters of a circle (with centre O) perpendicular to each other
and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Answer

Radius of larger circle, R = 7 cm
Radius of smaller circle, r = 7/2 cm
Height of ΔBCA = OC = 7 cm
Base of ΔBCA = AB = 14 cm
Area of ΔBCA = 1/2 × AB × OC = 1/2 × 7 × 14 = 49 cm2 
Area of larger circle = πR2 = 22/7 × 72 = 154 cm2 
Area of larger semicircle = 154/2 cm2 = 77 cm2 
Area of smaller circle = πr2 = 22/7 × 7/2 × 7/2 = 77/2 cm2
 
Area of the shaded region = Area of larger circle – Area of triangle – Area of larger semicircle + Area   of smaller circle
Area of the shaded region = (154 – 49 – 77 + 77/2) cm2
                                                        = 133/2 cm2 = 66.5 cm2 
 
10. The area of an equilateral triangle ABC is 17320.5 cm2. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)
 
Answer
 
ABC is an equilateral triangle.
∴ ∠A = ∠B = ∠C = 60°
There are three sectors each making 60°.
Area of ΔABC = 17320.5 cm2
⇒ √3/4 × (side)2  = 17320.5
⇒ (side)2  = 17320.5 × 4/1.73205
⇒ (side)2  = 4 × 104 
⇒ side  = 200 cm
Radius of the circles = 200/2 cm = 100 cm
Area of the sector = (60°/360°) × π r2 cm2
                                       = 1/6 × 3.14 × (100)2 cm2
                                       = 15700/3 cm2
Area of 3 sectors = 3 × 15700/3 = 15700 cm2 =

Area of the shaded region = Area of equilateral triangle ABC – Area of 3 sectors
                                          = 17320.5 – 15700 cm2 = 1620.5 cm2
 
11. On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.
 
Answer
Number of circular design = 9
Radius of the circular design = 7 cm
There are three circles in one side of square handkerchief.
∴ Side of the square = 3 × diameter of circle = 3 × 14 = 42 cm
Area of the square = 42 × 42 cm2 = 1764 cm2
 
Area of the circle = π r2 = 22/7 × 7 × 7 = 154 cm2
Total area of the design = 9 × 154 = 1386 cm2
 
Area of the remaining portion of the handkerchief = Area of the square – Total area of the design
                                                                                 = 1764 – 1386 = 378 cm2
 
12. In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
 (i) quadrant OACB,                       (ii) shaded region.
Answer
 
Radius of the quadrant = 3.5 cm = 7/2 cm
 
(i) Area of quadrant OACB = (πR2)/4 cm2
                                             = (22/7 × 7/2 × 7/2)/4 cm2
                                                            = 77/8 cm2
(ii) Area of triangle BOD = 1/2 × 7/2 × 2 cm2
                                                       = 7/2 cm2
Area of shaded region = Area of quadrant – Area of triangle BOD
                                    = (77/8 – 7/2) cm2 = 49/8 cm2 
                                                = 6.125 cm2
 
13. In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)
 
Answer
Side of square = OA = AB = 20 cm
Radius of the quadrant = OB
OAB is right angled triangle
By Pythagoras theorem in ΔOAB ,
OB2 = AB2  + OA2 
⇒ OB2 = 202  + 202
⇒ OB2 = 400  + 400
⇒ OB2 = 800
⇒ OB = 20√2 cm
Area of the quadrant = (πR2)/4 cm2 = 3.14/4 × (20√2)2 cm2 = 628 cm2
Area of the square = 20 × 20 = 400  cm2
 
Area of the shaded region = Area of the quadrant – Area of the square
                                          = 628 – 400 cm2 = 228 cm2
 
14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig. 12.32). If ∠AOB = 30°, find the area of the shaded region.
 
Answer
 
Radius of the larger circle, R = 21 cm
Radius of the smaller circle, r = 7 cm
Angle made by sectors of both concentric circles =  30°
Area of the larger sector = (30°/360°) × π R2 cm2
                                                     = 1/12 × 22/7 × 212 cm2
                                                     = 231/2 cm2
Area of the smaller circle = (30°/360°) × π r2 cm2
                                                     = 1/12 × 22/7 × 72 cm2
                                                     = 77/6 cm2
 
Area of the shaded region = 231/2 – 77/6 cm2
                                                        = 616/6 cm2 = 308/3 cm2
 
15. In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC
as diameter. Find the area of the shaded region.
 
Answer
 
Radius of the the quadrant ABC of circle = 14 cm
AB = AC = 14 cm
BC is diameter of semicircle.
ABC is right angled triangle.
By Pythagoras theorem in ΔABC,
BC2 = AB2  + AC2 
⇒ BC2 = 142  + 142
⇒ BC = 14√2 cm
Radius of semicircle = 14√2/2 cm = 7√2 cm

Area of ΔABC = 1/2 × 14 × 14 = 98 cm2
Area of quadrant = 1/4 × 22/7 × 14 × 14 = 154 cm2
Area of the semicircle = 1/2 × 22/7 × 7√2 × 7√2 = 154 cm2
 
Area of the shaded region = Area of the semicircle + Area of ΔABC – Area of quadrant
                                          = 154 + 98 – 154 cm2 = 98 cm2

Page No: 238
 
16. Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.
 
Answer
AB = BC = CD = AD = 8 cm
Area of ΔABC = Area of ΔADC = 1/2 × 8 × 8 = 32 cm2
Area of quadrant AECB = Area of quadrant AFCD = 1/4 × 22/7 × 82
                                                     = 352/7 cm2
 
Area of shaded region = (Area of quadrant AECB – Area of ΔABC) +                                           (Area of quadrant AFCD – Area of ΔADC)
                                    = (352/7 – 32) + (352/7 -32) cm2
                                                = 2 × (352/7 -32) cm2
                                                 =  256/7 cm2

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NCERT Solutions for Class 10 Maths Chapter 11 Construction | EduGrown

In This Post we are  providing Chapter 11 Construction NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Construction Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Construction NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 11 Construction

Exercise 11.1

In each of the following, give the justification of the construction also:

1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.

Answer

Steps of Construction:
Step I: AB = 7.6 cm is drawn.
Step II: A ray AX making an acute angle with  AB is drawn.
Step III: After that, a ray BY is drawn parallel to AX making an equal acute angle as in the previous step.
Step IV: Point A1, A2, A3, A4 and A5 is marked on AX and point B1, B2…. to B8 is marked on BY such that AA1 = A1A2 = A2A3 =….BB1= B1B2 = …. B7B8
Step V: A5 and B8 are joined and it intersected AB at point C diving it in the ratio 5:8.


AC : CB = 5 : 8

Justification:
ΔAA5C ~ ΔBB8C
∴ AA5/BB8 = AC/BC = 5/8

 
2.  Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.
 
Answer

Steps of Construction:
Step I: AB = 6 cm is drawn.
Step II: With A as a centre and radius equal to 4cm, an arc is draw.
Step III: Again, with B as a centre and radius equal to 5 cm an arc is drawn on same  side of AB intersecting previous arc at C.
Step IV: AC and BC are joined to form ΔABC.
Step V: A ray AX is drawn making an acute  angle with AB below it.
Step VI: 5 equal points (sum of the ratio = 2 + 3 =5) is marked on AX as A1 A2….A5
Step VII: A5B is joined. A2B’ is drawn parallel to A5B and B’C’ is drawn parallel to BC.
ΔAB’C’ is the required triangle
 

Justification:
∠A(Common)
∠C = ∠C’ and ∠B = ∠ B’ (corresponding angles)
Thus ΔAB’C’ ~  ΔABC by AAA similarity condition
From the figure,
AB’/AB = AA2/AA5 = 2/3
AB’ =2/3 AB
AC’ = 2/3 AC

3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are
7/5 of the corresponding sides of the first triangle.

Answer

Steps of Construction:
Step I: A triangle ABC with sides 5 cm, 6 cm and 7 cm is drawn.
Step II: A ray BX making an acute angle with BC is drawn opposite to vertex A.
Step III: 7 points as B1 B2 B3 B4 B5 B6 and B7 are marked on BX.
Step IV; Point B5 is joined with C to draw B5C.
Step V: B7C’ is drawn parallel to B5C and C’A’ is parallel to CA.
Thus A’BC’ is the required triangle.

Justification
ΔAB’C’ ~  ΔABC by AAA similarity condition
∴ AB/A’B = AC/A’C’ = BC/BC’
and BC/BC’ = BB5/BB7 = 5/7
∴A’B/AB = A’C’/AC = = BC’/BC = 7/5

4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.

Answer

Steps of Construction:
Step I: BC = 5 cm is drawn.
Step II: Perpendicular bisector of BC is drawn and it intersects BC at O.
Step III: At a distance of 4 cm, a point A is marked on the perpendicular bisector of BC.
Step IV: AB and AC is joined to form ΔABC.
Step V: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step VI: 3 points B1 B2 and B3 is marked BX.
Step VII: B2 is joined with C to form B2C.
Step VIII: B3C’ is drawn parallel to B2C and C’A’ is drawn parallel to CA.
Thus, A’BC’ is the required triangle formed.

Justification:

ΔAB’C’ ~  ΔABC by AA similarity condition.
∴ AB/AB’ = BC/B’C’ = AC/AC’
also,
AB/AB’ = AA2/AA3 = 2/3
⇒ AB’ = 3/2 AB, B’C’ = 3/2 BC and AC’ = 3/2 AC
 
 
5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
 
Answer

Steps of Construction:
Step I: BC = 6 cm is drawn.
Step II: At point B, AB = 5 cm is drawn making an
            ∠ABC = 60° with BC.
Step III: AC is joined to form ΔABC.
Step IV: A ray BX is drawn making an acute angle with BC  opposite to vertex A.
Step V: 4 points B1 B2 B3 and B4 at equal distance is marked on BX.
Step VII: B3 is joined with C’ to form B3C’.
Step VIII: C’A’ is drawn parallel CA.
Thus, A’BC’ is the required triangle.

Justification:

∠A = 60° (Common)
∠C = ∠C’
ΔAB’C’ ~  ΔABC by AA similarity condition.
∴ AB/AB’ = BC/B’C’ = AC/AC’
also,
AB/AB’ = AA3/AA4 = 4/3
⇒ AB’ = 3/4 AB, B’C’ = 3/4 BC and AC’ = 3/4 AC

6. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Δ ABC.

Answer

Sum of all side of triangle = 180°
∴ ∠A  + ∠B  + ∠C  = 180°
∠C = 180° – 150° = 30°
Steps of Construction:
Step I: BC = 7 cm is drawn.
Step II: At B, a ray is drawn making an angle of  45° with BC.
 
Step III: At C, a ray making an angle of 30° with  BC is drawn intersecting the previous ray at A. Thus, ∠A = 105°.
Step IV: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step V: 4 points B1 B2 B3 and B4 at equal distance is marked on BX.
Step VI: B3C is joined and B4C’ is made parallel to B3C.
Step VII: C’A’ is made parallel CA.
Thus, A’BC’ is the required triangle.

Justification:

∠B = 45° (Common)
∠C = ∠C’
ΔAB’C’ ~  ΔABC by AA similarity condition.
∴ BC/BC’ = AB/A’B’ = AC/A’C’
also,
BC/BC’ = BB3/BB4 = 34
⇒ AB = 4/3 AB’, BC = 4/3 BC’ and AC = 4/3 A’C’

7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Answer

Steps of Construction:
Step I: BC = 3 cm is drawn.
Step II: At B, A ray making an angle of 90° with BC is drawn.
Step III: With B as centre and radius equal to 4 cm, an arc is made on the previous ray intersecting it at point A.
Step IV: AC is joined to form ΔABC.
Step V: A ray BX is drawn making an acute angle with BC  opposite to vertex A.
Step VI: 5 points B1 B2 B3 B4 and B5 at equal distance is marked on BX.
Step VII: B3C is joined B5C’ is made parallel to B3C.
Step VIII: A’C’ is joined together.
Thus, ΔA’BC’ is the required triangle.

Justification:
As in the previous question 6
Exercise 11.2

In each of the following, give also the justification of the construction:

1.  Draw a circle of a radius of 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Answer

Steps of Construction:
Step I: With O as a centre and radius equal to 6 cm, a circle is drawn.
Step II: A point P at a distance of 10 cm from the centre O is taken. OP is joined.
Step III: Perpendicular bisector OP is drawn and let it intersected at M.
Step IV: With M as a centre and OM as a radius, a circle is drawn intersecting the previous circle at Q and R.
Step V: PQ and PR are joined.
Thus, PQ and PR are the tangents to the circle.
On measuring the length, tangents are equal to 8 cm.
PQ = PR = 8cm.

Justification:
OQ is joined.
∠PQO = 90° (Angle in the semi-circle)
∴ OQ ⊥ PQ
Therefore, OQ is the radius of the circle then PQ has to be a tangent of the circle. 
Similarly, PR is a tangent of the circle.
 
2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
 
Answer
 
Steps of Construction:
Step I: With O as a centre and radius equal to 4 cm, a circle is drawn.
Step II: With O as a centre and radius equal to 6 cm, a concentric circle is drawn.
Step III: P be any point on the circle of radius 6 cm and OP is joined.
Step IV: Perpendicular bisector of OP is drawn which cuts it at                 M
 
Step V: With M as a centre and OM as a radius, a circle is  drawn which intersect the circle of radius 4 cm at Q and R
Step VI: PQ and PR are joined.
Thus, PQ and PR are the two tangents.

Measurement:
OQ = 4 cm (Radius of the circle)
PQ = 6 cm ( Radius of the circle)
∠PQO = 90° (Angle in the semi circle)
Applying Pythagoras theorem in ΔPQO,
PQ2 + QO2 = PO2
⇒ PQ2 + 42= 62
⇒ PQ2 + 16 = 36
⇒ PQ2 = 36 – 16
⇒ PQ2 = 20
⇒ PQ = 2√5

Justification:

∠PQO = 90° (Angle in the semi circle)
∴ OQ ⊥ PQ
Therefor, OQ is the radius of the circle then PQ has to be a tangent of the circle. 
Similarly, PR is a tangent of the circle.

3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

Answer

Steps of Construction:
Step I: With O as a centre and radius equal to 3 cm, a                   circle is drawn.
Step II: The diameter of the circle is extended both sides and an arc is made to cut it at 7 cm.
Step III: Perpendicular bisector of OP and OQ is drawn and x and y be its mid-point.
Step IV: With O as a centre and Ox be its radius, a circle is drawn which intersected the previous circle at M and N.
Step V: Step IV is repeated with O as centre and Oy as radius and it intersected the circle at R and T.
Step VI: PM and PN are joined also QR and QT are joined.
Thus,  PM and PN are tangents to the circle from P and QR and QT are tangents to the circle from point Q.

Justification:

∠PMO = 90° (Angle in the semi-circle)
∴ OM ⊥ PM
Therefore, OM is the radius of the circle then PM has to be a tangent of the circle. 
Similarly, PN, QR and QT are tangents of the circle.

4. Draw a pair of tangents to a circle of radius 5 cm which is inclined to each other at an angle of 60°.

Answer

We know that radius of the circle is perpendicular to the tangents.
Sum of all the 4 angles of quadrilateral = 360°
∴ Angle between the radius (∠O)  = 360° – (90° + 90° + 60°) = 120°
Steps of Construction:
Step I: A point Q is taken on the circumference of the circle and OQ is joined. OQ is the radius of the circle.
Step II: Draw another radius OR making an angle equal to 120° with the previous one.
Step III: A point P is taken outside the circle. QP and PR are joined which is perpendicular OQ and OR.
Thus, QP and PR are the required tangents inclined to each other at an angle of 60°.

Justification:

Sum of all angles in the quadrilateral PQOR = 360°
∠QOR + ∠ORP + ∠OQR + ∠RPQ = 360°
⇒ 120° + 90° + 90° + ∠RPQ = 360°
⇒∠RPQ = 360° – 300°
⇒∠RPQ = 60°
Hence, QP and PR are tangents inclined to each other at an angle of 60°.

5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Answer

Steps of Construction:
Step I: A line segment AB of 8 cm is drawn.
Step II: With A as centre and radius equal to 4 cm, a circle is drawn which cut the line at point O.
Step III: With B as a centre and radius equal to  3 cm, a circle is drawn.
Step IV: With O as a centre and OA as a   radius, a circle is drawn which intersect the previous two circles at P, Q and R, S.
Step V: AP, AQ, BR and BS are joined.
Thus, AP, AQ, BR and BS are the required tangents.

Justification:
∠BPA = 90° (Angle in the semi circle)
∴ AP ⊥ PB
Therefor, BP is the radius of the circle then AP has to be a tangent of the circle. 
Similarly,  AQ, BR and BS are tangents of the circle.

6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.

Answer

Steps of Construction:
Step I: A ΔABC is drawn.
Step II: Perpendicular to AC is drawn to point B which intersected it at D.
Step III: With O as a centre and OC as a radius, a   circle is drawn. The circle through B, C,D is drawn.
Step IV: OA is joined and a circle is drawn with diameter OA which intersected the previous circle at B and E.
Step V: AE is joined.
Thus, AB and AE are the required tangents to the circle from A.

Justification:

∠OEA = 90° (Angle in the semi-circle)
∴ OE ⊥ AE
Therefore, OE is the radius of the circle then AE has to be a tangent of the circle. 
Similarly,  AB is the tangent of the circle.

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NCERT Solutions for Class 10 Maths Chapter 10 Circle | EduGrown

In This Post we are  providing Chapter 8 Circle NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Circle Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Circle NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 10 Circle

Exercise: 10.1
 
1. How many tangents can a circle have?
 
Answer
 
A circle can have infinite tangents.
 
2.  Fill in the blanks :
(i) A tangent to a circle intersects it in …………… point(s).
(ii) A line intersecting a circle in two points is called a ………….
(iii) A circle can have …………… parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called …………
 
Answer
 
(i) one
(ii) secant
(iii) two
(iv) point of contact
 
3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at
a point Q so that OQ = 12 cm. Length PQ is :
(A) 12 cm
(B) 13 cm
(C) 8.5 cm 
(D) √119 cm
 
Answer
 

The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ OP ⊥ PQ

By Pythagoras theorem in ΔOPQ,
OQ2 = OP2 + PQ2
⇒ (12)2 = 52 + PQ2
⇒PQ2 = 144 – 25
⇒PQ2 = 119
⇒PQ = √119 cm
(D) is the correct option.
 
4. Draw a circle and two lines parallel to a given line such that one is a tangent and the
other, a secant to the circle.
 
Answer
 
AB and XY are two parallel lines where AB is the tangent to the circle at point C while XY is the secant to the circle.
Exercise: 10.2
 
In Q.1 to 3, choose the correct option and give justification.
 
1.  From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A)  7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
 
Answer

The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ OP ⊥ PQ
also, ΔOPQ is right angled.
OQ = 25 cm and PQ = 24 cm (Given)
By Pythagoras theorem in ΔOPQ,
OQ2 = OP2 + PQ2
⇒ (25)2 = OP2 + (24)2
⇒ OP2 = 625 – 576
⇒ OP2 = 49
⇒ OP = 7 cm
The radius of the circle is option (A) 7 cm.


2.  In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(A) 60°
(B) 70° 
(C) 80° 
(D) 90°
 
Answer

OP and OQ are radii of the circle to the tangents TP and TQ respectively.
∴ OP ⊥ TP and,
∴ OQ ⊥ TQ
∠OPT = ∠OQT = 90°
In quadrilateral POQT,
Sum of all interior angles = 360°
∠PTQ + ∠OPT + ∠POQ + ∠OQT  = 360°
⇒ ∠PTQ + 90° + 110° + 90°  = 360°
⇒ ∠PTQ = 70°
∠PTQ is equal to option (B) 70°.
 
3.  If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠ POA is equal to
(A) 50°  
(B) 60° 
(C) 70°
(D) 80°

Answer
 
OA and OB are radii of the circle to the tangents PA and PB respectively.
∴ OA ⊥ PA and,
∴ OB ⊥ PB
∠OBP = ∠OAP = 90°
In quadrilateral AOBP,
Sum of all interior angles = 360°
∠AOB + ∠OBP + ∠OAP + ∠APB  = 360°
⇒ ∠AOB + 90° + 90° + 80°  = 360°
⇒ ∠AOB = 100°
Now,
In ΔOPB and ΔOPA,
AP = BP (Tangents from a point are equal)
OA = OB (Radii of the circle)
OP = OP (Common side)
∴ ΔOPB ≅ ΔOPA (by SSS congruence condition)
Thus ∠POB = ∠POA
∠AOB = ∠POB + ∠POA
⇒ 2 ∠POA = ∠AOB
⇒ ∠POA = 100°/2 = 50°
∠POA is equal to option  (A) 50°
 
Page No: 214
 
4.  Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
 
Answer
 
Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.
Radii of the circle to the tangents will be perpendicular to it.
∴ OB ⊥ RS and,
∴ OA ⊥ PQ
∠OBR = ∠OBS = ∠OAP = ∠OAQ = 90º
From the figure,
∠OBR = ∠OAQ (Alternate interior angles)
∠OBS = ∠OAP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.
Hence Proved that the tangents drawn at the ends of a diameter of a circle are parallel.
 
5.  Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
 
Answer
 
Let AB be the tangent to the circle at point P with centre O.
We have to prove that PQ passes through the point O.
Suppose that PQ doesn’t passes through point O. Join OP.
Through O, draw a straight line CD parallel to the tangent AB.
PQ intersect CD at R and also intersect AB at P.
AS, CD // AB PQ is the line of intersection,
∠ORP = ∠RPA (Alternate interior angles)
but also,
∠RPA = 90° (PQ ⊥ AB) 
⇒ ∠ORP  = 90°
∠ROP + ∠OPA = 180° (Co-interior angles)
⇒∠ROP + 90° = 180°
⇒∠ROP = 90°
Thus, the ΔORP has 2 right angles i.e. ∠ORP  and ∠ROP which is not possible.
Hence, our supposition is wrong. 
∴ PQ passes through the point O.
 
6.  The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
 
Answer
 
AB is a tangent drawn on this circle from point A.
∴ OB ⊥ AB
OA = 5cm and AB = 4 cm (Given)
In ΔABO,
By Pythagoras theorem in ΔABO,
OA2 = AB2 + BO2
⇒ 52 = 42 + BO2
⇒ BO2 = 25 – 16
⇒ BO2 = 9
⇒ BO = 3
∴ The radius of the circle is 3 cm.

7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the
larger circle which touches the smaller circle.

Answer

Let the two concentric circles with centre O.
AB be the chord of the larger circle which touches the smaller circle at point P. 
∴ AB is tangent to the smaller circle to the point P.
 
⇒ OP ⊥ AB
By Pythagoras theorem in ΔOPA,
OA2 =  AP2 + OP2
⇒ 52 = AP2 + 32
⇒ AP2 = 25 – 9
⇒ AP = 4
In ΔOPB,
Since OP ⊥ AB,
AP = PB (Perpendicular from the center of the circle bisects the chord)
AB = 2AP = 2 × 4 = 8 cm
 ∴ The length of the chord of the larger circle is 8 cm.

8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

Answer


From the figure we observe that,
DR = DS (Tangents on the circle from point D) … (i)
AP = AS (Tangents on the circle from point A) … (ii)
BP = BQ (Tangents on the circle from point B) … (iii)
CR = CQ (Tangents on the circle from point C) … (iv)
Adding all these equations,
DR + AP + BP + CR = DS + AS + BQ + CQ
⇒ (BP + AP) + (DR + CR)  = (DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC
 
9. In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.
 
Answer
 
We joined O and C
A/q,
In ΔOPA and ΔOCA,
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
∴ ΔOPA ≅ ΔOCA (SSS congruence criterion)
⇒ ∠POA = ∠COA … (i)
Similarly,
 ΔOQB  ≅ ΔOCB
∠QOB = ∠COB … (ii)
Since POQ is a diameter of the circle, it is a straight line.
∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 º
From equations (i) and (ii),
2∠COA + 2∠COB = 180º
⇒ ∠COA + ∠COB = 90º
⇒ ∠AOB = 90°
 
10.  Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
 
Answer
Consider a circle with centre O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.
It can be observed that
OA ⊥ PA
∴ ∠OAP = 90°
Similarly, OB ⊥ PB
∴ ∠OBP = 90°
In quadrilateral OAPB,
Sum of all interior angles = 360º
∠OAP +∠APB +∠PBO +∠BOA = 360º
⇒ 90º + ∠APB + 90º + ∠BOA = 360º
⇒ ∠APB + ∠BOA = 180º
∴ The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
 
11. Prove that the parallelogram circumscribing a circle is a rhombus.
 
Answer
 
ABCD is a parallelogram,
∴ AB = CD … (i)
∴ BC = AD … (ii)
From the figure, we observe that,
DR = DS (Tangents to the circle at D)
CR = CQ (Tangents to the circle at C)
BP = BQ (Tangents to the circle at B)
AP = AS (Tangents to the circle at A)
Adding all these,
DR + CR + BP + AP = DS + CQ + BQ + AS
⇒ (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC … (iii)
Putting the value of (i) and (ii) in equation (iii) we get,
2AB = 2BC
⇒ AB = BC … (iv)
By Comparing equations (i), (ii), and (iv) we get,
AB = BC = CD = DA
∴ ABCD is a rhombus.
 
12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.
 
Answer
 
In ΔABC,
Length of two tangents drawn from the same point to the circle are equal,
∴ CF = CD = 6cm
∴ BE = BD = 8cm
∴ AE = AF = x
We observed that,
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x
Now semi perimeter of triangle (s) is,
⇒ 2s = AB + BC + CA
= x + 8 + 14 + 6 + x
= 28 + 2x
⇒s = 14 + x
Area of ΔABC = √s (s – a)(s – b)(s – c)
 
= √(14 + x) (14 + x – 14)(14 + x – x – 6)(14 + x – x – 8)
= √(14 + x) (x)(8)(6)
= √(14 + x) 48 x … (i)
also, Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB)
= 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]
= 2×1/2 (4x + 24 + 32) = 56 + 4x … (ii)
Equating equation (i) and (ii) we get,
√(14 + x) 48 x = 56 + 4x
Squaring both sides,
48x (14 + x) = (56 + 4x)2
⇒ 48x = [4(14 + x)]2/(14 + x)
⇒ 48x = 16 (14 + x)
⇒ 48x = 224 + 16x
⇒ 32x = 224
⇒ x = 7 cm
Hence, AB = x + 8 = 7 + 8 = 15 cm
CA = 6 + x = 6 + 7 = 13 cm
 
13.  Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. 
 
Answer

Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S. Join the vertices of the quadrilateral ABCD to the center of the circle.
In ΔOAP and ΔOAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the circle)
OA = OA (Common side)
ΔOAP ≅ ΔOAS (SSS congruence condition)
∴ ∠POA = ∠AOS
⇒∠1 = ∠8
Similarly we get,
∠2 = ∠3
∠4 = ∠5
∠6 = ∠7
Adding all these angles,
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 +∠8 = 360º
⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º
⇒ 2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360º
⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º
⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180º
⇒ ∠AOB + ∠COD = 180º
Similarly, we can prove that ∠ BOC + ∠ DOA = 180º
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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NCERT Solutions for Class 10 Maths Chapter 9 Some Application of Trigonometry | EduGrown

In This Post we are  providing Chapter 8 Some Application Of Trigonometry NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These  Introduction to Trigonometry  Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Some Application Of Trigonometry NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 9 Some Application Of Trigonometry

Exercise: 9.1
 
Page No: 203
 
 
1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11). 
 
Answer

Let AB be the vertical pole Ac be 20 m  long rope tied to point C.
In  right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = AB/20
⇒ AB = 20/2
⇒ AB = 10
The height of the pole is 10 m.
 
2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
 
Answer
 
Let AC be the broken part of the tree.
∴ Total height of the tree = AB+AC
In  right ΔABC,
cos 30° = BC/AC
⇒ √3/2 = 8/AC
⇒ AC = 16/√3
Also,
tan 30° = AB/BC
⇒ 1/√3 = AB/8
⇒ AB = 8/√3
Total height of the tree = AB+AC = 16/√3 + 8/√3 = 24/√3
 
3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
 
Answer
 
There are two slides of height 1.5 m and 3 m. (Given)
Let AB is 1.5 m and PQ be 3 m slides. 
ABC is the slide inclined at 30° with length AC and PQR is the slide inclined at 
60° with length PR.
A/q,
In  right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = 1.5/AC
⇒ AC = 3m
also,
In  right ΔPQR,
sin 60° = PQ/PR
⇒ √3/2 = 3/PR
⇒ PR = 2√3 m
Hence, length of the slides are 3 m and 2√3 m respectively.
 
Page No: 204
 
4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
 
Answer
 
Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.
A/q,
In  right ΔABC,
tan 30° = AB/BC
⇒ 1/√3 = AB/30
⇒ AB = 10√3
Thus, the height of the tower is 10√3 m.

5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Answer

Let BC be the height of the kite from the ground,
AC be the inclined length of the string from the ground and A is the point where string of the kite is tied.
A/q,
In  right ΔABC,
sin 60° = BC/AC
⇒ √3/2 = 60/AC
⇒ AC = 40√3 m
Thus, the length of the string from the ground is 40√3 m.

6.  A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer

Let the boy initially standing at point Y with inclination 30° and then he approaches the building to
the point X with inclination 60°.
∴ XY is the distance he walked towards the building.
also, XY = CD.
Height of the building = AZ = 30 m
AB = AZ – BZ = (30 – 1.5) = 28.5 m
A/q,
In  right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = 28.5/BD
⇒ BD = 28.5√3 m
also,
In  right ΔABC,
tan 60° = AB/BC
⇒ √3 = 28.5/BC
⇒ BC = 28.5/√3 = 28.5√3/3 m
∴ XY = CD = BD – BC = (28.5√3 – 28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 m.
Thus, the distance boy walked towards the building is 57/√3 m.

7.  From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer

Let BC be the 20 m high building.
D is the point on the ground from where the elevation is taken.
Height of transmission tower = AB = AC – BC
A/q,
In  right ΔBCD,
tan 45° = BC/CD
⇒ 1 = 20/CD
⇒ CD = 20 m
also,
In  right ΔACD,
tan 60° = AC/CD
⇒ √3 = AC/20
⇒ AC = 20√3 m
Height of transmission tower = AB = AC – BC = (20√3 – 20) m = 20(√3 – 1) m.

8.  A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Answer

Let AB be the height of statue.
D is the point on the ground from where the elevation is taken.
Height of pedestal = BC = AC – AB
A/q,
In  right ΔBCD,
tan 45° = BC/CD
⇒ 1 =  BC/CD
⇒ BC = CD.
also,
In  right ΔACD,
tan 60° = AC/CD
⇒ √3 = AB+BC/CD
⇒ √3CD = 1.6 m + BC
⇒ √3BC = 1.6 m + BC
⇒ √3BC – BC = 1.6 m
⇒ BC(√3-1) = 1.6 m
⇒ BC = 1.6/(√3-1) m
⇒ BC = 0.8(√3+1) m
Thus, the height of the pedestal is 0.8(√3+1) m.

9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answer

Let CD be the height of the tower equal to 50 m (Given)
Let AB be the height of the building.
BC be the distance between the foots of the building and the tower.
Elevation is 30° and 60° from the tower and the building respectively.
A/q,
In  right ΔBCD,
tan 60° = CD/BC
⇒ √3 = 50/BC
⇒ BC = 50/√3
also,
In  right ΔABC,
tan 30° = AB/BC
⇒ 1/√3 = AB/BC
⇒ AB = 50/3
Thus, the height of the building is 50/3.

10. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Answer

Let AB and CD be the poles of equal height.
O is the point between them from where the height of elevation taken.
BD is the distance between the poles.
A/q,
AB = CD,
OB + OD = 80 m
Now,
In  right ΔCDO,
tan 30° = CD/OD
⇒ 1/√3 = CD/OD
⇒ CD = OD/√3 … (i)
also,
In  right ΔABO,
tan 60° = AB/OB
⇒ √3 = AB/(80-OD)
⇒ AB = √3(80-OD)
AB = CD (Given)
⇒ √3(80-OD) = OD/√3
⇒ 3(80-OD) = OD
⇒ 240 – 3 OD = OD
⇒ 4 OD = 240
⇒ OD = 60
Putting the value of OD in equation (i)
CD = OD/√3 ⇒ CD = 60/√3 ⇒ CD = 20√3 m
also,
OB + OD = 80 m ⇒ OB = (80-60) m = 20 m
Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and 60 m respectively.


11.  A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

Answer

Here, AB is the height of the tower.
CD = 20 m (given)
A/q,
In  right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = AB/(20+BC)
⇒ AB = (20+BC)/√3 … (i)
also,
In  right ΔABC,
tan 60° = AB/BC
⇒ √3 = AB/BC
⇒ AB = √3 BC … (ii)
From eqn (i) and (ii)
AB = √3 BC = (20+BC)/√3
⇒ 3 BC = 20 + BC
⇒ 2 BC = 20 ⇒ BC = 10 m
Putting the value of BC in eqn (ii)
AB = 10√3 m
Thus, the height of the tower 10√3 m and the width of the canal is 10 m.

12.  From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer

Let AB be the building of height 7 m and EC be the height of tower.
A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°
EC = DE + CD
also, CD = AB = 7 m.
and BC = AD
A/q,
In  right ΔABC,
tan 45° = AB/BC
⇒ 1= 7/BC
⇒ BC = 7 m = AD
also,
In  right ΔADE,
tan 60° = DE/AD
⇒ √3 = DE/7
⇒ DE = 7√3 m
Height of the tower = EC =  DE + CD
                                = (7√3 + 7) m = 7(√3+1) m.

13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer


Let AB be the lighthouse of height 75 m.
Let C and D be the positions of the ships.
30° and 45° are the angles of depression from the lighthouse.
A/q,
In  right ΔABC,
tan 45° = AB/BC
⇒ 1= 75/BC
⇒ BC = 75 m
also,
In  right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = 75/BD
⇒ BD = 75√3  m
 
The distance between the two ships = CD = BD – BC = (75√3 – 75) m = 75(√3-1) m.

Page No: 205

 
14.  A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m
from the ground. The height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

Answer

Let the initial position of the balloon be A and final position be B.
Height of balloon above the girl height = 88.2 m – 1.2 m = 87 m
Distance travelled by the balloon =
 DE = CE – CD
A/q,
In  right ΔBEC,
tan 30° = BE/CE
⇒ 1/√3= 87/CE
⇒ CE = 87√3 m
also,
In  right ΔADC,
tan 60° = AD/CD
⇒ √3= 87/CD
⇒ CD = 87/√3 m = 29√3 m
Distance travelled by the balloon =  DE = CE – CD = (87√3 – 29√3) m = 29√3(3 – 1) m = 58√3 m.

15.  A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer

Let AB be the tower.
D is the initial and C is the final position of the car respectively.
Angles of depression are measured from A.
BC is the distance from the foot of the tower to the car.
A/q,
In  right ΔABC,
tan 60° = AB/BC
⇒ √3 = AB/BC
⇒ BC = AB/√3 m
also,
In  right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = AB/(BC + CD)
⇒ AB√3 = BC + CD
⇒ AB√3 = AB/√3 + CD
⇒ CD = AB√3 – AB/√3
⇒ CD = AB(√3 – 1/√3)
⇒ CD = 2AB/√3
Here, distance of BC is half of CD. Thus, the time taken is also half.
Time taken by car to travel distance CD = 6 sec.
Time taken by car to travel BC = 6/2 = 3 sec.

16.  The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer

Let AB be the tower.
C and D be the two points with distance 4 m and 9 m from the base respectively.
A/q,
In  right ΔABC,
tan x = AB/BC
⇒ tan x = AB/4
⇒ AB = 4 tan x … (i)
also,
In  right ΔABD,
tan (90°-x) = AB/BD
⇒ cot x = AB/9
⇒ AB = 9 cot x  … (ii)
Multiplying  eqn (i) and (ii)
AB2 = 9 cot x × 4 tan x
⇒ AB2 = 36
⇒ AB = ± 6
Height cannot be negative. Therefore, the height of the tower is 6 m. Hence, Proved.

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry | EduGrown

In This Post we are  providing Chapter 8 Introduction to Trigonometry NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These  Introduction to Trigonometry  Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths  Introduction to Trigonometry  NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Exercise 8.1
 
1. In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A
(ii) sin C, cos C

Answer

In Δ ABC,∠B = 90º
By Applying Pythagoras theorem , we get
AC2 = AB2 + BC2 = (24)2 + 72 = (576+49) cm2 = 625 cm2
⇒ AC = 25
 
(i) sin A = BC/AC = 7/25    cos A = AB/AC = 24/25
(ii) sin C = AB/AC = 24/25
     cos C = BC/AC = 7/25

2.  In Fig. 8.13, find tan P – cot R.

Answer

By Applying Pythagoras theorem in ΔPQR , we get
PR2 = PQ2 + QR2 = (13)2 = (12)2 + QR2 = 169 = 144  + QR2
⇒  QR2 = 25 ⇒  QR = 5 cm
Now,
tan P = QR/PQ = 5/12
cot R = QR/PQ = 5/12
A/q
tan P – cot R = 5/12 – 5/12 = 0

3. If sin A =3/4, calculate cos A and tan A.

Answer
 
Let ΔABC be a right-angled triangle, right-angled at B.
We know that sin A = BC/AC = 3/4

Let BC be 3k and AC will be 4k where k is a positive real number.

By Pythagoras theorem we get,
AC2 = AB2 + BC2 
(4k)2 = AB2 + (3k)2
16k2 – 9k2 = AB2
AB2 = 7k2
AB = √7 k

cos A = AB/AC = √7 k/4k = √7/4
tan A = BC/AB = 3k/√7 k = 3/√7
 

4. Given 15 cot A = 8, find sin A and sec A.

 
Answer

Let ΔABC be a right-angled triangle, right-angled at B.
We know that cot A = AB/BC = 8/15   (Given)
Let AB be 8k and BC will be 15k where k is a positive real number. 
By Pythagoras theorem we get,
AC2 = AB2 + BC2 
AC2 = (8k)2 + (15k)2
AC2 = 64k2 + 225k2
AC2 = 289k2
AC = 17 k
sin A = BC/AC = 15k/17k = 15/17
sec A = AC/AB = 17k/8 k = 17/8
 
 
5. Given sec θ = 13/12, calculate all other trigonometric ratios.
 
Answer
 
Let ΔABC be a right-angled triangle, right-angled at B.
We know that sec θ = OP/OM = 13/12   (Given)
Let OP be 13k and OM will be 12k where k is a positive real number. 
By Pythagoras theorem we get,
OP2 = OM2 + MP2 
(13k)2 = (12k)2 + MP2 
169k2 – 144k2 = MP2
MP2 = 25k2

MP = 5

 
Now,
sin θ = MP/OP = 5k/13k = 5/13
cos θ = OM/OP = 12k/13k = 12/13
tan θ = MP/OM = 5k/12k = 5/12
cot θ = OM/MP = 12k/5k = 12/5
cosec θ = OP/MP = 13k/5k = 13/5
 
6.  If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
 
Answer

Let ΔABC in which CD ⊥ AB.

A/q,

cos A = cos B
⇒ AD/AC = BD/BC
⇒ AD/BD = AC/BC
Let AD/BD = AC/BC = k
⇒ AD = kBD  …. (i)
⇒ AC = kBC  …. (ii)

By applying Pythagoras theorem in ΔCAD and ΔCBD we get,
CD2 = AC2 – AD2 …. (iii)
and also CD2 = BC2 – BD2 …. (iv)
From equations (iii) and (iv) we get,
AC2 – AD2 = BC2 – BD2
⇒ (kBC)2 – (k BD)2 = BC2 – BD2
⇒ k2 (BC2 – BD2) = BC2 – BD2
⇒ k2 = 1
⇒ k = 1
Putting this value in equation (ii), we obtain
AC = BC
⇒ ∠A = ∠B  (Angles opposite to equal sides of a triangle are equal-isosceles triangle)

 
7. If cot θ =7/8, evaluate : 
(i)(1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ)
(ii) cot2θ
 
Answer
 
Let ΔABC in which ∠B = 90º and ∠C = θ
A/q,
cot θ = BC/AB = 7/8

Let BC = 7k and AB = 8k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC2 = AB2 + BC2 
AC2 = (8k)2 + (7k)2
AC2 = 64k2 + 49k2
AC2 = 113k2
AC = √113 k

sin θ = AB/AC = 8k/√113 k = 8/√113 
and cos θ = BC/AC = 7k/√113 k = 7/√113 
(i) (1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ) = (1-sin2θ)/(1-cos2θ) = {1 – (8/√113)2}/{1 – (7/√113)2}
     = {1 – (64/113)}/{1 – (49/113)} = {(113 – 64)/113}/{(113 – 49)/113} = 49/64
(ii) cot2θ = (7/8)2 = 49/64
 
8.  If 3cot A = 4/3 , check whether (1-tan2A)/(1+tan2A) = cos2A – sin2A or not.

Answer


Let ΔABC in which ∠B = 90º,
A/q,
cot A = AB/BC = 4/3
Let AB = 4k and BC = 3k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC2 = AB2 + BC2 
AC2 = (4k)2 + (3k)2
AC2 = 16k2 + 9k2
AC2 = 25k2
AC = 5k
tan A = BC/AB = 3/4
sin A = BC/AC = 3/5
cos A = AB/AC = 4/5
L.H.S. = (1-tan2A)/(1+tan2A) = 1- (3/4)2/1+ (3/4)2 = (1- 9/16)/(1+ 9/16) = (16-9)/(16+9) = 7/25
R.H.S. = cos2A – sin2A = (4/5)2 – (3/4)2 = (16/25) – (9/25) = 7/25
R.H.S. = L.H.S.
Hence,  (1-tan2A)/(1+tan2A) = cos2A – sin2A

9. In triangle ABC, right-angled at B, if tan A =1/√3 find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C

Answer

Let ΔABC in which ∠B = 90º,
A/q,
tan A = BC/AB = 1/√3
Let AB = √3 k and BC = k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC2 = AB2 + BC2 
AC2 = (√3 k)2 + (k)2
AC2 = 3k2 + k2
AC2 = 4k2
AC = 2k
sin A = BC/AC = 1/2                   cos A = AB/AC = √3/2 ,
sin C = AB/AC = √3/2                   cos A = BC/AC = 1/2
(i) sin A cos C + cos A sin C = (1/2×1/2) + (√3/2×√3/2) = 1/4+3/4 = 4/4 = 1
(ii) cos A cos C – sin A sin C = (√3/2×1/2) – (1/2×√3/2) = √3/4 – √3/4 = 0

10. In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Answer

Given that, PR + QR = 25 , PQ = 5
Let PR be x.  ∴ QR = 25 – x

By Pythagoras theorem ,
PR2 = PQ2 + QR2
x2 = (5)2 + (25 – x)2
x2 = 25 + 625 + x2 – 50x
50x = 650
x = 13
∴ PR = 13 cm
QR = (25 – 13) cm = 12 cm

sin P = QR/PR = 12/13
cos P = PQ/PR = 5/13
tan P = QR/PQ = 12/5
 
11.  State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = 4/3 for some angle θ.
 
Answer
 
(i) False.
In ΔABC in which ∠B = 90º,
     AB = 3, BC = 4 and AC = 5
Value of tan A = 4/3 which is greater than.
The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as
it will follow the Pythagoras theorem.
AC2 = AB2 + BC2 
52 = 32 + 42
25 = 9 + 16
25 = 25

(ii) True.
Let a ΔABC in which ∠B = 90º,AC be 12k and AB be 5k, where k is a positive real number.
By Pythagoras theorem we get,
AC2 = AB2 + BC2 
(12k)2 = (5k)2 + BC2 
BC2 + 25k2 = 144k2
BC2 = 119k2

Such a triangle is possible as it will follow the Pythagoras theorem.


(iii) False.
Abbreviation used for cosecant of angle A is cosec A.cos A is the abbreviation used for cosine of angle A.

(iv) False.
cot A is not the product of cot and A. It is the cotangent of ∠A.

(v) False.
sin θ = Height/Hypotenuse
We know that in a right angled triangle, Hypotenuse is the longest side. 
∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.
Exercise 8.2

1. Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan245° + cos230° – sin260°
(iii) cos 45°/(sec 30° + cosec 30°)    (iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
(v) (5cos260° + 4sec230° – tan245°)/(sin230° + cos230°)

Answer

(i) sin 60° cos 30° + sin 30° cos 60°
     =  (√3/2×√3/2) + (1/2×1/2) = 3/4 + 1/4 = 4/4 = 1

(ii) 2 tan245° + cos230° – sin260°
     = 2×(1)2 + (√3/2)2 – (√3/2)2 = 2

(iii) cos 45°/(sec 30° + cosec 30°)
     = 1/√2/(2/√3 + 2) = 1/√2/{(2+2√3)/√3)
     = √3/√2×(2+2√3) = √3/(2√2+2√6)
     = √3(2√6-2√2)/(2√6+2√2)(2√6-2√2)
     = 2√3(√6-√2)/(2√6)2-(2√2)2
       =  2√3(√6-√2)/(24-8) =  2√3(√6-√2)/16
     = √3(√6-√2)/8 = (√18-√6)/8 = (3√2-√6)/8

(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
      = (1/2+1-2/√3)/(2/√3+1/2+1)
      = (3/2-2/√3)/(3/2+2/√3)
      = (3√3-4/2√3)/(3√3+4/2√3)
      = (3√3-4)/(3√3+4)
      = (3√3-4)(3√3-4)/(3√3+4)(3√3-4)
      = (3√3-4)2/(3√3)2-(4)2
        = (27+16-24√3)/(27-16)
      = (43-24√3)/11]

(v) (5cos260° + 4sec230° – tan245°)/(sin230° + cos230°)
     = 5(1/2)2+4(2/√3)2-12/(1/2)2+(√3/2)2
      = (5/4+16/3-1)/(1/4+3/4)
    = (15+64-12)/12/(4/4)
    = 67/12

2. Choose the correct option and justify your choice :
(i) 2tan 30°/1+tan230° =
     (A) sin 60°            (B) cos 60°          (C) tan 60°            (D) sin 30°
(ii) 1-tan245°/1+tan245° =
     (A) tan 90°            (B) 1                    (C) sin 45°            (D) 0
(iii)  sin 2A = 2 sin A is true when A =
     (A) 0°                   (B) 30°                  (C) 45°                 (D) 60°
(iv) 2tan30°/1-tan230° =
     (A) cos 60°          (B) sin 60°             (C) tan 60°           (D) sin 30°

Answer

(i) (A) is correct.
2tan 30°/1+tan230° = 2(1/√3)/1+(1/√3)2
= (2/√3)/(1+1/3) = (2/√3)/(4/3)
= 6/4√3 = √3/2 = sin 60°

(ii)  (D) is correct.
1-tan245°/1+tan245° = (1-12)/(1+12)
= 0/2 = 0

(iii) (A) is correct.
sin 2A = 2 sin A is true when A =
= As sin 2A = sin 0° = 0
2 sin A = 2sin 0° = 2×0 = 0
or,
sin 2A = 2sin A cos A
⇒2sin A cos A = 2 sin A
⇒ 2cos A = 2 ⇒ cos A = 1
⇒ A = 0°

(iv) (C) is correct.
2tan30°/1-tan230° =  2(1/√3)/1-(1/√3)2
= (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°
 
3. If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.
 
Answer
 
tan (A + B) = √3
⇒ tan (A + B) = tan 60°
⇒ (A + B) = 60° … (i)
 tan (A – B) = 1/√3
⇒ tan (A – B) = tan 30°
⇒ (A – B) = 30° … (ii)
Adding (i) and (ii), we get
A + B + A – B = 60° + 30°
2A = 90°
A= 45°
Putting the value of A in equation (i)
45° + B = 60°
⇒ B = 60° – 45°
⇒ B = 15°
Thus, A = 45° and B = 15°
 
4. State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
 
Answer
 
(i) False.
Let A = 30° and B = 60°, then
sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,
sin A + sin B = sin 30° + sin 60°
= 1/2 + √3/2 = 1+√3/2

(ii) True.
sin 0° = 0
sin 30° = 1/2
sin 45° = 1/√2
sin 60° = √3/2
sin  90° = 1
Thus the value of sin θ increases as θ increases.

(iii) False.
cos 0° = 1
cos 30° = √3/2
cos 45° = 1/√2
cos 60° = 1/2
cos 90° = 0
Thus the value of cos θ decreases as θ increases.

(iv) True.
cot A = cos A/sin A
cot 0° = cos 0°/sin 0° = 1/0 = undefined.
Excercise 8.3
 
 
 
1. Evaluate :
(i) sin 18°/cos 72°        (ii) tan 26°/cot 64°        (iii)  cos 48° – sin 42°       (iv)  cosec 31° – sec 59°
 
Answer
 
(i) sin 18°/cos 72°
    = sin (90° – 18°) /cos 72° 
    = cos 72° /cos 72° = 1

(ii) tan 26°/cot 64°
    = tan (90° – 36°)/cot 64°
    = cot 64°/cot 64° = 1

(iii) cos 48° – sin 42°
      = cos (90° – 42°) – sin 42°
      = sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°
     = cosec (90° – 59°) – sec 59°
     = sec 59° – sec 59° = 0
 
2.  Show that :
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
 
Answer
 

(i) tan 48° tan 23° tan 42° tan 67°
    = tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°
    = cot 42° cot 67° tan 42° tan 67°
    = (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

(ii) cos 38° cos 52° – sin 38° sin 52°
    = cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°
    = sin 52° sin 38° – sin 38° sin 52° = 0

 
3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
 
Answer 
 

A/q,
tan 2A = cot (A- 18°)
⇒ cot (90° – 2A) = cot (A -18°)
Equating angles,
⇒ 90° – 2A = A- 18° ⇒ 108° = 3A
⇒ A = 36°

 
4.  If tan A = cot B, prove that A + B = 90°.
 
Answer
 
A/q, 

tan A = cot B
⇒ tan A = tan (90° – B)
⇒ A = 90° – B
⇒ A + B = 90°

 
5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
 
Answer
 

A/q,
sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A – 20°)

Equating angles,
90° – 4A= A- 20°
⇒ 110° = 5A
⇒ A = 22°
 
Page No : 190
 
6. If A, B and C are interior angles of a triangle ABC, then show that
    sin (B+C/2) = cos A/2
 
Answer
 
In a triangle, sum of all the interior angles
A + B + C = 180°
⇒ B + C = 180° – A
⇒ (B+C)/2 = (180°-A)/2
⇒ (B+C)/2 = (90°-A/2)
⇒ sin (B+C)/2 = sin (90°-A/2)
⇒ sin (B+C)/2 = cos A/2
 
7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
 
Answer
 

sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°

Excercise 8.4
 
1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
 
Answer
 
cosec2A – cot2A = 1
⇒ cosec2A = 1 + cot2A
⇒ 1/sin2A = 1 + cot2A
⇒sin2A = 1/(1+cot2A)

Now,
sin2A = 1/(1+cot2A)
⇒ 1 – cos2A = 1/(1+cot2A)
⇒cos2A = 1 – 1/(1+cot2A)
⇒cos2A = (1-1+cot2A)/(1+cot2A)
⇒ 1/sec2A = cot2A/(1+cot2A)
⇒ secA = (1+cot2A)/cot2A

also,
tan A = sin A/cos A and cot A = cos A/sin A
⇒ tan A = 1/cot A

2. Write all the other trigonometric ratios of ∠A in terms of sec A.

Answer

We know that,
sec A = 1/cos A
⇒ cos A = 1/sec A
also,
cos2A + sin2A = 1
⇒  sin2A = 1 – cos2A
⇒  sin2A = 1 – (1/sec2A)
⇒  sin2A = (sec2A-1)/sec2A


also,
sin A = 1/cosec A
⇒cosec A = 1/sin A

Now,
sec2A – tan2A = 1
⇒ tan2A = sec2A + 1

also,
tan A = 1/cot A
⇒ cot A = 1/tan A


3. Evaluate :
(i) (sin263° + sin227°)/(cos217° + cos273°)
(ii)  sin 25° cos 65° + cos 25° sin 65°

Answer

(i) (sin263° + sin227°)/(cos217° + cos273°)
   = [sin2(90°-27°) + sin227°]/[cos2(90°-73°) + cos273°)]
   = (cos227° + sin227°)/(sin227° + cos273°)
   = 1/1 =1                       (∵ sin2A + cos2A = 1)

(ii) sin 25° cos 65° + cos 25° sin 65°
   = sin(90°-25°) cos 65° + cos(90°-65°) sin 65°
   = cos 65° cos 65° + sin 65° sin 65°
   = cos265° + sin265° = 1

4. Choose the correct option. Justify your choice.
(i) 9 sec2A – 9 tan2A =
      (A) 1                 (B) 9              (C) 8                (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
      (A) 0                 (B) 1              (C) 2                (D) – 1
(iii) (secA + tanA) (1 – sinA) =
      (A) secA           (B) sinA        (C) cosecA      (D) cosA
(iv) 1+tan2A/1+cot2A = 
      (A) sec2A                 (B) -1              (C) cot2A                (D) tan2A
 
Answer
 
(i) (B) is correct.
9 sec2A – 9 tan2A

= 9 (sec2A – tan2A)
= 9×1 = 9             (∵ sec2 A – tan2 A = 1)


(ii) (C) is correct
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)   
= (1 + sin θ/cos θ + 1/cos θ) (1 + cos θ/sin θ – 1/sin θ)
= (cos θ+sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ
= (cos θ+sin θ)2-12/(cos θ sin θ)
= (cos2θ + sin2θ + 2cos θ sin θ -1)/(cos θ sin θ)
= (1+ 2cos θ sin θ -1)/(cos θ sin θ)
= (2cos θ sin θ)/(cos θ sin θ) = 2

(iii) (D) is correct.
(secA + tanA) (1 – sinA)
= (1/cos A + sin A/cos A) (1 – sinA)
= (1+sin A/cos A) (1 – sinA)
= (1 – sin2A)/cos A
= cos2A/cos A = cos A

(iv) (D) is correct.
1+tan2A/1+cot2A 
= (1+1/cot2A)/1+cot2A
= (cot2A+1/cot2A)×(1/1+cot2A)
= 1/cot2A = tan2A
 
5. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.
(i) (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ)
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
     [Hint : Write the expression in terms of sin θ and cos θ]
(iv) (1 + sec A)/sec A = sin2A/(1-cos A)  
     [Hint : Simplify LHS and RHS separately]
(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec2A = 1+cot2A.
 
 
 
 
(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
     [Hint : Simplify LHS and RHS separately]
(x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A

Answer

(i) (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ)
L.H.S. =  (cosec θ – cot θ)2
           = (cosec2θ + cot2θ – 2cosec θ cot θ)
           = (1/sin2θ + cos2θ/sin2θ – 2cos θ/sin2θ)
           = (1 + cos2θ – 2cos θ)/(1 – cos2θ)
           = (1-cos θ)2/(1 – cosθ)(1+cos θ)
           = (1-cos θ)/(1+cos θ) = R.H.S.

(ii)  cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
 L.H.S. = cos A/(1+sin A) + (1+sin A)/cos A
            = [cos2A + (1+sin A)2]/(1+sin A)cos A
            = (cos2A + sin2A + 1 + 2sin A)/(1+sin A)cos A
            = (1 + 1 + 2sin A)/(1+sin A)cos A
            = (2+ 2sin A)/(1+sin A)cos A
            = 2(1+sin A)/(1+sin A)cos A
            = 2/cos A = 2 sec A = R.H.S.

(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)
           = [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]
           = [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ]
           = sin2θ/[cos θ(sin θ-cos θ)] + cos2θ/[sin θ(cos θ-sin θ)]
           = sin2θ/[cos θ(sin θ-cos θ)] – cos2θ/[sin θ(sin θ-cos θ)]
           = 1/(sin θ-cos θ) [(sin2θ/cos θ) – (cos2θ/sin θ)]
           = 1/(sin θ-cos θ) × [(sin3θ – cos3θ)/sin θ cos θ]
           = [(sin θ-cos θ)(sin2θ+cos2θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]
           = (1 + sin θ cos θ)/sin θ cos θ
           = 1/sin θ cos θ + 1
           = 1 + sec θ cosec θ = R.H.S.

(iv)  (1 + sec A)/sec A = sin2A/(1-cos A)
L.H.S. = (1 + sec A)/sec A
           = (1 + 1/cos A)/1/cos A
           = (cos A + 1)/cos A/1/cos A
           = cos A + 1
R.H.S. = sin2A/(1-cos A)
            = (1 – cos2A)/(1-cos A)
            = (1-cos A)(1+cos A)/(1-cos A)
            = cos A + 1
L.H.S. = R.H.S.

(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec2A = 1+cot2A.
L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)
           Dividing Numerator and Denominator by sin A,
           = (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A
           = (cot A – 1 + cosec A)/(cot A+ 1 – cosec A)
           = (cot A – cosec2A + cot2A + cosec A)/(cot A+ 1 – cosec A) (using cosec2A – cot2A = 1)
           = [(cot A + cosec A) – (cosec2A – cot2A)]/(cot A+ 1 – cosec A)
           = [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]/(1 – cosec A + cot A)
           =  (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)
           =  cot A + cosec A = R.H.S.


Dividing Numerator and Denominator of L.H.S. by cos A,

= sec A + tan A = R.H.S.

(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
L.H.S. = (sin θ – 2sin3θ)/(2cos3θ – cos θ)
           = [sin θ(1 – 2sin2θ)]/[cos θ(2cos2θ- 1)]
           = sin θ[1 – 2(1-cos2θ)]/[cos θ(2cos2θ -1)]
          = [sin θ(2cos2θ -1)]/[cos θ(2cos2θ -1)]
          = tan θ = R.H.S.

(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A
L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2
               = (sin2A + cosec2A + 2 sin A cosec A) + (cos2A + sec2A + 2 cos A sec A)
           = (sin2A + cos2A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan2A + 1 + cot2A
           = 1 + 2 + 2 + 2 + tan2A + cot2A
           = 7+tan2A+cot2A = R.H.S.

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
L.H.S. = (cosec A – sin A)(sec A – cos A)
           = (1/sin A – sin A)(1/cos A – cos A)
           = [(1-sin2A)/sin A][(1-cos2A)/cos A]
           = (cos2A/sin A)×(sin2A/cos A)
           = cos A sin A
R.H.S. = 1/(tan A+cotA)
            = 1/(sin A/cos A +cos A/sin A)
            = 1/[(sin2A+cos2A)/sin A cos A]
            = cos A sin A
L.H.S. = R.H.S.

(x)  (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A
L.H.S. = (1+tan2A/1+cot2A)
           = (1+tan2A/1+1/tan2A)
           = 1+tan2A/[(1+tan2A)/tan2A]
           = tan2A

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NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry | EduGrown

In This Post we are  providing Chapter 7Coordinate Geometry NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Coordinate Geometry  Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Coordinate Geometry  NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Exercise 7.1

1. Find the distance between the following pairs of points:
(i) (2, 3), (4, 1) (ii) (−5, 7), (−1, 3) (iii) (a, b), (− a, − b)

Answer

(i) Distance between the points is given by


(ii) Distance between (−5, 7) and (−1, 3) is given by


(iii) Distance between (a, b) and (− a, − b) is given by


2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

 
Answer
 

Distance between points (0, 0) and (36, 15)

Yes, Assume town A at origin point (0, 0).
Therefore, town B will be at point (36, 15) with respect to town A.
And hence, as calculated above, the distance between town A and B will be 39 km.

 
3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
 
Answer
 

Let the points (1, 5), (2, 3), and (- 2,-11) be representing the vertices A, B, and C of the given triangle respectively.Let A = (1, 5), B = (2, 3) and C = (- 2,-11)

Since AB + BC ≠ CA
Therefore, the points (1, 5), (2, 3), and ( – 2, – 11) are not collinear.

 
4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
 
Answer
 

Let the points (5, – 2), (6, 4), and (7, – 2) are representing the vertices A, B, and C of the given triangle respectively.

Therefore, AB = BC
As two sides are equal in length, therefore, ABC is an isosceles triangle.

 
5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees.
Using distance formula, find which of them is correct.
Answer
 
Clearly from the figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6, 1). 
By using distance formula, we get

It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are of the same length.
Therefore, ABCD is a square and hence, Champa was correct


6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
 
Answer
 
Let the points ( – 1, – 2), (1, 0), ( – 1, 2), and ( – 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square.

(ii) Let the points ( – 3, 5), (3, 1), (0, 3), and ( – 1, – 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
It can be observed that all sides of this quadrilateral are of different lengths. Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc.

(iii) Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
It can be observed that opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.


7. Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).

Answer

We have to find a point on x-axis. Therefore, its y-coordinate will be 0.
Let the point on x-axis be (x,0)
(x – 2)2 + 25 = (x – 2)2 + 81
x2 + 4 – 4x + 25 = x2 + 4 + 4x + 81
8x = 25 -81
8x = -56
x = -7
Therefore, the point is ( – 7, 0).

8. Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.

Answer

It is given that the distance between (2, – 3) and (10, y) is 10.
64 + (y + 3)2 = 100
(y +3)2 = 36
y + 3 = ±6
y + 3 = +6 or y + 3 = -6
Therefore, y = 3 or -9

9. If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also find the distance QR and PR.

Answer

PQ = QR

41 = x2 + 25
16 = x2
x = ±4
Therefore, point R is (4, 6) or ( – 4, 6).
When point R is (4, 6),


10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).

Answer

Point (x, y) is equidistant from (3, 6) and ( – 3, 4).
x2 + 9 – 6x + y2 + 36 – 12y = x2 + 9 + 6x + y2 + 16 – 8y
36 – 16 = 6x + 6x + 12y – 8y
20 = 12x + 4y
3x + y = 5
3x + y – 5 = 0

Exercise 7.2

1. Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2:3.

Answer

Let P(x, y) be the required point. Using the section formula, we get
Therefore, the point is (1, 3).
 
2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

Answer

Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB
Therefore, point P divides AB internally in the ratio 1:2.


3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flagexactly halfway between the line segment joining the two flags, where should she post her flag?
 
Answer
 
It can be observed that Niharika posted the green flag at 1/4th of the distance AD i.e., (1×100/4)m = 25m from the starting point of 2nd line. Therefore, the coordinates of this point G is (2, 25).
Similarly, Preet posted red flag at 1/5 of the distance AD i.e., (1×100/5) m = 20m from the starting point of 8th line. Therefore, the coordinates of this point R are (8, 20).
Distance between these flags by using distance formula = GR

Therefore, Rashmi should post her blue flag at 22.5m on 5th line.
 
4. Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6).

Answer

Let the ratio in which the line segment joining ( -3, 10) and (6, -8) is divided by point ( -1, 6) be k:1.
Therefore, -1 = 6k-3/k+1
–k – 1 = 6k -3
7k = 2
k = 2/7
Therefore, the required ratio is 2:7.

5. Find the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Answer

Let the ratio in which the line segment joining A (1, – 5) and B ( – 4, 5) is divided by x-axis be k:1.
Therefore, the coordinates of the point of division is (-4k+1/k+1, 5k-5/k+1).
We know that y-coordinate of any point on x-axis is 0.
∴ 5k-5/k+1 = 0
Therefore, x-axis divides it in the ratio 1:1.

6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Answer

Let A,B,C and D be the points (1,2) (4,y), (x,6) and (3,5) respectively.
Mid point of diagonal AC is 
and Mid point of Diagonal BD is 
Since the diagonals of a parallelogram bisect each other, the mid point of AC and BD are same.
∴ x+1/2 = 7/2 and 4 = 5+y/2
⇒ x + 1 = 7 and 5 + y = 8
⇒ x = 6 and y = 3
 
7. Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, – 3) and B is (1, 4).
 
Answer
 
Let the coordinates of point A be (x, y).
Mid-point of AB is (2, – 3), which is the center of the circle.

⇒ x + 1 = 4 and y + 4 = -6
⇒ x = 3 and y = -10
Therefore, the coordinates of A are (3,-10).

8. If A and B are (–2, –2) and (2, –4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.

Answer

The coordinates of point A and B are (-2,-2) and (2,-4) respectively.
Since AP = 3/7 AB
Therefore, AP:PB = 3:4
Point P divides the line segment AB in the ratio 3:4.
 
9. Find the coordinates of the points which divide the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.
 
Answer
 
From the figure, it can be observed that points X, Y, Z are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.


10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order. [Hint: Area of a rhombus = 1/2(product of its diagonals)]
Answer

Let (3, 0), (4, 5), ( – 1, 4) and ( – 2, – 1) are the vertices A, B, C, D of a rhombus ABCD.

Exercises 7.3

1. Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)
 
Answer
 
(i) Area of a triangle is given by
Area of triangle = 1/2 {x1 (y2 – y3)+ x2 (y3 – y1)+ x3 (y1 – y2)}
Area of the given triangle = 1/2 [2 { 0- (-4)} + (-1) {(-4) – (3)} + 2 (3 – 0)]
                                          = 1/2 {8 + 7 + 6}
                                          = 21/2 square units.

(ii) Area of the given triangle = 1/2 [-5 { (-5)- (4)} + 3(2-(-1)) + 5{-1 – (-5)}]
                                                = 1/2{35 + 9 + 20}
                                                = 32 square units

2. In each of the following find the value of ‘k‘, for which the points are collinear.
(i) (7, -2), (5, 1), (3, –k) 
(ii) (8, 1), (k, -4), (2, -5)


Answer

(i) For collinear points, area of triangle formed by them is zero.
Therefore, for points (7, -2) (5, 1), and (3, k), area = 0
1/2 [7 { 1- k} + 5(k-(-2)) + 3{(-2) + 1}] = 0
7 – 7k + 5k +10 -9 = 0
-2k + 8 = 0
k = 4

(ii) For collinear points, area of triangle formed by them is zero.
Therefore, for points (8, 1), (k, – 4), and (2, – 5), area = 0
1/2 [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0
8 – 6k + 10 = 0
6k = 18
k = 3

3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Answer

 

Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by

D = (0+2/2 , -1+1/2) = (1,0)
E = (0+0/2 , -3-1/2) = (0,1)
F = (2+0/2 , 1+3/2) = (1,2)
Area of a triangle = 1/2 {x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)}
Area of ΔDEF = 1/2 {1(2-1) + 1(1-0) + 0(0-2)}
                        = 1/2 (1+1) = 1 square units
Area of ΔABC = 1/2 [0(1-3) + 2{3-(-1)} + 0(-1-1)]
                         = 1/2 {8} = 4 square units
Therefore, the required ratio is 1:4.

4. Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).


Answer

Let the vertices of the quadrilateral be A ( – 4, – 2), B ( – 3, – 5), C (3, – 2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.

Area of a triangle = 1/2 {x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)}
Area of ΔABC = 1/2 [(-4) {(-5) – (-2)} + (-3) {(-2) – (-2)} + 3 {(-2) – (-5)}]
                         =  1/2 (12+0+9)
                         = 21/2 square units
Area of ΔACD = 1/2 [(-4) {(-2) – (3)} + 3{(3) – (-2)} + 2 {(-2) – (-2)}]
                         = 1/2 (20+15+0)
                         = 35/2 square units
Area of ☐ABCD  = Area of ΔABC + Area of ΔACD
                             = (21/2 + 35/2) square units = 28 square units

5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).


Answer

Let the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2).
Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.

Coordinates of point D = (3+5/2, -2+2/2) = (4,0)
Area of a triangle = 1/2 {x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)}
Area of ΔABD = 1/2 [(4) {(-2) – (0)} + 3{(0) – (-6)} + (4) {(-6) – (-2)}]

                       = 1/2 (-8+18-16)
                       = -3 square units
However, area cannot be negative. Therefore, area of ΔABD is 3 square units.
Area of ΔABD = 1/2 [(4) {0 – (2)} + 4{(2) – (-6)} + (5) {(-6) – (0)}]
                         = 1/2 (-8+32-30)
                         = -3 square units
However, area cannot be negative. Therefore, area of ΔABD is 3 square units.
The area of both sides is same. Thus, median AD has divided ΔABC in two triangles of equal areas.

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NCERT Solutions for Class 10 Maths Chapter 6 Triangle | EduGrown

In This Post we are  providing Chapter 6 Triangle NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Triangle Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Triangle NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 6 Triangle

Exercise 6.1

1. Fill in the blanks using correct word given in the brackets:-

(i) All circles are __________. (congruent, similar)
► Similar

(ii) All squares are __________. (similar, congruent)

► Similar

(iii) All __________ triangles are similar. (isosceles, equilateral)
► Equilateral

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

► (a) Equal, (b) Proportional

2. Give two different examples of pair of
(i) Similar figures
(ii) Non-similar figures

Answer

(i) Two twenty-rupee notes, Two two rupees coins.
(ii) One rupee coin and five rupees coin, One rupee not and ten rupees note.

3. State whether the following quadrilaterals are similar or not:

Triangles Exercise 6.1 Question No. 3
 
Answer
 
The given two figures are not similar because their corresponding angles are not equal.
Exercise 6.2
 

1. In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Triangles Exercise 6.2 Question No. 1
 
Answer
 
(i) In △ ABC, DE∥BC (Given)
∴ AD/DB = AE/EC [By using Basic proportionality theorem]

⇒ 1.5/3 = 1/EC

⇒ Σ EC = 3/1.5

EC = 3×10/15 = 2 cm
Hence, EC = 2 cm.

(ii) In △ ABC, DE∥BC (Given)
∴ AD/DB = AE/EC [By using Basic proportionality theorem]
⇒ AD/7.2 = 1.8/5.4
⇒ AD = 1.8×7.2/5.4 = 18/10 × 72/10 × 10/54 = 24/10
⇒ AD = 2.4
Hence, AD = 2.4 cm.

2. E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm


Answer

Triangles Exercise 6.2 Answer 3

 

 
In ΔPQR, E and F are two points on side PQ and PR respectively.
(i) PE = 3.9 cm, EQ = 3 cm (Given)
PF = 3.6 cm, FR = 2,4 cm (Given)
∴ PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3 [By using Basic proportionality theorem]
And, PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5
So, PE/EQ ≠ PF/FR
Hence, EF is not parallel to QR.
 
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8cm, RF = 9cm
∴ PE/QE = 4/4.5 = 40/45 = 8/9 [By using Basic proportionality theorem]
And, PF/RF = 8/9
So, PE/QE = PF/RF
Hence, EF is parallel to QR.
 
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm (Given)
Here, EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm
So, PE/EQ = 0.18/1.10 = 18/110 = 9/55 … (i)
And, PE/FR = 0.36/2.20 = 36/220 = 9/55 … (ii)
∴ PE/EQ = PF/FR.
Hence, EF is parallel to QR.
 

3. In the fig 6.18, if LM || CB and LN || CD, prove that AM/MB = AN/AD

Triangles Exercise 6.2 Question No. 3
Answer
 
In the given figure, LM || CB

By using basic proportionality theorem, we get,

AM/MB = AL/LC … (i)
Similarly, LN || CD
∴ AN/AD = AL/LC … (ii)
 

From (i) and (ii), we get
AM/MB = AN/AD

4. In the fig 6.19, DE||AC and DF||AE. Prove that

BF/FE = BE/EC
 
Answer
 

In ΔABC, DE || AC (Given)

∴ BD/DA = BE/EC …(i) [By using Basic Proportionality Theorem]

 

In  ΔABC, DF || AE (Given)

∴ BD/DA = BF/FE …(ii) [By using Basic Proportionality Theorem]

From equation (i) and (ii), we get
BE/EC = BF/FE

5. In the fig 6.20, DE||OQ and DF||OR, show that EF||QR.

Triangles Exercise 6.2 Question No. 5
 
Answer
 
In ΔPQO, DE || OQ (Given)
∴ PD/DO = PE/EQ …(i) [By using Basic Proportionality Theorem]
 
In ΔPQO, DE || OQ (Given)
∴ PD/DO = PF/FR …(ii) [By using Basic Proportionality Theorem]
 
From equation (i) and (ii), we get
PE/EQ = PF/FR
 
In ΔPQR, EF || QR. [By converse of Basic Proportionality Theorem]
 

6. In the fig 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Triangles Exercise 6.2 Question No. 6
 
Answer
 
In ΔOPQ, AB || PQ (Given)
∴ OA/AP = OB/BQ …(i) [By using Basic Proportionality Theorem]
 
In ΔOPR, AC || PR (Given)
∴ OA/AP = OC/CR …(ii) [By using Basic Proportionality Theorem]
 
From equation (i) and (ii), we get
OB/BQ = OC/CR
 
In ΔOQR, BC || QR. [By converse of Basic Proportionality Theorem].
 

7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Answer

Triangles Exercise 6.2 Answer 7
Given: ΔABC in which D is the mid point of AB such that AD=DB.
A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.
 
To Prove: E is the mid point of AC.
 
Proof: D is the mid-point of AB.
∴ AD=DB

⇒ AD/BD = 1 … (i)

In ΔABC, DE || BC,

Therefore, AD/DB = AE/EC [By using Basic Proportionality Theorem]

⇒1 = AE/EC [From equation (i)]
∴ AE =EC
Hence, E is the mid point of AC.

8. Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Answer

Triangles Exercise 6.2 Answer 8
 

Given: ΔABC in which D and E are the mid points of AB and AC respectively such that AD=BD and AE=EC.

To Prove: DE || BC

Proof: D is the mid point of AB (Given)

∴ AD=DB

⇒ AD/BD = 1 … (i)

Also, E is the mid-point of AC (Given)
∴ AE=EC

⇒AE/EC = 1 [From equation (i)]

From equation (i) and (ii), we get
AD/BD = AE/EC
Hence, DE || BC [By converse of Basic Proportionality Theorem]

9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

Answer

Triangles Exercise 6.2 Answer 9


Given: ABCD is a trapezium in which AB || DC in which diagonals AC and BD intersect each other at O.

To Prove: AO/BO = CO/DO

Construction: Through O, draw EO || DC || AB

Proof: In ΔADC, we have
OE || DC (By Construction)

∴ AE/ED = AO/CO  …(i) [By using Basic Proportionality Theorem]

In ΔABD, we have
OE || AB (By Construction)

∴ DE/EA = DO/BO …(ii) [By using Basic Proportionality Theorem]

From equation (i) and (ii), we get
AO/CO = BO/DO
⇒  AO/BO = CO/DO

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

Answer

Triangles Exercise 6.2 Answer 10
 
Given: Quadrilateral ABCD in which diagonals AC and BD intersects each other at O such that AO/BO = CO/DO.
 
To Prove: ABCD is a trapezium
 
Construction: Through O, draw line EO, where EO || AB, which meets AD at E.
 
Proof: In ΔDAB, we have
EO || AB
∴ DE/EA = DO/OB …(i) [By using Basic Proportionality Theorem]
Also,  AO/BO = CO/DO (Given)
⇒ AO/CO = BO/DO
⇒ CO/AO = BO/DO
⇒ DO/OB = CO/AO …(ii) 
 
From equation (i) and (ii), we get
DE/EA = CO/AO
Therefore, By using converse of Basic Proportionality Theorem, EO || DC also EO || AB
⇒ AB || DC.
Hence, quadrilateral ABCD is a trapezium with AB || CD.
Exercise 6.3
 

1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

Triangles Exercise 6.3 Question No. 1
 
Answer
 
(i) In  ΔABC and ΔPQR, we have
∠A = ∠P = 60° (Given)

∠B = ∠Q = 80° (Given)
∠C = ∠R = 40° (Given)
∴ ΔABC ~ ΔPQR (AAA similarity criterion)

(ii) In  ΔABC and ΔPQR, we have
AB/QR = BC/RP = CA/PQ
∴  ΔABC ~ ΔQRP (SSS similarity criterion)

(iii) In ΔLMP and ΔDEF, we have
LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
MP/DE = 2/4 = 1/2
PL/DF = 3/6 = 1/2
LM/EF= 2.7/5 = 27/50
Here, MP/DE = PL/DF ≠ LM/EF
Hence, ΔLMP and ΔDEF are not similar.

(iv) In ΔMNL and ΔQPR, we have
MN/QP = ML/QR = 1/2
∠M = ∠Q = 70°
∴ ΔMNL ~ ΔQPR (SAS similarity criterion)

(v) In ΔABC and ΔDEF, we have
AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°
Here, AB/DF = 2.5/5 = 1/2
And, BC/EF = 3/6 = 1/2
⇒ ∠B ≠ ∠F
Hence, ΔABC and ΔDEF are not similar.

(vi) In ΔDEF,we have
∠D + ∠E + ∠F = 180° (sum of angles of a triangle)
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 180° – 70° – 80°
⇒ ∠F = 30°
In PQR, we have
∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 180° – 80° -30°
⇒ ∠P = 70°
In ΔDEF and ΔPQR, we have
∠D = ∠P = 70°
∠F = ∠Q = 80°
∠F = ∠R = 30°
Hence, ΔDEF ~ ΔPQR (AAA similarity criterion)

Page No: 139

2.  In the fig 6.35, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

Triangles Exercise 6.3 Question No. 2

Answer

DOB is a straight line.
Therefore, ∠DOC + ∠ COB = 180°
⇒ ∠DOC = 180° – 125°
= 55°

In ΔDOC,
∠DCO + ∠ CDO + ∠ DOC = 180°
(Sum of the measures of the angles of a triangle is 180º.)
⇒ ∠DCO + 70º + 55º = 180°
⇒ ∠DCO = 55°
It is given that ΔODC ~ ΔOBA.

∴ ∠OAB = ∠OCD [Corresponding angles are equal in similar triangles.]
⇒ ∠ OAB = 55°
∴ ∠OAB = ∠OCD [Corresponding angles are equal in similar triangles.]

⇒ ∠OAB = 55°


3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD

Answer

Triangles Exercise 6.3 Answer 3

In ΔDOC and ΔBOA,
∠CDO = ∠ABO [Alternate interior angles as AB || CD]
∠DCO = ∠BAO [Alternate interior angles as AB || CD]
∠DOC = ∠BOA [Vertically opposite angles]
∴ ΔDOC ~ ΔBOA [AAA similarity criterion]

∴ DO/BO = OC/OA  [ Corresponding sides are proportional]
⇒ OA/OC = OB/OD
 

Page No: 140

4. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
Triangles Exercise 6.3 Question No. 4
 

Answer

In ΔPQR, ∠PQR = ∠PRQ
∴ PQ = PR …(i)
Given,QR/QS = QT/PR
Using (i), we get
QR/QS = QT/QP …(ii)
In ΔPQS and ΔTQR,
QR/QS = QT/QP [using (ii)]
∠Q = ∠Q
∴ ΔPQS ~ ΔTQR [SAS similarity criterion]

5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.

Answer

Triangles Exercise 6.3 Answer 5

In ΔRPQ and ΔRST,
∠RTS = ∠QPS (Given)
∠R = ∠R (Common angle)
∴ ΔRPQ ~ ΔRTS (By AA similarity criterion)

 
6. In the fig 6.37, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
Triangles Exercise 6.3 Question No. 6
 
Answer
 

It is given that ΔABE ≅ ΔACD.
∴ AB = AC [By cpct] …(i)
And, AD = AE [By cpct] …(ii)
In ΔADE and ΔABC,

AD/AB = AE/AC [Dividing equation (ii) by (i)]

∠A = ∠A [Common angle]
∴ ΔADE ~ ΔABC [By SAS similarity criterion]

 
7. In the fig 6.38, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
Triangles Exercise 6.3 Question No. 7

(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC


Answer


(i) In ΔAEP and ΔCDP,
∠AEP = ∠CDP (Each 90°)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔCDP

(ii) In ΔABD and ΔCBE,
∠ADB = ∠CEB (Each 90°)

∠ABD = ∠CBE (Common)
Hence, by using AA similarity criterion,
ΔABD ~ ΔCBE


(iii) In ΔAEP and ΔADB,
∠AEP = ∠ADB (Each 90°)

∠PAE = ∠DAB (Common)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔADB



(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC (Each 90°)
∠PCD = ∠BCE (Common angle)
Hence, by using AA similarity criterion,
ΔPDC ~ ΔBEC

8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

Answer

Triangles Exercise 6.3 Answer 8

In ΔABE and ΔCFB,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ΔABE ~ ΔCFB (By AA similarity criterion)

 
9. In the fig 6.39, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:
Triangles Exercise 6.3 Question No. 9
(i) ΔABC ~ ΔAMP
(ii) CA/PA = BC/MP
 
Answer
 
(i) In ΔABC and ΔAMP, we have
∠A = ∠A (common angle)
∠ABC = ∠AMP = 90° (each 90°)
∴  ΔABC ~ ΔAMP (By AA similarity criterion)
 
(ii) As, ΔABC ~ ΔAMP (By AA similarity criterion)
If two triangles are similar then the corresponding sides are equal,
Hence, CA/PA = BC/MP
 
10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:

(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF

Answer

Triangles Exercise 6.3 Answer 10
 

(i) It is given that ΔABC ~ ΔFEG.
∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE
∠ACB = ∠FGE
∴ ∠ACD = ∠FGH (Angle bisector)
And, ∠DCB = ∠HGE (Angle bisector)
In ΔACD and ΔFGH,
∠A = ∠F (Proved above)
∠ACD = ∠FGH (Proved above)
∴ ΔACD ~ ΔFGH (By AA similarity criterion)

⇒ CD/GH = AC/FG
 
(ii) In ΔDCB and ΔHGE,
∠DCB = ∠HGE (Proved above)
∠B = ∠E (Proved above)
∴ ΔDCB ~ ΔHGE (By AA similarity criterion)
(iii) In ΔDCA and ΔHGF,
∠ACD = ∠FGH (Proved above)
∠A = ∠F (Proved above)
∴ ΔDCA ~ ΔHGF (By AA similarity criterion)


Page No: 141

11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

Triangles Exercise 6.3 Question No. 11


Answer

It is given that ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ABD = ∠ECF
In ΔABD and ΔECF,
∠ADB = ∠EFC (Each 90°)
∠BAD = ∠CEF (Proved above)
∴ ΔABD ~ ΔECF (By using AA similarity criterion)

 
12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.
Triangles Exercise 6.3 Question No. 12
Answer
 
Given: ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR
i.e., AB/PQ = BC/QR = AD/PM
 
To Prove: ΔABC ~ ΔPQR
 
Proof: AB/PQ = BC/QR = AD/PM
⇒ AB/PQ = BC/QR = AD/PM (D is the mid-point of BC. M is the mid point of QR)
⇒ ΔABD ~ ΔPQM [SSS similarity criterion]
∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]
⇒ ∠ABC = ∠PQR
In ΔABC and ΔPQR
AB/PQ = BC/QR …(i)
∠ABC = ∠PQR …(ii)
From equation (i) and (ii), we get
ΔABC ~ ΔPQR [By SAS similarity criterion]
 

13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD

Answer

Triangles Exercise 6.3 Answer 13

In ΔADC and ΔBAC,
∠ADC = ∠BAC (Given)
∠ACD = ∠BCA (Common angle)
∴ ΔADC ~ ΔBAC (By AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.

∴ CA/CB =CD/CA
⇒ CA2 = CB.CD.
 

14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.

Answer

Triangles Exercise 6.3 Answer 14
Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that AB/PQ = AC/PR = AD/PM
 
To Prove: ΔABC ~ ΔPQR
 
Construction: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.
 
Proof: In ΔABD and ΔCDE, we have
AD = DE  [By Construction]
BD = DC [∴ AP is the median]
and, ∠ADB = ∠CDE [Vertically opp. angles]
∴ ΔABD ≅ ΔCDE [By SAS criterion of congruence]
⇒ AB = CE [CPCT] …(i)
Also, in ΔPQM and ΔMNR, we have
PM = MN [By Construction]
QM = MR [∴ PM is the median]
and, ∠PMQ = ∠NMR [Vertically opposite angles]
∴ ΔPQM = ΔMNR [By SAS criterion of congruence]
⇒ PQ = RN [CPCT] …(ii)
Now, AB/PQ = AC/PR = AD/PM
⇒ CE/RN = AC/PR = AD/PM …[From (i) and (ii)]
⇒ CE/RN = AC/PR = 2AD/2PM
⇒ CE/RN = AC/PR = AE/PN [∴ 2AD = AE and 2PM = PN]
∴ ΔACE ~ ΔPRN [By SSS similarity criterion]
Therefore, ∠2 = ∠4
Similarly, ∠1 = ∠3
∴ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠A = ∠P …(iii)
Now, In ΔABC and ΔPQR, we have
AB/PQ = AC/PR (Given)
∠A = ∠P [From (iii)]
∴ ΔABC ~ ΔPQR [By SAS similarity criterion]
 

15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Answer

Triangles Exercise 6.3 Answer 14
Length of the vertical pole = 6m (Given)
Shadow of the pole = 4 m (Given)
Let Height of tower = h m
Length of shadow of the tower = 28 m (Given)
In ΔABC and ΔDEF,
∠C = ∠E (angular elevation of sum)
∠B = ∠F = 90°
∴ ΔABC ~ ΔDEF (By AA similarity criterion)
∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)
∴ 6/h = 4/28
⇒ h = 6×28/4
⇒ h = 6 × 7
⇒ h = 42 m
Hence, the height of the tower is 42 m.
 

16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.

 
Answer
 
 
Triangles Exercise 6.3 Answer 16

It is given that ΔABC ~ ΔPQR
We know that the corresponding sides of similar triangles are in proportion.∴ AB/PQ = AC/PR = BC/QR …(i)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R …(ii)
Since AD and PM are medians, they will divide their opposite sides.∴ BD = BC/2 and QM = QR/2 …(iii)
From equations (i) and (iii), we get
AB/PQ = BD/QM …(iv)
In ΔABD and ΔPQM,
∠B = ∠Q [Using equation (ii)]
AB/PQ = BD/QM [Using equation (iv)]
∴ ΔABD ~ ΔPQM (By SAS similarity criterion)⇒ AB/PQ = BD/QM = AD/PM.

Exercise 6.4

1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

Answer

It is given that,
Area of ΔABC = 64 cm2
Area of ΔDEF = 121 cm2
EF = 15.4 cm
and, ΔABC ~ ΔDEF
∴ Area of ΔABC/Area of ΔDEF = AB2/DE2
= AC2/DF2 = BC2/EF2 …(i)
[If two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides]
∴ 64/121 = BC2/EF2
⇒ (8/11)2 = (BC/15.4)2
⇒ 8/11 = BC/15.4
⇒ BC = 8×15.4/11
⇒ BC = 8 × 1.4
⇒ BC = 11.2 cm

2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Answer

Triangles Exercise 6.4 Answer 2
ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.
In ΔAOB and ΔCOD, we have
∠1 = ∠2 (Alternate angles)
∠3 = ∠4 (Alternate angles)
∠5 = ∠6 (Vertically opposite angle)
∴ ΔAOB ~ ΔCOD [By AAA similarity criterion]
Now, Area of (ΔAOB)/Area of (ΔCOD)
= AB2/CD2 [If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides]
= (2CD)2/CD2 [∴ AB = CD]
∴ Area of (ΔAOB)/Area of (ΔCOD)
= 4CD2/CD = 4/1
Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1

3. In the fig 6.53, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.
Triangles Exercise 6.4 Question No. 3

Answer

Triangles Exercise 6.4 Answer 3
Given: ABC and DBC are triangles on the same base BC. Ad intersects BC at O.
 
To Prove: area (ΔABC)/area (ΔDBC) = AO/DO.
 
Construction: Let us draw two perpendiculars AP and DM on line BC.
 
Proof: We know that area of a triangle = 1/2 × Base × Height

In ΔAPO and ΔDMO,
∠APO = ∠DMO (Each equals to 90°)
∠AOP = ∠DOM (Vertically opposite angles)
∴ ΔAPO ~ ΔDMO (By AA similarity criterion)∴ AP/DM = AO/DO
⇒ area (ΔABC)/area (ΔDBC) = AO/DO.

4. If the areas of two similar triangles are equal, prove that they are congruent.

Answer

Triangles Exercise 6.4 Answer 4

Given: ΔABC and ΔPQR are similar and equal in area.

To Prove: ΔABC ≅ ΔPQR

Proof: Since, ΔABC ~ ΔPQR
∴ Area of (ΔABC)/Area of (ΔPQR) = BC2/QR2
⇒ BC2/QR2 =1 [Since, Area(ΔABC) = (ΔPQR)
⇒ BC2/QR2
⇒ BC = QR
Similarly, we can prove that
AB = PQ and AC = PR
Thus, ΔABC ≅ ΔPQR [BY SSS criterion of congruence]

5. D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.
 
Answer
Triangles Exercise 6.4 Answer 5
Given: D, E and F are the mid-points of the sides AB, BC and CA respectively of the ΔABC.

To Find: area(ΔDEF) and area(ΔABC)

Solution: In ΔABC, we have
F is the mid point of AB (Given)
E is the mid-point of AC (Given)
So, by the mid-point theorem, we have
FE || BC and FE = 1/2BC
⇒ FE || BC and FE || BD [BD = 1/2BC]
∴ BDEF is parallelogram [Opposite sides of parallelogram are equal and parallel]
Similarly in ΔFBD and ΔDEF, we have
FB = DE (Opposite sides of parallelogram BDEF)
FD = FD (Common)
BD = FE (Opposite sides of parallelogram BDEF)
∴ ΔFBD ≅ ΔDEF
Similarly, we can prove that
ΔAFE ≅ ΔDEF
ΔEDC ≅ ΔDEF
If triangles are congruent,then they are equal in area.
So, area(ΔFBD) = area(ΔDEF) …(i)
area(ΔAFE) = area(ΔDEF) …(ii)
and, area(ΔEDC) = area(ΔDEF) …(iii)
Now, area(ΔABC) = area(ΔFBD) + area(ΔDEF) + area(ΔAFE) + area(ΔEDC) …(iv)
area(ΔABC) = area(ΔDEF) + area(ΔDEF) + area(ΔDEF) + area(ΔDEF)
⇒ area(ΔDEF) = 1/4area(ΔABC) [From (i), (ii) and (iii)]
⇒ area(ΔDEF)/area(ΔABC) = 1/4
Hence, area(ΔDEF):area(ΔABC) = 1:4

6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Answer
Triangles Exercise 6.4 Answer 6
Given: AM and DN are the medians of triangles ABC and DEF respectively and ΔABC ~ ΔDEF.
 
To Prove: area(ΔABC)/area(ΔDEF) = AM2/DN2

Proof: ΔABC ~ ΔDEF (Given)
∴ area(ΔABC)/area(ΔDEF) = (AB2/DE2) …(i)
and, AB/DE = BC/EF = CA/FD …(ii)

In ΔABM and ΔDEN, we have
∠B = ∠E [Since ΔABC ~ ΔDEF]
AB/DE = BM/EN [Prove in (i)]
∴ ΔABC ~ ΔDEF [By SAS similarity criterion]
⇒ AB/DE = AM/DN …(iii)
∴ ΔABM ~ ΔDEN
As the areas of two similar triangles are proportional to the squares of the corresponding sides.
∴ area(ΔABC)/area(ΔDEF) = AB2/DE2 = AM2/DN2

7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Answer
Triangles Exercise 6.4 Answer 7

Given: ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

To Prove: area(ΔBQC) = 1/2area(ΔAPC)

Proof: ΔAPC and ΔBQC are both equilateral triangles (Given)
∴ ΔAPC ~ ΔBQC [AAA similarity criterion]
∴ area(ΔAPC)/area(ΔBQC) = (AC2/BC2) = AC2/BC2
⇒ area(ΔAPC) = 2 × area(ΔBQC)
⇒ area(ΔBQC) = 1/2area(ΔAPC)

Tick the correct answer and justify:

8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4

Answer

Triangles Exercise 6.4 Answer 8
ΔABC and ΔBDE are two equilateral triangle. D is the mid point of BC.
∴ BD = DC = 1/2BC
Let each side of triangle is 2a.
As, ΔABC ~ ΔBDE
∴ area(ΔABC)/area(ΔBDE) = AB2/BD2 = (2a)2/(a)2 = 4a2/a2 = 4/1 = 4:1
Hence, the correct option is (C).
 
9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81

Answer
Triangles Exercise 6.4 Answer 9
Let ABC and DEF are two similarity triangles ΔABC ~ ΔDEF (Given)
and,  AB/DE = AC/DF = BC/EF = 4/9 (Given)
∴ area(ΔABC)/area(ΔDEF) = AB2/DE2 [the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides]
∴ area(ΔABC)/area(ΔDEF) = (4/9)2 = 16/81 = 16:81
Hence, the correct option is (D).
Exercise 6.5
 

1.  Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm

Answer

(i) Given that the sides of the triangle are 7 cm, 24 cm, and 25 cm.
Squaring the lengths of these sides, we will get 49, 576, and 625.
49 + 576 = 625
(7)2 + (24)2 = (25)2
The sides of the given triangle are satisfying Pythagoras theorem.Hence, it is right angled triangle.
Length of Hypotenuse = 25 cm

(ii) Given that the sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will get 9, 64, and 36.
However, 9 + 36 ≠ 64
Or, 32 + 62 ≠ 82
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.

(iii) Given that sides are 50 cm, 80 cm, and 100 cm.
Squaring the lengths of these sides, we will get 2500, 6400, and 10000.
However, 2500 + 6400 ≠ 10000
Or, 502 + 802 ≠ 1002
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.
Hence, it is not a right triangle.

(iv) Given that sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will get 169, 144, and 25.
Clearly, 144 +25 = 169

Or, 122 + 52 = 132
The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
Length of the hypotenuse of this triangle is 13 cm.


2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM × MR.

Answer

Given: ΔPQR is right angled at P is a point on QR such that PM ⊥QR.
 
To prove: PM2 = QM × MR
 
Proof: In ΔPQM, we have
PQ2 = PM2 + QM2 [By Pythagoras theorem]
Or, PM2 = PQ2 – QM2 …(i)
In ΔPMR, we have
PR2 = PM2 + MR2 [By Pythagoras theorem]
Or, PM2 = PR2 – MR2 …(ii)
Adding (i) and (ii), we get
2PM2 = (PQ2 + PM2) – (QM2 + MR2)
= QR2 – QM2 – MR2        [∴ QR2 = PQ2 + PR2]
= (QM + MR)2 – QM2 – MR2
= 2QM × MR
∴ PM2 = QM × MR
 

3. In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB2 = BC × BD
(ii) AC2 = BC × DC
(iii) AD2 = BD × CD

Triangles Exercise 6.5 Question No. 3


Answer

(i) In ΔADB and ΔCAB, we have
∠DAB = ∠ACB (Each equals to 90°)
∠ABD = ∠CBA (Common angle)
∴ ΔADB ~ ΔCAB [AA similarity criterion]
⇒ AB/CB = BD/AB
⇒ AB2 = CB × BD

(ii) Let ∠CAB = x
In ΔCBA,
∠CBA = 180° – 90° – x
∠CBA = 90° – x
Similarly, in ΔCAD
∠CAD = 90° – ∠CBA = 90° – x
∠CDA = 180° – 90° – (90° – x)
∠CDA = x
In ΔCBA and ΔCAD, we have
∠CBA = ∠CAD
∠CAB = ∠CDA
∠ACB = ∠DCA (Each equals to 90°)
∴ ΔCBA ~ ΔCAD [By AAA similarity criterion]
⇒ AC/DC = BC/AC
⇒ AC2 =  DC × BC

(iii) In ΔDCA and ΔDAB, we have
∠DCA = ∠DAB (Each equals to 90°)
∠CDA = ∠ADB (common angle)
∴ ΔDCA ~ ΔDAB [By AA similarity criterion]
⇒ DC/DA = DA/DA
⇒ AD2 = BD × CD

4. ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2 .

Answer

Triangles Exercise 6.5 Answer 4
Given that ΔABC is an isosceles triangle right angled at C.
In ΔACB, ∠C = 90°
AC = BC (Given)
AB2 = AC2 + BC2 ([By using Pythagoras theorem]
= AC2 + AC2 [Since, AC = BC]
AB2 = 2AC2
 

5. ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.

Answer

Triangles Exercise 6.5 Answer 5
Given that ΔABC is an isosceles triangle having AC = BC and AB2 = 2AC2
In ΔACB,
AC = BC (Given)
AB2 = 2AC2 (Given)
AB2 = AC2 + AC2
= AC2 + BC2 [Since, AC = BC]
Hence, By Pythagoras theorem ΔABC is right angle triangle.
 
6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.
 
Answer
Triangles Exercise 6.5 Answer 6
ABC is an equilateral triangle of side 2a. 
Draw, AD ⊥ BC
In ΔADB and ΔADC, we have
AB = AC [Given]
AD = AD [Given]
∠ADB = ∠ADC [equal to 90°]
Therefore, ΔADB ≅ ΔADC by RHS congruence.
Hence, BD = DC [by CPCT]
In right angled ΔADB, 
AB2 = AD2 + BD2
(2a)2 = AD2 + a2 
⇒ AD2  = 4a2 – a2
⇒ AD2  = 3a2 
⇒ AD  = √3a 
 
7. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.
 
Answer
Triangles Exercise 6.5 Answer 7
ABCD is a rhombus whose diagonals AC and BD intersect at O. [Given]
We have to prove that, 
AB2 + BC2 + CD2 + AD2 = AC2 + BD2

Since, the diagonals of a rhombus bisect each other at right angles.

Therefore, AO = CO and BO = DO
In ΔAOB,
∠AOB = 90°
AB2 = AO2 + BO2 … (i) [By Pythagoras]
Similarly, 
AD2 = AO2 + DO2 … (ii)
DC2 = DO2 + CO2 … (iii)
BC2 = CO2 + BO2 … (iv)
Adding equations (i) + (ii) + (iii) + (iv)  we get,
AB2 + AD2 + DC2 + BC2  =  2(AO2 + BO2 + DO2 + CO2 )

= 4AO2 + 4BO2 [Since, AO = CO and BO =DO]
= (2AO)2 + (2BO)2 = AC2 + BD2

Page No: 151

8. In Fig. 6.54, O is a point in the interior of a triangle

Triangles Exercise 6.5 Question No. 8

ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 ,
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.

Answer

Join OA, OB and OC

Triangles Exercise 6.5 Answer 8
(i) Applying Pythagoras theorem in ΔAOF, we have
OA2 = OF2 + AF2

Similarly, in ΔBOD
OB2 = OD2 + BD2
Similarly, in ΔCOE
OC2 = OE2 + EC2
Adding these equations,
OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE2 + EC2
OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2.

(ii) AF2 + BD2 + EC2 = (OA2 – OE2) + (OC2 – OD2) + (OB2 – OF2)
∴ AF2 + BD2 + CE2 = AE2 + CD2 + BF2.

9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Answer

Triangles Exercise 6.5 Answer 9
Let BA be the wall and Ac be the ladder,
Therefore, by Pythagoras theorem,we have
AC2 = AB2 + BC2
102 = 82 + BC2
BC2 = 100 – 64
BC2 = 36
BC = 6m

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.

10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?

Answer

Triangles Exercise 6.5 Answer 10

Let AB be the pole and AC be the wire.
By Pythagoras theorem,

AC2 = AB2 + BC2
242 = 182 + BC2
BC2 = 576 – 324
BC2 = 252
BC = 6√7m
Therefore, the distance from the base is 6√7m.


11. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after hours?

Answer

Triangles Exercise 6.5 Answer 11
Speed of first aeroplane = 1000 km/hr
Distance covered by first aeroplane due north in  hours (OA) = 100 × 3/2 km = 1500 km
Speed of second aeroplane = 1200 km/hr
Distance covered by second aeroplane due west in hours (OB) = 1200 × 3/2 km = 1800 km
In right angle ΔAOB, we have
AB2 = AO2 + OB2
⇒ AB2 = (1500)2 + (1800)2
⇒ AB = √2250000 + 3240000
= √5490000
⇒ AB = 300√61 km
Hence, the distance between two aeroplanes will be 300√61 km.
 

12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Answer

Triangles Exercise 6.5 Answer 12

 

 

Let CD and AB be the poles of height 11 m and 6 m.
Therefore, CP = 11 – 6 = 5 m
From the figure, it can be observed that AP = 12m
Applying Pythagoras theorem for ΔAPC, we get

AP2 = PC2 + AC2
(12m)2 + (5m)2 = (AC)2
AC2 = (144+25)m2 = 169 m2
AC = 13m
Therefore, the distance between their tops is 13 m.


13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.

Answer

Triangles Exercise 6.5 Answer 13
Applying Pythagoras theorem in ΔACE, we get
AC2 + CE2 = AE2 ….(i)
Applying Pythagoras theorem in ΔBCD, we get
BC2 + CD2 = BD2 ….(ii)
Using equations (i) and (ii), we get
AC2 + CE2 + BC2 + CD2 = AE2 + BD2 …(iii)
Applying Pythagoras theorem in ΔCDE, we get
DE2 = CD2 + CE2
Applying Pythagoras theorem in ΔABC, we get
AB2 = AC2 + CB2

Putting these values in equation (iii), we get
DE2 + AB2 = AE2 + BD2.

14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Fig. 6.55). Prove that 2AB2 = 2AC2 + BC2.

Triangles Exercise 6.5 Question No. 14
Answer
 
Given that in ΔABC, we have
AD ⊥BC and BD = 3CD
In right angle triangles ADB and ADC, we have
AB2 = AD2 + BD2 …(i)
AC2 = AD2 + DC2 …(ii) [By Pythagoras theorem]
Subtracting equation (ii) from equation (i), we get
AB2 – AC2 = BD2 – DC2

= 9CD2 – CD2  [∴ BD = 3CD]               
= 9CD2 = 8(BC/4)2 [Since, BC = DB + CD = 3CD + CD = 4CD]
Therefore, AB2 – AC2 = BC2/2
⇒ 2(AB2 – AC2) = BC2
⇒ 2AB2 – 2AC2 = BC2
∴ 2AB2 = 2AC2 + BC2.

15.  In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 = 7AB2.

Answer

Triangles Exercise 6.5 Answer 15

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2
And, AE = a√3/2
Given that, BD = 1/3BC

∴ BD = a/3

DE = BE – BD = a/2 – a/3 = a/6

Applying Pythagoras theorem in ΔADE, we get
AD2 = AE2 + DE2 

⇒ 9 AD2 = 7 AB2


16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Answer

Triangles Exercise 6.5 Answer 16

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
Applying Pythagoras theorem in ΔABE, we get
AB2 = AE2 + BE2

4AE2 = 3a2
⇒ 4 × (Square of altitude) = 3 × (Square of one side)


17. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120°

(B) 60°
(C) 90° 
(D) 45°


Answer

Triangles Exercise 6.5 Answer 17

Given that, AB = 6√3 cm, AC = 12 cm, and BC = 6 cm
We can observe that
AB2 = 108
AC2 = 144
And, BC2 = 36
AB2 + BC2 = AC2
The given triangle, ΔABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
∴ ∠B = 90°
Hence, the correct option is (C).

Exercise 6.6

1. In Fig. 6.56, PS is the bisector of ∠QPR of ΔPQR. Prove that QS/SR = PQ/PR

 
Answer
 
Given, in figure, PS is the bisector of ∠QPR of ∆PQR. 
Now, draw RT SP || to meet QP produced in T.
 
Proof:
 
∵ RT SP || and transversal PR intersects them
∴ ∠1 = ∠2 (Alternate interior angle)…(i)
∴ RT SP || and transversalQT intersects them
∴ ∠3 = ∠4 (Corresponding angle) …(ii)
But ∠1 = ∠3 (Given)
∴ ∠2 = ∠4 [From Eqs. (i) and (ii)]
∴ PT = PR …(iii) (∵ Sides opposite to equal angles of a triangle are equal)
Now, in ∆QRT,
PS || RT (By construction)
∴ QS/SR = PQ/PT (By basic proportionally theorem)
⇒ QS/SR = PQ/PR [From Eq. (iii)]


2. In Fig. 6.57, D is a point on hypotenuse AC of ΔABC, such that BD ⊥ AC, DM ⊥ BC and
DN ⊥ AB. Prove that :
(i) DM2 = DN.MC
(ii) DN2 = DM.AN

Answer

Given that, D is a point on hypotenuse AC of ∆ABC, DM ⊥ BC and DN ⊥ AB.
Now, join NM.
Let BD and NM intersect at O.

(i) In ∆DMC and ∆NDM,
∠DMC = ∠NDM (Each equal to 90°)
∠MCD = ∠DMN
Let MCD = ∠1
Then, ∠MDC = 90° − (90°-∠1)
= ∠1 (∵∠MCD + ∠MDC + ∠DMC = 180°) 
∴ ∠ODM = 90° − (90° − ∠1)
= ∠1
⇒ ∠DMN = ∠1
∴ ∆DMO ~ ∆NDM  (AA similarity criterion)
∴ DM/ND = MC/DM
(Corresponding sides of the similar triangles are proportional)
⇒ DM2 = MC ND
 
(ii) In ∆DNM and ∆NAD,
∠NDM = ∠AND (Each equal to 90°)
∠DNM = ∠NAD
Let ∠NAD = ∠2
Then, ∠NDA = 90° − ∠2
∵∠NDA + ∠DAN + ∠DNA = 180°
∴ ∠ODN = 90° − (90° − ∠2) = ∠2
∴ ∠DNO = ∠2
∴ ∆DNM ~ ∆NAD  (AA similarity criterion)
∴ DN/NA = DM/ND
⇒ DN/AN = DM/DN
⇒ DN2 = DM×AN


3. In Fig. 6.58, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2 BC.BD.

 
Answer
 
Given that, in figure, ABC is a triangle in which ∠ABC> 90° and AD ⊥ CB produced. 
 
Proof :
 
In right ADC, 
∠D = 90° 
AC2 = AD2 + DC2  (By Pythagoras theorem) 
= AD2 + (BD + BC)2 [∵DC = DB + BC] 
= (AD2 + DB2) + BC2 + 2BD.BC [∵ (a + b)2 = a2 + b2 + 2ab] 
= AB2 + BC2 + 2BC.BD 
 
[∵In right ADB with ∠D = 90°, AB2 = AD2 + DB2] (By Pythagoras theorem) 


4. In Fig. 6.59, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 – 2BC.BD.

 

Answer

Given that, in figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC.

Proof:

In right △ADC,
∠D = 90°

AC2 = AD2 + DC2  (By Pythagoras theorem) 

= AD2 + (BC – BD)2  [∵BC = BD + DC]
= AD2 + (BC – BD)2  (BC = BD + DC)
= AD2 + BC2 + BD2 – 2BC.BD [∵ (a + b)2 = a2 + b2 + 2ab]
= (AD2 + BD2) + BC2 – 2BC . BD
= AB2 + BC2 – 2BC . BD
{In right △ADB with ∠D = 90°, AB2 = AD2 + BD2} (By Pythagoras theorem) 

5. In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :
(i) AC2 = AD2 + BC.DM + (BC/2)2
(ii) AB2 = AD2 – BC.DM + (BC/2)2
(iii) AC2 + AB2 = 2 AD2 + 1/2 BC2


Answer

Given that, in figure, AD is a median of a ∆ABC and AM ⊥ BC.
Proof:
(i) In right ∆AMC,

 
6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.


Answer

Given that, ABCD is a parallelogram whose diagonals are AC and BD.

Now, draw AM⊥DC and BN⊥D  (produced).
Proof:
In right ∆AMD and ∆BNC,

AD = BC (Opposite sides of a parallelogram)
AM = BN (Both are altitudes of the same parallelogram to the same base) ,
△AMD ⩭ △BNC (RHS congruence criterion)
MD = NC (CPCT) —(i)

In right △BND,
∠N = 90°
BD2 = BN2 + DN2 (By Pythagoras theorem)
= BN2 + (DC + CN)2 (∵ DN = DC + CN) 
= BN2 + DC2 + CN2 + 2DC.CN [∵ (a + b)2 = a2 + b2 + 2ab]
= (BN2 + CN2) + DC2 + 2DC.CN
= BC2 + DC2 + 2DC.CN — (ii) (∵In right △BNC with ∠N = 90°)
BN2 + CN2 = BC2  (By Pythagoras theorem)

In right △AMC,
∠M = 90°
AC2 = AM2 + MC2 (∵MC = DC – DM)
= AM2 + (DC – DM)2 [∵ (a + b)2 = a2 + b2 + 2ab]
= AM2 + DC2 + DM2 – 2DC.DM
(AM2 + DM2) + DC2 – 2DC.DM
= AD2 + DC2 – 2DC.DM
[∵ In  right triangle AMD with ∠M = 90°, AD2 = AM2 + DM2 (By Pythagoras theorem)]
= AD2 + AB2 = 2DC.CN  — (iii)
[∵ DC = AB, opposite sides of parallelogram and BM = CN from eq (i)]
Now, on adding Eqs. (iii) and (ii), we get
AC2 + BD2 = (AD2 + AB2) + (BC2 + DC2)
= AB2 + BC2 + CD2 + DA2

7. In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that :
(i) Δ APC ~ Δ DPB 
(ii) AP . PB = CP . DP

Answer

Given that, in figure, two chords AB and CD intersects each other at the point P.

Proof:

(i) ∆APC and ∆DPB
∠APC = ∠DPB (Vertically opposite angles)
∠CAP = ∠BDP (Angles in the same segment)
∴ ∆APC ~ ∆DPB (AA similarity criterion)

(ii) ∆APC ~ ∆DPB [Proved in (i)]
∴ AP/DP = CP/BP
(∴ Corresponding sides of two similar triangles are proportional)
⇒ AP.BP = CP.DP
⇒ AP.PB = CP.DP

 

8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that
(i) Δ PAC ~ Δ PDB
(ii) PA.PB = PC.PD


Answer

Given that, in figure, two chords AB and CD of a circle intersect each other at the point P (when produced) out the circle.

Proof:

(i) We know that, in a cyclic quadrilaterals, the exterior angle is equal to the interior opposite angle.
Therefore, ∠PAC = ∠PDB …(i)
and ∠PCA = ∠PBD …(ii)
In view of Eqs. (i) and (ii), we get
∆PAC ~ ∆PDB  (∵ AA similarity criterion)

(ii) ∆PAC ~ ∆PDB  [Proved in (i)]
∴ PA/PD = PC/PB
(∵ Corresponding sides of the similar triangles are proportional)
⇒ PA.PB = PC.PD

9. In Fig. 6.63, D is a point on side BC of ΔABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠BAC.

Answer
 
Given that,D is a point on side BC of ∆ABC such that BD/CD = AB/AC
Now, from BA produce cut off AE = A. JoinCE.


10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?


Answer

Length of the string that she has out

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions | EduGrown

In This Post we are  providing Chapter 5 Arithmetic Progressions NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Arithmetic Progressions Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Arithmetic Progressions  NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Exercise 5.1

1. In which of the following situations, does the list of numbers involved make as arithmetic progression and why?

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

Answer

It can be observed that
Taxi fare for 1st km = 15
Taxi fare for first 2 km = 15 + 8 = 23
Taxi fare for first 3 km = 23 + 8 = 31
Taxi fare for first 4 km = 31 + 8 = 39

Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Answer

Let the initial volume of air in a cylinder be V litres. In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time. In other words, after every stroke, only 1 – 1/4 = 3/4th part of air will remain.
Therefore, volumes will be V, 3V/4 , (3V/4)2 , (3V/4)3…
Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.

Answer

Cost of digging for first metre = 150
Cost of digging for first 2 metres = 150 + 50 = 200
Cost of digging for first 3 metres = 200 + 50 = 250
Cost of digging for first 4 metres = 250 + 50 = 300
Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.

(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Answer

We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

 

2. Write first four terms of the A.P. when the first term a and the common differenced are given as follows
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = – 3
(iv) a = -1 d = 1/2
(v) a = – 1.25, d = – 0.25

Answer

(i) a = 10, d = 10
Let the series be a1, a2, a3, a4, a5 …
a1 = a = 10
a2 = a1 + d = 10 + 10 = 20
a3 = a2 + d = 20 + 10 = 30
a4 = a3 + d = 30 + 10 = 40
a5 = a4 + d = 40 + 10 = 50
Therefore, the series will be 10, 20, 30, 40, 50 …
First four terms of this A.P. will be 10, 20, 30, and 40.

(ii) a = – 2, d = 0
Let the series be a1, a2, a3, a4 …
a1 = a = -2
a2 = a1 + d = – 2 + 0 = – 2
a3 = a2 + d = – 2 + 0 = – 2
a4 = a3 + d = – 2 + 0 = – 2
Therefore, the series will be – 2, – 2, – 2, – 2 …
First four terms of this A.P. will be – 2, – 2, – 2 and – 2.

(iii) a = 4, d = – 3
Let the series be a1, a2, a3, a4 …
a1 = a = 4
a2 = a1 + d = 4 – 3 = 1
a3 = a2 + d = 1 – 3 = – 2
a4 = a3 + d = – 2 – 3 = – 5
Therefore, the series will be 4, 1, – 2 – 5 …
First four terms of this A.P. will be 4, 1, – 2 and – 5.

(iv) a = – 1, d = 1/2
Let the series be a1, a2, a3, a4 …a1 = a = -1
a2 = a1 + d = -1 + 1/2 = -1/2
a3 = a2 + d = -1/2 + 1/2 = 0
a4 = a3 + d = 0 + 1/2 = 1/2
Clearly, the series will be-1, -1/2, 0, 1/2
First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v) a = – 1.25, d = – 0.25
Let the series be a1, a2, a3, a4 …
a1 = a = – 1.25
a2 = a1 + d = – 1.25 – 0.25 = – 1.50
a3 = a2 + d = – 1.50 – 0.25 = – 1.75
a4 = a3 + d = – 1.75 – 0.25 = – 2.00
Clearly, the series will be 1.25, – 1.50, – 1.75, – 2.00 ……..
First four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.

3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1, – 1, – 3 …
(ii) -5, – 1, 3, 7 …
(iii) 1/3, 5/3, 9/3, 13/3 ….
(iv) 0.6, 1.7, 2.8, 3.9 …

Answer

(i) 3, 1, – 1, – 3 …
Here, first term, a = 3
Common difference, d = Second term – First term
= 1 – 3 = – 2

(ii) – 5, – 1, 3, 7 …
Here, first term, a = – 5
Common difference, d = Second term – First term
= ( – 1) – ( – 5) = – 1 + 5 = 4
(iii) 1/3, 5/3, 9/3, 13/3 ….
Here, first term, a = 1/3

Common difference, d = Second term – First term 
= 5/3 – 1/3 = 4/3


(iv) 0.6, 1.7, 2.8, 3.9 …
Here, first term, a = 0.6
Common difference, d = Second term – First term
= 1.7 – 0.6
= 1.1

4. Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ….
(iii) -1.2, -3.2, -5.2, -7.2 …
(iv) -10, – 6, – 2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
(vii) 0, – 4, – 8, – 12 …
(viii) -1/2, -1/2, -1/2, -1/2 ….
(ix) 1, 3, 9, 27 …
(x) a, 2a, 3a, 4a …
(xi) a, a2, a3, a4 …
(xii) √2, √8, √18, √32 …
(xiii) √3, √6, √9, √12 …
(xiv) 12, 32, 52, 72 …
(xv) 12, 52, 72, 73 …

Answer

(i) 2, 4, 8, 16 …
Here,
a2 – a1 = 4 – 2 = 2
a3 – a2 = 8 – 4 = 4
a4 – a3 = 16 – 8 = 8
⇒ an+1 – an is not the same every time.

Therefore, the given numbers are forming an A.P.

(ii) 2, 5/2, 3, 7/2 ….
Here,

a2 – a1 = 5/2 – 2 = 1/2
a3 – a2 = 3 – 5/2 = 1/2
a4 – a3 = 7/2 – 3 = 1/2
⇒ an+1 – an is same every time.
Therefore, d = 1/2 and the given numbers are in A.P.
Three more terms are
a5 = 7/2 + 1/2 = 4
a6 = 4 + 1/2 = 9/2
a7 = 9/2 + 1/2 = 5

(iii) -1.2, – 3.2, -5.2, -7.2 …
Here,
a2 – a1 = ( -3.2) – ( -1.2) = -2
a3 – a2 = ( -5.2) – ( -3.2) = -2
a4 – a3 = ( -7.2) – ( -5.2) = -2
⇒ an+1 – an is same every time.
Therefore, d = -2 and the given numbers are in A.P.
Three more terms are
a5 = – 7.2 – 2 = – 9.2
a6 = – 9.2 – 2 = – 11.2
a7 = – 11.2 – 2 = – 13.2

(iv) -10, – 6, – 2, 2 …
Here,
a2 – a1 = (-6) – (-10) = 4
a3 – a2 = (-2) – (-6) = 4
a4 – a3 = (2) – (-2) = 4
⇒ an+1 – an is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Three more terms are
a5 = 2 + 4 = 6
a6 = 6 + 4 = 10
a7 = 10 + 4 = 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
Here,
a2 – a1 = 3 + √2 – 3 = √2
a3 – a2 = (3 + 2√2) – (3 + √2) = √2
a4 – a3 = (3 + 3√2) – (3 + 2√2) = √2
⇒ an+1 – an is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a5 = (3 + √2) + √2 = 3 + 4√2
a6 = (3 + 4√2) + √2 = 3 + 5√2
a7 = (3 + 5√2) + √2 = 3 + 6√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….
Here,
a2 – a1 = 0.22 – 0.2 = 0.02
a3 – a2 = 0.222 – 0.22 = 0.002
a4 – a3 = 0.2222 – 0.222 = 0.0002
⇒ an+1 – an is not the same every time.

Therefore, the given numbers are forming an A.P.

(vii) 0, -4, -8, -12 …
Here,
a2 – a1 = (-4) – 0 = -4
a3 – a2 = (-8) – (-4) = -4
a4 – a3 = (-12) – (-8) = -4
⇒ an+1 – an is same every time.
Therefore, d = -4 and the given numbers are in A.P.
Three more terms are
a5 = -12 – 4 = -16
a6 = -16 – 4 = -20
a7 = -20 – 4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ….
Here,
a2 – a1 = (-1/2) – (-1/2) = 0
a3 – a2 = (-1/2) – (-1/2) = 0
a4 – a3 = (-1/2) – (-1/2) = 0
⇒ an+1 – an is same every time.
Therefore, d = 0 and the given numbers are in A.P.
Three more terms are
a5 = (-1/2) – 0 = -1/2
a6 = (-1/2) – 0 = -1/2
a7 = (-1/2) – 0 = -1/2

(ix) 1, 3, 9, 27 …
Here,
a2 – a1 = 3 – 1 = 2
a3 – a2 = 9 – 3 = 6
a4 – a3 = 27 – 9 = 18
⇒ an+1 – an is not the same every time.
Therefore, the given numbers are forming an A.P.

(x) a, 2a, 3a, 4a …
Here,
a2 – a1 = 2a – a = a
a3 – a2 = 3a – 2a = a
a4 – a3 = 4a – 3a = a
⇒ an+1 – an is same every time.
Therefore, d = a and the given numbers are in A.P.
Three more terms are
a5 = 4a + a = 5a
a6 = 5a + a = 6a
a7 = 6a + a = 7a

(xi) a, a2, a3, a4 …
Here,
a2 – a1 = a2 – a = (a – 1)
a3 – a2 = a3 – a2 = a2 (a – 1)
a4 – a3 = a4 – a3 = a3(a – 1)
⇒ an+1 – an is not the same every time.
Therefore, the given numbers are forming an A.P.

(xii) √2, √8, √18, √32 …
Here,
a2 – a1 = √8 – √2  = 2√2 – √2 = √2
a3 – a2 = √18 – √8 = 3√2 – 2√2 = √2
a4 – a3 = 4√2 – 3√2 = √2
⇒ an+1 – an is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a5 = √32  + √2 = 4√2 + √2 = 5√2 = √50
a6 = 5√2 +√2 = 6√2 = √72
a7 = 6√2 + √2 = 7√2 = √98


(xiii) √3, √6, √9, √12 …
Here,
a2 – a1 = √6 – √3 = √3 × 2 -√3 = √3(√2 – 1)
a3 – a2 = √9 – √6 = 3 – √6 = √3(√3 – √2)
a4 – a3 = √12 – √9 = 2√3 – √3 × 3 = √3(2 – √3)
⇒ an+1 – an is not the same every time.

Therefore, the given numbers are forming an A.P.

(xiv) 12, 32, 52, 72 …

Or, 1, 9, 25, 49 …..
Here,
a2 − a1 = 9 − 1 = 8
a3 − a2 = 25 − 9 = 16
a4 − a3 = 49 − 25 = 24
⇒ an+1 – an is not the same every time.

Therefore, the given numbers are forming an A.P.


(xv) 12, 52, 72, 73 …
Or 1, 25, 49, 73 …
Here,
a2 − a1 = 25 − 1 = 24
a3 − a2 = 49 − 25 = 24
a4 − a3 = 73 − 49 = 24
i.e., ak+1 − ak is same every time.
⇒ an+1 – an is same every time.
Therefore, d = 24 and the given numbers are in A.P.
Three more terms are
a5 = 73+ 24 = 97
a6 = 97 + 24 = 121
a7 = 121 + 24 = 145

Exercise 5.2

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the A.P.

 
a
d
n
an
(i)
7
3
8
……
(ii)
− 18
…..
10
0
(iii)
…..
− 3
18
− 5
(iv)
− 18.9
2.5
…..
3.6
(v)
3.5
0
105
…..

Answer

(i) a = 7, d = 3, n = 8, an = ?
We know that,
For an A.P. an = a + (n − 1) d
= 7 + (8 − 1) 3
= 7 + (7) 3
= 7 + 21 = 28
Hence, an = 28

(ii) Given that
a = −18, n = 10, an = 0, d = ?
We know that,
an = a + (n − 1) d
0 = − 18 + (10 − 1) d
18 = 9d
d = 18/9 = 2
Hence, common difference, d = 2

(iii) Given that
d = −3, n = 18, an = −5
We know that,
an = a + (n − 1) d
−5 = a + (18 − 1) (−3)
−5 = a + (17) (−3)
−5 = a − 51
a = 51 − 5 = 46
Hence, a = 46

(iv) a = −18.9, d = 2.5, an = 3.6, n = ?
We know that,
an = a + (n − 1) d
3.6 = − 18.9 + (n − 1) 2.5
3.6 + 18.9 = (n − 1) 2.5
22.5 = (n − 1) 2.5
(n – 1) = 22.5/2.5
n – 1 = 9
n = 10
Hence, n = 10

(v) a = 3.5, d = 0, n = 105, an = ?
We know that,
an = a + (n − 1) d
an = 3.5 + (105 − 1) 0
an = 3.5 + 104 × 0
an = 3.5
Hence, an = 3.5

Page No: 106

Choose the correct choice in the following and justify
(i) 30th term of the A.P: 10, 7, 4, …, is
(A) 97 (B) 77 (C) −77 (D.) −87

(ii) 11th term of the A.P. -3, -1/2, ,2 …. is
(A) 28 (B) 22 (C) – 38 (D)

Answer

(i) Given that
A.P. 10, 7, 4, …
First term, a = 10
Common difference, d = a2 − a1 = 7 − 10 = −3
We know that, an = a + (n − 1) d
a30 = 10 + (30 − 1) (−3)
a30 = 10 + (29) (−3)
a30 = 10 − 87 = −77
Hence, the correct answer is option C.

(ii) Given that A.P. is -3, -1/2, ,2 …
First term a = – 3
Common difference, d = a2 − a1 = (-1/2) – (-3)
= (-1/2) + 3 = 5/2
We know that, an = a + (n − 1) d
a11 = 3 + (11 -1)(5/2)
a11 = 3 + (10)(5/2)
a11 = -3 + 25
a11 = 22
Hence, the answer is option B.

3. In the following APs find the missing term in the boxes.
 
Answer
 
(i) For this A.P.,
a = 2
a3 = 26
We know that, an = a + (n − 1) d
a3 = 2 + (3 – 1) d
26 = 2 + 2d
24 = 2d
d = 12
a2 = 2 + (2 – 1) 12
= 14
Therefore, 14 is the missing term.

(ii) For this A.P.,
a2 = 13 and
a4 = 3
We know that, an = a + (n − 1) d
a2 = a + (2 – 1) d
13 = a + d … (i)
a4 = a + (4 – 1) d
3 = a + 3d … (ii)
On subtracting (i) from (ii), we get
– 10 = 2d
d = – 5
From equation (i), we get
13 = a + (-5)
a = 18
a3 = 18 + (3 – 1) (-5)
= 18 + 2 (-5) = 18 – 10 = 8
Therefore, the missing terms are 18 and 8 respectively.

(iii) For this A.P.,
a = 5 and
a4 = 19/2
We know that, an = a + (n − 1) d
a4 = a + (4 – 1) d
19/2 = 5 + 3d
19/2 – 5 = 3d3d = 9/2
d = 3/2

a2 = a + (2 – 1) d
a2 = 5 + 3/2
a2 = 13/2

a3 = a + (3 – 1) d
a3 = 5 + 2×3/2
a3 = 8
Therefore, the missing terms are 13/2 and 8 respectively.

(iv) For this A.P.,
a = −4 and
a6 = 6
We know that,
an = a + (n − 1) d
a6 = a + (6 − 1) d
6 = − 4 + 5d
10 = 5d
d = 2
a2 = a + d = − 4 + 2 = −2
a3 = a + 2d = − 4 + 2 (2) = 0
a4 = a + 3d = − 4 + 3 (2) = 2
a5 = a + 4d = − 4 + 4 (2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.

(v)
For this A.P.,
a2 = 38
a6 = −22
We know that
an = a + (n − 1) d
a2 = a + (2 − 1) d
38 = a + d … (i)
a6 = a + (6 − 1) d
−22 = a + 5d … (ii)
On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d
d = −15
a = a2 − d = 38 − (−15) = 53
a3 = a + 2d = 53 + 2 (−15) = 23
a4 = a + 3d = 53 + 3 (−15) = 8
a5 = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.

4. Which term of the A.P. 3, 8, 13, 18, … is 78?

Answer

3, 8, 13, 18, …
For this A.P.,
a = 3
d = a2 − a1 = 8 − 3 = 5
Let nth term of this A.P. be 78.
an = a + (n − 1) d
78 = 3 + (n − 1) 5
75 = (n − 1) 5
(n − 1) = 15
n = 16
Hence, 16th term of this A.P. is 78.

5. Find the number of terms in each of the following A.P.
(i) 7, 13, 19, …, 205
(ii) 18,, 13,…., -47

Answer

(i) For this A.P.,
a = 7
d = a2 − a1 = 13 − 7 = 6
Let there are n terms in this A.P.
an = 205
We know that
an = a + (n − 1) d
Therefore, 205 = 7 + (n − 1) 6
198 = (n − 1) 6
33 = (n − 1)
n = 34
Therefore, this given series has 34 terms in it.

(ii) For this A.P.,
a = 18

Let there are n terms in this A.P.
an = 205
an = a + (n − 1) d
-47 = 18 + (n – 1) (-5/2)
-47 – 18 = (n – 1) (-5/2)
-65 = (n – 1)(-5/2)
(n – 1) = -130/-5
(n – 1) = 26
n = 27
Therefore, this given A.P. has 27 terms in it.
 
6. Check whether -150 is a term of the A.P. 11, 8, 5, 2, …
 
Answer

For this A.P.,
a = 11
d = a2 − a1 = 8 − 11 = −3
Let −150 be the nth term of this A.P.
We know that,
an = a + (n − 1) d
-150 = 11 + (n – 1)(-3)
-150 = 11 – 3n + 3
-164 = -3n
n = 164/3
Clearly, n is not an integer.
Therefore, – 150 is not a term of this A.P.
 
7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.

Answer

Given that,
a11 = 38
a16 = 73
We know that,
an = a + (n − 1) d
a11 = a + (11 − 1) d
38 = a + 10d … (i)
Similarly,
a16 = a + (16 − 1) d
73 = a + 15d … (ii)
On subtracting (i) from (ii), we get
35 = 5d
d = 7
From equation (i),
38 = a + 10 × (7)
38 − 70 = a
a = −32
a31 = a + (31 − 1) d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, 31st term is 178.

8. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Answer

Given that,
a3 = 12
a50 = 106
We know that,
an = a + (n − 1) d
a3 = a + (3 − 1) d
12 = a + 2d … (i)
Similarly, a50 = a + (50 − 1) d
106 = a + 49d … (ii)
On subtracting (i) from (ii), we get
94 = 47d
d = 2
From equation (i), we get
12 = a + 2 (2)
a = 12 − 4 = 8
a29 = a + (29 − 1) d
a29 = 8 + (28)2
a29 = 8 + 56 = 64
Therefore, 29th term is 64.

9. If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.

Answer

Given that,
a3 = 4
a9 = −8
We know that,
an = a + (n − 1) d
a3 = a + (3 − 1) d
4 = a + 2d … (i)
a9 = a + (9 − 1) d
−8 = a + 8d … (ii)
On subtracting equation (i) from (ii), we get,
−12 = 6d
d = −2
From equation (i), we get,
4 = a + 2 (−2)
4 = a − 4
a = 8
Let nth term of this A.P. be zero.
an = a + (n − 1) d
0 = 8 + (n − 1) (−2)
0 = 8 − 2n + 2
2n = 10
n = 5
Hence, 5th term of this A.P. is 0.

10. If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.

Answer

We know that,
For an A.P., an = a + (n − 1) d
a17 = a + (17 − 1) d
a17 = a + 16d
Similarly, a10 = a + 9d
It is given that
a17 − a10 = 7
(a + 16d) − (a + 9d) = 7
7d = 7
d = 1
Therefore, the common difference is 1.

11. Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?

Answer

Given A.P. is 3, 15, 27, 39, …
a = 3
d = a2 − a1 = 15 − 3 = 12
a54 = a + (54 − 1) d
= 3 + (53) (12)
= 3 + 636 = 639
132 + 639 = 771
We have to find the term of this A.P. which is 771.
Let nth term be 771.
an = a + (n − 1) d
771 = 3 + (n − 1) 12
768 = (n − 1) 12
(n − 1) = 64
n = 65
Therefore, 65th term was 132 more than 54th term.

Or

Let nth term be 132 more than 54th term.
n = 54 + 132/2
= 54 + 11 = 65th term

12. Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?

Answer

Let the first term of these A.P.s be a1 and a2 respectively and the common difference of these A.P.s be d.
For first A.P.,
a100 = a1 + (100 − 1) d
= a1 + 99d
a1000 = a1 + (1000 − 1) d
a1000 = a1 + 999d
For second A.P.,
a100 = a2 + (100 − 1) d
= a2 + 99d
a1000 = a2 + (1000 − 1) d
= a2 + 999d
Given that, difference between
100th term of these A.P.s = 100
Therefore, (a1 + 99d) − (a2 + 99d) = 100
a1 − a2 = 100 … (i)
Difference between 1000th terms of these A.P.s
(a1 + 999d) − (a2 + 999d) = a1 − a2
From equation (i),
This difference, a1 − a2 = 100
Hence, the difference between 1000th terms of these A.P. will be 100.

13. How many three digit numbers are divisible by 7?

Answer

First three-digit number that is divisible by 7 = 105
Next number = 105 + 7 = 112
Therefore, 105, 112, 119, …
All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.
The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.
The series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
a = 105
d = 7
an = 994
n = ?
an = a + (n − 1) d
994 = 105 + (n − 1) 7
889 = (n − 1) 7
(n − 1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7.

Or

Three digit numbers which are divisible by 7 are 105, 112, 119, …. 994 .
These numbers form an AP with a = 105 and d = 7.
Let number of three-digit numbers divisible by 7 be n, an = 994
⇒ a + (n – 1) d = 994
⇒ 105 + (n – 1) × 7 = 994
⇒7(n – 1) = 889
⇒ n – 1 = 127
⇒ n = 128

14. How many multiples of 4 lie between 10 and 250?

Answer

First multiple of 4 that is greater than 10 is 12. Next will be 16.
Therefore, 12, 16, 20, 24, …
All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows.
12, 16, 20, 24, …, 248
Let 248 be the nth term of this A.P.
a = 12
d = 4
an = 248
an = a + (n – 1) d
248 = 12 + (n – 1) × 4
236/4 = n – 1
59  = n – 1
n = 60
Therefore, there are 60 multiples of 4 between 10 and 250.

Or

Multiples of 4 lies between 10 and 250 are 12, 16, 20, …., 248.
These numbers form an AP with a = 12 and d = 4.
Let number of three-digit numbers divisible by 4 be n, an = 248
⇒ a + (n – 1) d = 248
⇒ 12 + (n – 1) × 4 = 248
⇒4(n – 1) = 248
⇒ n – 1 = 59
⇒ n = 60

15. For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal?

Answer

63, 65, 67, …
a = 63
d = a2 − a1 = 65 − 63 = 2
nth term of this A.P. = an = a + (n − 1) d
an= 63 + (n − 1) 2 = 63 + 2n − 2
an = 61 + 2n … (i)
3, 10, 17, …
a = 3
d = a2 − a1 = 10 − 3 = 7
nth term of this A.P. = 3 + (n − 1) 7
an = 3 + 7n − 7
an = 7n − 4 … (ii)
It is given that, nth term of these A.P.s are equal to each other.
Equating both these equations, we obtain
61 + 2n = 7n − 4
61 + 4 = 5n
5n = 65
n = 13
Therefore, 13th terms of both these A.P.s are equal to each other.

16. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.

Answer

a3 = 16
a + (3 − 1) d = 16
a + 2d = 16 … (i)
a7 − a5 = 12
[a+ (7 − 1) d] − [a + (5 − 1) d]= 12
(a + 6d) − (a + 4d) = 12
2d = 12
d = 6
From equation (i), we get,
a + 2 (6) = 16
a + 12 = 16
a = 4
Therefore, A.P. will be
4, 10, 16, 22, …

Page No: 107

17. Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.

Answer

Given A.P. is
3, 8, 13, …, 253
Common difference for this A.P. is 5.
Therefore, this A.P. can be written in reverse order as
253, 248, 243, …, 13, 8, 5
For this A.P.,
a = 253
d = 248 − 253 = −5
n = 20
a20 = a + (20 − 1) d
a20 = 253 + (19) (−5)
a20 = 253 − 95
a = 158
Therefore, 20th term from the last term is 158.

18. The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.

Answer

We know that,
an = a + (n − 1) d
a4 = a + (4 − 1) d
a4 = a + 3d
Similarly,
a8 = a + 7d
a6 = a + 5d
a10 = a + 9d
Given that, a4 + a8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12 … (i)
a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 … (ii)
On subtracting equation (i) from (ii), we get,
2d = 22 − 12
2d = 10
d = 5
From equation (i), we get
a + 5d = 12
a + 5 (5) = 12
a + 25 = 12
a = −13
a2 = a + d = − 13 + 5 = −8
a3 = a2 + d = − 8 + 5 = −3
Therefore, the first three terms of this A.P. are −13, −8, and −3.

19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Answer

It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.
Therefore, the salaries of each year after 1995 are
5000, 5200, 5400, …
Here, a = 5000
d = 200
Let after nth year, his salary be Rs 7000.
Therefore, an = a + (n − 1) d
7000 = 5000 + (n − 1) 200
200(n − 1) = 2000
(n − 1) = 10
n = 11
Therefore, in 11th year, his salary will be Rs 7000.

20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her week, her weekly savings become Rs 20.75, find n.
Answer

Given that,
a = 5
d = 1.75
an = 20.75
n = ?
an = a + (n − 1) d
20.75 = 5 + (n – 1) × 1.75
15.75 = (n – 1) × 1.75
(n – 1) = 15.75/1.75 = 1575/175
= 63/7 = 9
n – 1 = 9
n = 10
Hence, n is 10.
Exercise 5.3

1. Find the sum of the following APs.
(i) 2, 7, 12 ,…., to 10 terms.
(ii) − 37, − 33, − 29 ,…, to 12 terms
(iii) 0.6, 1.7, 2.8 ,…….., to 100 terms
(iv) 1/15, 1/12, 1/10, …… , to 11 terms

Answer

(i) 2, 7, 12 ,…, to 10 terms
For this A.P.,
a = 2
d = a2 − a1 = 7 − 2 = 5
n = 10
We know that,
Sn = n/2 [2a + (n – 1) d]
S10 = 10/2 [2(2) + (10 – 1) × 5]
= 5[4 + (9) × (5)]
= 5 × 49 = 245

(ii) −37, −33, −29 ,…, to 12 terms
For this A.P.,
a = −37
d = a2 − a1 = (−33) − (−37)
= − 33 + 37 = 4
n = 12
We know that,
Sn = n/2 [2a + (n – 1) d]
S12 = 12/2 [2(-37) + (12 – 1) × 4]
= 6[-74 + 11 × 4]
= 6[-74 + 44]
= 6(-30) = -180

(iii) 0.6, 1.7, 2.8 ,…, to 100 terms
For this A.P.,
a = 0.6
d = a2 − a1 = 1.7 − 0.6 = 1.1
n = 100
We know that,
Sn = n/2 [2a + (n – 1) d]
S12 = 50/2 [1.2 + (99) × 1.1]
= 50[1.2 + 108.9]
= 50[110.1]
= 5505


(iv) 1/15, 1/12, 1/10, …… , to 11 terms
For this A.P.,


2. Find the sums given below
(i) 7 + + 14 + ……………… +84
(ii)+ 14 + ………… + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)

Answer

(i) For this A.P.,
a = 7
l = 84
d = a2 − a1 = – 7 = 21/2 – 7 = 7/2
Let 84 be the nth term of this A.P.
l = a (n – 1)d
84 = 7 + (n – 1) × 7/2
77 = (n – 1) × 7/2
22 = n − 1
n = 23
We know that,
Sn = n/2 (a + l)
Sn = 23/2 (7 + 84)
= (23×91/2) = 2093/2
= 


(ii) 34 + 32 + 30 + ……….. + 10
For this A.P.,
a = 34
d = a2 − a1 = 32 − 34 = −2
l = 10
Let 10 be the nth term of this A.P.
l = a + (n − 1) d
10 = 34 + (n − 1) (−2)
−24 = (n − 1) (−2)
12 = n − 1
n = 13
Sn = n/2 (a + l)
= 13/2 (34 + 10)
= (13×44/2) = 13 × 22
= 286

(iii) (−5) + (−8) + (−11) + ………… + (−230) For this A.P.,
a = −5
l = −230
d = a2 − a1 = (−8) − (−5)
= − 8 + 5 = −3
Let −230 be the nth term of this A.P.
l = a + (n − 1)d
−230 = − 5 + (n − 1) (−3)
−225 = (n − 1) (−3)
(n − 1) = 75
n = 76
And,
Sn = n/2 (a + l)
= 76/2 [(-5) + (-230)]
= 38(-235)
= -8930

3. In an AP
(i) Given a = 5, d = 3, an = 50, find n and Sn.
(ii) Given a = 7, a13 = 35, find d and S13.
(iii) Given a12 = 37, d = 3, find a and S12.
(iv) Given a3 = 15, S10 = 125, find d and a10.
(v) Given d = 5, S9 = 75, find a and a9.
(vi) Given a = 2, d = 8, Sn = 90, find n and an.
(vii) Given a = 8, an = 62, Sn = 210, find n and d.
(viii) Given an = 4, d = 2, Sn = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x) Given l = 28, S = 144 and there are total 9 terms. Find a.

Answer

(i) Given that, a = 5, d = 3, an = 50
As an = a + (n − 1)d,
⇒ 50 = 5 + (n – 1) × 3
⇒ 3(n – 1) = 45
⇒ n – 1 = 15
⇒ n = 16
Now, Sn = n/2 (a + an)
Sn = 16/2 (5 + 50) = 440


(ii) Given that, a = 7, a13 = 35
As an = a + (n − 1)d, ⇒ 35 = 7 + (13 – 1)d
⇒ 12d = 28
⇒ d = 28/12 = 2.33
Now, Sn = n/2 (a + an)
S13 = 13/2 (7 + 35) = 273

(iii)Given that, a12 = 37, d = 3 As an = a + (n − 1)d,
⇒ a12 = a + (12 − 1)3
⇒ 37 = a + 33
⇒ a = 4
Sn = n/2 (a + an)
Sn = 12/2 (4 + 37)
= 246

(iv) Given that, a3 = 15, S10 = 125
As an = a + (n − 1)d,
a3 = a + (3 − 1)d
15 = a + 2d … (i)
Sn = n/2 [2a + (n – 1)d]
S10 = 10/2 [2a + (10 – 1)d]
125 = 5(2a + 9d)
25 = 2a + 9d … (ii)
On multiplying equation (i) by (ii), we get
30 = 2a + 4d … (iii)
On subtracting equation (iii) from (ii), we get
−5 = 5d
d = −1
From equation (i),
15 = a + 2(−1)
15 = a − 2
a = 17
a10 = a + (10 − 1)d
a10 = 17 + (9) (−1)
a10 = 17 − 9 = 8

(v) Given that, d = 5, S9 = 75
As Sn = n/2 [2a + (n – 1)d]
S9 = 9/2 [2a + (9 – 1)5]
25 = 3(a + 20)
25 = 3a + 60
3a = 25 − 60
a = -35/3
an = a + (n − 1)d
a9 = a + (9 − 1) (5)
= -35/3 + 8(5)
= -35/3 + 40
= (35+120/3) = 85/3

(vi) Given that, a = 2, d = 8, Sn = 90
As Sn = n/2 [2a + (n – 1)d]
90 = n/2 [2a + (n – 1)d]
⇒ 180 = n(4 + 8n – 8) = n(8n – 4) = 8n2 – 4n
⇒ 8n2 – 4n – 180 = 0
⇒ 2n2 – n – 45 = 0
⇒ 2n2 – 10n + 9n – 45 = 0
⇒ 2n(n -5) + 9(n – 5) = 0
⇒ (2n – 9)(2n + 9) = 0
So, n = 5 (as it is positive integer)
∴ a5 = 8 + 5 × 4 = 34

(vii) Given that, a = 8, an = 62, Sn = 210
As Sn = n/2 (a + an)
210 = n/2 (8 + 62)
⇒ 35n = 210
⇒ n = 210/35 = 6
Now, 62 = 8 + 5d
⇒ 5d = 62 – 8 = 54
⇒ d = 54/5 = 10.8

(viii) Given that, an = 4, d = 2, Sn = −14
an = a + (n − 1)d
4 = a + (n − 1)2
4 = a + 2n − 2
a + 2n = 6
a = 6 − 2n … (i)
Sn = n/2 (a + an)
-14 = n/2 (a + 4)
−28 = n (a + 4)
−28 = n (6 − 2n + 4) {From equation (i)}
−28 = n (− 2n + 10)
−28 = − 2n2 + 10n
2n2 − 10n − 28 = 0
n2 − 5n −14 = 0
n2 − 7n + 2n − 14 = 0
n (n − 7) + 2(n − 7) = 0
(n − 7) (n + 2) = 0
Either n − 7 = 0 or n + 2 = 0
n = 7 or n = −2
However, n can neither be negative nor fractional.
Therefore, n = 7
From equation (i), we get
a = 6 − 2n
a = 6 − 2(7)
= 6 − 14
= −8

(ix) Given that, a = 3, n = 8, S = 192
As Sn = n/2 [2a + (n – 1)d]
192 = 8/2 [2 × 3 + (8 – 1)d]
192 = 4 [6 + 7d]
48 = 6 + 7d
42 = 7d
d = 6

(x) Given that, l = 28, S = 144 and there are total of 9 terms.
Sn = n/2 (a + l)
144 = 9/2 (a + 28)
(16) × (2) = a + 28
32 = a + 28
a = 4

Page No: 113

4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Answer

Let there be n terms of this A.P.
For this A.P., a = 9
d = a2 − a1 = 17 − 9 = 8
As Sn = n/2 [2a + (n – 1)d]
636 = n/2 [2 × a + (8 – 1) × 8]
636 = n/2 [18 + (n– 1) × 8]
636 = n [9 + 4n − 4]
636 = n (4n + 5)
4n2 + 5n − 636 = 0
4n2 + 53n − 48n − 636 = 0
n (4n + 53) − 12 (4n + 53) = 0
(4n + 53) (n − 12) = 0
Either 4n + 53 = 0 or n − 12 = 0
n = (-53/4) or n = 12
n cannot be (-53/4). As the number of terms can neither be negative nor fractional, therefore, n = 12 only.

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Answer

Given that,
a = 5
l = 45
Sn = 400
Sn = n/2 (a + l)
400 = n/2 (5 + 45)
400 = n/2 (50)
n = 16
l = a + (n − 1) d
45 = 5 + (16 − 1) d
40 = 15d
d = 40/15 = 8/3

6. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Answer

Given that,

a = 17
l = 350
d = 9
Let there be n terms in the A.P.
l = a + (n − 1) d
350 = 17 + (n − 1)9
333 = (n − 1)9
(n − 1) = 37
n = 38
Sn = n/2 (a + l)
S38 = 13/2 (17 + 350)
= 19 × 367
= 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Answer

d = 7
a22 = 149
S22 = ?
an = a + (n − 1)d
a22 = a + (22 − 1)d
149 = a + 21 × 7
149 = a + 147
a = 2
Sn = n/2 (a + an)
= 22/2 (2 + 149)
= 11 × 151
= 1661

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Answer

Given that,
a2 = 14
a3 = 18
d = a3 − a2 = 18 − 14 = 4
a2 = a + d
14 = a + 4
a = 10
Sn = n/2 [2a + (n – 1)d]
S51 = 51/2 [2 × 10 + (51 – 1) × 4]
= 51/2 [2 + (20) × 4]
= 51×220/2
= 51 × 110
= 5610

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Answer

Given that,

S7 = 49
S17 = 289
S7  = 7/2 [2a + (n – 1)d]
S7 = 7/2 [2a + (7 – 1)d]
49 = 7/2 [2a + 16d]
7 = (a + 3d)
a + 3d = 7 … (i)
Similarly,
S17 = 17/2 [2a + (17 – 1)d]
289 = 17/2 (2a + 16d)
17 = (a + 8d)
a + 8d = 17 … (ii)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
From equation (i),
a + 3(2) = 7
a + 6 = 7
a = 1
Sn = n/2 [2a + (n – 1)d]
= n/2 [2(1) + (n – 1) × 2]
= n/2 (2 + 2n – 2)
= n/2 (2n)
= n2

10. Show that a1, a2 … , an , … form an AP where an is defined as below
(i) an = 3 + 4n
(ii) an = 9 − 5n
Also find the sum of the first 15 terms in each case.

Answer

(i) an = 3 + 4n
a1 = 3 + 4(1) = 7
a2 = 3 + 4(2) = 3 + 8 = 11
a3 = 3 + 4(3) = 3 + 12 = 15
a4 = 3 + 4(4) = 3 + 16 = 19
It can be observed that
a2 − a1 = 11 − 7 = 4
a3 − a2 = 15 − 11 = 4
a4 − a3 = 19 − 15 = 4
i.e., ak + 1 − ak is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
Sn = n/2 [2a + (n – 1)d]
S15 = 15/2 [2(7) + (15 – 1) × 4]
= 15/2 [(14) + 56]
= 15/2 (70)
= 15 × 35
= 525

(ii) an = 9 − 5n
a1 = 9 − 5 × 1 = 9 − 5 = 4
a2 = 9 − 5 × 2 = 9 − 10 = −1
a3 = 9 − 5 × 3 = 9 − 15 = −6
a4 = 9 − 5 × 4 = 9 − 20 = −11
It can be observed that
a2 − a1 = − 1 − 4 = −5
a3 − a2 = − 6 − (−1) = −5
a4 − a3 = − 11 − (−6) = −5
i.e., ak + 1 − ak is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.
Sn = n/2 [2a + (n – 1)d]
S15 = 15/2 [2(4) + (15 – 1) (-5)]
= 15/2 [8 + 14(-5)]
= 15/2 (8 – 70)
= 15/2 (-62)
= 15(-31)
= -465

11. If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the10th and the nth terms.

Answer

Given that,
Sn = 4n − n2
First term, a = S1 = 4(1) − (1)2 = 4 − 1 = 3
Sum of first two terms = S2
= 4(2) − (2)2 = 8 − 4 = 4
Second term, a2 = S2 − S1 = 4 − 3 = 1
d = a2 − a = 1 − 3 = −2
an = a + (n − 1)d
= 3 + (n − 1) (−2)
= 3 − 2n + 2
= 5 − 2n
Therefore, a3 = 5 − 2(3) = 5 − 6 = −1
a10 = 5 − 2(10) = 5 − 20 = −15
Hence, the sum of first two terms is 4. The second term is 1. 3rd, 10th, and nth terms are −1, −15, and 5 − 2n respectively.

12. Find the sum of first 40 positive integers divisible by 6.

Answer

The positive integers that are divisible by 6 are
6, 12, 18, 24 …
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
a = 6
d = 6
S40 = ?
Sn = n/2 [2a + (n – 1)d]
S40 = 40/2 [2(6) + (40 – 1) 6]
= 20[12 + (39) (6)]
= 20(12 + 234)
= 20 × 246
= 4920

13. Find the sum of first 15 multiples of 8.

Answer

The multiples of 8 are
8, 16, 24, 32…
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S15 = ?
Sn = n/2 [2a + (n – 1)d]
S15 = 15/2 [2(8) + (15 – 1)8]
= 15/2[6 + (14) (8)]
= 15/2[16 + 112]
= 15(128)/2
= 15 × 64
= 960

14. Find the sum of the odd numbers between 0 and 50.

Answer

The odd numbers between 0 and 50 are
1, 3, 5, 7, 9 … 49
Therefore, it can be observed that these odd numbers are in an A.P.
a = 1
d = 2
l = 49
l = a + (n − 1) d
49 = 1 + (n − 1)2
48 = 2(n − 1)
n − 1 = 24
n = 25
Sn = n/2 (a + l)
S25 = 25/2 (1 + 49)
= 25(50)/2
=(25)(25)
= 625

15. A contract on construction job specifies a penalty for delay of completion beyond a certain dateas follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

Answer

It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.
a = 200
d = 50
Penalty that has to be paid if he has delayed the work by 30 days = S30
= 30/2 [2(200) + (30 – 1) 50]

= 15 [400 + 1450]
= 15 (1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty.

16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

Answer

Let the cost of 1st prize be P.
Cost of 2nd prize = P − 20
And cost of 3rd prize = P − 40
It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and first term as P.
a = P
d = −20
Given that, S7 = 700
7/2 [2a + (7 – 1)d] = 700

a + 3(−20) = 100
a − 60 = 100
a = 160
Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

Answer

It can be observed that the number of trees planted by the students is in an AP.
1, 2, 3, 4, 5………………..12
First term, a = 1
Common difference, d = 2 − 1 = 1
Sn = n/2 [2a + (n – 1)d]
S12 = 12/2 [2(1) + (12 – 1)(1)]
= 6 (2 + 11)
= 6 (13)
= 78
Therefore, number of trees planted by 1 section of the classes = 78
Number of trees planted by 3 sections of the classes = 3 × 78 = 234


Therefore, 234 trees will be planted by the students.



18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)

Answer

perimeter of semi-circle = πr
P1 = π(0.5) = π/2 cm
P2 = π(1) = π cm
P3 = π(1.5) = 3π/2 cm
P1, P2, P3 are the lengths of the semi-circles

π/2, π, 3π/2, 2π, ….
P1 = π/2 cm
P2 = π cm
d = P2- P1 = π – π/2 = π/2
First term = P1 = a = π/2 cm
Sn = n/2 [2a + (n – 1)d]
Therefor, Sum of the length of 13 consecutive circles
S13 = 13/2 [2(π/2) + (13 – 1)π/2]
=  13/2 [π + 6π]
=13/2 (7π)  = 13/2 × 7 × 22/7
= 143 cm

Page No: 114

19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Answer

It can be observed that the numbers of logs in rows are in an A.P.
20, 19, 18…
For this A.P.,
a = 20
d = a2 − a1 = 19 − 20 = −1
Let a total of 200 logs be placed in n rows.
Sn = 200
Sn = n/2 [2a + (n – 1)d]
S12 = 12/2 [2(20) + (n – 1)(-1)]
400 = n (40 − n + 1)
400 = n (41 − n)
400 = 41n − n2
n2 − 41n + 400 = 0
n2 − 16n − 25n + 400 = 0
n (n − 16) −25 (n − 16) = 0
(n − 16) (n − 25) = 0
Either (n − 16) = 0 or n − 25 = 0
n = 16 or n = 25
an = a + (n − 1)d
a16 = 20 + (16 − 1) (−1)
a16 = 20 − 15
a16 = 5
Similarly,
a25 = 20 + (25 − 1) (−1)
a25 = 20 − 24
= −4
Clearly, the number of logs in 16th row is 5. However, the number of logs in 25th row is negative, which is not possible.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5.

20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

 

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 ×(5 + 3)]

Answer

The distances of potatoes from the bucket are 5, 8, 11, 14…
Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept because first she has to first pick the potato and again return back to the same place in order to start picking the second potato.. Therefore, distances to be run are
10, 16, 22, 28, 34,……….
a = 10
d = 16 − 10 = 6
S10 =?
S10 = 10/2 [2(20) + (n – 1)(-1)]
= 5[20 + 54]
= 5 (74)
= 370
Therefore, the competitor will run a total distance of 370 m.

Exercise 5.4 (Optional)

1. Which term of the AP : 121, 117, 113, . . ., is its first negative term?
[Hint : Find n for an < 0]

Answer

We have the A.P. having a = 121 and d = 117 – 121 = – 4
∴  an = a + (n – 1) d
= 121 + (n – 1) × (- 4)
= 121 – 4n + 4
= 125 – 4n
For the first negative term, we have
an < 0
⇒ (125 – 4n) < 0
⇒ 125 < 4n
⇒  125/4 <n
⇒ n > 31 1⁄4

Thus, the first negative term is 32nd term.

 
2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
 
Answer
 
 
3. A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 21⁄2 m apart, what is the length of the wood required for the rungs?
[Hint : Number of rungs = 250/25 + 1]
 
Answer
 
 
4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint : Sx-1 = S49 – Sx]
 
Answer
 
 
5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1⁄4 m and a tread of 1⁄2 m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace.
[Hint : Volume of concrete required to build the first step = 1/4 × 1/2 × 50m3]
 
Answer
 

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations | EduGrown

In This Post we are  providing Chapter 4 Quadratic Equation NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Quadratic Equation Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Quadratic Equation  NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equation

Exercise 4.1

1. Check whether the following are quadratic equations:
(i) (x + 1)2 = 2(x – 3)
(ii) x2 – 2x = (-2)(3 – x)
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
(iv) (x – 3)(2x + 1) = x(x + 5)
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
(vi) x2 + 3x + 1 = (x – 2)2
(vii) (x + 2)3 = 2x(x2 – 1)
(viii) x3 – 4x2 – x + 1 = (x – 2)3

Answer

(i) (x + 2)2 = 2(x – 3)
⇒ x2 + 2x + 1 = 2x – 6
⇒ x2 + 7 = 0
It is of the form ax2 + bx + c = 0.
Hence, the given equation is quadratic equation.

(ii) x2 – 2x = (-2)(3 – x)
⇒ x2 – 2x = –6 + 2x 
⇒ x2 – 4x + 6 = 0
It is of the form ax2 + bx + c = 0.
Hence, the given equation is quadratic equation.

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
⇒ x2 – x – 2 = x2 + 2x – 3
⇒ 3x – 1 =0
It is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.

(iv) (x – 3)(2x + 1) = x(x + 5)
⇒ 2x2 – 5x – 3 = x2 + 5x
⇒  x2 – 10x – 3 = 0
It is of the form ax2 + bx + c = 0.
Hence, the given equation is quadratic equation.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
⇒ 2x2 – 7x + 3 = x2 + 4x – 5
⇒ x2 – 11x + 8 = 0
It is of the form ax2 + bx + c = 0.
Hence, the given equation is quadratic equation.

(vi) x2 + 3x + 1 = (x – 2)2
⇒ x2 + 3x + 1 = x2 + 4 – 4x
⇒ 7x – 3 = 0
It is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.

(vii) (x + 2)3 = 2x(x2 – 1)
⇒ x3 + 8 + x2 + 12x = 2x3 – 2x
⇒ x3 + 14x – 6x2 – 8 = 0
It is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.

(viii) x3 – 4x2 – x + 1 = (x – 2)3
⇒  x3 – 4x2 – x + 1 = x3 – 8 – 6x2  + 12x
⇒ 2x2 – 13x + 9 = 0
It is of the form ax2 + bx + c = 0.
Hence, the given equation is quadratic equation.

2. Represent the following situations in the form of quadratic equations.


(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
 
Answer

Let the breadth of the rectangular plot = x m
Hence, the length of the plot is (2x + 1) m.
Formula of area of rectangle = length × breadth = 528 m2
Putting the value of length and width, we get
(2x + 1) × x = 528
⇒ 2x2 + x =528
⇒ 2x2 + x – 528 = 0

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

Answer

Let the first integer number = x
Next consecutive positive integer will = x + 1
Product of both integers = x × (x +1) = 306
⇒ x2 + x = 306
⇒ x2 + x – 306 = 0

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

Answer

Let take Rohan’s age = x years
Hence, his mother’s age = x + 26
3 years from now
Rohan’s age = x + 3
Age of Rohan’s mother will = x + 26 + 3 = x + 29
The product of their ages 3 years from now will be 360 so that
(x + 3)(x + 29) = 360
⇒ x2 + 29x + 3x + 87 = 360
⇒ x2 + 32x + 87 – 360 = 0
⇒ x2 + 32x – 273 = 0

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Answer

Let the speed of train be x km/h.
Time taken to travel 480 km = 480/x km/h

In second condition, let the speed of train = (x – 8) km/h
It is also given that the train will take 3 hours to cover the same distance.
Therefore, time taken to travel 480 km = (480/x + 3) km/h

Speed × Time = Distance

(x – 8)(480/x + 3) = 480
⇒ 480 + 3x – 3840/x – 24 = 480
⇒ 3x – 3840/x = 24
⇒ 3x2 – 8x – 1280 = 0

Exercise 4.2

1. Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) √2 x2 + 7x + 5√2 = 0
(iv) 2x2 – x + 1/8 = 0
(v) 100x2 – 20x + 1 = 0

Answer

(i) x2 – 3x – 10
= x2 – 5x + 2x – 10
= x(x – 5) + 2(x – 5)
= (x – 5)(x + 2)
Roots of this equation are the values for which (x – 5)(x + 2) = 0
∴ x – 5 = 0 or x + 2 = 0
⇒ x = 5 or x = -2

(ii) 2x2 + x – 6
= 2x2 + 4x – 3x – 6
= 2x(x + 2) – 3(x + 2)
= (x + 2)(2x – 3)
Roots of this equation are the values for which (x + 2)(2x – 3) = 0
∴ x + 2 = 0 or 2x – 3 = 0
⇒ x = -2 or x = 3/2
(iii) √2 x2 + 7x + 5√2
= √2 x2 + 5x + 2x + 5√2
= x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)
Roots of this equation are the values for which (√2x + 5)(x + √2) = 0
∴ √2x + 5 = 0 or x + √2 = 0
⇒ x = -5/√2 or x = -√2

(iv) 2x2 – x + 1/8
= 1/8 (16x2  – 8x + 1)
= 1/8 (16x2  – 4x -4x + 1)
= 1/8 (4x(4x  – 1) -1(4x – 1))
= 1/8(4x – 1)2
Roots of this equation are the values for which (4x – 1)2 = 0
∴ (4x – 1) = 0 or (4x – 1) = 0
⇒ x = 1/4 or x = 1/4

(v) 100x2 – 20x + 1
= 100x2 – 10x – 10x + 1
= 10x(10x – 1) -1(10x – 1)
= (10x – 1)2
Roots of this equation are the values for which (10x – 1)2 = 0
∴ (10x – 1) = 0 or (10x – 1) = 0
⇒ x = 1/10 or x = 1/10

2. (i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. Find out how many marbles they had to start with.
 
Answer
 
Let the number of John’s marbles be x.
Therefore, number of Jivanti’s marble = 45 – x
After losing 5 marbles,
Number of John’s marbles = x – 5
Number of Jivanti’s marbles = 45 – x – 5 = 40 – x
It is given that the product of their marbles is 124.
∴ (x – 5)(40 – x) = 124
⇒ x2 – 45x + 324 = 0
⇒ x2 – 36x – 9x + 324 = 0
⇒ x(x – 36) -9(x – 36) = 0
⇒ (x – 36)(x – 9) = 0
Either x – 36 = 0 or x – 9 = 0
⇒ x = 36 or x = 9
If the number of John’s marbles = 36,
Then, number of Jivanti’s marbles = 45 – 36 = 9
If number of John’s marbles = 9,
Then, number of Jivanti’s marbles = 45 – 9 = 36

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs 750. Find out the number of toys produced on that day.

Answer

Let the number of toys produced be x.
∴ Cost of production of each toy = Rs (55 – x)
It is given that, total production of the toys = Rs 750
∴ x(55 – x) = 750
⇒ x2 – 55x + 750 = 0
⇒ x2 – 25x – 30x + 750 = 0
⇒ x(x – 25) -30(x – 25) = 0
⇒ (x – 25)(x – 30) = 0
Either, x -25 = 0 or x – 30 = 0
⇒ x = 25 or x = 30
Hence, the number of toys will be either 25 or 30.

3. Find two numbers whose sum is 27 and product is 182.

Answer

Let the first number be x and the second number is 27 – x.
Therefore, their product = x (27 – x)
It is given that the product of these numbers is 182.
Therefore, x(27 – x) = 182
⇒ x2 – 27x – 182 = 0
⇒ x2 – 13x – 14x + 182 = 0
⇒ x(x – 13) -14(x – 13) = 0
⇒ (x – 13)(x -14) = 0
Either x = -13 = 0 or x – 14 = 0
⇒ x = 13 or x = 14
If first number = 13, then
Other number = 27 – 13 = 14
If first number = 14, then
Other number = 27 – 14 = 13
Therefore, the numbers are 13 and 14.

4. Find two consecutive positive integers, sum of whose squares is 365.

Answer

Let the consecutive positive integers be x and x + 1.
Therefore, x2 + (x + 1)2 = 365
⇒ x2 + x2 + 1 + 2x = 365
⇒ 2x2 + 2x – 364 = 0
⇒ x2 + x – 182 = 0
⇒ x2 + 14x – 13x – 182 = 0
⇒ x(x + 14) -13(x + 14) = 0
⇒ (x + 14)(x – 13) = 0
Either x + 14 = 0 or x – 13 = 0,
⇒ x = – 14 or x = 13
Since the integers are positive, x can only be 13.
∴ x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers will be 13 and 14.

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Answer

Let the base of the right triangle be x cm.
Its altitude = (x – 7) cm
From Pythagoras theorem, we have
Base2 + Altitude2 = Hypotenuse2
∴ x2 + (x – 7)2 = 132
⇒ x2 + x2 + 49 – 14x = 169
⇒ 2x2 – 14x – 120 = 0
⇒ x2 – 7x – 60 = 0
⇒ x2 – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12)(x + 5) = 0
Either x – 12 = 0 or x + 5 = 0,
⇒ x = 12 or x = – 5
Since sides are positive, x can only be 12.
Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm = 5 cm.

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.

Answer

Let the number of articles produced be x.
Therefore, cost of production of each article = Rs (2x + 3)
It is given that the total production is Rs 90.∴ x(2x + 3) = 0
⇒ 2x2 + 3x – 90 = 0
⇒ 2x2 + 15x -12x – 90 = 0
⇒ x(2x + 15) -6(2x + 15) = 0
⇒ (2x + 15)(x – 6) = 0
Either 2x + 15 = 0 or x – 6 = 0
⇒ x = -15/2 or x = 6
As the number of articles produced can only be a positive integer, therefore, x can only be 6.
Hence, number of articles produced = 6
Cost of each article = 2 × 6 + 3 = Rs 15.
Exercise 4.3

1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x2 – 7x +3 = 0
(ii) 2x2 + x – 4 = 0
(iii) 4x2 + 4√3x + 3 = 0
(iv) 2x2 + x + 4 = 0

Answer

(i) 2x2 – 7x + 3 = 0
⇒ 2x2 – 7x = – 3
On dividing both sides of the equation by 2, we get
⇒ x2 – 7x/2  = -3/2
⇒ x2 – 2 × x ×  7/4 = -3/2
On adding (7/4)2 to both sides of equation, we get
⇒ (x)2 – 2 × x × 7/4 + (7/4)2 = (7/4)2 – 3/2
⇒ (x – 7/4)2 = 49/16 – 3/2
⇒ (x – 7/4)2 = 25/16
⇒ (x – 7/4) = ± 5/4
⇒ x = 7/4 ± 5/4
⇒ x = 7/4 + 5/4 or x = 7/4 – 5/4
⇒ x = 12/4 or x = 2/4
⇒ x = 3 or 1/2

(ii) 2x2 + x – 4 = 0
⇒ 2x2 + x = 4
On dividing both sides of the equation, we get
⇒ x2 + x/2 = 2
On adding (1/4)2 to both sides of the equation, we get
⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = 2 + (1/4)2
⇒ (x + 1/4)2 = 33/16
⇒ x + 1/4 = ± √33/4
⇒ x = ± √33/4 – 1/4
⇒ x = ± √33-1/4
⇒ x = √33-1/4 or x = -√33-1/4

(iii) 4x2 + 4√3x + 3 = 0
⇒ (2x)2 + 2 × 2x × √3 + (√3)2 = 0
⇒ (2x + √3)2 = 0
⇒ (2x + √3) = 0 and (2x + √3) = 0
⇒ x = -√3/2 or x = -√3/2

(iv) 2x2 + x + 4 = 0
⇒ 2x2 + x = -4
On dividing both sides of the equation, we get
⇒ x2 + 1/2x = 2
⇒ x2 + 2 × x × 1/4 = -2
On adding (1/4)2 to both sides of the equation, we get
⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = (1/4)2 – 2
⇒ (x + 1/4)2 = 1/16 – 2
⇒ (x + 1/4)2 = -31/16
However, the square of number cannot be negative.
Therefore, there is no real root for the given equation.

2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

Answer

(i) 2x2 – 7x + 3 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = -7 and c = 3
By using quadratic formula, we get
x = –b±√b2 – 4ac/2a
⇒ x = 7±√49 – 24/4
⇒ x = 7±√25/4
⇒ x = 7±5/4
⇒ x = 7+5/4 or x = 7-5/4
⇒ x = 12/4 or 2/4
∴  x = 3 or 1/2

(ii) 2x2 + x – 4 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = 1 and c = -4
By using quadratic formula, we get
x = –b±√b2 – 4ac/2a
⇒x = -1±√1+32/4
⇒x = -1±√33/4
∴ x = -1+√33/4 or x = -1-√33/4

(iii) 4x2 + 4√3x + 3 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 4, b = 4√3 and c = 3
By using quadratic formula, we get
x = –b±√b2 – 4ac/2a
⇒ x = -4√3±√48-48/8
⇒ x = -4√3±0/8
∴ x = -√3/2 or x = -√3/2

(iv) 2x2 + x + 4 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = 1 and c = 4
By using quadratic formula, we get
x = –b±√b2 – 4ac/2a
⇒ x = -1±√1-32/4
⇒ x = -1±√-31/4
The square of a number can never be negative.
∴There is no real solution of this equation.

Page No: 88

3. Find the roots of the following equations:
(i) x-1/x = 3, x ≠ 0
(ii) 1/x+4 – 1/x-7 = 11/30, x = -4, 7

Answer

(i) x-1/x = 3
⇒ x2 – 3x -1 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = -3 and c = -1
By using quadratic formula, we get
x = –b±√b2 – 4ac/2a
⇒ x = 3±√9+4/2
⇒ x = 3±√13/2
∴ x = 3+√13/2 or x = 3-√13/2

(ii) 1/x+4 – 1/x-7 = 11/30
⇒ x-7-x-4/(x+4)(x-7) = 11/30
⇒ -11/(x+4)(x-7) = 11/30
⇒ (x+4)(x-7) = -30
⇒ x2 – 3x – 28 = 30
⇒ x2 – 3x + 2 = 0
⇒ x2 – 2x – x + 2 = 0
⇒ x(x – 2) – 1(x – 2) = 0
⇒ (x – 2)(x – 1) = 0
⇒ x = 1 or 2

4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Answer

Let the present age of Rehman be x years.
Three years ago, his age was (x – 3) years.
Five years hence, his age will be (x + 5) years.
It is given that the sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is 1/3.

∴ 1/x-3 + 1/x-5 = 1/3
x+5+x-3/(x-3)(x+5) = 1/3
2x+2/(x-3)(x+5) = 1/3
⇒ 3(2x + 2) = (x-3)(x+5)
⇒ 6x + 6 = x2 + 2x – 15
⇒ x2 – 4x – 21 = 0
⇒ x2 – 7x + 3x – 21 = 0
⇒ x(x – 7) + 3(x – 7) = 0
⇒ (x – 7)(x + 3) = 0
⇒ x = 7, -3
However, age cannot be negative.
Therefore, Rehman’s present age is 7 years.

 

5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Answer

Let the marks in Maths be x.
Then, the marks in English will be 30 – x.
According to the question,
(x + 2)(30 – x – 3) = 210
(x + 2)(27 – x) = 210
⇒ –x2 + 25x + 54 = 210
⇒ x2 – 25x + 156 = 0
⇒ x2 – 12x – 13x + 156 = 0
⇒ x(x – 12) -13(x – 12) = 0
⇒ (x – 12)(x – 13) = 0
⇒ x = 12, 13
If the marks in Maths are 12, then marks in English will be 30 – 12 = 18
If the marks in Maths are 13, then marks in English will be 30 – 13 = 17

 
6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
 
Answer
 

Let the shorter side of the rectangle be x m.
Then, larger side of the rectangle = (x + 30) m

⇒ x2 + (x + 30)2 = (x + 60)2
⇒ x2 + x2 + 900 + 60x = x2 + 3600 + 120x
⇒ x2 – 60x – 2700 = 0
⇒ x2 – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30(x -90)
⇒ (x – 90)(x + 30) = 0
⇒ x = 90, -30

However, side cannot be negative. Therefore, the length of the shorter side will be 90 m.
Hence, length of the larger side will be (90 + 30) m = 120 m.

 
7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
 
Answer
 

Let the larger and smaller number be x and y respectively.
According to the question,
x2 – y2 = 180 and y2 = 8x
⇒ x2 – 8x = 180
⇒ x2 – 8x – 180 = 0
⇒ x2 – 18x + 10x – 180 = 0
⇒ x(x – 18) +10(x – 18) = 0
⇒ (x – 18)(x + 10) = 0
⇒ x = 18, -10

However, the larger number cannot be negative as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.
Therefore, the larger number will be 18 only.
x = 18
∴ y2 = 8x = 8 × 18 = 144
⇒ y = ±√44 = ±12
∴ Smaller number = ±12
Therefore, the numbers are 18 and 12 or 18 and – 12.

8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Answer

Let the speed of the train be x km/hr.
Time taken to cover 360 km = 360/x hr.According to the question,
⇒ (x + 5)(360-1/x) = 360
⇒ 360 – x + 1800-5/x = 360
⇒ x2 + 5x + 10x – 1800 = 0
⇒ x(x + 45) -40(x + 45) = 0
⇒ (x + 45)(x – 40) = 0
⇒ x = 40, -45
However, speed cannot be negative.
Therefore, the speed of train is 40 km/h.

 

9. Two water taps together can fill a tank inhours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer

Let the time taken by the smaller pipe to fill the tank be x hr.
Time taken by the larger pipe = (x – 10) hr
Part of tank filled by smaller pipe in 1 hour = 1/x

Part of tank filled by larger pipe in 1 hour = 1/x – 10

It is given that the tank can be filled in = 75/8 hours by both the pipes together. Therefore,
1/x + 1/x-10 = 8/75
x-10+x/x(x-10) = 8/75
⇒ 2x-10/x(x-10) = 8/75
⇒ 75(2x – 10) = 8x2 – 80x
⇒ 150x – 750 = 8x2 – 80x
⇒ 8x2 – 230x +750 = 0
⇒ 8x2 – 200x – 30x + 750 = 0
⇒ 8x(x – 25) -30(x – 25) = 0
⇒ (x – 25)(8x -30) = 0
⇒ x = 25, 30/8
Time taken by the smaller pipe cannot be 30/8 = 3.75 hours. As in this case, the time taken by the larger pipe will be negative, which is logically not possible.

Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours respectively.

10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Answer

Let the average speed of passenger train be x km/h.
Average speed of express train = (x + 11) km/h
It is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.

 
⇒ 132 × 11 = x(x + 11)
⇒ x2 + 11x – 1452 = 0
⇒ x2 +  44x -33x -1452 = 0
⇒ x(x + 44) -33(x + 44) = 0
⇒ (x + 44)(x – 33) = 0
⇒ x = – 44, 33

Speed cannot be negative.

Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.

 

11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Answer

Let the sides of the two squares be x m and y m. Therefore, their perimeter will be 4x and 4y respectively and their areas will be x2 and y2 respectively.
It is given that
4x – 4y = 24
x – y = 6
x = y + 6
Also, x2 + y2 = 468
⇒ (6 + y2) + y2 = 468
⇒ 36 + y2 + 12y + y2 = 468
⇒ 2y2 + 12y + 432 = 0
⇒ y2 + 6y – 216 = 0
⇒ y2 + 18y – 12y – 216 = 0
⇒ y(y +18) -12(y + 18) = 0
⇒ (y + 18)(y – 12) = 0
⇒ y = -18, 12
However, side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.

Exercise 4.4

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;
(i) 2x2 – 3x + 5 = 0
(ii) 3x2 – 4√3x + 4 = 0
(iii) 2x2 – 6x + 3 = 0

Answer

(i) Consider the equation
x2 – 3x + 5 = 0
Comparing it with ax2 + bx + c = 0, we get
a = 2, b = -3 and c = 5
Discriminant = b2 – 4ac
= ( – 3)2 – 4 (2) (5) = 9 – 40
= – 31
As b2 – 4ac < 0,
Therefore, no real root is possible for the given equation.

(ii) 3x2 – 4√3x + 4 = 0
Comparing it with ax2 + bx + c = 0, we get
a = 3, b = -4√3 and c = 4
Discriminant = b2 – 4ac
= (-4√3)2 – 4(3)(4)
= 48 – 48 = 0
As b2 – 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be –b/2a and –b/2a.-b/2a = -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3
Therefore, the roots are 2/√3 and 2/√3.

 
(iii) 2x2 – 6x + 3 = 0

Comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = -6, c = 3
Discriminant = b2 – 4ac

= (-6)2 – 4 (2) (3)
= 36 – 24 = 12
As b2 – 4ac > 0,
Therefore, distinct real roots exist for this equation:

2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2 + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0


Answer

(i) 2x2 + kx + 3 = 0
Comparing equation with ax2 + bx + c = 0, we get
a = 2, b = k and c = 3
Discriminant = b2 – 4ac

= (k)2 – 4(2) (3)
= k2 – 24
For equal roots,
Discriminant = 0
k2 – 24 = 0
k2 = 24
k = ±√24 = ±2√6

(ii) kx(x – 2) + 6 = 0
or kx2 – 2kx + 6 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = k, b = – 2k and c = 6
Discriminant = b2 – 4ac
= ( – 2k)2 – 4 (k) (6)
= 4k2 – 24k
For equal roots,
b2 – 4ac = 0
4k2 – 24k = 0
4k (k – 6) = 0
Either 4k = 0
or k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms ‘x2‘ and ‘x‘.
Therefore, if this equation has two equal roots, k should be 6 only.
 
3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

Answer

Let the breadth of mango grove be l.
Length of mango grove will be 2l.
Area of mango grove = (2l) (l)= 2l2
2l2 = 800
l2 = 800/2 = 400
l2 – 400 =0
Comparing this equation with al2 + bl + c = 0, we get
a = 1, b = 0, c = 400
Discriminant = b2 – 4ac
= (0)2 – 4 × (1) × ( – 400) = 1600
Here, b2 – 4ac > 0
Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.
l = ±20
However, length cannot be negative.
Therefore, breadth of mango grove = 20 m
Length of mango grove = 2 × 20 = 40 m


4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Answer

Let the age of one friend be x years.
then the age of the other friend will be (20 – x) years.
4 years ago,

Age of 1st friend = (x – 4) years
Age of 2nd friend = (20 – x – 4) = (16 – x) years
A/q we get that,
(x – 4) (16 – x) = 48

16x – x2 – 64 + 4x = 48
 – x2 + 20x – 112 = 0
x2 – 20x + 112 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = -20 and c = 112
Discriminant = b2 – 4ac
= (-20)2 – 4 × 112
= 400 – 448 = -48
b2 – 4ac < 0
Therefore, there will be no real solution possible for the equations. Such type of condition doesn’t exist.

5. Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth.

Answer

Let the length and breadth of the park be l and b.
Perimeter = 2 (l + b) = 80
l + b = 40
Or, b = 40 – l
Area = l×b = l(40 – l) = 40l – l240l –  l2 = 400
l2 –  40l + 400 = 0
Comparing this equation with al2 + bl + c = 0, we get
a = 1, b = -40, c = 400
Discriminant = b2 – 4ac
(-40)2 – 4 × 400
= 1600 – 1600 = 0
b2 – 4ac = 0
Therefore, this equation has equal real roots. And hence, this situation is possible.
Root of this equation,l = –b/2a
l = (40)/2(1) = 40/2 = 20
Therefore, length of park, l = 20 m
And breadth of park, b = 40 – l = 40 – 20 = 20 m.

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