Author Archives: Swati

Chapter 18 – Surface Areas and Volume of a Cuboid and Cube Exercise Ex. 18.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with coloured paper with picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively. How many square sheets of paper of side 40 cm would she require?

Solution 4

Question 5

Solution 5

Question 6Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9Hameed has built a cubical water tank with lid for his house, with each outer edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm. Find how much he would spend for the titles, if the cost of tiles is Rs. 360 per dozen.Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Ravish wanted to make a temporary shelter for his car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m x 3m?Solution 12Length of shelter = 4 m
Breadth of shelter = 3 m
Height of shelter = 2.5 m

The tarpaulin will be required for top and four sides of the shelter.
Area of Tarpaulin required = 2(lh + bh) + lb
= [2(4  2.5 + 3  2.5) + 4  3] m²
= [2(10 + 7.5) + 12] m²
= 47 m²

Question 13

An open box is made of wood 3 cm thick. Its external length, breadth and height are 1.48m, 1.16 m and 8.3 dm. Find the cost of painting the inner surface at Rs 50 per sq meter.Solution 13

Question 14

Solution 14

Question 15The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?Solution 15Total surface area of one brick = 2(lb + bh + lh)
 = [2(22.5 × 10 + 10 × 7.5 + 22.5 × 7.5)]cm²
 = 2(225 + 75 + 168.75) 
          = (2 × 468.75) cm²
 = 937.5 cm²
Let n number of bricks be painted by the container.
Area of n bricks = 937.5n cm²
Area that can be painted by the container = 9.375 m² = 93750 cm²
 93750 = 937.5n
n = 100
Thus, 100 bricks can be painted out by the container.

Question 16

Solution 16

Question 17

The cost of preparing the walls of a room 12 m long at the rate of Rs 1.35 per square meter is Rs 340.20 and the cost of matting the floor at 85 paise per square meter is Rs 91.80. Find the height of the room.Solution 17

Question 18

Solution 18

Question 19A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm. The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm² and the rate of painting is 10 paise per cm², find the total expenses required for polishing and painting the surface of the bookshelf. 

                                           Solution 19    External length (l) of bookshelf = 85 cm
    External breadth (b) of bookshelf = 25 cm
    External height (h) of bookshelf = 110 cm
    External surface area of shelf while leaving front face of shelf
                                                  = lh + 2 (lb + bh)
                                                  = [85  110 + 2 (85  25 + 25  110)] cm²
                                                  = 19100 cm²
    Area of front face = [85  110 – 75  100 + 2 (75  5)] cm²
                                                  = 1850 + 750 cm²
                                                  = 2600 cm²
    Area to be polished = (19100 + 2600) cm² = 21700 cm²
    Cost of polishing 1 cm² area = Rs 0.20
    Cost of polishing 21700 cm² area = Rs (21700  0.20) = Rs 4340    

    Now, length (l), breadth (b) height (h) of each row of bookshelf is 75 cm, 20 cm, and    30cm  respectively.
    Area to be painted in 1 row = 2 (l + h) b + lh
                                           = [2 (75 + 30)  20 + 75  30] cm²
                                           = (4200 + 2250) cm²
                                           = 6450 cm²
    Area to be painted in 3 rows = (3  6450) cm² = 19350 cm²
    Cost of painting 1 cm² area = Rs 0.10
    Cost of painting 19350 cm² area = Rs (19350  0.10) = Rs 1935    
    Total expense required for polishing and painting the surface of the bookshelf                                             = Rs(4340 + 1935) = Rs 6275

Chapter 18 – Surface Areas and Volume of a Cuboid and Cube Exercise Ex. 18.2

Question 1A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How, many litres of water can it holds? Solution 1Volume of tank = l  b  h = (6  5  4.5) m³ = 135 m³  It is given that:
 1 m³ = 1000 litres


 Thus, the tank can hold 135000 litres of water.

Question 2A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?   Solution 2Let height of cuboidal vessel be h.
Length (l) of vessel = 10 m
Width (b) of vessel = 8 m
Volume of vessel = 380 m³
l  b  h = 380        
 10  8  h = 380
 h = 4.75

 Thus, the height of the vessel should be 4.75 m.    

Question 3Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs 30 per m³.

Solution 3Length (l) of the cuboidal pit = 8 m
Width (b) of the cuboidal pit = 6 m
 Depth (h) of the cuboidal pit = 3 m
Volume of the cuboidal pit = l  b  h = (8  6  3)  = 144 m³
Cost of digging 1 m³ = Rs 30
Cost of digging 144 m³ = Rs (144 30) = Rs 4320

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?Solution 16Length (l) of the cuboidal tank = 20 m
Breadth (b) of the cuboidal tank = 15 m
Height (h) of the cuboidal tank = 6 m

Capacity of tank = l × b × h    = (20 × 15 × 6) m³ = 1800 m³ = 1800000 litres

Water consumed by people of village in 1 day = 4000 × 150 litres = 600000 litres

Let water of this tank lasts for n days.
Water consumed by all people of village in n days = capacity of tank
n × 600000 = 1800000
n = 3    
Thus, the water of tank will last for 3 days.

Question 17

A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in fig. If the edge of each cube is 3 cm, find the volume of the structure built by the child

Solution 17

Question 18A godown measures 40 m  25 m  10 m. Find the maximum number of wooden crates each measuring 1.5 m  1.25 m  0.5 m that can be stored in the godown.    

Solution 18Length  of the godown = 40 m
    Breadth  of the godown = 25 m
    Height  of the godown = 10 m

Volume of godown = l₁ b₁ h₁ = (40  25  10)  = 10000 

    Length  of a wooden crate = 1.5 m
    Breadth  of a wooden crate = 1.25 m
    Height  of a wooden crate = 0.5 m

Volume of a wooden crate =  = (1.5  1.25  0.5) m3 = 0.9375 

Let n wooden crates be stored in the godown.
Volume of n wooden crates = volume of godown
0.9375  n = 10000

Thus, 10666 wooden crates can be stored in godown.

Question 19

A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm x 12 cm x 8 cm, how many bricks would be required?Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?Solution 22Rate of water flow = 2 km per hour  
Depth (h) of river = 3 m
Width (b) of river = 40 m
Volume of water flowed in 1 min 
Thus, in 1 minute 4000  = 4000000 litres of water will fall into the sea.    

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

A rectangular container, whose base is a square of side 5 cm, stands on a horizontal table, and holds water upto 1 cm from the top. When a cube is placed in the water it is completely submerged, the water rises to the top and 2 cubic cm of water overflows. Calculate the volume of the cube and also the length of its edge.Solution 26

Question 27

Solution 27

Question 28

Solution 28

Chapter 17 – Heron’s Formula Exercise Ex. 17.1

Question 1

Find the area of the triangle whose sides are respectively 150 cm, 120 cm and 200 cm.Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5The perimeter of triangular field is 540 m and its sides are in the ratio 25:17:12. Find the area of the triangle.Solution 5The sides of the triangular field are in the ratio 25:17:12.
Let the sides of triangle be 25x, 17x, and 12x.
  Perimeter of this triangle = 540 m
             25x + 17x + 12x = 540 m
                         54x = 540 m
                            x = 10 m
  Sides of triangle will be 250 m, 170 m, and 120 m. Semi-perimeter (s) =  By Heron’s formula:

    So, area of the triangle is 9000 m².Question 6

The perimeter of right triangle is 300m. If its sides are in the ratio 3 : 5 : 7. Find the area of the triangle.Solution 6

Question 7

The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.Solution 7

Question 8

A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.Solution 8

Question 9

The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side.Solution 9

Question 10

The perimeter of an isosceles triangle is 42 cm and its base is (3/2) times each of the equal sides. Find the length of each side of the triangle, area of the triangle and the height of the triangle.Solution 10

Question 11

Find the area of the shaded region in fig.12.12

Solution 11

Chapter 17 – Heron’s Formula Exercise Ex. 17.2

Question 1Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.Solution 1

For ABC
AC² = AB² + BC²
(5)² = (3)² + (4)²
So, ABC is a right angle triangle, right angled at point B.
Area of ABCFor ADC
Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm
            s = 7 cm
By Heron’s formula
Area of triangle 

Area of ABCD = Area of ABC + Area of ACD
    = (6 + 9.166) cm² = 15.166 cm² = 15.2 cm² (approximately)

Question 2

Solution 2

Question 3

Solution 3

Question 4A park, in the shape of a quadrilateral ABCD, has  = 90^o, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?Solution 4Let us join BD.
In BCD applying Pythagoras theorem
BD² = BC² + CD²
       = (12)² + (5)²
       = 144 + 25
BD² = 169
  BD = 13 m

Area of BCD

                  For ABD

                    By Heron’s formula Area of triangle  

                               Area of park = Area of ABD + Area of BCD
                         = 35.496 + 30 m²                         = 65.496 m²                         = 65. 5 m² (approximately)

Question 5

Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.Solution 5

Question 6

A rhombus sheet, whose perimeter is 32 m and whose one diagonal is 10 m long, is painted on both sides at the rate of Rs 5 per m². Find the cost of painting.Solution 6

Question 7

Solution 7

Question 8

Find the area of a quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.Solution 8

Question 09

The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.Solution 09

Question 10

Find the area of the blades of the magnetic compass shown in fig.

(Take √11 = 3.32)

Solution 10

Question 11

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

A hand fan is made by stitching 10 equal size triangular strips of two different types of paper as shown in fig., The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each type of paper needed to make the hand fan.

Solution 14

Chapter 16 – Constructions Exercise Ex. 16.1

Question 1

Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.Solution 1

Question 2

Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this segment.Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Draw a line segment AB and bisect it. Bisect one of the equal parts of obtain a line segment of length  (AB)Solution 6

Question 7

Solution 7

Chapter 16 – Constructions Exercise Ex. 16.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Using ruler and compasses only, draw a right angle.Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10(i)

Solution 10(i)

Question 10(ii)

Solution 10(ii)

Question 11(i)

Solution 11(i)

Question 11(ii)

Solution 11(ii)

Question 11(iii)

Solution 11(iii)

Question 11(iv)

Solution 11(iv)

Question 11(v)

Construct an angle of 15^o.Solution 11(v)

Question 11(vi)

Solution 11(vi)

Chapter 16 – Constructions Exercise Ex. 16.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Construct a right angled triangle whose perimeter is equal to 10 cm and one actute angle equal to 60^o.Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 15 – Circles Exercise Ex. 15.1

Question 1

Fill in the blanks:

(i) All points lying inside/outside a circle are called …… points/ … points.

(ii) Circles having the same centre and different radii are called … circles.

(iii) A point whose distance from the centre of a circle is greater than its radius lies in … of the circle.

(iv) A continuous piece of a circle is … of the circle.

(v) The longest chord of a circle is a … of the circle.

(vi) An arc is a … when its ends are the ends of a diameter.

(vii) Segment of a circle is the region between an arc and … of the circle.

(viii) A circle divides the plane, on which it lies, in …. parts.Solution 1

(i) interior/exterior

(ii) concentric

(iii) the exterior

(iv) arc

(v) diameter

(vi) semi-circle

(vii) centre

(viii) threeQuestion 2

Write the truth value (T/F) of the following with suitable reasons:

(i) A circle is a plane figure.

(ii) Line segment joining the centre to any point on the circle is a radius of the circle.

(iii) If a circle is divided into three equal arcs each is a major arc.

(iv) A circle has only finite number of equal chords.

(v) A chord of a circle, which is twice as long is its radius is a diameter of the circle.

(vi) Sector is the region between the chord and its corresponding arc.

(vii) The degree measure of an arc is the complement of the central angle containing the arc.

(viii) The degree measure of a semi-circle is 180^o.Solution 2

(i) T

(ii) T

(iii) T

(iv) F

(v) T

(vi) T

(vii) F

(viii) T

Chapter 15 – Circles Exercise Ex. 15.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Give a method to find the centre of a given circle.Solution 4

Steps of construction:

(1) Take three point A, B and C on the given circle.

(2) Join AB and BC.

(3) Draw the perpendicular bisectors of chord AB and BC which interesect each other at O.

(4) Point O will be the required circle because we know that the perpendicular bisector of a chord always passes through the centre.Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord form the centre?Solution 11

                                              Distance of smaller chord AB from centre of circle = 4 cm.
OM = 4 cm

In OMB

In ONDOD=OB=5cm             (radii of same circle)

 So, distance of bigger chord from centre is 3 cm.Question 12

Solution 12

Question 13

Solution 13

Question 14

Prove that two different circles cannot intersect each other at more than two points.Solution 14

Suppose two different circles can intersect each other at three points then they will pass through the three common points but we know that there is one and only one circle with passes through three non-collinear points, which contradicts our supposition.

Hence, two different circles cannot intersect each other at more than two points.Question 15Two chords AB and CD of lengths 5 cm and 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.Solution 15Draw OM  AB and ON  CD. Join OB and OD 

                     (Perpendicular from centre bisects the chord)

Let ON be x, so OM will be 6 – x
In MOB

In NOD

 We have OB = OD             (radii of same circle)
So, from equation (1) and (2) 

From equation (2) 

So, radius of circle is found to be  cm.

Chapter 15 – Circles Exercise Ex. 15.3

Question 1

Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha, Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha?Solution 1

Question 2

A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.Solution 2

Chapter 15 – Circles Exercise Ex. 15.4

Question 1

In fig., O is the centre of the circle. If ∠APB = 50°, find ∠AOB and ∠OAB.

Solution 1

Question 2

In fig., O is the centre of the circle. Find ∠BAC.

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 3(vii)

If O is the centre of the circle. Find the value of x in the following figure:

Solution 3(vii)

Question 3(viii)

Solution 3(viii)

Question 3(ix)

If O is the centre of the circle. Find the value of x in the following figure:

Solution 3(ix)

Question 3(x)

If O is the centre of the circle. Find the value of x in the following figure:

Solution 3(x)

Question 3(xi)

If O is the centre of the circle. Find the value of x in the following figure:

Solution 3(xi)

Question 3(xii)

Solution 3(xii)

Question 4

Solution 4

Question 5

In fig., O is the centre of the circle, BO is  the bisector of ∠ABC. Show that AB = AC.

Solution 5

Question 6

In fig., O and O’ are centres of two circles intersecting at B and C. ABD is straight line, find x.

Solution 6

Question 7

In fig., if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.

Solution 7

Question 8

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.Solution 8

Question 9

In fig., it is given given that O is the centre of the circle and ∠AOC = 150°. Find ∠ABC.

Solution 9

Question 10

In fig., O is the centre of the circle, prove that ∠x = ∠y + ∠z.

Solution 10

Question 11

in fig., O is the centre of a circle and PQ is a diameter. If ∠ROS = 40°, find. ∠RTS.

Solution 11

Chapter 15 – Circles Exercise Ex. 15.5

Question 1

In fig., ΔABC is an equilateral triangle. Find m∠BEC.

Solution 1

Question 2

In fig., ΔPQR is an isosceles triangle with PQ = PR and m∠PQR = 35°. find m∠QSR and m∠QTR.

Solution 2

Question 3

In fig., O is the centre of the circle. If ∠BOD = 160°, find the values of x and y.

Solution 3

Question 4

In fig., ABCD is a cyclic qudrilateral. If ∠BCD = 100° and ABD = 70°, find ∠ADB.

Solution 4

Question 5

If ABCD is a cyclic quadrilateral in which AD ∥ BC. Prove that ∠B = ∠C.

Solution 5

Question 6

In fig., O is the centre of the circle. find ∠CBD.

Solution 6

Question 7

In fig., AB and CD are diameters of a circle with centre O. If ∠OBD = 50°, find ∠AOC.

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

In fig., O is the centre of the circle and DAB = 50. calculate the values of x and y.

Solution 11

Question 12

In fig., if ∠BAC = 60°, and ∠BCA = 20°, find ∠ADC.

Solution 12

Question 13

In fig., if ABC is an equilateral triangle. Find ∠BDC and ∠BEC.

Solution 13

Question 14

In fig., O is the centre of the circle. If ∠CEA = 30°, find the values of x, y and z.

Solution 14

Question 15

In fig., ∠BAD = 78°, ∠DCF = x° and DEF = y° find the values of x and y.

Solution 15

Question 16

Solution 16

Question 17

In fig., ABCD is cyclic qudrilateral. Find the value of x.

Solution 17

Question 18(i)

Solution 18(i)

Question 18(ii)

Solution 18(ii)

Question 18(iii)

Solution 18(iii)

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

In fig., ABCD is cyclic quadrilaterial in which AC an BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.

Solution 23

Question 24

Solution 24

Question 25

Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.Solution 25

Let O be the centre of the circle circumscribing the cyclic rectangle ABCD. Since ABC = 90^o and AC is a chord of the circle, so, AC is a diameter of the circle. Similarly, BD is a diameter.

Hence, point of intersection of AC and BD is the centre of the circle.Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half of the hypotenuse.Solution 29

Chapter 14 – Areas of Parallelograms and Triangles Exercise Ex. 14.1

Question 1

Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:

Solution 1

(i) ΔAPB and trapezium ABCD are on the same base AB and between the same parallels AB and CD.

(ii) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.

(iii) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC but they are not on the same base.

(iv) ΔQRT and parallelogram PQRS are on the same base QR and between the same parallels QR and PS.

(v) Parallelogram PQRS and trapezium SMNR are on the same base SR but they are not between the same parallels.

(vi) Parallelograms PQRS, AQRD, BQRC are between the same parallels. Also, parallelograms PQRS, BPSC and APSD are between the same parallels.

Chapter 14 – Areas of Parallelograms and Triangles Exercise Ex. 14.2

Question 1In the given figure, ABCD is parallelogram, AE  DC and CF  AD. If AB = 16 cm. AE = 8 cm and CF = 10 cm, find AD.  

Solution 1In parallelogram ABCD, CD = AB = 16 cm     [Opposite sides of a parallelogram are equal]
We know that,
Area of parallelogram = Base x corresponding attitude
Area of parallelogram ABCD = CD x AE = AD x CF  
16 cm x 8 cm = AD x 10 cmAD =  cm = 12.8 cm.Thus, the length of AD is 12.8 cm.Question 2

In Q. No. 1, if AD = 6 cm, CF = 10 cm, and AE = 8, find AB.Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 14 – Areas of Parallelograms and Triangles Exercise Ex. 14.3

Question 1

In fig., compute the area of quadrilateral ABCD.

Solution 1

Question 2

In the fig., PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of ΔOTS if PQ = 8 cm.

Solution 2

Question 3

Compute the area of trapezium PQRS in fig.

Solution 3

Question 4

In fig., ∠AOB = 90, AC = BC, OA = 12 cm and OC = 6.5 cm. find the area of ΔAOB

Solution 4

Question 5

In fig., ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

Solution 5

Question 6

Solution 6

Question 7

In fig., ABCD is a trapezium in which AB ∥ DC. PRove that ar (ΔAOD) = ar (ΔBOC)

Solution 7

Question 8

Solution 8

Question 9

In fig., ABC and ABD are two triangles on the base Ab. If the line segment CD is bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD).

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

In fig., D and E are two points on BC such that BD = DE = EC. Show that ar (ΔABD) = ar (ΔADE) = ar(ΔAEC).

Solution 15

Draw a line l through A parallel to BC.

Given that, BD = DE = EC.

We observe that the triangles ABD, ADE and AEC are on the equal bases and between the same parallels l and BC. Therefore, their areas are equal.

Hence, ar (ΔABD) = ar (ΔADE) = ar(ΔAEC).Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

(i) 
(ii) 

(iii)

Question 20

Solution 20

Question 21

In fig., CD ∥ AE and CY ∥ BA.

(i) Name a triangle equal in area of ΔCBX 

(ii) Prove that ar (ΔZDE) = ar (ΔCZA)

(iii) Prove that ar (BCZY) = ar (ΔEDZ)

Solution 21

Question 22

In fig., PSDA is a parallelogram in which PQ = QR = RS and AP ∥ BQ ∥CR. Prove that ar(Δ PQE) = ar (Δ CFD).

Solution 22

Question 23

In fig., ABCD is a trapezium in which AB DC and DC = 40 cm and AB = 60 cm. If X and Y are, respectively, the mid – points of AD and BC, prove that:

(i) XY = 50 cm

(ii) DCYX is a trapezium

(iii) ar (trap. DCYX) = (9/11)ar (trap.(XYBA)

Solution 23

Question 24

Solution 24

Question 25

In fig., X and Y are the mid-points of AC and AB respectively, QP ∥ BC and CYQ and BXP are straight lines. Prove that ar(Δ ABP) = ar (Δ ACQ)

Solution 25

Question 26

In fig., ABCD and AEFD are two parallelograms. Prove that

(i) PE = FQ

(ii) ar(ΔAPE) : ar(ΔPFA) = ar Δ(QFD) : ar (ΔPFD)

(iii) ar(ΔPEA) = ar (ΔQFD)

Solution 26

Question 27

In fig. ABCD is a ∥^gm. O is any point on AC. PQ ∥ AB and LM ∥ AD. Prove that ar(∥^gm DLOP) = ar (∥^gm BMOQ).

Solution 27

Question 28

Solution 28

Question 29

In fig., ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F. Prove that

(i) ar (BDE) = ar (ABC)

(ii) ar(BDE) = ar(BAE)

(iii) ar (BFE) = ar(AFD)

(iv) ar(ABC) = 2 ar(BEC)

(v) ar (FED) = ar(AFC)

(vi) ar(BFE) = 2 ar (EFD)

Solution 29

Given, ABC and BDE are two equilateral triangles.

Let AB = BC = CA = x. Then, BD = = DE = BE

(i) We have,

ar(ABC) = x²

ar (BDE) = 

ar(BDE) = ar (ABC)

(ii) It is given that triangles ABC and BED are equilateral triangles.

ACB = DBE = 60^o

BE||AC(Since, alternate angles are equal)

Triangles BAE and BEC are on the same base BE and between the same parallels BE and AC.

ar (BAE) = ar(BEC)

ar (BAE) =2 ar (BDE)

[ ED is a median of EBC ar(BEC) = 2ar(BDE)]

ar (BDE) = ar(BAE)

(iii) Since ABC and BDE are equilateral triangles.

ABC = 60^o and BDE = 60^o

ABC = BDE

AB||DE(Since, alternate angles are equal)

Triangles BED and AED are on the same base ED and between the same parallels AB and DE.

ar (BED) = ar(AED)

ar (BED) ar(EFD) = ar(AED) ar(EFD)

ar(BEF) = ar(AFD)

(iv) Since ED is a median of BEC

ar (BEC) = 2 ar (BDE)

ar (BEC) = ar (ABC)[From (i), ar (BDE) = ar (ABC)]

ar(BEC) = ar (ABC)

ar (ABC) = 2 ar (BEC)

(v) Let h be the height of vertex E, corresponding to the side BD in triangle BDE.

Let H be the height of vertex A, corresponding to the side BC in triangle ABC.

From part (i),

ar(BDE) = ar (ABC)

From part (iii),

ar (BFE) = ar (AFD)

(vi) ar (AFC) = ar (AFD) + ar (ADC)

= ar (BFE) + ar (ABC)

(Using part (iii); and AD is the median of ABC)

= ar (BFE) + 4 ar (BDE)(Using part (i))

= ar (BFE) + 2 ar (BDE) (2)

Now, from part (v),

ar (BFE) = 2ar (FED) (3)

ar (BDE) = ar (BFE) + ar (FED)

= 2 ar (FED) + ar (FED)

= 3 ar (FED) (4)

From (2), (3) and (4), we get,

ar (AFC) = 2ar (FED) + 2 3 ar (FED) = 8 ar (FED)

Hence, ar (FED) = ar(AFC)Now, fromQuestion 30

If fig., ABC is a right triangle right angled at A, BCED, ACFG and ABMN are square on the sides BC, CA and AB respectively. Line segment AX DE meets BC at Y. Show that

(i) MBC ABD

(ii) ar (BYXD) = 2ar(MBC)

(iii) ar(BYXD) = ar(ABMN)

(iv) FCB ACE

(v) ar(CYXE) = 2ar (FCB)

(vi) ar(CYXE) = ar (ACFG)

(vii) ar(BCED) = ar (ABMN) + ar (ACFG)

Solution 30

(i) In MBC and ABD, we have

MB = AB

BC = BD

And MBC = ABD

[MBC and ABD are obtained by adding ABC to a right angle]

So, by SAS congruence criterion, we have

MBC ABD

ar (MBC) = ar(ABD) (1)

(ii) Clearly, triangle ABD and rectangle BYXD are on the same base BD and between the same parallels AX and BD.

ar(ABD) = ar (rect. BYXD)

ar (rect. BYXD) = 2 ar(ABD)

ar (rect. BYXD) = 2 ar (MBC)…(2)

[ ar (ABD) = ar (MBC), from (1)]

(iii) Since triangle MBC and square MBAN are on the same base MB and between the same parallels MB and NC.

2 ar (MBC) = ar (MBAN) (3)

From (2) and (3), we have

ar (sq. MBAN) = ar(rec. BYXD)

(iv) In triangles FCB and ACE, we have

FC = AC

CB = CE

And, FCB = ACE

[FCB and ACE are obtained by adding ACB to a right angle]

So, by SAS congruence criterion, we have

FCB ACE

(v) We have,

FCB ACE

ar (FCB) = ar (ACE)

Clearly, ACE and rectangle CYXE are on the same base CE ad between the same parallels CE and AX.

2 ar (ACE) = ar (CYXE)

2 ar (FCB) = ar (CYXE) (4)

(vi) Clearly, FCB and rectangle FCAG are on the same base FC and between the same parallels FC and BG.

2ar (FCB) = ar(FCAG) (5)

From (4) and (5), we get

ar(CYXE) = ar (ACFG)

(vii) Applying Pythagoras theorem in ACB, we have

BC² = AB² + AC²

Chapter 13 – Quadrilaterals Exercise Ex. 13.1

Question 1

Solution 1

Question 2

Solution 2

Question 3The angles of quadrilateral are in the ratio 3: 5: 9: 13, Find all the angles of the quadrilateral.Solution 3Let the common ratio between the angles is x. So, the angles will be 3x, 5x, 9x and 13x respectively.
Since the sum of all interior angles of a quadrilateral is 360^o.
 3x + 5x + 9x + 13x = 360^o
30x = 360^o
    x = 12^o
Hence, the angles are
3x = 3  12 = 36^o
5x = 5  12 = 60^o
9x = 9  12 = 108^o
13x = 13  12 = 156^o

Question 4

Solution 4

Chapter 13 – Quadrilaterals Exercise Ex. 13.2

Question 1

Two opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of each of the parallelogram.Solution 1

Question 2

If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.Solution 2

Question 3

Find the measure of all the angles of a parallelogram, if one angle is 24^o less than twice the smallest angle.Solution 3

Question 4

The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm what is the measure of the shorter side?Solution 4

Question 5

In a parallelogram ABCD, ∠D = 135°, determine the measures of ∠A and ∠B.Solution 5

Question 6

ABCD is a parallelogram in which ∠A = 70. Compute ∠B, ∠C and ∠D.Solution 6

Question 7

In fig., ABCD is a parallelogram in which ∠DAB = 75° and ∠DBC = 60°. Compute ∠CDB and ∠ADB.

Solution 7

Question 8

Solution 8

i. F

ii. T

iii. F

iv. F

v. T

vi. F

vii. F

viii. TQuestion 9

In fig., ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠b meet at P, prove that AD = DP, PC = BC and DC = 2AD.

Solution 9

Question 10

In fig., ABCD is a parallelogram and E is the mid-point of side BC. IF DE and AB when produced meet at F, prove that AF = 2AB.

Solution 10

Chapter 13 – Quadrilaterals Exercise Ex. 13.3

Question 1

In a parallelogram ABCD, determine sum of angles ∠C and ∠D.Solution 1

C and D are cosecutive interior angles on the same side of the transversal CD. Therefore, 

C + D = 180^oQuestion 2

In a parallelogram ABCD, if ∠B = 135°, determine the measures of its other angles.Solution 2

Question 3

ABCD is a square. AC and BD intersect at O. State the measure of AOB.Solution 3

Since, diagonals of a square bisect each other at right angle. Therefore, AOB = 90^oQuestion 4

Solution 4

Question 5

The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.Solution 5

Question 6

P and Q are the points of trisection of the diagonal BD of a parallelogram ABCD. Prove that CQ is parallel to AP. Prove also that AC bisects PQ.Solution 6

Question 7

ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is square.Solution 7

Question 8

ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.Solution 8

Question 9

ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F. Prove that BF = BC.Solution 9

Chapter 13 – Quadrilaterals Exercise Ex. 13.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

In fig., triangle ABC is right-angled at B. Give that AB = 9 cm, AC = 15 cm and D, E are the mid-points of the sides AB and AC respectively, calculate

(i) The length of BC

(ii) The area of ADE.

Solution 7

Question 8

In fig., M, N, and P are the mid-points of AB, AC and BC respectively. If MN = 3 cm, Np = 3.5 cm and MP = 2.5 cm, Calculate BC, AB and AC.

Solution 8

Question 9

In fig., AB = AC and CP ∥ BA and AP is the bisector of exterior ∠CAD of ΔABC. Prove that (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram.

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.Solution 12

                                          Let ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively.
Join PQ, QR, RS, SP and BD.
In ABD, S and P are mid points of AD and AB respectively.
So, By using mid-point theorem, we can say that
SP || BD and SP =   BD             … (1)  
Similarly in BCD
QR || BD and QR =  BD               … (2)
From equations (1) and (2), we have
SP || QR and SP = QR  
As in quadrilateral SPQR one pair of opposite sides are equal and parallel to
each other.
So, SPQR is a parallelogram.Since, diagonals of a parallelogram bisect each other.
Hence, PR and QS bisect each other.

Question 13

Fill in the blanks to make the following statements correct:

(i) The triangle formed by joining the mid-points of the sides of an isosceles traingle is ______.

(ii) The triangle formed by joining the mid-points of the sides of a right triangle is ______ .

(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is ______ .Solution 13

(i) isosceles

(ii) right triangle

(iii) parallelogramQuestion 14

Solution 14

Question 15

In fig., BE ⊥ AC. AD is any line from A to BC interesting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°.

Solution 15

Question 16

Solution 16

Question 17

In fig., ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = (1/4)AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.

Solution 17

Question 18

In fig., ABCD and PQRC are rectangle and Q is the mid-point of AC. Prove that 

i. DP = PC ii. PR = (1/2) AC

Solution 18

Question 19

ABCD is a parallelogram, E and F are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC and G, P and H respectively. Prove that GP = PH.Solution 19

Question 20

BM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. If L is the mid-point of BC, prove that LM = LN.Solution 20

Chapter 12 – Congruent Triangles Exercise Ex. 12.1

Question 1

In fig., the sides BA and CA have been produced such that BA = AD and CA = AE.

Prove that segment DE || BC

Solution 1

Question 2

Solution 2

Question 3

Prove that the medians of an equilateral triangle are equal.Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

The vertical angle of an isosceles triangle is 100^o. Find its base angles.Solution 6

Question 7

 In fig., AB = Ac and ∠ACD = 105°, find ∠BAC. 

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

In fig., AB =AC and DB = DC, find the ratio ∠ABD = ∠ACD. 

Solution 10

Question 11

Determine the measure of each of the equal angles of a right-angled isosceles triangle.

                                                                                   OR

ABC is a right-angled triangle in which A = 90^o and AB = AC. Find B and C.Solution 11

Question 12

Solution 12

Question 13

AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See fig.). Show that the line PQ is perpendicular bisector of AB.

Solution 13

Chapter 12 – Congruent Triangles Exercise Ex. 12.2

Question 1

Solution 1

Question 2

In fig., it is given RT = TS, ∠1 = 2∠2 and ∠4 = 2∠3 prove that ΔRBT ≅ ΔSAT.

Solution 2

Question 3

Solution 3

Chapter 12 – Congruent Triangles Exercise Ex. 12.3

Question 1

In two right triangles one side and acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.Solution 1

Let ABC and DEF be two right triangles.

Question 2

Solution 2

Question 3

Solution 3

Question 4Show that the angles of an equilateral triangle are 60^o each.Solution 4

 Let us consider that ABC is an equilateral triangle.
So, AB = BC = AC
Now, AB = AC⇒ ∠C = ∠B         (angles opposite to equal sides of a triangle are equal)

We also have
AC = BC    
⇒ ∠B = ∠A             (angles opposite to equal sides of a triangle are equal)

So, we have
∠A = ∠B = ∠C
    Now, in ΔABC
∠A + ∠B + ∠C = 180^o
⇒ ∠A + ∠A + ∠A = 180^o
⇒ 3∠A = 180^o
⇒ ∠A = 60^o
⇒ ∠A = ∠B = ∠C = 60^o
Hence, in an equilateral triangle all interior angles are of 60^o.

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 12 – Congruent Triangles Exercise Ex. 12.4

Question 1

In fig., it is given that Ab = CD and AD = BC. prove that ΔADC ≅ ΔCBA

Solution 1

Question 2

Solution 2

Chapter 12 – Congruent Triangles Exercise Ex. 12.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

In fig., AD ⊥ CD and CB ⊥ CD. If AQ = BP an DP = CQ, prove that ∠DAQ = ∠CBP.

Solution 4

Question 5

Which of the following statements are True (T) and which are False (f):

(i) Sides opposite to equal angles of a triangle may be unequal.

(ii) Angles opposite to equal sides of a triangle are equal.

(iii) The measure of each angle of an equilaterial triangle is 60^o.

(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isoscles.

(v) The bisectors of two equal angles of a traingle are equal.

(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.

(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.

(viii) If any two sides of a right triangle are respectively equal to two sides of other right triagnle, then the two triangles are congruent.

(ix) Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle.Solution 5

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) False

(vii) False

(viii) False

(ix) TrueQuestion 6

Solution 6

(i) equal

(ii) equal

(iii) equal

(iv) BC

(v) AC

(vi) equal to

(vii) EFDQuestion 7

Solution 7

Chapter 12 – Congruent Triangles Exercise Ex. 12.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Is it possible to draw a triangle with sides of length 2cm, 3cm and 7 cm?Solution 4

Here, 2 + 3 < 7

Hence, it is not possible because triangle can be drawn only if the sum of any two sides is greater than third side.Question 5

Solution 5

Question 6

Solution 6

Question 7

In fig., prove that:

i. CD + DA + AB + BC > 2AC

ii. CD + DA + AB > BC

Solution 7

Question 8

Which of the following statements are true (T) and which are false (F)?

(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.

(ii) Sum of any two sides of a triangle is greater than twice the median drawn to the third side.

(iii) Sum of any two sides of a triangle is greater than the third side.

(iv) Difference of any two sides of a triangle is equal to the third side.

(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it.

(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.Solution 8

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) TrueQuestion 9

Solution 9

(i) largest

(ii) less

(iii) greater

(iv) smaller

(v) less

(vi) greaterQuestion 10

Solution 10

Question 11

Solution 11

Chapter 11 Triangle and its Angles Exercise Ex. 11.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Two angles of a triangle are equal and the third angle is greater than each of those angles by 30^o. Determine all the angles of the triangle.Solution 4

Question 5

Solution 5

Question 6

Can a triangle have:

(i) Two right angles?

(ii) Two obtuse angles?

(iii) Two acute angles?

(iv) All angles more than 60^o?

(v) All angles less than 60^o?

(vi) All angles equal to 60^o?

Justify your answer in each case.Solution 6

(i) No

As two right angles would sum up to 180^o, and we know that the sum of all three angles of a triangle is 180^o, so the third angle will become zero. This is not possible, so a triangle cannot have two right angles.

(ii) No

A triangle cannot have 2 obtuse angles, since then the sum of those two angles will be greater than 180^o which is not possible as the sum of all three angles of a triangle is 180^o.

(iii) Yes

A triangle can have 2 acute angles.

(iv) No

The sum of all the internal angles of a triangle is 180^o. Having all angles more than 60^o will make that sum more than 180^o, which is impossible.

(v) No

The sum of all the internal angles of a triangle is 180^o. Having all angles less than 60^o will make that sum less than 180^o, which is impossible.

(vi) Yes

The sum of all the internal angles of a triangle is 180^o.  So, a triangle can have all angles as 60^o. Such triangles are called equilateral triangles.Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 11 Triangle and its Angles Exercise Ex. 11.2

Question 1

The exterior angles, obtained on producing both the base of a triangle both ways are 104^o and 136^o. Find all the angles of the triangle.

Solution 1

Question 2

In fig., the sides BC, CA and AB of a ΔABC have been produced to D, E, and F respectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the ΔABC.

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4

In fig., AC ⊥ CE and ∠A : ∠B : ∠C = 3 : 2 : 1, find the value of ∠ECD.

Solution 4

Question 5

In fig. AB ∥ DE. Find ∠ACD.

Solution 5

Question 6

Which of the following statements are true (T) and which are false (F):

Solution 6

Question 7

Fill in the blanks to make the following statements true:

(i) Sum of the angle of triangle is ______ .

(ii) An exterior angle of a triangle is equal to the two ______ opposite angles.

(iii) An exterior angle of a traingle is always _______ than either of the interior oppsite angles.

(iv) A traingle cannot have more than ______ right angles.

(v) A triangles cannot have more than ______ obtuse angles.Solution 7

(i) 180^o

(ii) interior

(iii) greater

(iv) one

(v) oneQuestion 8

Solution 8

Question 9

Solution 9

Question 10

In fig., AB divides ∠DAC in the ratio 1 : 3 and AB = DB. Determine the value of x.

Solution 10

Question 11

Solution 11

Question 12

In fig., AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find ∠MAN.

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

In fig. AE bisects ∠CAD and ∠B = ∠C. Prove that AE ∥ BC.

Solution 15

Chapter 10 – Lines and Angles Exercise Ex. 10.1

Question 1

Write the complement of each of the following angles:

(i) 20^o

(ii) 35^o

(iii) 90^o

(iv) 77^o

(v) 30^oSolution 1

Question 2

Write the supplement of each of the following angles:

(i) 54^o

(ii) 132^o

(iii) 138^oSolution 2

(i) 54°

Since, the sum of an angle and its supplement is 180°

∴Its supplement will be 180° – 54° = 126°.

(ii) 132°

Since, the sum of an angle and its supplement is 180°

∴Its supplement will be 180° – 132° = 48°.

(iii) 138°

Since, the sum of an angle and its supplement is 180°

∴Its supplement will be 180° – 138° = 42°.Question 3

If an angle is 28^o less than its complement, find its measure.Solution 3

Question 4

If an angle is 30^o more than one half of its complement, find the measure of the angle.Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

If the complement of an angle is equal to the supplement of the thrice of it. Find the measure of angle.Solution 13

Let the measure of the angle be x^o.

Its complement will be (90^o – x^o) and its supplement will be (180^o – x^o).

Supplement of thrice of the angle = (180^o – 3x^o)

According to the given information:

(90^o – x^o) = (180^o – 3x^o)

3x – x = 180 – 90

2x = 90

x = 45

Thus, the measure of the angle is 45^o.

The measure of the angle is 45^oQuestion 14

Solution 14

Chapter 10 – Lines and Angles Exercise Ex. 10.2

Question 1

In fig., OA and OB are opposite rays:

(i) If x = 25°, what is the value of y?

(ii) if y = 35°, what is the value of x?

Solution 1

Question 2

In fig., write all pairs of adjacent angles and all the linear pairs.

Solution 2

Question 3

In fig., find x. further find ∠BOC, ∠COD and ∠AOD

Solution 3

Question 4

In fig., rays OA, OB, OC, OD and OE have the common end point O. Show that  ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360ºSolution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

How many pairs of adjacent angles, in all, can you name in fig.

Solution 7

Question 8

In fig., determine the value of x.

Solution 8

Question 9

In fig., AOC is a line, find x.

Solution 9

Question 10

In Fig., POS is a line, find x.

Solution 10

Question 11

In fig., ACB is a line such that ∠DCA = 5x and ∠DCB = 4x. Find the values of ∠DCA and∠DCB

Solution 11

Question 12

Give POR = 3x and QOR = 2x + 10, find the value of x for which POQ will be aline.

Solution 12

Question 13

What value of y would make AOB a line in fig., if ∠AOC = 4y and ∠BOC = (6y + 30)?

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

In Fig., Lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 5:7, find all the angles.

Solution 16

Question 17

In Fig. If a greater than b by one third of a right-angle. find the values of a and b.

Solution 17

Question 18

Solution 18

Question 19In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

                                    Solution 19Given that OR ⊥ PQ   ∴ POR = 90^o        ⇒ ∠POS  + ∠SOR = 90^o ∠ROS = 90^o  – ∠POS                … (1)     ∠QOR = 90^o                     (As OR ⊥ PQ)     ∠QOS – ∠ROS = 90^o     ∠ROS = ∠QOS – 90^o             … (2)     On adding equations (1) and (2), we have     2 ∠ROS = ∠QOS – ∠POS

Chapter 10 – Lines and Angles Exercise Ex. 10.3

Question 1

In fig., lines l₁ aans l₂ intersect at O, forming angles as shown in the figure. If x = 45, find the values of y, z and u.

Solution 1

Question 2

In fig., three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.

Solution 2

Question 3

In fig. , find the values of x, y and z.

Solution 3

Question 4

In Fig., find the value of X.

Solution 4

Question 5

Solution 5

Question 6

In fig., rays AB and CD intersect at O.

(i) Determine y when x = 60^o

(ii) Determine x when y = 40

Solution 6

Question 7

In fig., lines AB, CD and EF intersect at O. Find the measures of ∠AOC, ∠COF, ∠DOE and ∠BOF.

Solution 7

Question 8

Solution 8

Question 9

In fig., lines AB, and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution 9

Question 10

Which of the following statements are true (T) and which are false (F)?

(i) Angles forming a linear pair are supplementary.

(ii) If two adjacent angles are equal, then each angle measures 90^o.

(iii) Angles forming a linear pair can both be acute angles.

(iv) If angles forming a linear pair are equal, then each of these angles is of measure 90^o.Solution 10

(i) True

(ii) False

(iii) False

(iv) TrueQuestion 11

Fill in the blanks so as to make the following statements true:

(i) If one angle of a linear pair is acute, then its other angle will be ________.

(ii) A ray stands on a line, then the sum of the two adjacent angles so formed is _______.

(iii) If the sum of two adjacent angles is 180^o, then the ______ arms of the two angles are opposite rays.Solution 11

(i) obtuse.

(ii) 180^o

(iii) uncommonQuestion 12

Solution 12

Question 13

Solution 13

Chapter 10 – Lines and Angles Exercise Ex. 10.4

Question 1

In fig., AB ∥ CD and ∠1 and ∠2 are in the ratio 3:2. determine all angles from 1 to 8.

Solution 1

Question 2

In fig., l, m and n are parallel lines intersected by transversal p at x, y and z respectively. find ∠1, ∠2, ∠3.

Solution 2

Question 3

In fig., if AB ∥ CD and CD ∥ EF, find ∠ACE.

Solution 3

Question 4

In fig., state which lines are parallel and why.

Solution 4

Question 5

In fig. if l ∥ m, n ∥ p and ∠1 = 85°, find ∠2.

Solution 5

Question 6

If two straight lines are perpendicular to the same line, prove that they are parallel to each other.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Let AB and CD be perpendicuar to line MN.

Question 9

In fig., ∠1 = 60° and ∠2 = (2/3)^rd of a right angle. prove that l ∥ m

Solution 9

Question 10

In fig., if l ∥ m ∥ n and ∠1 = 60°, find ∠2.

Solution 10

Question 11

Solution 11

Let AB and CD be perpendicuar to line MN.

Question 12

Solution 12

Question 13

Solution 13

Question 14

In fig., p is transversal to lines m and n, ∠2 = 120° and ∠5 = 60°. Prove that m ∥ n. 

Solution 14

Question 15

In fig., transceral l intersects two lines m and n, ∠4 = 110° and ∠7 = 65°. is m ∥ n ?

Solution 15

Question 16

Which pair of lines in Fig., are parallel? give reasons.

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

In fig., AB ∥ CD ∥ EF and GH ∥ KL. Find ∠HKL

Solution 20

Question 21

In fig., show that AB ∥ EF.

Solution 21

Question 22

In Fig., PQ ∥ AB and PR BC. IF ∠QPR = 102º, determine ∠ABC. Give reasons.

Solution 22

Question 23

Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.Solution 23

Consider the angles AOB and ACB.

Question 24

In fig.,  lines AB and CD are parallel and p is any point as shown in the figure. Show that ∠ABP + ∠CDP = ∠DPB.

Solution 24

Question 25

In fig., AB ∥ CD and P is any point shown in the figure. Prove that:

∠ABP + ∠BPD + ∠CDP = 360°

Solution 25

Question 26

In fig., arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC = ∠DEF

 Solution 26

Question 27

 Solution 27

Chapter 9 Introduction to Euclid’s Geometry Exercise Ex. 9.1

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 4

Write the truth value (T/F) of each of the following statements:

(i) Two lines intersect in a point.

(ii) Two lines may intersect in two points.

(iii) A segment has no length.

(iv) Two distinct points always determine a line.

(v) Every ray has a finite length.

(vi) A ray has one end-point only.

(vii) A segment has one end-point only.

(viii) The ray AB is same as ray BA.

(ix) Only a single line may pass through a given point.

(x) Two lines are coincident if they have only one point in common.Solution 4

(i) False

(ii) False

(iii) False

(iv) True

(v) False

(vi) True

(vii) False

(viii) False

(ix) False

(x) FalseQuestion 5

In fig., name the following:

(i) Five line segments.

(ii) Five rays.

(iii) Four collinear points.

(iv) Two pairs of non-intersecting line segments.Solution 5

Question 6

Fill in the blanks so as to make the following statements true:

(i) Two distinct points in a plane determine a ______ line.

(ii) Two distinct ______ in a plane cannot have more than one point in common.

(iii) Given a line and a point, not on the line, there is one and only ______ line which passes through the given point and is ______ to the given line.

(iv) A line separates a plane into ______ parts namely the ______ and the _______ itself.Solution 6

(i) unique

(ii) lines

(iii) perpendicular, perpendicular

(iv) three, two half planes, line.