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Areas Related to Circles Class 10 Important Questions

The start of class 10 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 10 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

Question 1.
In figure, ABCD is a square of side 14 cm. Semi-circles are drawn with each side of square as diameter. Find the area of the shaded region.

Solution:
Area of the square ABCD = 14 x 14 = 196 cm²
Area of semicircle AOB=1/2 x πr²
=1/2×22/7x7x7
Similarly, area of semicircle DOC = 77 cm²
Hence, the area of shaded region (Part W and Part Y) = Area of square -Area of two semicircles AOB and COD
= 196 – 154 = 42 cm²
Therefore, area of four shaded parts (i.e. X, Y, W, Z) = (2 x 42) cm² = 84 cm²

Question 2.
In figure, are shown two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre O and radius OP while arc PBQ is a semicircle drawn on PQ as diameter with centre M. If OP = PQ = 10 cm, show that area of shaded region is
25(√3- π /6)cm²

Solution:
OP = OQ = 10 cm
PQ = 10 cm
So, ΔOPQ is an equilateral triangle
∠POQ = 60°
Area of segment PAQM = Area of sector OPAQ – Area of ΔOPQ
=60/360 x π x 10 x 10-√3/4 x 10 x 10
=(100π/6-100√3/4)cm²
Area of semicircle =1/2 x π x 5 x 5 = 25/2π cm²
Area of the shaded region =25/2π-(100π/6-100√3/4)=25π/2-50π/3+25√3
=75π-100π/6+25√3=25√3-25π/6
=25(√3-π/6)cm²

Question 3.
In figure, O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region.

Solution:
Here, BC² = AB²-AC²
= 169-144 = 25 BC = 5
Area of shaded region = Area of semicircle – Area of right triangle ABC
=1/2 x πr²-1/2AC x BC
=1/2 x 3.14(13/2)²-1/2 x 12 x 5
=66.33 – 30 = 36.33 cm²

Question 4.
In figure, find the area of the shaded region, enclosed between two concentric circles of radii 7 cm and 14 cm where∠AOC = 40°.

Solution:
Area of shaded region=360°-θ/360° x π(R²-r²)
=320°/360° x π[(14)²-(7)²]
=8/9 x 22/7(196-49)=8/9 x 22/7 x 147
=1232/3=410.67 cm²

Question 5.
Find the area of shaded region in figure, where a circle of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm

Solution:
Area of  ΔOAB=√3/4(side)²=√3/4 x (12)²
=36√3=36 x 1.73
=62.28 cm²
area of circle with center O =πr²=3.14 x (6)²
=3.14 x 36 =113.04 cm²
area of sector(OLQP)=πr² x θ/360°=3.14 x 6² x 60°/360°
=3.14 x 36 x 1/6 =18.84 cm²
area of shaded region=area of ΔOAB+area of circle-2 area of sector OLQP
=(62.28+113.04-2 x 18.84)cm²
=137.64cm²

Question 6.
In figure, is a chord AB of a circle, with centre O and radius 10 cm, that subtends a right angle at the centre of the circle. Find the area of the minor segment AQBP. Hence, find the area of major segment ALBQA

Solution:
Area of minor segment APBQ=θ/360° x πr²-r²sin45°cos45°
=3.14 x 100/4-100 x 1/√2 x 1/√2
=(78.5-50)cm²=28.5 cm²
Area of major segment ALBQA =πr²-area of minor segment
=3.14 x (10)²-28.5
=(314-28.5)cm²=285.5 cm²

Long Answer Type Questions [4 Marks]

Question 7.
An elastic belt is placed around the rim of a pulley of radius 5 cm. From one point C on the belt, the elastic belt is pulled directly away from the centre O of the pulley until it is at P, 10 cm from the point O. Find the length of the belt that is still in contact with the pulley.Also find the shaded area, (use π = 3.14 and √3 = 1.73)

Solution:

Question 8.
In figure, is shown a sector OAP of a circle with centre O, containing Z0. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of shaded region is r[tanθ+secθ+πθ/180-1]

Solution:

Question 9.
Find the area of the shaded region in figure, where APD, AQB, BRC and CSD are semi-circles of diameter
14 cm, 3.5 cm, 7 cm and 3.5 cm respectively

Solution:
Area of shaded region
= Area of semicircle APD + Area of semicircle BRC – 2 x Area of semicircle AQB

2015

Short Answer Type Questions II [3 Marks

Question 10.
In figure, APB and AQO are semicircle, and AO = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region.

Solution:
Let r be the radius of the semicircle APB, i.e. OB = OA = r, then r/2 is the radius of the semicircle AQO.

Question 11.
In figure, find the area of the shaded region

Solution:

Question 12.
Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60°. Also find the area of the corresponding major segment
Solution:

Question 13.
All the vertices of a rhombus lie on a circle.Find the area of rhombus,if the area of circle is 1256 cm²
Solution:

Question 14.
The long and short hand of a clock are 6cm and 4 cm long respectively,Find the sum of the distance travelled by their tips in 24hrs.
Solution:

Question 15.

Solution:

Important Links

NCERT Quick Revision Notes- Areas Related to Circles

NCERT Solution- Areas Related to Circles

Important MCQs- Areas Related to Circles

For Free, Video Lectures Click here

Constructions Class 10 Important Questions

The start of class 10 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 10 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

Question 1.
Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to smaller circle from a point on the larger circle. Also measure its length
Solution:

Now after measuring, PA and PB comes out to be 4 cm.
Steps of construction of tangents:

1. Take point O. Draw 2 concentric circles of radii 3 cm and 5 cm respectively.
2. Locate point P on the circumference of larger circle.
3. Join OP and bisect it. Let M be mid-point of OP.
4. Taking M as centre and MP as radius, draw an arc intersecting smaller circle at A and B.
5. Join PA and PB. Thus, PA, PB are required tangents

Question 2.
Construct a triangle ABC in which BC = 6 cm, AB = 5 cm and∠ABC = 60°. Then construct another triangle whose sides are 3/4 times the corresponding sides of ΔABC.
Solution:

Steps of construction:

1. Draw ΔABC with side BC = 6 cm, AB = 5 cm, ∠ABC = 60°.
2. Draw ray BX making an acute angle with BC on opposite side of vertex A.
3. Locate 4 points P1 P2, P3, P4 on line segment BY.
4. Join P4C and draw a line through P3, parallel to P4C intersecting BC at C’.
5. Draw a line through C’ parallel to AC intersecting AB at A’. ΔA’BC’ is the required triangle

Question 3.
Draw a triangle ABC with BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct a triangle whose sides are 4/5 times the corresponding sides of ΔABC.
Solution:

Given, ∠B = 45°,∠A = 105°
Sum of all interior angles in Δ = 180°
∠A +∠B + ∠C = 180°
∠C = 30°
Steps of construction:

1. Draw ΔABC with side BC = 7 cm,∠B = 45°, ∠C = 30°.
2. Draw a ray BX making an acute angle with BC on opposite side of vertex A.
3. Locate 5 points P1, P2, P3, P4, P5 on BZ.
4. Join P5C. Draw line through P4 parallel to P5C intersecting BC at C’.
5. Through C’, draw line parallel to AC intersecting AB at A’. ΔA’BC’ is the required triangle

Question 4.
Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other
Solution:

Steps of construction:
1. Draw a circle of radius 4 cm with centre O.
2. Take point A on circle. Join OA.
3- Draw line AP perpendicular to radius OA.
4. Draw ∠AOB = 120° at O.
5. Join A and B at P, to get 2 tangents. Here ∠APB = 60°.

Question 5.
Draw an isosceles ΔABC in which BC = 5.5 cm and altitude AL = 3 cm. Then construct another triangle whose sides are  3/4 of the corresponding sides of ΔABC
Solution:

Steps of construction:

1. Draw BC = 5.5 cm.
2. Construct AP the perpendicular bisector of BC meeting BC at L.
3. Along LP cut off LA = 3 cm.
4. Join BA and CA. Then ΔABC so obtained is the required ΔABC.
5. Draw an acute angle CBY and cut 4 equal lengths as BA1 = A1A2 = A2A3 = A3A4 and join CA4.
6. Now draw a line through A3 parallel to CA4 intersecting BC at C’.
7. Draw a line through C’ and parallel to AC intersecting AB at A’. BA’C’ is the required triangle.

Question 6.
Draw a triangle with sides 5 cm, 6 cm and 7 cm. Then draw another triangle whose 4/5 sides are y of the corresponding sides of first triangle
Solution:

1. Draw a line segment AB of length 7 cm.
   Then using A as centre and distance 5 cm draw an arc C.
   Also draw an arc using B as centre and with distance 6 cm, which intersect earlier drawn arc at C. Join AC and BC.
2. Draw an acute angle BAZ and cut AZ as AA1 = A1A2 = A2A3 = A3A4 = A4A5 and join BA5.
3. Through A4 draw a line parallel BA5 intersecting AB at B’.
4. Through B’ draw a line parallel to BC intersecting AC at C’. AAB’C’ is the required triangle.

Question 7.
Draw a ΔABC in which AB = 4 cm, BC = 5 cm and AC = 6 cm. Then construct another triangle whose sides are 3/5 of the corresponding sides of ΔABC
Solution:

Steps of construction:

1. Draw a line segment AC = 6 cm.
2. Draw an arc with A as centre and radius equal to 5 cm.
3. Draw an arc with C as centre and radius equal to 4 cm intersecting the previous drawn arc at B.
4. Join AB and CB, then ΔABC is required triangle.
5. Below AC make an acute angle CAX.
6. Along AX mark of 5 points A1, A2, A3, A4, A5 such that AA1= A1A2 = A2A3= A3A4 = A4A5.
7. Join A5C.
8. From A3 draw A3D | | A3C meeting AC at D.
9. From D, draw ED | | BC meeting AB at E. Then we have ΔEDA which is the required triangle.

Question 8.
Draw a triangle with sides 4 cm, 5 cm and 6 cm. Then construct another triangle whose sides are 2/5 of the corresponding sides of given (first) triangle
Solution:

Steps of construction:

1. Draw a line segment BC = 6 cm.
2. Draw an arc with B as centre and radius equal to 5 cm.
3. Draw an arc with C as centre and radius equal to 4 cm intersecting the previous one at A.
4. Join AB and AC, then ΔABC is the required triangle.
5. Below BC, make an acute angle CBX.
6. Along BX, mark off 5 points B1, B2, B3, B4, B5 such that BB1 = B1B2 = B2B3 = B3B4 = B4B5.
7. Join B5C.
8. From B2, draw B2D ¦ ¦B5C, meeting BC at D.
9. From D, draw ED ¦ ¦ AC, meeting BA at E. Then we have ΔEDB which is the required triangle.

Question 9.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are 3/4 times the sides of first triangle
Solution:

Steps of construction:

1. Draw line BC = 8 cm then at B draw a line making angle of 90°.
2. Cut a length of 6 cm and name it A. Join AC. ΔABC is the right triangle.
3. Below BC make an acute angle ∠CBX.
4. Along BX mark off 4 points B1, B2, B3, B4 such that BB1=B1B2 = B2B3 = B3B4.
5. Join B4C.
6. From B3 draw B3D ¦ ¦ B4C meeting BC at D.
7. From D draw ED 11 AC meeting BA at E. Now we have ΔEBD which is the required triangle whose sides are 3/4 of the corresponding sides of ΔABC

2015

Long Answer Type Questions [4 Marks]

Question 10.
Construct a triangle ABC with BC = 7 cm, ∠B = 60° and AB = 6 cm. Construct another triangle whose sides are 3/4 times the corresponding sides of ΔABC
Solution:

Steps of construction:

1. Draw a line segment BC = 7 cm.
2. Draw ∠B = 60° at point B. Thus∠XBC = 60°
3. Take an arc of 6 cm, with B as centre mark an arc on BX to get point A.
4. Join AC.
5. ΔABC is constructed triangle.
6. Draw an acute angle CBY below BC.
7. Take points P1, P2, P3, P4, at BY such that BP1 = P1P2=P2P3=P3P4
8. Join P4C with line.
9. Draw a line parallel to P4C through the point P3 which intersects BC at C’.
10. Join P3C’ with line
11. Draw a line parallel to AC through the point C’ which intersects AB at point A’. ΔA’BC’ is the required triangle whose sides are 3/4 times the corresponding sides of ΔABC.

Question 11.
Construct an isosceles triangle whose base is 6 cm and altitude 4 cm. Then construct another triangle whose sides are 3/4 times the corresponding sides of the isosceles triangle.
Solution:

Steps of construction:

1. Draw a line segment BC = 6 cm.
2. Draw perpendicular bisector of BC which intersects BC at point D.
3. Take an arc of 4 cm, with D as centre mark on arc ⊥ bisector as point A.
4. Join AB and AC. ΔABC is constructed isosceles Δ.
5. Draw an acute angle CBY below BC.
6. Take points P1,P2,P3, P4 at BY such that BP1 = P1P2 = P2P3=P3P4
7. Join P4C with dotted line.
8. Draw a line parallel to P4C through the point P3 which intersects BC at C’.
9. Join P3C’ with dotted line.
10. Draw a line parallel to AC through the point C’ which intersects AB at A’.
    ΔABC is the required triangle whose sides are 3/4 times the corresponding sides of ΔABC.

Question 12.
Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle
Solution:

Required tangents are

1. BP and BQ
2. AR and AS.
   Steps of construction:

• Draw AB = 7 cm. Taking A and B as centres, draw two circles of 3 cm and 2 cm radius.
• Bisect line AB. Let mid-point of AB be C.
• Taking C as centre, draw circle of AC radius which will intersect circles at P, Q, R, S. Join BP, BQ, AR, AS

Question 13.
Construct a ΔABC in which AB = 6 cm, ∠A = 30° and ∠B = 60°. Construct another ΔAB’C’ similar to ΔABC with base AB’ = 8 cm
Solution:

Steps of construction:

1. Draw AB = 6 cm and make an angle of 60° at point B and 30° at point A.
2. Make any acute angle BAX at point A.
3. Cut four arcs A1,A2, A3, A4 on line AX such that AA1=A1A2=A2A3=A3A4
4. Join B to A3.
5. Draw line from A4 parallel to A3B cutting AB extended to B’.
6. Draw line from B’ | | BC cuts AC at C’.

Question 14.
Construct a right triangle ABC with AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD, the perpendicular from B on AC. Draw the circle through B, C and D and construct the tangents from A to this circle
Solution:

Thus, AP and AB are the required tangents
Steps of construction:

1. Draw BC = 8 cm, ∠B = 90°.
2. Take an arc of 6 cm, with B as centre, mark an arc on point A. Join AB.
3. Draw BD ⊥ AC. Bisect line BC at E as mid-point of BC.
4. Taking E as centre and EC as its radius, draw circle which will intersect AC at D. Join BD.
5. Mark point P on circle. Join A to P.

Question 15.
Construct a triangle ABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Now construct another triangle whose sides are 5/7  times the corresponding sides of ΔABC.
Solution:
pt>
Steps of construction:

1. Draw BC = 6 cm.
2. Take any radius (less than half of BC) and centre B, draw an arc intersecting BC at P. With same radius and centre P, draw another arc intersecting previous arc at Q.
3. Join BQ, extend it to D.
4. Take radius = 5 cm and centre B, we draw arc intersecting BD at A. Join AC, get ΔABC.
5. Draw line BX, as ∠CBX is any acute angle. Draw 7 equal radius arcs on line BX intersecting at B1, B2, B3, B4, B5, B6, B7 as BB1=B1B2=B2B3=B3B4=B4B5=B5B6=B6B7
6. Join B7 to C. Draw line from B5 as parallel to B7C intersecting BC at C’.
7. Draw line from C’as parallel to AC intersecting AB at A’.

Question 16.
Draw a circle of radius 3 cm. From a point P, 7 cm away from its centre draw two tangents to the circle. Measure the length of each tangent.
Solution:

We know, radius perpendicular to tangent with OA = 3 cm, OP = 7 cm
In right ΔOAP, (OP)² = (OA)² + (PA)²
PB = PA = 2√10 cm
Steps of construction:

1. Take point O as centre, draw circle of radius 3 cm. Locate point P, 7 cm away from its centre O. Join OP.
2. Bisect OP. Let Q be mid-point of PO.
3. Taking Q as centre and QO as radius, draw a circle.
4. Let this circle intersect previous circle at A and B.
5. Join AP, BP which are required tangents.
   .’. AP = 6.3 cm (approx.)

Question 17.
To a circle of radius 4 cm, draw two tangents which are inclined to each other at an angle of 60°.
Solution:
Refer to Ans 4.

Question 18.
Draw a circle of radius 3.5 cm. Draw two tangents to the circle which are perpendicular to each other
Solution:

Steps of construction:

1. Draw a circle of radius 3.5 cm with centre O.
2. Take point A on circle. Join OA.
3. Draw perpendicular to OA at A.
4. Draw radius OB, making an angle of 90° with OA.
5. Draw perpendicular to OB at point B. Let these perpendiculars intersect at C.
   Hence, CA and CB are required tangents inclined at angle of 90°.

Important Links

NCERT Quick Revision Notes- Constructions

NCERT Solution- Constructions

Important MCQs- Constructions

For Free, Video Lectures Click here

Circles Class 10 Important Questions

The start of class 10 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 10 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

Question 4.
In given figure, a circle is inscribed in a ΔABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. If the lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF

Solution:

Question 5.
If given figure, AP and BP are tangents to a circle with centre O, such that AP = 5 cm and ∠APB = 60°. Find the length of chord AB.

Solution:

Question 6.
In figure, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that AB + CD = BC + DA.

Solution:

Question 7.
In given figure, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r.If PO = 2r, show that ∠OTS =∠OST = 30°.

Solution:

Question 8.
In given figure, from a point P, two tangents PT and PS are drawn to a circle with centre O such that ∠SPT = 120°, Prove that OP = 2PS

Solution:

Question 9.
In given figure, there are two concentric circles of radii 6 cm and 4 cm with centre O. If AP is a tangent to the larger circle and BP to the smaller circle and length of AP is 8cm, find the length of BP

Solution:

Long Answer Type Questions [4 Marks]

Question 10.
Prove that the lengths of tangents drawn from an external point to a circle are equal
Solution:

Question 11.


Solution:

Question 12.
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact
Solution:

Question 13.
In given figure, two equal circles, with centres O and O’, touch each other at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of DO’/CO.

Solution:

Question 14.
In given figure, AB is a chord of a circle, with centre O, such that AB = 16 cm and radius of circle is 10 cm. Tangents at A and B intersect each other at P. Find the length of PA

Solution:

Question 18.
In figure, AB is the diameter of a circle with centre O and AT is a tangent. If ∠AOQ = 58°, find ∠ATQ

Solution:

Question 19.
From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of the line segment PQ.
Solution:

Question 20.
In figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120°, then prove that OR = PR + RQ.

Solution:

Question 21.
In figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively of lengths 6 cm and 9 cm. If the area of ΔABC is 54 cm², then find the lengths of sides AB and AC

Solution:

Question 22.
In figure, O is the centre of a circle. PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70°, find ∠TRQ

Solution:

Question 23.
In figure, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the lengths of TP and TQ

Solution:

Long Answer Type Questions [4 Marks]

Question 24.
In figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS

Solution:

Question 25.
Prove that the tangent at any point of a circle is perpendicular to the radius throug the point of contact
Solution:
Refer to Ans. 12

Question 26.
Prove that the lengths of the tangents drawn from an external point to a circle are equal
Solution:
Refer to Ans. 10.

Question 27.
Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.
Solution:

Question 28.
In figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30°, prove that BA: AT = 2:1

Solution:

Important Links

NCERT Quick Revision Notes- Circles

NCERT Solution- Circles

Important MCQs- Circles

For Free, Video Lectures Click here

Some Applications of Trigonometry Class 10 Important Questions

The start of class 10 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 10 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

Short Answer Type Questions II [3 Marks]

Question 4.
The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building,
Solution:

Question 5.
A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of hill as 30°. Find the distance of the hill from the ship and the height of the hill.
Solution:

Question 6.
Two men on either side of a 75 m high building and in line with base of building observe the angles of elevation of the top of the building as 30° and 60°. Find the distance between the two men
Solution:

Question 7.
A 7 m long flagstaff is fixed on the top of a tower standing on the horizontal plane. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are 60° and 45° respectively. Find the height of the tower correct to one place of decimal

Solution:

Question 8.
An aeroplane, when flying at a height of 4000 m  from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the aeroplanes at that instant
Solution:

Long Answer Type Questions [4 Marks]

Question 9.
A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Find the speed of flying of the bird.
Solution:

Question 10.
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 60° and 30° respectively. Find the height of the tower.
Solution:

Question 11.
The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX.
Solution:

Question 12.
As observed from the top of a light house, 100 m high above sea level, the angles of depression of a ship, sailing directly towards it, changes from 30° to 60°. Find the distance travelled by the ship during the period of observation.
Solution:

Question 13.
From a point on the ground, the angle of elevation of the top of a tower is observed to be 60°. From a point 40 m vertically above the first point of observation, the angle of elevation of the top of the tower is 30°. Find the height of the tower and its horizontal distance from the point of observation.
Solution:

Question 14.

Solution:

Important Links

NCERT Quick Revision Notes- Some Applications of Trigonometry

NCERT Solution- Some Applications of Trigonometry

Important MCQs- Some Applications of Trigonometry

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Introduction to Trigonometry Class 10 Important Questions

The start of class 10 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 10 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

Question 1.
Given tan A = 5/12 , find the other trigonometric ratios of the angle A.
Solution:

Question 2.
Prove that 1/sec A – tan A-1/cosA=1/ cos A -1/sec A + tan A
Solution:

Question 3.
If sin θ = 12/13, 0° < θ < 90°, find the value of: sin² θ- cos² θ /2 sin θ. cos θ  x 1/tan² θ
Solution:

Long Answer Type Questions [4 Marks]

Question 4.
If sin (A + B) = 1 and tan (A – B) = 1/√3, find the value of:

1. tan A + cot B
2. sec A – cosec B

Solution:

Question 9.
If sec A = x + 1/4x, prove that sec A + tan A = 2x or 1/2x
Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.
If sec θ – tan θ = x, show that: sec θ=1/2(x+1/x) and tan θ=1/2(1/x-x)
Solution:

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NCERT Quick Revision Notes- Coordinate Geometry

NCERT Solution- Coordinate Geometry

Important MCQs- Coordinate Geometry

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Coordinate Geometry Class 10 Important Questions

The start of class 10 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 10 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

Question 1.
Find the value of ‘k”, for which the points are collinear: (7, -2), (5, 1), (3, k).
Solution:
Let the given points be
A (x₁, y₁) = (7, -2), B (x₂, Y₂) = (5, 1) and C (x₃, y₃) = (3, k)
Since these points are collinear therefore area (∆ABC) = 0
⇒ \(\frac{1}{2}\) [x₁(y₂ – Y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
⇒ x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) = 0
⇒ 7(1 – k) + 5(k + 2) + 3(-2 -1) = 0
⇒ 7 – 7k + 5k + 10 – 9 = 0
⇒ -2k + 8 = 0
⇒ 2k = 8
⇒ k = 4
Hence, given points are collinear for k = 4.

Question 2.
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.
Solution:
Let A (x₁, y₁) = (0, -1), B (x₂, y₂) = (2, 1), C (x₃, y₃) = (0, 3) be the vertices of ∆ABC.
Now, let P, Q, R be the mid-points of BC, CA and AB, respectively.
So, coordinates of P, Q, R are

Ratio of ar (∆PQR) to the ar (∆ABC) = 1 : 4.

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

Let A(4, -2), B(-3, -5), C(3, -2) and D(2, 3) be the vertices of the quadrilateral ABCD.
Now, area of quadrilateral ABCD
= area of ∆ABC + area of ∆ADC

Question 4.
A median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A (4,-6), B (3, -2) and C (5, 2).
Solution:
Since AD is the median of ∆ABC, therefore, D is the mid-point of BC.

Hence, the median divides it into two triangles of equal areas.

Question 5.
Find the ratio in which the point P (x, 2), divides the line segment joining the points A (12, 5) and B (4, -3). Also find the value of x.
Solution:

The ratio in which p divides the line segment is \(\frac{3}{5}\), i.e., 3 : 5.

Question 6.
If A (4, 2), B (7, 6) and C (1, 4) are the vertices of a ∆ABC and AD is its median, prove that the median AD divides into two triangles of equal areas.
Solution:
Given: AD is the median on BC.
⇒ BD = DC
The coordinates of midpoint D are given by.

Hence, AD divides ∆ABC into two equal areas.

Question 7.
If the point A (2, -4) is equidistant from P (3, 8) and Q (-10, y), find the values of y. Also find distance PQ.
Solution:
Given points are A(2, 4), P(3, 8) and Q(-10, y)
According to the question,
PA = QA

Question 8.
The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point Care (0, -3). The origin is the mid-point of the base. Find the coordinates of the points A and B. Also find the coordinates of another point D such that BACD is a rhombus.
Solution:
∵ O is the mid-point of the base BC.
∴ Coordinates of point B are (0, 3). So,
BC = 6 units Let the coordinates of point A be (x, 0).
Using distance formula,

∴ Coordinates of point A = (x, 0) = (3√3, 0)
Since BACD is a rhombus.
∴ AB = AC = CD = DB
∴ Coordinates of point D = (-3√3, 0).

Question 9.
Prove that the area of a triangle with vertices (t, t-2), (t + 2, t + 2) and (t + 3, t) is independent of t.
Solution:
Area of a triangle = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Area of the triangle = \(\frac{1}{2}\)[t + 2 – t) + (t + 2) (t – t + 2) + (t + 3) (t – 2 – t – 2)]
= \(\frac{1}{2}\) [2t + 2t + 4 – 4t – 12 ]
= 4 sq. units
which is independent of t.
Hence proved.

Question 10.
The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is (\(\frac{7}{2}\), y), find the value of y.
Solution:
Given: ar(∆ABC) = 5 sq. units

Question 11.
The coordinates of the points A, B and Care (6, 3), (-3,5) and (4,-2) respectively. P(x, y) is any point in the plane. Show that \(\frac { ar(∆PBC) }{ ar(∆ABC) } \) = \(\frac{x+y-2}{7}\)
Solution:
P(x, y), B(-3, 5), C(4, -2), A(6, 3)

Question 12.
In Fig. 6.32, the vertices of ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line-segment DE is drawn to intersect the sides AB and AC at D and E respectively such that \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{3}\) Calculate the area of ∆ADE and compare it with area of ∆ABC.
Solution:

\

Question 13.
If a = b = 0, prove that the points (a, a²), (b, b²) (0, 0) will not be collinear.
Solution:
∵ We know that three points are collinear if area of triangle = 0
∴ Area of triangle with vertices (a, a²), (b, b²) and (0, 0)

∵ Area of ∆ ≠ 0
∴ Given points are not collinear.

Coordinate Geometry Class 10 Extra Questions HOTS

Question 1.
The line joining the points (2, 1) and (5, -8) is trisected by the points P and Q. If the point P lies on the line 2x – y + k = 0, find the value of k.
Solution:

As line segment AB is trisected by the points P and Q.
Therefore,
Case I: When AP : PB = 1 : 2.

⇒ P (3, -2)
Since the point P (3,-2) lies on the line
2x – y + k = 0 =
⇒ 2 × 3-(-2) + k = 0
⇒ k = -8

Case II: When AP : PB = 2 : 1.

Coordinates of point P are

Since the point P(4, -5) lies on the line
2x – y + k = 0
∴ 2 × 4-(-5) + k = 0
∴ k = -13

Question 2.
Prove that the diagonals of a rectangle bisect each other and are equal.
Solution:

Let OACB be a rectangle such that OA is along x-axis and OB is along y-axis. Let OA = a and OB = b.
Then, the coordinates of A and B are (a,0) and (0, b) respectively.
Since, OACB is a rectangle. Therefore,
AC = OB
⇒ AC = b
Also, OA = a
⇒ BC = a
So, the coordinates of Care (a, b).

∴ OC = AB.

Question 3.
In what ratio does the y-axis divide the line segment joining the point P (4, 5) and Q (3, -7)?
Also, find the coordinates of the point of intersection.
Solution:
Suppose y-axis divides PQ in the ratio k : 1. Then, the coordinates of the point of division are|

Question 4.
Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).
Solution:
Let O(x, y) be the centre of circle. Given points are A(6, -6), B(3, -7) and C(3, 3).

Question 5.
If the coordinates of the mid-points of the sides of a triangle are (1, 1), (2, – 3) and (3, 4). Find its centroid.
Solution:

Let P(1, 1), Q(2, -3), R(3, 4) be the mid-points of sides AB, BC and CA respectively, of triangle ABC. Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of triangle ABC. Then, P is the mid-point of AB.

Question 6.
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, -2) and B(3, 7).
Solution:
Let P(x₁, y₁) be common point of both lines and divide the line segment joining A(2, -2) and B(3, 7) in ratio k : 1.

Question 7.
Show that ∆ABC with vertices A (-2, 0), B (2, 0) and C (0, 2) is similar to ∆DEF with vertices
D(4, 0) E (4, 0) and F (0, 4).
Solution:
Given vertices of ∆ABC and ∆DEF are

A(-2, 0), B(2, 0), C(0, 2), D(-4, 0), E(4, 0) and F(0, 4)


Here, we see that sides of ∆DEF are twice the sides of a ∆ABC.
Hence, both triangles are similar.

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Important MCQs- Coordinate Geometry

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Triangles Class 10 Important Questions

The start of class 10 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 10 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

Question 18.
In the given figure, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and EC = 2 cm. (2012)

Solution:
In ∆ABL, CD || LA

Question 19.
If a line segment intersects sides AB and AC of a ∆ABC at D and E respectively and is parallel to BC, prove that ADAB=AEAC. (2013)
Solution:
Given. In ∆ABC, DE || BC

To prove. ADAB=AEAC
Proof.
In ∆ADE and ∆ABC
∠1 = ∠1 … Common
∠2 = ∠3 … [Corresponding angles
∆ADE ~ ∆ABC …[AA similarity
∴ ADAB=AEAC
…[In ~∆s corresponding sides are proportional

Question 20.
In a ∆ABC, DE || BC with D on AB and E on AC. If ADDB=34 , find BCDE. (2013)
Solution:
Given: In a ∆ABC, DE || BC with D on AB and E
on AC and ADDB=34
To find: BCDE
Proof. Let AD = 3k,

DB = 4k
∴ AB = 3k + 4k = 7k
In ∆ADE and ∆ABC,
∠1 = ∠1 …[Common
∠2 = ∠3 … [Corresponding angles
∴ ∆ADE ~ ∆ABC …[AA similarity

Question 21.
In the figure, if DE || OB and EF || BC, then prove that DF || OC. (2014)

Solution:
Given. In ∆ABC, DE || OB and EF || BC
To prove. DF || OC
Proof. In ∆AOB, DE || OB … [Given

Question 22.
If the perimeters of two similar triangles ABC and DEF are 50 cm and 70 cm respectively and one side of ∆ABC = 20 cm, then find the corresponding side of ∆DEF. (2014)
Solution:

Given. ∆ABC ~ ∆DEF,
Perimeter(∆ABC) = 50 cm
Perimeter(∆DEF) = 70 cm
One side of ∆ABC = 20 cm
To Find. Corresponding side of ∆DEF (i.e.,) DE. ∆ABC ~ ∆DEF …[Given

∴ The corresponding side of ADEF = 28 cm

Question 23.
A vertical pole of length 8 m casts a shadow 6 cm long on the ground and at the same time a tower casts a shadow 30 m long. Find the height of tower. (2014)
Solution:

Let BC be the pole and EF be the tower Shadow AB = 6 m and DE = 30 m.
In ∆ABC and ∆DEF,
∠2 = ∠4 … [Each 90°
∠1 = ∠3 … [Sun’s angle of elevation at the same time
∆ABC ~ ∆DEF …[AA similarity
ABDE=BCEF … [In -As corresponding sides are proportional
⇒ 630=8EF ∴ EF = 40 m

Question 24.
In given figure, EB ⊥ AC, BG ⊥ AE and CF ⊥ AE (2015)
Prove that:
(a) ∆ABG ~ ∆DCB
(b) BCBD=BEBA

Solution:

Given: EB ⊥ AC, BG ⊥ AE and CF ⊥ AE.
To prove: (a) ∆ABG – ∆DCB,
(b) BCBD=BEBA
Proof: (a) In ∆ABG and ∆DCB,
∠2 = ∠5 … [each 90°
∠6 = ∠4 … [corresponding angles
∴ ∆ABG ~ ∆DCB … [By AA similarity
(Hence Proved)
∴ ∠1 = ∠3 …(CPCT … [In ~∆s, corresponding angles are equal

(b) In ∆ABE and ∆DBC,
∠1 = ∠3 …(proved above
∠ABE = ∠5 … [each is 90°, EB ⊥ AC (Given)
∆ABE ~ ∆DBC … [By AA similarity
BCBD=BEBA
… [In ~∆s, corresponding sides are proportional
∴ BCBD=BEBA (Hence Proved)

Question 25.
∆ABC ~ ∆PQR. AD is the median to BC and PM is the median to QR. Prove that ABPQ=ADPM. (2017D)
Solution:

∆ABC ~ ∆PQR … [Given
∠1 = ∠2 … [In ~∆s corresponding angles are equal

Question 26.
State whether the given pairs of triangles are similar or not. In case of similarity mention the criterion. (2015)

Solution:

(b) In ∆PQR, ∠P + ∠Q + ∠ZR = 180° …[Angle-Sum Property of a ∆
45° + 78° + ∠R = 180°
∠R = 180° – 45° – 78° = 57°
In ∆LMN, ∠L + ∠M + ∠N = 180° …[Angle-Sum Property of a ∆
57° + 45° + ∠N = 180°
∠N = 180° – 57 – 45° = 78°
∠P = ∠M … (each = 45°
∠Q = ∠N … (each = 78°
∠R = ∠L …(each = 57°
∴ ∆PQR – ∆MNL …[By AAA similarity theorem

Question 27.
In the figure of ∆ABC, D divides CA in the ratio 4 : 3. If DE || BC, then find ar (BCDE) : ar (∆ABC). (2015)

Solution:
Given:
D divides CA in 4 : 3
CD = 4K
DA = 3K
DE || BC …[Given

In ∆AED and ∆ABC,
∠1 = ∠1 …[common
∠2 = ∠3 … corresponding angles
∴ ∆AED – ∆ABC …(AA similarity
⇒ ar(△AED)ar(△ABC)=(ADAC)2
… [The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides
⇒ frac(3K)2(7K)2=9K249K2=ar(△AED)ar(△ABC)=949
Let ar(∆AED) = 9p
and ar(∆ABC) = 49p
ar(BCDE) = ar (∆ABC) – ar (∆ADE)
= 49p – 9p = 40p
∴ ar(BCDE)ar(△ABC)=40p49p
∴ ar (BCDE) : ar(AABC) = 40 : 49

Question 28.
In the given figure, DE || BC and AD : DB = 7 : 5, find \frac { ar\left( \triangle DEF \right) }{ ar\left( \triangle CFB \right) } [/latex] (2017OD)

Solution:
Given: In ∆ABC, DE || BC and AD : DB = 7 : 5.
To find: ar(△DEF)ar(△CFB) = ?

Proof: Let AD = 7k
and BD = 5k then
AB = 7k + 5k = 12k
In ∆ADE and ∆ABC,
∠1 = ∠1 …(Common
∠2 = ∠ABC … [Corresponding angles

Question 29.
In the given figure, the line segment XY is parallel to the side AC of ∆ABC and it divides the triangle into two parts of equal areas. Find the ratio AXAB. (2017OD)

Solution:
We have XY || AC … [Given
So, ∠BXY = ∠A and ∠BYX = ∠C …[Corresponding angles
∴ ∆ABC ~ ∆XBY …[AA similarity criterion

Question 30.
In the given figure, AD ⊥ BC and BD = 13CD. Prove that 2AC² = 2AB² + BC². (2012)

Solution:
BC = BD + DC = BD + 3BD = 4BD
∴ BC4 = BD
In rt. ∆ADB, AD² = AB² – BD² ….(ii)
In rt. ∆ADC, AD² = AC² – CD² …(iii)
From (ii) and (iii), we get
AC² – CD² = AB² – BD²
AC² = AB² – BD² + CD²

∴ 2AC² = 2AB² + BC² (Hence proved)

Question 31.
In the given figure, ∆ABC is right-angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE. (2012, 2017D)

Solution:
Given: ∆ABC is rt. ∠ed at C and DE ⊥ AB.
AD = 3 cm, DC = 2 cm, BC = 12 cm
To prove:
(i) ∆ABC ~ ∆ADE; (ii) AE = ? and DE = ?
Proof. (i) In ∆ABC and ∆ADE,
∠ACB = ∠AED … [Each 90°
∠BAC = ∠DAE …(Common .
∴ ∆ABC ~ ∆ADE …[AA Similarity Criterion

(ii) ∴ ABAD=BCDE=ACAE … [side are proportional
AB3=12DE=3+2AE
…..[In rt. ∆ACB, … AB² = AC² + BC² (By Pythagoras’ theorem)
= (5)² + (12)² = 169
∴ AB = 13 cm

Question 32.
In ∆ABC, if AP ⊥ BC and AC² = BC² – AB², then prove that PA² = PB × CP. (2015)
Solution:

AC² = BC² – AB² …Given
AC² + AB² = BC²
∴ ∠BAC = 90° … [By converse of Pythagoras’ theorem
∆APB ~ ∆CPA
[If a perpendicular is drawn from the vertex of the right angle of a triangle to the hypotenuse then As on both sides of the perpendicular are similar to the whole triangle and to each other.
∴ APCP=PBPA … [In ~∆s, corresponding sides are proportional
∴ PA² = PB. CP (Hence Proved)

Question 33.
ABCD is a rhombus. Prove that AB² + BC² + CD² + DA² = AC² + BD². (2013)
Solution:
Given. In rhombus ABCD, diagonals AC and BD intersect at O.

To prove: AB² + BC² + CD² + DA² = AC² + BD²
Proof: AC ⊥ BD [∵ Diagonals of a rhombus bisect each other at right angles
∴ OA = OC and
OB = OD
In rt. ∆AOB,
AB² = OA² + OB² … [Pythagoras’ theorem
AB² = (AC2)2+(BD2)2
AB² = (AC2)2+(BD2)2
4AB² = AC² + BD²
AB² + AB² + AB² + AB² = AC² + BD²
∴ AB² + BC² + CD² + DA² = AC² + BD²
…[∵ In a rhombus, all sides are equal

Question 34.
The diagonals of trapezium ABCD intersect each other at point o. If AB = 2CD, find the ratio of area of the ∆AOB to area of ∆COD. (2013)
Solution:
In ∆AOB and ∆COD, … [Alternate int. ∠s
∠1 = ∆3
∠2 = ∠4

Question 35.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO=CODO. Show that ABCD is a trapezium. (2014)
Solution:
1st method.
Given: Quadrilateral ABCD in which
AC and BD intersect each other at 0.
Such that AOBO=CODO
To prove: ABCD is a trapezium
Const.: From O, draw OE || CD.


But these are alternate interior angles
∴ AB || DC Quad. ABCD is a trapezium.

Triangles Class 10 Important Questions Long Answer (4 Marks).

Question 36.
In a rectangle ABCD, E is middle point of AD. If AD = 40 m and AB = 48 m, then find EB. (2014D)
Solution:

E is the mid-point of AD …[Given
AE = 402 = 20 m
∠A = 90° …[Angle of a rectangle
In rt. ∆BAE,
EB² = AB² + AE² …[Pythagoras’ theorem
= (48)² + (20)²
= 2304 + 400 = 2704
∴ EB = 2704−−−−√ = 52 m

Question 37.
Let ABC be a triangle and D and E be two points on side AB such that AD = BE. If DP || BC and EQ || AC, then prove that PQ || AB. (2013)
Solution:

In ∆ABC,
DP || BC
and EQ || AC … [Given

Now, in ∆ABC, P and Q divide sides CA and CB respectively in the same ratio.
∴ PQ || AB

Question 38.
In the figure, ∠BED = ∠BDE & E divides BC in the ratio 2 : 1.
Prove that AF × BE = 2 AD × CF. (2015)

Solution:
Construction:
Draw CG || DF
Proof: E divides
BC in 2 : 1.
BEEC=21 …(i)

Question 39.
In the given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, then find the pair of parallel lines and hence their lengths. (2015)

Solution:

Question 40.
If sides AB, BC and median AD of AABC are proportional to the corresponding sides PQ, QR and median PM of PQR, show that ∆ABC ~ ∆PQR. (2017OD)
Solution:

Question 41.
Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. (2012)
Solution:
Given: ∆ABC ~ ∆DEF

Question 42.
State and prove converse of Pythagoras theorem. Using the above theorem, solve the following: In ∆ABC, AB = 63–√ cm, BC = 6 cm and AC = 12 cm, find ∠B. (2015)
Solution:
Part I:
Statement: Prove that, in a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

To prove: ∠ABC = 90°
Const.: Draw a right angle ∆DEF in which DE = BC and EF = AB.
Proof: In rt. ∆ABC,
AB² + BC² = AC² …(i) Given
In rt. ∆DEF
DE² + EF² = DF² … [By Pythagoras’ theorem
BC² + AB² = DF²…(ii)…[∵ DE = BC; EF = AB
From (i) and (ii), we get
AC² = DF² = AC = DF
Now, DE = BC …[By construction
EF = AB …[By construction
DF = AC … [Proved above :
∴ ∆DEF = ∆ABC … (SSS congruence :
∴ ∠DEF = ∠ABC …[c.p.c.t.
∵ ∠DEF = 90° ∴ ∠ABC = 90°
Given: In rt. ∆ABC,
AB² + BC² = AC²
AB² + BC² = (63–√)² + (6)²
= 108 + 36 = 144 = (12)²
AB² + BC² = AC² ∴ ∠B = 90° … [Above theorem

Question 43.
In the given figure, BL and CM are medians of a triangle ABC, right angled at A. Prove that: 4(BL² + CM²) = 5BC² (2012)

Solution:
Given: BL and CM are medians of ∆ABC, right angled at A.
To prove: 4(BL² + CM²) = 5 BC²
Proof: In ∆ABC, BC² = BA² + CA² …(i)
In ∆BAL,
BL² = BA² + AL² …[Pythagoras’ theorem
BL² = BA² + (CA2)2
BL² = BA²+ CA24
⇒ 4BL² = 4BA² + CA² …(ii)
Now, In ∆MCA,
MC² = CA² + MA² …[Pythagoras’ theorem
MC² = CA²2 + (BA2)2
MC² = CA² + BA24
4MC² = 4CA² + BA²
Adding (ii) and (iii), we get
4BL² + 4MC² = 4BA² + CA² + 4CA²+ BA² …[From (ii) & (iii)
4(BL² + MC²) = 5BA² + 5CA²
4(BL² + MC²) = 5(BA² + CA²)
∴ 4(BL² + MC²) = 5BC² … [Using (1)
Hence proved.

Question 44.
In the given figure, AD is median of ∆ABC and AE ⊥ BC. (2013)
Prove that b² + c² = 2p² + 12 a².

Solution:

Proof. Let ED = x
BD = DC = BC2=a2 = …[∵ AD is the median
In rt. ∆AEC, AC² = AE² + EC² …..[By Pythagoras’ theorem
b² = h² + (ED + DC)²
b² = (p² – x²) + (x = a2)²
…[∵ In rt. ∆AED, x² + h² = p² ⇒ h² = p² – x² …(i)
b² = p² – x² + x² + (a2)2²+ 2(x)(a2)
b² = p² + ax + a24 …(ii)
In rt. ∆AEB, AB² = AE² + BE² … [By Pythagoras’ theorem

Question 45.
In a ∆ABC, the perpendicular from A on the side BC of a AABC intersects BC at D such that DB = 3 CD. Prove that 2 AB² = 2 AC² + BC². (2013; 2017OD)
Solution:
In rt. ∆ADB,
AD² = AB² – BD² …(i) [Pythagoras’ theorem
In rt. ∆ADC,
AD² = AC² – DC² …(ii) [Pythagoras’ theorem

From (i) and (ii), we get
AB² – BD² = AC² – DC²
AB² = AC² + BD² – DC²
Now, BC = BD + DC
= 3CD + CD = 4 CD …[∵ BD = 3CD (Given)
⇒ BC² = 16 CD² …(iv) [Squaring
Now, AB² = AC² + BD² – DC² …[From (iii)
= AC² + 9 DC² – DC² ….[∵ BD = 3 CD ⇒ BD² = 9 CD²
= AC² + 8 DC²
= AC² + 16DC22
= AC² + BC22 … [From (iv)
∴ 2AB² = 2AC² + BC² … [Proved

Question 46.
In ∆ABC, altitudes AD and CE intersect each other at the point P. Prove that: (2014)
(i) ∆APE ~ ∆CPD
(ii) AP × PD = CP × PE
(iii) ∆ADB ~ ∆CEB
(iv) AB × CE = BC × AD
Solution:

Given. In ∆ABC, AD ⊥ BC & CE ⊥ AB.
To prove. (i) ∆APE ~ ∆CPD
(ii) AP × PD = CP × PE
(iii) ∆ADB ~ ∆CEB
(iv) AB × CE = BC × AD
Proof: (i) In ∆APE and ∆CPD,
∠1 = ∠4 …[Each 90°
∠2 = ∠3 …[Vertically opposite angles
∴ ∆APE ~ ∆CPD …[AA similarity
(ii) APCP=PEPD … [In ~ ∆s corresponding sides are proportional
∴ AP × PD = CP × PE
(iii) In ∆ADB and ∆CEB,
∠5 = ∠7 …[Each 90°
∠6 = ∠6 …(Common
∴ ∆ADB ~ ∆CEB …[AA similarity
(iv) ∴ ABCB=ADCE … [In ~ ∆s corresponding sides are proportional
∴ AB × CE = BC × AD

Question 47.
In the figure, PQR and QST are two right triangles, right angled at R and T resepctively. Prove that QR × QS = QP × QT. (2014)

Solution:
Given: Two rt. ∆’s PQR and QST.

To prove: QR × QS = QP × QT
Proof: In ∆PRQ and ∆STQ,
∠1 = ∠1 … [Common
∠2 = ∠3 … [Each 90°
∆PRQ ~ ∆STO …(AA similarity
∴ QRQT=QPQS ..[In -∆s corresponding sides are proportional
∴ QR × QS = QP × QT (Hence proved)

Question 48.
In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that ar(ABC)ar(DBC)=AODO. (2012)

Solution:
Given: ABC and DBC are two As on the same base BC. AD intersects BC at O.
To prove:

Question 49.
Hypotenuse of a right triangle is 25 cm and out of the remaining two sides, one is longer than the other by 5 cm. Find the lengths of the other
two sides. (2013)
Solution:
Let Base, AB = x cm
Then altitude, BC = (x + 5) cm
In rt. ∆,
By Pythagoras’ theorem

AB² + BC² = AC²
⇒ (x)² + (x + 5)² = 25²
⇒ x²2 + x² + 10x + 25 – 625 = 0
⇒ 2x² + 10x – 600 = 0
⇒ x² + 5x – 300 = 0 … [Dividing both sides by 2
⇒ x² + 20x – 15x – 300 = 0
⇒ x(x + 20) – 15(x + 20) = 0
(x – 15)(x + 20) = 0
x – 15 = 0 or x + 20 = 0
x = 15 or x = -20
Base cannot be -ve
∴ x = 15 cm
∴ Length of the other side = 15 + 5 = 20 cm
Two sides are = 15 cm and 20 cm

Question 50.
In Figure, AB ⊥ BC, FG ⊥ BC and DE ⊥ AC. Prove that ∆ADE ~ ∆GCF. (2016 OD)

Solution:
In rt. ∆ABC,
∠A + ∠C = 90° …(i)
In rt. ∆AED,
∠A + ∠2 = 90°
From (i) and (ii), ∠C = ∠2
Similarly, ∠A = ∠1
Now in ∆ADE & ∆GCF
∠A = 1 … [Proved
∠C = 2 … [Proved
∠AED = ∠GFC … [rt. ∠s
∴ ∆ADE – ∆GCF …(Hence Proved)

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Arithmetic Progressions Class 10 Important Questions

The start of class 10 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 10 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

Question 1.
If the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P
Solution:

Question 2.
If the ratio of the sum of first n terms of two A.P.’s is (7n + 1): (4n + 27), find the ratio of their mth terms.
Solution:

Question 3.
The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
Solution:

Question 4.
The sums of first n terms of three arithmetic progressions are S1, S2 and S3 respectively. The first term of each A.P. is 1 and their common differences are 1,2 and 3 respectively. Prove that S2 + S3 = 2Sr
Solution:

Question 5.
Divide 56 in four parts in A.R such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is 5 : 6.
Solution:

Question 6.
The pih, 9th and rth terms of an A.P. are a, b and c respectively. Show that a(q – r) + b(r-p) + c(p – q) = 0
Solution:

Question 7.
The sums of first n terms of three A.Ps’ are S1, S2 and S3. The first term of each is 5 and their common differences are 2,4 and 6 respectively. Prove that S1 + S3 = 2Sr
Solution:

Question 8.
A thief runs with a uniform speed of 100 m/minute. After one minute a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief.
Solution:

Question 9.
A thief, after committing a theft, runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 m/minute every succeeding minute. After how many minutes, the policeman will catch the thief?
Solution:

Question 10.
The sum of three numbers in A.P. is 12 and sum of their cubes is 288, Find the numbers.
Solution:

Question 11.
The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses proceding the house numbered X is equal to sum of the numbers of houses following X.
Solution:

Question 12.
Reshma wanted to save at least ? 6,500 for sending her daughter to school next year (after 12 months). She saved ? 450 in the first month and raised her savings by ? 20 every next month. How much will she be able to save in next 12 months? Will she be able to send her daughter to the school next year?
Solution:

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Quadratic Equations Class 10 Important Questions

The start of class 10 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 10 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

Question 1.
Find the values of k for which the quadratic equation (3k + 1)x² + 2(k + 1)x + 1 = 0 has equal roots. Also find the roots. (2014D)
Solution:
(3k + 1)x² + 2(k + 1) + 1 = 0
Here, a = 3k + 1, b = 2(k + 1), c = 1
D = 0 …[∵ Roots are equal
As b² – 4ac = 0
∴ [2(k + 1)]² – 4(3k + 1)(1) = 0
4(k + 1)² – 4(3k + 1) = 0
4(k² + 2k + 1 – 3k – 1) = 0
(k² – k) = 04 ⇒ k(k – 1) = 0
k = 0 or k – 1 = 0
∴ k = 0 or k = 1
Roots are x = −b2a ..[As equal roots (Given)
x = −2(k+1)2(3k+1) ⇒ x = −(k+1)(3k+1)
When k = 0, x = −(0+1)0+1 = -1
∴ Equal roots are -1 and -1
When k = 1, x = −(1+1)3+1
x= −24=−12
∴ Equal roots are −12 and −12

Question 2.
Find the value of p for which the quadratic equation (2p + 1)x² – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots. (2014D)
Solution:
(2p + 1)x² – (7p + 2)x + (7p – 3) = 0
Here, a = 2p + 1, b = -(7p + 2), c = 7p – 3
D = 0 …[∵ Equal roots As h2 – 4ac = 0
∴ [-(7p + 2)]² – 4(2p + 1)(7p – 3) = 0
⇒ (7p + 2)² – 4(14p² – 6p + 7p – 3) = 0
⇒ 49p² + 28p + 4 – 56p² + 24p – 28p + 12 = 0
⇒ -7p² + 24p + 16 = 0
⇒ 7p² – 24p – 16 = 0 … [Dividing both sides by -1
⇒ 7p² – 28p + 4p – 16 = 0
⇒ 7p(p – 4) + 4(p – 4) = 0
⇒ (p – 4) (7p + 4) = 0
⇒ p – 4 = 0 or 7p + 4 = 0
⇒ p = 4 or p = −47


∴ Equal roots are 7 and 7.

Question 3.
Find the roots of the equation 1x+4−1x−7=1130, x ≠ -4, 7 (2011D)
Solution:

⇒ (x + 4)(x – 7) = – 30
⇒ x² – 7x + 4x – 28 + 30 = 0
⇒ x² – 3x + 2 = 0
⇒ x² – x – 2x + 2 = 0
⇒ x(x – 1) – 2(x – 1) = 0
⇒ (x – 1)(x – 2) = 0
⇒ x – 1 = 0 or x – 2 = 0
∴ x = 1 or x = 2

Question 4.
Find the roots of the equation: 12x−3+1x−5=1, x ≠ 32, 5. (2011OD)
Solution:

Question 5.
Solve for x:
1x−3+2x−2=8x; x ≠ 0, 2, 3 (2013OD)
Solution:


⇒ 8(x – 3)(x – 2) = x(3x – 8)
⇒ 8(x² – 3x – 2x +6) = 3x² – 8x
⇒ 8x² – 24x – 16x + 48 – 3x² + 8x = 0
⇒ 5x² – 32x + 48 = 0
⇒ 5x² – 20x – 12x + 48 = 0
⇒ 5x(x – 4) – 12(x – 4) = 0
⇒ (x – 4)(5x – 12) = 0
⇒ x – 4 = 0 or 5x – 12 = 0
x = 4 or x = 125

Question 6.
Solve for x: 4x−3=52x+3; x ≠ 0, −32 (2013OD)
Solution:

⇒ 5x = (2x + 3 (4 – 3x)
⇒ 5x = 8x – 6x² + 12 – 9x
⇒ 5x – 8x + 6x² – 12 + x = 0
⇒ 6x + 6x – 12 = 0
⇒ x² + x – 2 = 0 …[Dividing by 6
⇒ x² + 2x – x – 2 = 0
⇒ x(x + 2) – 1(x + 2) = 0
⇒ (x – 1)(x + 2) = 0
⇒ x – 1 = 0 or x + 2 = 0
∴ x = 1 or x = -2

Question 7.
Solve for x: x−2x−3+x−4x−5=103; x ≠ 3, 5 (2014OD)
Solution:

⇒ 4(x² – 8x + 15) = (6x – 24)
⇒ 4x² – 32x + 60 – 6x + 24 = 0
⇒ 4x² – 38x + 84 = 0
⇒ 2x² – 19x + 42 = 0 …[Dividing by 2
⇒ 2x² – 12x – 7x + 42 = 0
⇒ 2x(x -6) – 7(x – 6) = 0
⇒ (x – 6) (2x – 7) = 0
⇒ x – 6 = 0 or 2x – 7 = 0
∴ x = 6 or x = 72

Question 8.
Solve for x: 3x+1+4x−1=294x−1;x≠1,−1,14 (2015D)
Solution:

⇒ [3(x – 1) + 4(x + 1)] [4x – 1] = 29(x² – 1)
⇒ (3x – 3 + 4x + 4) [4x – 1] = 29(x² – 1)
⇒ (7x + 1) (4x – 1) = 29x² – 29
⇒ 28x² – 3x – 1 = 29x² – 29
⇒ x² + 3x – 28 = 0
⇒ x² + 7x – 4x – 28 = 0
⇒ x(x + 7) – 4(x + 7) = 0
⇒ (x + 7) (x – 4) = 0
x = -7 or x = 4
∴ x = -7 and 4

Question 9.
Solve for x: 2x+1+32(x−2)=235x,x≠0,−1,2 (2015D)
Solution:

⇒ 5x[4 (x – 2) + 3x + 3) = 46(x + 1) (x – 2)
⇒ 5x[4x – 8 + 3x + 3) = 46[x² – 1x – 2]
⇒ 5x (7x – 5) = 46 (x² – x – 2)
⇒ 35x² – 25x = 46x² – 46x – 92
⇒ 35x² – 46x² – 25x + 46x + 92 = 0
⇒ 11x² – 21x – 92 = 0
Here, a = 11, b = -21, c = -92
D = b² – 4ac
= (-21)² – 4 × 11 × (-92)
= 441 + 4048 = 4489

Question 10.
Find x in terms of a, b and c: ax−a+bx−b=2cx−c, x ≠ a, b, c (2016D)
Solution:

⇒ (x – c)[ax – ab + bx – ab] = 2c(x – a)(x – b)
⇒ (x – c)(ax + bx – 2ab) = 2c(x² – bx – ax + ab)
⇒ ax² + bx² – 2abx – acx – bcx + 2abc = 2cx² – 2bcx – 2cax + 2abc
⇒ ax² + bx² – 2abx – acx – bcx – 2cx² + 2bcx + 2cax = 0
⇒ ax² + bx² – 2cx² – 2abx + bcx + cax = 0
⇒ x²(a + b – 2c) + x(-2ab + bc + ca) = 0
⇒ x[x (a + b – 2c) + (-2ab + bc + ca)] = 0
⇒ x = 0 or x (a + b – 2c) + (-2ab + bc + ca) = 0
⇒ x = 0 or x (a + b – 2c) = 2ab – bc – ca = 0
∴ x = 2ab−bc−cda+b−2c

Question 61.
Solve the following for x:  (2013D)
Solution:

⇒ 2x² + 2ax + bx + ab = 0
⇒ 2x (x + a) + b(x + a) = 0
⇒ (x + a) (2x + b) = 0
⇒ x + a = 0 or 2x + b = 0
⇒ x = -a or x = −b2

Question 11.
A shopkeeper buys some books for 80. If he had bought 4 more books for the same amount, each book would have cost ₹1 less. Find the number of books he bought. (2012D)
Solution:
Let the number of books he bought = x
Increased number of books he had bought = x +4
Total amount = ₹80
According to the problem,

⇒ x(x + 4) = 320
⇒ x² + 4x – 320 = 0
⇒ x² + 20x – 16x – 320 = 0
⇒ x(x + 20) – 16(x + 20) = 0
⇒ (x + 20) (x – 16) = 0
⇒ x + 20 = 0 or x – 16 = 0
⇒ x = -20 … (neglected) or x = 16
∴ Number of books he bought = 16

Question 12.
Sum of the areas of two squares is 400 cm². If the difference of their perimeters is 16 cm, find the sides of the two squares. (2013D)
Solution:
Let the side of Large square = x cm
Let the side of small square = y cm
According to the Question,
x² + y² = 400… (i) …[∵ area of square = (side)²
4x – 4y = 16 …[∵ Perimeter of square = 4 sides
⇒ x – y = 4 … [Dividing both sides by 4
⇒ x = 4 + y …(ii)
Putting the value of x in equation (i),
(4 + y)² + y2² = 400
⇒ y² + 8y + 16 + y² – 400 = 0
⇒ 2y² + 8y – 384 = 0
⇒ y² + 4y – 192 = 0 … [Dividing both sides by 2
⇒ y² + 16y – 12y – 192 = 0
⇒ y(y + 16) – 12(y + 16) = 0
⇒ (y – 12)(y + 16) = 0
⇒ y – 12 = 0 or y + 16 = 0
⇒ y = 12 or y = -16 … [Neglecting negative value
∴ Side of small square = y = 12 cm
and Side of large square = x = 4 + 12 = 16 cm

Question 13.
The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field. (2015OD)
Solution:
Let the length of shorter side be x m.
∴ length of diagonal = (x + 16) m
and length of longer side = (x + 14) m
Using pythagoras theorem,
(l)² + (b)² = (h)²
∴ x² + (x + 14)2² = (x + 16)²
⇒ x² + x² + 196 + 28x = x² + 256 + 32x
⇒ x² – 4x – 60 = 0
⇒ x² – 10x + 6x – 60 = 0
⇒ x(x – 10) + 6(x – 10) = 0
⇒ (x – 10) (x + 6) = 0
⇒ x – 10 = 0 or x + 6 = 0
⇒ x = 10 or x = -6 (Reject)
⇒ x = 10 m …[As length cannot be negative
Length of shorter side = x = 10 m
Length of diagonal = (x + 16) m = 26 m
Length of longer side = (x + 14)m = 24m
∴ Length of sides are 10 m and 24 m.

Question 14.
The sum of three numbers in A.P. is 12 and sum of their cubes is 288. Find the numbers. (2016D)
Solution:
Let three numbers in A.P. are a – d, a, a + d.
a – d + a + a + d = 12
⇒ 3a = 12 ⇒a = 4
(a – d)³ + (a)³ + (a + d)³ = 288
⇒ a³ – 3a²d + 3ad² – d³ + a³ + a³ + 3a²d + 3ad² + d³ = 288
⇒ 3a³ + 6ad² = 288
⇒ 3a(a² + 2d²) = 288
⇒ 3 × 4(4² + 2d²) = 288
⇒ (16 + 2d²) = 28812
⇒ 2d² = 24 – 16 = 8
⇒ d² = 4 ⇒ d = ± 2
When, a = 4, d = 2, numbers are –
a – d, a, a + d, i.e., 2, 4, 6
When, a = 4, d = -2, numbers are –
a – d, a, a + d, i.e., 6, 4, 2

Question 15.
The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find the area of the triangle. (2016D)
Solution:

Perimeter of right ∆ = 60 cm …[Given
a + b + c = 60
a + b + 25 = 60
a + b = 60 – 25 = 35 …(i)
In rt. ∆ACB, AC² + BC² = AB²
b² + a² = (25)² …[Pythagoras’ theorem
a² + b² = 625 ….(ii)
From (i), a + b = 35
(a + b)² = (35) … [Squaring both sides
a² + b² + 2ab = 1225
625 + 2ab = 1225 … [From (ii)
2ab = 1225 – 625 = 600 ⇒ ab = 300 … (iii)
Area of ∆ = 12 × base × corresponding altitude
= 12 × b × a = 12 (300) ..[From (iii)
= 150 cm²

Question 16.
The sum of two numbers is 9 and the sum of their reciprocals is 12. Find the numbers. (2012D)
Solution:
Let the numbers be x and 9 – x.
According to the Question,
1x+19−x=12
9−x+xx(9−x)=12
⇒ 18 = 9x – x²
⇒ x² – 9x + 18 = 0
⇒ x² – 3x – 6x + 18 = 0
⇒ x(x – 3) – 6(x – 3) = 0
⇒ (x – 3) (x – 6) = 0
⇒ x – 3 = 0 or x – 6 = 0
⇒ x = 3 or x = 6
When x = 3, nos. are 3 and 6.
When x = 6, nos. are 6 and 3.

Question 17.
The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by 115. Find the fraction. (20120D)
Solution:
Let the denominator be x and the numerator be x – 3.
∴ Fraction =x−3x
New denominator = x + 1
According to the Question,

⇒ 15x² – 45x = 14x² – 45x + 14x – 45
⇒ 15x² – 14x² – 14x + 45 = 0
⇒ x² – 14x + 45 = 0
⇒ x² – 5x – 9x + 45 = 0
⇒ x(x – 5) – 9(x – 5) = 0
⇒ (x – 5) (x – 9) = 0
⇒ x – 5 = 0 or x – 9 = 0
⇒ x = 5 or x = 9
When x = 5, fraction = 5−35=25
When x = 9, fraction = 9−39=69=23
∴ Fraction = 25 or 23

Question 18.
The difference of two natural numbers is 5 and the difference of their reciprocals is 110. Find the numbers. (2014D)
Solution:
Let the larger natural number be x and the smaller natural number be y.
According to the Question,

Question 19.
The difference of two natural numbers is 5 and the difference of their reciprocals is 514. Find the numbers. (2014D)
Solution:
Let the larger number be x and the smaller number be y.
According to the Question,

⇒ xy = 14
(5 + y)y = 14 … [From (i)
y² + 5y – 14 = 0
⇒ y² + 7y – 2y – 14 = 0
y(y + 7) – 2(y + 7) = 0
(y – 2) (y + 7) = 0
y – 2 = 0 or y + 7 = 0
y = 2 or y = -7
When y = 2, x = 5 + 2 = 7 …[From (1)
∴ Numbers are 7 and 2.
When y = -7, x = 5 + (-7) = -2 …[From (i)
∴ Numbers are -2 and (-7).

Question 20.
The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original fraction is a 2920. Find the original fraction. (2015D)
Solution: .
Let the denominator and numerator of the
fraction be x and x – 3 respectively.
Let the fraction be x−3x.
By the given condition,

⇒ 20[(x – 3) (x + 2) + x(x – 1)] = 29(x² + 2x)
⇒ 20(x² – x – 6 + x² – x) = 29x² + 58x
⇒ 20(2x² – 2x – 6) = 29x² + 58x
⇒ 40x² – 29x² – 40x – 58x = 120
⇒ 11x² – 98x – 120 = 0
⇒ 11x² – 110x + 12x – 120 = 0
⇒ 11x(x – 10) + 12(x – 10) = 0
⇒ (11x + 12) (x – 10) = 0
⇒ 11x + 12 = 0 or x – 10 = 0
⇒ x = −1211 (Reject) or x = 10
Now denominator (x) = 10
then, numerator = x – 3 = 7
∴ The fraction is 1o.

Question 21.
A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find the length and breadth of the rectangular park. (2016OD)
Solution:
Let length of the rectangular park = x m,
breadth of the rectangular park = (x -3)m
∴ Area of the rectangular park = x(x – 3)m²… (i)
Base of an isosceles triangle = (x – 3)m
Altitude of an isosceles triangle = 12 m
∴ Area of isosceles triangle
= 1/2 × base × altitude
= 1/2 × (x – 3) × 12
= 6(x – 3) …(ii)
According to the question,
Ar.(rectangle) – Ar.(isosceles ∆) = 4 m²
⇒ x(x – 3) – 6(x – 3) = 4 … [From (i) & (ii)
⇒ x² – 3x – 6x + 18 – 4 = 0
⇒ x² – 9x + 14 = 0
⇒ x² – 7x – 2x + 14 = 0
⇒ x(x – 7) – 2(x – 7) = 0
⇒ (x – 2) (x – 7) = 0
⇒ x – 2 = 0 or x – 7 = 0
⇒ x = 2 or x = 7
When x = 2, breadth of rectangle becomes -ve, so this is not possible.
∴ Length of the rectangular park, x = 7 m
and Breadth = (x – 3) = 4 m.

Question 22.
A train travels 180 km at a uniform speed. If the speed had been 9 km/hour more, it would have taken 1 hour less for the same journey. Find the speed of the train. (2011OD)
Solution:
Let the speed of the train = x km/hr
Let the increased speed of the train = (x + 9) km/hr
According to the question,

⇒ x(x + 9) = 1620
⇒ x² + 9x – 1620 = 0
⇒ x² + 45x – 36x – 1620 = 0
⇒ x(x + 45) – 36(x + 45) = 0
⇒ (x – 36) (x + 45) = 0
⇒ x – 36 = 0 or x + 45 = 0
⇒ x = 36 or x = -45 ….[Rejecting negative value as the speed cannot be -ve
∴ Speed of the train = 36 km/hr

Question 23.
In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by 100 km/h and time increased by 30 minutes. Find the original duration of the flight. (2012OD)
Solution:
Let the average speed of the aircraft = x km/hr
the reduced speed of the aircraft = (x – 100 km/hr
Then Distance = 2800 km
According to the Question,

⇒ x² – 100x = 560000
⇒ x² – 100x – 560000 = 0
⇒ x² – 800x + 700x – 560000 = 0
⇒ x(x – 800) + 700(x – 800) = 0
⇒ (x – 800) (x + 700) = 0
⇒ x – 800 = 0 or x + 700 = 0
⇒ x = 800 or x = -700
As speed of the aircraft cannot be -ve.
∴ Speed = 800 km/hr
∴ Original duration/time = Distance  Speed 
= 2800800=72
= 31/2 hrs. or 3 hrs. 30 mins. or 210 mins.

Question 24.
While boarding an aeroplane, a passenger got hurt. The pilot, showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 minutes. To reach the destination, 1500 km away in time, the pilot increased the speed by 100 km/hour. Find the original speed/hour of the plane. (2013OD)
Solution:
Let the original speed of the aeroplane = x km/hr
The increased speed of the aeroplane = (x + 100) km/hr
Given: Distance = 1500 km
According to the Question,

⇒ x(x + 100) = 300000
⇒ x² + 100x – 300000 = 0
⇒ x² + 600x – 500x – 300000 = 0
⇒ x(x + 600) – 500(x + 600) = 0
⇒ (x – 500) (x + 600) = 0
⇒ x – 500 = 0 or x + 600 = 0
⇒ x = 500 or x = -600 (rejected)
Since speed cannot be negative.
∴ Original speed of the aeroplane = 500 km/hr
Original time =  Distance  Speed =1,500500 = 3 hrs.

Question 25.
A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed? (2015OD)
Solution:
Let the original average speed of (first) train be x km/hr.
Now, new speed will be = (x + 6) km/hr.
We know,

⇒ 54x + 324 + 63x = 3x (x + 6)
⇒ 117x + 324 = 3x² + 18x
⇒ 3x² + 18x – 117x – 324 = 0
⇒ 3x² – 99x – 324 = 0
⇒ x² – 33x – 108 = 0
⇒ x² – 36x + 3x – 108 = 0
⇒ x(x – 36) + 3(x – 36) = 0
⇒ (x – 36) (x + 3) = 0
⇒ x – 36 = 0 or x + 3 = 0
x = 36 or x = -3 (Reject)
∴ First speed of train = 36 km/h.

Question 26.
A bus travels at a certain average speed for a distance of 75 km and then travels a distance of 90 km at an average speed of 10 km/h more than the first speed. If it takes 3 hours to complete the total journey, find its first speed. (2015OD)
Solution:
Let the first average speed of the bus be x km/hr.
Now, new speed of bus = (x + 10) km/hr.


⇒ 75x + 750 + 90x = 3(x² + 10x)
⇒ 165x + 750 – 3x² – 30x = 0
⇒ 3x² – 135x – 750 = 0
⇒ x² – 45x – 250 = 0
⇒ x² – 50x + 5x – 250 = 0
⇒ x(x – 50) + 5(x – 50) = 0
⇒ (x – 50) (x + 5) = 0
∴ x = 50 or x = -5 (Reject)
∴ Speed = 50 km/h

Question 27.
A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed which is 20 km per hour more than the first speed. If the truck covers the total distance in 5 hours, find the first speed of the truck. (2015OD)
Solution:
Let the first average speed of truck be x km/hr.

⇒ 150x + 3000 + 200x = 5(x² + 20x)
⇒ 350x + 3000 – 5x² – 100x = 0
⇒ x² – 50x – 600 = 0
⇒ x² – 60x + 10x – 600 = 0
⇒ x(x – 60) + 10(x – 60) = 0
⇒ (x – 60) (x + 10) = 0
⇒ x – 60 = 0 or x + 10 = 0
⇒ x = 60 or x = -10 (Reject)
∴ Speed = 60 km/hr.

Question 28.
Two pipes running together can fill a tank in 1119 minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately. (2016OD)
Solution:
Let the quicker pipe take to fill the cistern = x minutes
Then the slower pipe takes to fill the cistern = (x + 3) minutes
According to Question,

13xv + 39x = 80x + 120
13x² + 39x – 80x – 120 = 0
13x² – 41x – 120 = 0
13x² – 65x + 24x – 120 = 0
13x(x – 5) + 24(x – 5) = 0
(x – 5)(13x + 24) = 0
x – 5 = 0 or 13x + 24 = 0
x = 5 or x = −2413 (rejected) …[∵ x > 0
∴ x = 5 Hence, the faster pipe fills the cistern in 5 minutes, and the slower pipe fills the cistern in 8(5 + 3) minutes.

Question 29.
A motor boat whose speed is 20 km/h in still water, takes 1 hour more to go 48 km upstream than to return downstream to the same spot. Find the speed of the stream. (2011D)
Solution:
Let the speed of the stream be x km/hr
∴ Speed of the boat upstream = (20 – x) km/hr
and speed of the boat downstream = (20 + x) km/hr
Given, Distance = 48 km
According to the Question,

⇒ 96x = 400 – x²
⇒ x² + 96x – 400 = 0
⇒ x² + 100x – 4x – 400 = 0
⇒ x (x + 100) – 4 (x + 100) = 0
⇒ (x – 4) (x + 100) = 0
⇒ x – 4 = 0 or x + 100 = 0
⇒ x = 4 or x = – 100 ….[Rejecting negative value as the speed cannot be -ve
∴ Speed of the stream = 4 km/hr

Question 30.
A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot.
Find the speed of the stream. (2014OD)
Solution:
Let speed of the stream be x km/hr,
Speed of the boat upstream = (18 – x) km/hr
and Speed of the boat downstream = (18 + x) km/hr
Given, Distance = 24 km
According to the Question,

⇒ 24(2x)324−x2=1
⇒ 48x = 324 – x²
⇒ x² + 48x – 324 = 0
⇒ x² + 54x – 6x – 324 = 0
⇒ x(x + 54) – 6(x + 54) = 0
⇒ (x – 6) (x + 54) = 0
x – 6 = 0 or x + 54 = 0
x = 6 or x= -54 (rejected)
Since speed cannot be negative
∴ Speed of stream, x = 6 km/hr

Question 31.
The time taken by a person to cover 150 km was 2 hours more than the time taken in the return journey. If he returned at a speed of 10 km/hour more than the speed while going, find the speed per hour in each direction. (2016D)
Solution:
Let the speed of a person while going = x km/hr Then the speed of a person while returning = (x + 10) km/hr
Given, Distance = 150 km

⇒ 5x(x + 10) = 3,000
⇒ x(x + 10) = 600 …[Dividing both sides by 5
⇒ x² + 10x – 600 = 0
⇒ x² + 30x – 20x – 600 = 0
⇒ x(x + 30) – 20(x + 30) = 0
⇒ (x + 30) (x – 20) = 0
⇒ x + 30 = 0 or x – 20 = 0
⇒ x = -30 (rejected) or x = 20
Since, speed can not be negative.
∴ Speed x = 20 km/hr.
∴ Speed while going = x = 20 km/hr
and Speed while returning
= (x + 10) = 20 + 10 = 30 km/hr

Question 32.
To fill a swimming pool two pipes are to be used. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool. (2015D)
Solution:
Let the bigger pipe fill the tank in x hrs.
∴ the smaller pipe fills the tanks in (x + 10) hrs.

⇒ 2(13x + 40) = x² + 10x
⇒ 26x + 80 = x² + 10x
⇒ x² + 10x – 26x = 80
⇒ x² – 16x – 80 = 0
⇒ x² – 20x + 4x – 80 = 0
⇒ x(x – 20) + 4(x – 20) = 0
⇒ (x – 20) (x + 4) = 0
⇒ x – 20 = 0 or x + 4 = 0
x = 20 x = -4 (Reject)
Hence, the pipe with larger diameter fills the tank in 20 hours.
and the pipe with smaller diameter fills the tank in 30 hours.

Important Links

NCERT Quick Revision Notes- Quadratic Equations

NCERT Solution- Quadratic Equations

Important MCQs- Quadratic Equations

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Pair of Linear Equations in Two Variables Class 10 Important Questions

The start of class 10 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 10 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

Question 1.
The age of the father is twice the sum of the ages of his 2 children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father. (2012)
Solution:
Let the present ages of his children be x years and y years.
Then the present age of the father = 2(x + y) …(i)
After 20 years, his children’s ages will be
(x + 20) and (y + 20) years
After 20 years, father’s age will be 2(x + y) + 20
According to the Question,
⇒ 2(x + y) + 20 = x + 20 + y + 20
⇒ 2x + 2y + 20 = x + y + 40
⇒ 2x + 2y – x – y = 40 – 20
⇒ x + y = 20 …[From (i)
∴ Present age of father = 2(20) = 40 years

Question 2.
A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number. (2013)
Solution:
Let unit’s place digit be x and ten’s place digit bey.
Then original number = x + 10y
and reversed number = 10x + y
According to the Question,
x + 10y = 7(x + y)
x + 10y = 7x + 7y
⇒ 10y – 7y = 7x – x
⇒ 3y = 6x ⇒ y = 2x …(i)
(x + 10y) – (10x + y) = 18
x + 10y – 10x – y = 18
⇒ 9y – 9x = 180
⇒ y – x = 2 …[Dividing by 9
⇒ 2x – x = 2 …[From (i)
∴ x = 2
Putting the value of ‘x’ in (i), we get y = 2(2) = 4
∴ Required number = x + 10y
= 2 + 10(4) = 42

Question 3.
Sita Devi wants to make a rectangular pond on the road side for the purpose of providing drinking water for street animals. The area of the pond will be decreased by 3 square feet if its length is decreased by 2 ft. and breadth is increased by 1 ft. Its area will be increased by 4 square feet if the length is increased by 1 ft. and breadth remains same. Find the dimensions of the pond. (2014)
Solution:
Let length of rectangular pond = x
and breadth of rectangular pond = y
Area of rectangular pond = xy
According to Question,

∴Length of rectangular pond = 7 ft.
Breadth of rectangular pond = 4 ft.

Question 4.
On reversing the digits of a two digit number, number obtained is 9 less than three times the original number. If difference of these two numbers is 45, find the original number. (2014)
Solution:
Let unit’s place digit be x and ten’s place digit bey.
∴ Original number = x + 10y Reversed number = 10x + y
According to the Question,
10x + y = 3(x + 10y) – 9
⇒ 10x + y = 3x + 30y – 9
⇒ 10x + y – 3x – 30y = -9
⇒ 7x – 29y = -9 …(i)
10x + y – (x + 10y) = 45
⇒ 9x – 9y = 45
⇒ x – y = 5 …[Dividing both sides by 9
⇒ x – 5 + y …(ii)
Solving (i),
7x – 29y = -9
7(5 + y) – 29y = -9 …[From (ii)
35+ 7y – 29y = -9
-22y = -9 – 35
-22y = -44 ⇒ y = 4422 = 2
Putting the value of y in (ii),
x = 5 + 2 = 7
∴ Original number = x + 10y
= 7 + 10(2) = 27

Question 5.
Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream. 2017D
Solution:
Let the speed of the stream = x km/hr
Speed of the boat in still water = 15 km/hr
then, the speed of the boat upstream = (15 – x) km/hr
and the speed of the boat downstream = (15 + x) km/hr

∴ Speed of stream = 5 km/hr

Question 6.
The owner of a taxi company decides to run all the taxis on CNG fuel instead of petrol/diesel. The taxi charges in city comprises of fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is 789 and for journey of 20 km, the charge paid is ₹145.
What will a person have to pay for travelling a distance of 30 km? (2014)
Solution:
Let the fixed charges = 7x
and the charge per km = ₹y
According to the Question,

Putting the value of y in (i), we get
x + 12(7) = 89
x + 84 = 89 ⇒ x = 89 – 84 = 5
Total fare for 30 km = x + 30y = 5 + 30(7)
= 5 + 210 = ₹215

Question 7.
A boat takes 4 hours to go 44 km downstream and it can go 20 km upstream in the same time. Find the speed of the stream and that of the boat in still water. (2015)
Solution:
Let the speed of the stream = y km/hr
Let the speed of boat in still water = x km/hr
then, the speed of the boat in downstream = (x + y) km/hr
and, the speed of the boat in upstream = (x – y) km/hr

From (i), x = 11 – 3 = 8
∴ Speed of the stream, y =3 km/hr
Speed of the boat in still water, x = 8 km/hr

Question 8.
A man travels 300 km partly by train and partly by car. He takes 4 hours if the travels 60 km by train and the rest by car. If he travels 100 km by train and the remaining by car, he takes 10 minutes longer. Find the speeds of the train and the car separately. (2017D)
Solution:
Let the speed of the train = x km/hr
Let the speed of the car = y km/ hr
According to the Question,

∴ Speed of the train = 60 km/hr
and Speed of the car = 80 kn/hr

Question 9.
The owner of a taxi company decides to run all the taxis on CNG fuel instead of petrol/diesel. The taxi charges in city comprises of fixed charges together with the charge for the distance covered. For a journey of 13 km, the charge paid is ₹129 and for a journey of 22 km, the charge paid is ₹210.
What will a person have to pay for travelling a distance of 32 km? (2014 )
Solution:
Let fixed charge be ₹x and the charge for the distance = ₹y per km
According to the Question,
For a journey of 13 km,
x + 13y = 129 ⇒ x = 129 – 13y …(/)
For a journey of 22 km, x + 22y = 210 …(ii)
⇒ 129 – 13y + 22y = 210 …[From (i)
⇒ 9y = 210 – 129 = 81
⇒ 9y = 81 ⇒ y = 9
From (i), x = 129 – 13(9)
= 129 – 117 = 12
∴ Fixed charge, x = ₹12
∴ The charge for the distance, y = ₹9 per km
To pay for travelling a distance of 32 km
= x + 32y = 12 + 32(9) = 12 + 288 = ₹300

Question 10.
Solve the following pair of linear equations graphically:
x + 3y = 6 ; 2x – 3y = 12
Also find the area of the triangle formed by the lines representing the given equations with y-axis. (2012, 2015)
Solution:

By plotting points and joining them, the lines intersesct at A(6, 0)
∴ x = 6, y = 0
Line x + 3y = 6 intersects Y-axis at B(0, 2) and Line 2x – 3y = 12 intersects Y-axis at C(0, -4). Therefore, Area of triangle formed by the lines with y-axis.
Area of triangle
= 12 × base × corresponding altitude
= 12 × BC × AO = 12 × 6 × 6 = 18 sq. units

Question 11.
Draw the graphs of following equations:
2x – y = 1; x + 2y = 13
Find the solution of the equations from the graph and shade the triangular region formed by the lines and the y-axis. (2013)
Solution:

By plotting the points and joining them, the lines intersect at A(3,5).
∴ x = 3, y = 5
Here ∆ABC is the required triangle.

Question 12.
Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and x-axis. (2012, 2017D)
Solution:

Lines intersect at (2, 3)
∴ x = 2, y = 3
Vertices of ∆ABC are A(2, 3), B(-1, 0) and C(4, 0)

Question 13.
Amit bought two pencils and three chocolates for ₹11 and Sumeet bought one pencil and two chocolates for ₹7. Represent this situation in the form of a pair of linear equations. Find the price of one pencil and that of one chocolate graphically. (2017OD)
Solution:
Let the price of one pencil = ₹x and the price of one chocolate = ₹y.
As per the Question,

Lines intersect at (1, 3).
∴ x = 1, y = 3
Therefore the price of one pencil = ₹1 and price of one chocolate = ₹3

Question 14.
7x – 5y – 4 = 0 is given. Write another linear equation, so that the lines represented by the pair are:
(i) intersecting
(ii) coincident
(iii) parallel (2015 OD)
Solution:
7x – 5y – 4 = 0

Important Links

NCERT Quick Revision Notes- Pair of Linear Equations in Two Variables

NCERT Solution-Pair of Linear Equations in Two Variables

Important MCQs- Pair of Linear Equations in Two Variables

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