Chapter 7 : Lines And Angles - R S Aggarwal And V Aggarwal Solutions for Class 9 Maths CBSE

If you love numbers & have a deep interest in statistics, then Mathematics is probably one of the scoring subject ever. A Class 9th CBSE Mathematics is one of the challenging subject one can undergo as, it involves a lot of crunching of complex numbers, geometrical formulae, diagrams & much more. Hence, to simplify the mathematical complexity, TopperLearning has framed a customise solution that involves Test preparation notes, Textbook solutions, Videos, & several other study material that help the student to memorise the concepts quickly. We have bifurcated our CBSE Class 9 Study Material (Solution) into 3 Different parts namely: 

- CBSE Class 9 Practice Test includes Revision notes, Question bank & Sample Paper 
- TopperLearning Classroom that includes Videos, Ask the Experts & Preparation Tips
- Text Book Solutions, etc

TopperLearning packages involves all the ingredients of CBSE Class 9 Study Material that includes Solved Question Papers, High Animated Videos, Solutions by SME (Subject Matter Expert), CBSE class 9 Preparation Tips, Update to CBSE 9th Mathematics syllabus, Practice books, Reference Materials, etc that help you to score great marks with glorious marks.

Getting a good score is now relatively easy if you prefer TopperLearning solutions for CBSE Class 9th Mathematics subject. By purchasing our package, you will be accessed to guaranteed success in your examination!

Read  more
Page / Exercise

Chapter 7 - Lines And Angles Excercise MCQ

Question 1

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is

(a) an isosceles triangle

(b) an obtuse triangle

(c) an equilateral triangle

(d) a right triangle

Solution 1

Correct option: (d)

In a right triangle, one angle is 90° and the sum of acute angles of a right triangle is 90°.

Question 2

An exterior angle of a triangle is 110° and its two interior opposite angles are equal. Each of these equal angles is

(a) 70° 

(b) 55° 

(c) 35° 

(d) 

Solution 2

Correct option: (b)

Let each interior opposite angle be x.

Then, x + x = 110° (Exterior angle property of a triangle)

2x = 110° 

x = 55° 

Question 3

The angles of a triangle are in the ratio 3:5:7 The triangle is

  1. Acute angled
  2. Obtuse angled
  3. Right angled
  4. an isosceles triangle
Solution 3

Question 4

If one of the angles of triangle is 130° then the angle between the bisector of the other two angles can be

(a) 50° 

(b) 65° 

(c) 90° 

(d) 155

Solution 4

Correct option: (d)

  

Let A = 130° 

In ΔABC, by angle sum property,

B + C + A = 180° 

B + C + 130° = 180° 

B + C = 50° 

Question 5

In the given figure, AOB is a straight line. The value of x is

  

 

(a) 12

(b) 15

(c) 20

(d) 25

Solution 5

Correct option: (b)

AOB is a straight line.

AOB = 180° 

60° + 5x° + 3x° = 180° 

60° + 8x° = 180° 

8x° = 120° 

x = 15° 

Question 6

The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle of the triangle is

(a) 120° 

(b) 100° 

(c) 80° 

(d) 60° 

Solution 6

Correct option: (c)

By angle sum property,

2x + 3x + 4x = 180° 

9x = 180° 

x = 20° 

Hence, largest angle = 4x = 4(20°) = 80° 

Question 7

In the given figure, OAB = 110° and BCD = 130° then ABC is equal to

 

  

(a) 40° 

(b) 50° 

(c) 60° 

(d) 70° 

Solution 7

Correct option: (c)

Through B draw YBZ OA CD.

  

Now, OA YB and AB is the transversal.

OAB + YBA = 180° (interior angles are supplementary)

110° + YBA = 180° 

YBA = 70° 

Also, CD BZ and BC is the transversal.

DCB + CBZ = 180° (interior angles are supplementary)

130° + CBZ = 180° 

CBZ = 50° 

Now, YBZ = 180° (straight angle)

YBA + ABC + CBZ = 180° 

70° + x + 50° = 180° 

x = 60° 

ABC = 60° 

Question 8

If two angles are complements of each other, then each angle is

  1. An acute angle
  2. An obtuse angle
  3. A right angle
  4. A reflex angle
Solution 8

Correct option: (a)

Two angles are said to be complementary, if the sum of their measures is 90°.

Clearly, the measures of each of the angles have to be less than 90°.

Hence, each angle is an acute angle.

Question 9

An angle which measures more than 180° but less than 360°, is called

  1. An acute angle
  2. An obtuse angle
  3. A straight angle
  4. A reflex angle
Solution 9

Correct option: (d)

An angle which measures more than 180o but less than 360o is called a reflex angle.

Question 10

The measure of an angle is five times its complement. The angle measures

  1. 25°
  2. 35°
  3. 65°
  4. 75°
Solution 10

Question 11

Two complementary angles are such that twice the measure of the one is equal to three times the measure of the other. The larger of the two measures

  1. 72°o
  2. 54°
  3. 63°
  4. 36°
Solution 11

Question 12

In the given figure, AOB is a straight line. If AOC = 4x° and BOC = 5x°, then AOC =?

Solution 12

Question 13

In the given figure, AOB is a straight line. If AOC = (3x + 10) ° and BOC = (4x - 26) °, then BOC =?

  1. 96°
  2. 86°
  3. 76°
  4. 106°

Solution 13

Question 14

In the given figure, AOB is a straight line. If AOC = (3x - 10) °, COD = 50° and BOD = (x +20) °, then AOC =?

  1. 40°
  2. 60°
  3. 80°
  4. 50°
Solution 14

Question 15

Which of the following statements is false?

  1. Through a given point, only one straight line can be drawn
  2. Through two given points, it is possible to draw one and only one straight line.
  3. Two straight lines can intersect only at one point
  4. A line segment can be produced to any desired length.
Solution 15

Correct option: (a)

Option (a) is false, since through a given point we can draw an infinite number of straight lines.

Question 16

An angle is one-fifth of its supplement. The measure of the angle is

  1. 15°
  2. 30°
  3. 75°
  4. 150°
Solution 16

Question 17

In the adjoining figure, AOB is straight line. If x:y:z = 4:5:6, then y = ?

  1. 60°
  2. 80°
  3. 48°
  4. 72°

Solution 17

Question 18

In the given figure, straight lines AB and CD intersect at O. If AOC =ϕ, BOC = θ and θ = 3 ϕ, then ϕ =?

  1. 30°
  2. 40°
  3. 45°
  4. 60°

Solution 18

Question 19

In the given figure, straight lines AB and CD intersect at O. If AOC + BOD = 130°, then AOD =?

  1. 65°
  2. 115°
  3. 110°
  4. 125°
Solution 19

Question 20

In the given figure AB is a mirror, PQ is the incident ray and and QR is the reflected ray. If PQR = 108°, then AQP =?

  1. 72°
  2. 18°
  3. 36°
  4. 54°

Solution 20

Question 21

In the given figure, AB CD, If BAO = 60° and OCD = 110° then AOC = ?

  

 

(a) 70° 

(b) 60° 

(c) 50° 

(d) 40° 

Solution 21

Correct option: (c)

Let AOC = x° 

Draw YOZ CD AB.

  

Now, YO AB and OA is the transversal.

YOA = OAB = 60° (alternate angles)

Again, OZ CD and OC is the transversal.

COZ + OCD = 180° (interior angles)

COZ + 110° = 180° 

COZ = 70° 

Now, YOZ = 180° (straight angle)

YOA + AOC + COZ = 180° 

60° + x + 70° = 180° 

x = 50° 

AOC = 50° 

Question 22

In the given figure, AB CD. If AOC = 30° and OAB = 100°, then OCD =?

  1. 130°
  2. 150°
  3. 80°
  4. 100°

Solution 22

Question 23

In the given figure, AB CD. If CAB = 80o and EFC= 25°, then CEF =?

  1. 65°
  2. 55°
  3. 45°
  4. 75°

Solution 23

Question 24

In the given figure, AB CD, CD EF and y:z = 3:7, then x = ?

  1. 108°
  2. 126°
  3. 162°
  4. 63°

Solution 24

Question 25

In the given figure, AB CD. If APQ = 70° and RPD = 120°, then QPR =?

  1. 50°
  2. 60°
  3. 40°
  4. 35°

Solution 25

  

Question 26

In the given figure AB CD. If EAB = 50° and ECD=60°, then AEB =?

  1. 50°
  2. 60°
  3. 70°
  4. 50°

Solution 26

Question 27

In the given figure, OAB = 75°, OBA=55° and OCD = 100°. Then ODC=?

  1. 20°
  2. 25°
  3. 30°
  4. 35°

Solution 27

Question 28

In the adjoining figure y =?

  1. 36°
  2. 54°
  3. 63°
  4. 72°

Solution 28

Chapter 7 - Lines And Angles Excercise Ex. 7A

Question 1

Define the following terms:

(i) Angle (ii) Interior of an angle (iii) Obtuse angle (iv) Reflex angle (v) Complementary angles (vi) Supplementary angles

Solution 1

(i) Angle: Two rays having a common end point form an angle.

(ii) Interior of an angle: The interior of AOB is the set of all points in its plane, which lie on the same side of OA as B and also on same side of OB as A.

(iii) Obtuse angle: An angle whose measure is more than 90o but less than 180o, is called an obtuse angle.

(iv) Reflex angle: An angle whose measure is more than 180o but less than 360o is called a reflex angle.

(v) Complementary angles: Two angles are said to be complementary, if the sum of their measures is 90o.

(vi) Supplementary angles: Two angles are said to be supplementary, if the sum of their measures is 180o.


Question 2

Find the complement of each of the following angles.

16o

Solution 2

Complement of 16o = 90 - 16o = 74o

Question 3

Find the complement of each of the following angles.

46o 30

Solution 3


Complement of 46o 30' = 90o - 46o 30' = 43o 30'

Question 4

Find the complement of each of the following angle:

55° 

Solution 4

Complement of 55° = 90° - 55° = 35°  

Question 5

Find the complement of each of the following angle:

90° 

Solution 5

Complement of 90° = 90° - 90° = 0° 

Question 6

Find the supplement of each of the following angles.

75o 36'

Solution 6

Supplement of 75o 36' = 180o - 75o 36' = 104o 24'

Question 7

Find the supplement of each of the following angle:

42° 

Solution 7

Supplement of 42° = 180° - 42° = 138° 

Question 8

Find the supplement of each of the following angle:

90° 

Solution 8

Supplement of 90° = 180° - 90° = 90° 

Question 9

Find the supplement of each of the following angle:

124° 

Solution 9

Supplement of 124° = 180° - 124° = 56° 

Question 10

Find the measure of an angle which is

(i) equal to its complement, (ii) equal to its supplement.

Solution 10

(i) Let the required angle be xo

Then, its complement = 90o - xo

The measure of an angle which is equal to its complement is 45o.

(ii) Let the required angle be xo

Then, its supplement = 180o - xo

The measure of an angle which is equal to its supplement is 90o.

Question 11

Find the measure of an angle which is 36o more than its complement.

Solution 11

Let the required angle be xo

Then its complement is 90o - xo

The measure of an angle which is 36o more than its complement is 63o.

Question 12

Find the measure of an angle which is 30° less than its supplement.

Solution 12

Let the measure of the required angle = x° 

Then, measure of its supplement = (180 - x)° 

It is given that

x° = (180 - x)° - 30° 

x° = 180° - x° - 30° 

2x° = 150° 

x° = 75° 

Hence, the measure of the required angle is 75°. 

Question 13

Find the angle which is four times its complement.

Solution 13

Let the required angle be xo

Then, its complement = 90o - xo

The required angle is 72o.

Question 14

Find the angle which is five times its supplement.

Solution 14

Let the required angle be xo

Then, its supplement is 180o - xo

The required angle is 150o.

Question 15

Find the angle whose supplement is four times its complement.

Solution 15

Let the required angle be xo

Then, its complement is 90o - xo and its supplement is 180o - xo

That is we have,


The required angle is 60o.


Question 16

Find the angle whose complement is one-third of its supplement.

Solution 16

Let the required angle be xo

Then, its complement is 90o - xo and its supplement is 180o - xo

The required angle is 45o.

Question 17

Two complementary angles are in the ratio 4: 5. Find the angles.

Solution 17

Let the two required angles be xo and 90o - xo.

Then

5x = 4(90 - x)

5x = 360 - 4x

5x + 4x = 360

9x = 360

Thus, the required angles are 40o and 90o - xo = 90 o - 40o = 50o.


Question 18

Find the value of x for which the angles (2x - 5)° and (x - 10)° are the complementary angles.

Solution 18

(2x - 5)° and (x - 10)° are complementary angles.

(2x - 5)° + (x - 10)° = 90° 

2x - 5° + x - 10° = 90° 

3x - 15° = 90° 

3x = 105° 

x = 35° 

Chapter 7 - Lines And Angles Excercise Ex. 7B

Question 1

In the given figure, AOB is a straight line. Find the value of x.

Solution 1

 

Since BOC and COA form a linear pair of angles, we have

BOC + COA = 180o

xo + 62o = 180o

x = 180 - 62

x = 118o

Question 2

In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find AOC and BOD.

 

  

Solution 2

AOB is a straight angle.

AOB = 180° 

AOC + COD + BOD = 180° 

(3x - 7)° + 55° + (x + 20)° = 180° 

4x + 68° = 180° 

4x = 112° 

x = 28° 

Thus, AOC = (3x - 7)° = 3(28°) - 7° = 84° - 7° = 77° 

And, BOD = (x + 20)° = 28° + 20° = 48° 

Question 3

In the given figure, AOB is a straight line. Find the value of x. Hence, find AOC, COD and BOD.

Solution 3

Since BOD and DOA from a linear pair of angles.

BOD + DOA = 180o

BOD + DOC + COA = 180o

xo + (2x - 19)o + (3x + 7)o = 180o

6x - 12 = 180

6x = 180 + 12 = 192


x = 32

AOC = (3x + 7)o = (3 32 + 7)o = 103o

COD = (2x - 19)o = (2 32 - 19)o = 45o

and BOD = xo = 32o


Question 4

In the given figure, x: y: z = 5: 4: 6. If XOY is a straight line, find the values of x, y and z.

Solution 4

x: y: z = 5: 4: 6

The sum of their ratios = 5 + 4 + 6 = 15

But x + y + z = 180o

[Since, XOY is a straight line]

So, if the total sum of the measures is 15, then the measure of x is 5.

If the sum of angles is 180o, then, measure of 

And, if the total sum of the measures is 15, then the measure of y is 4.

If the sum of the angles is 180o, then, measure of

And z = 180o - x - y

= 180o - 60o - 48o

= 180o - 108o = 72o

x = 60, y = 48 and z = 72.


Question 5

In the given figure, what value of x will make AOB, a straight line?

Solution 5

AOB will be a straight line, if two adjacent angles form a linear pair.

BOC + AOC = 180o

(4x - 36)o + (3x + 20)o = 180o

4x - 36 + 3x + 20 = 180

7x - 16 = 180o

7x = 180 + 16 = 196

The value of x = 28.

Question 6

Two lines AB and CD intersect at O. If AOC = 50o, find AOD, BOD and BOC.

Solution 6

Since AOC and AOD form a linear pair.

AOC + AOD = 180o

50o + AOD = 180o

AOD = 180o - 50o = 130o

AOD and BOC are vertically opposite angles.

AOD = BOC

BOC = 130o

BOD and AOC are vertically opposite angles.

BOD = AOC

BOD = 50o

Question 7

In the given figure, three coplanar lines AB,CD and EF intersect at a point O, forming angles as shown. Find the values of x,y,z and t.

Solution 7

Since COE and DOF are vertically opposite angles, we have,

COE = DOF

z = 50o

Also BOD and COA are vertically opposite angles.

So, BOD = COA

t = 90o

As COA and AOD form a linear pair,

COA + AOD = 180o

COA + AOF + FOD = 180o [t = 90o]

t + x + 50o = 180o

90o + xo + 50o = 180o

x + 140 = 180

x = 180 - 140 = 40

Since EOB and AOF are vertically opposite angles

So, EOB = AOF

y = x = 40

Thus, x = 40 = y = 40, z = 50 and t = 90

Question 8

In the given figure, three coplanar lines AB,CD and EF intersect at a point O. Find the value of x. Hence, find AOD, COE and AOE.

Solution 8

Since COE and EOD form a linear pair of angles.

COE + EOD = 180o

COE + EOA + AOD = 180o

5x + EOA + 2x = 180

5x + BOF + 2x = 180

[EOA and BOF are vertically opposite angles so, EOA = BOF]

5x + 3x + 2x = 180

10x = 180

x = 18

Now AOD = 2xo = 2 18o = 36o

COE = 5xo = 5 18o = 90o

and, EOA = BOF = 3xo = 3 18o = 54o

Question 9

Two adjacent angles on a straight line are in the ratio 5 : 4. Find the measure of each one of these angles.

Solution 9

Let the two adjacent angles be 5x and 4x.

Now, since these angles form a linear pair.

So, 5x + 4x = 180o

9x = 180o

The required angles are 5x = 5x = 5 20o = 100o

and 4x = 4 20o = 80o

Question 10

If two straight lines intersect each other in such a way that one of the angles formed measures 90o, show that each of the remaining angles measures 90o.

Solution 10

Let two straight lines AB and CD intersect at O and let AOC = 90o.

Now, AOC = BOD [Vertically opposite angles]

BOD = 90o

Also, as AOC and AOD form a linear pair.

90o + AOD = 180o

AOD = 180o - 90o = 90o

Since, BOC = AOD [Verticallty opposite angles]

BOC = 90o

Thus, each of the remaining angles is 90o.

Question 11

Two lines AB and CD intersect at a point O such that BOC +AOD = 280o, as shown in the figure. Find all the four angles.

Solution 11

Since, AOD and BOC are vertically opposite angles.

AOD = BOC

Now, AOD + BOC = 280o [Given]

AOD + AOD = 280o

2AOD = 280o

AOD =

BOC = AOD = 140o

As, AOC and AOD form a linear pair.

So, AOC + AOD = 180o

AOC + 140o = 180o

AOC = 180o - 140o = 40o

Since, AOC and BOD are vertically opposite angles.

AOC = BOD

BOD = 40o

BOC = 140o, AOC = 40o , AOD = 140o and BOD = 40o.

Question 12

Two lines AB and CD intersect each other at a point O such that AOC : AOD = 5 : 7. Find all the angles.

  

Solution 12

Let AOC = 5x and AOD = 7x

Now, AOC + AOD = 180° (linear pair of angles)

5x + 7x = 180° 

12x = 180° 

x = 15° 

AOC = 5x = 5(15°) = 75° and AOD = 7x = 7(15°) = 105° 

Now, AOC = BOD (vertically opposite angles)

BOD = 75° 

Also, AOD = BOC (vertically opposite angles)

BOC = 105° 

Question 13

In the given figure, three lines AB, CD and EF intersect at a point O such that AOE = 35° and BOD = 40°. Find the measure of AOC, BOF, COF and DOE.

  

Solution 13

BOD = 40° 

AOC = BOD = 40° (vertically opposite angles)

AOE = 35° 

BOF = AOE = 35° (vertically opposite angles)

AOB is a straight angle.

AOB = 180° 

AOE + EOD + BOD = 180° 

35° + EOD + 40° = 180° 

EOD + 75° = 180° 

EOD = 105° 

Now, COF = EOD = 105° (vertically opposite angles)

Question 14

In the given figure, the two lines AB and CD intersect at a point O such that BOC = 125°. Find the values of x, y and z.

 

  

Solution 14

AOC + BOC = 180° (linear pair of angles)

x + 125 = 180° 

x = 55° 

Now, AOD = BOC  (vertically opposite angles)

y = 125° 

Also, BOD = AOC (vertically opposite angles)

z = 55° 

Question 15

If two straight lines intersect each other then prove that the ray opposite to the bisector of one of the angles so formed bisects the vertically opposite angle.

Solution 15

Given : AB and CD are two lines which are intersecting at O. OE is a ray bisecting the BOD. OF is a ray opposite to ray OE.

To Prove: AOF = COF

Proof : Since are two opposite rays, is a straight line passing through O.

AOF = BOE

and COF = DOE

[Vertically opposite angles]

But BOE = DOE (Given)

AOF = COF

Hence, proved.

Question 16

Prove that the bisectors of two adjacent supplementary angles include a right angle.

Solution 16

Given: is the bisector of BCD and is the bisector of ACD.

To Prove: ECF = 90o

Proof: Since ACD and BCD forms a linear pair.

ACD + BCD = 180o

ACE + ECD + DCF + FCB = 180o

ECD + ECD + DCF + DCF = 180o

because ACE = ECD

and DCF = FCB

2(ECD) + 2 (CDF) = 180o

2(ECD + DCF) = 180o

ECD + DCF =

ECF = 90o (Proved)

Chapter 7 - Lines And Angles Excercise Ex. 7C

Question 1

In the given figure, l m and a transversal t cuts them. If 1 = 120°, find the measure of each of the remaining marked angles.

 

  

Solution 1

Given, 1 = 120° 

Now, 1 + 2 = 180° (linear pair)

120° + 2 = 180° 

2 = 60° 

1 = 3  (vertically opposite angles)

3 = 120° 

Also, 2 = 4  (vertically opposite angles)

4 = 60° 

Line l line m and line t is a transversal.

5 = 1 = 120° (corresponding angles)

 6 = 2 = 60° (corresponding angles)

 7 = 3 = 120° (corresponding angles)

 8 = 4 = 60° (corresponding angles)

Question 2

In the given figure, l m and a transversal t cuts them. If 7 = 80°, find the measure of each of the remaining marked angles.

  

Solution 2

Given, 7 = 80° 

Now, 7 + 8 = 180° (linear pair)

80° + 8 = 180° 

8 = 100° 

7 = 5 (vertically opposite angles)

5 = 80° 

Also, 6 = 8 (vertically opposite angles)

6 = 100° 

Line l line m and line t is a transversal.

1 = 5 = 80°  (corresponding angles)

 2 = 6 = 100° (corresponding angles)

 3 = 7 = 80°  (corresponding angles)

 4 = 8 = 100° (corresponding angles) 

Question 3

In the given figure, l m and a transversal t cuts them. If 1 : 2 = 2 : 3, find the measure of each of the marked angles.

  

Solution 3

Given, 1 : 2 = 2 : 3

Now, 1 + 2 = 180° (linear pair)

2x + 3x = 180° 

5x = 180° 

x = 36° 

1 = 2x = 72° and 2 = 3x = 108° 

1 = 3 (vertically opposite angles)

3 = 72° 

Also, 2 = 4  (vertically opposite angles)

4 = 108° 

Line l line m and line t is a transversal.

5 = 1 = 72°  (corresponding angles)

 6 = 2 = 108° (corresponding angles)

 7 = 3 = 72°  (corresponding angles)

 8 = 4 = 108° (corresponding angles) 

Question 4

For what value of x will the lines l and m be parallel to each other?

 

Solution 4

Lines l and m will be parallel if 3x - 20 = 2x + 10

[Since, if corresponding angles are equal, lines are parallel]

3x - 2x = 10 + 20

x = 30

Question 5

For what value of x will the lines l and m be parallel to each other?

 

  

*Question modified, back answer incorrect.

Solution 5

For lines l and m to be parallel to each other, the corresponding angles (3x + 5)° and (4x)° should be equal.

(3x + 5)° = 4x° 

x = 5° 

Question 6

In the given figure, AB || CD and BC || ED. Find the value of x.

Solution 6

Since AB || CD and BC is a transversal.

So, BCD = ABC = xo     [Alternate angles]

As BC || ED and CD is a transversal.

BCD + EDC = 180o    

  BCD + 75o =180o

BCD = 180o - 75o = 105o 

ABC = 105o                 [since BCD = ABC]

xo = ABC = 105o

Hence, x = 105.


Question 7

In the given figure, AB || CD || EF. Find the value of x.

Solution 7

Since AB || CD and BC is a transversal.

So, ABC = BCD                [atternate interior angles]

70o = xo + ECD(i)

Now, CD || EF and CE is transversal.

So,ECD + CEF = 180o    [sum of consecutive interior angles is 180o]

ECD + 130o = 180o

ECD = 180o - 130o = 50o

Putting ECD = 50o in (i) we get,

70o = xo + 50o

x = 70 - 50 = 20

Question 8

In the give figure, AB CD. Find the values of x, y and z.

  

Solution 8

AB CD and EF is transversal.

AEF = EFG (alternate angles)

Given, AEF = 75° 

EFG = y = 75° 

 

Now, EFC + EFG = 180° (linear pair)

x + y = 180° 

x + 75° = 180° 

x = 105° 

 

EGD = EFG + FEG (Exterior angle property)

125° = y + z

125° = 75° + z

z = 50° 

 

Thus, x = 105°, y = 75° and z = 50° 

Question 9

Ineach of the figures given below, AB || CD. Find the value of x in each case.

Solution 9

Through E draw EG || CD. Now since EG||CD and ED is a transversal.

So,GED = EDC = 65o[Alternate interior angles]

Since EG || CD and AB || CD,

EG||AB and EB is transversal.

So,BEG = ABE = 35o[Alternate interior angles]

So,DEB = xo

BEG + GED = 35o + 65o = 100o.

Hence, x = 100.

Question 10

Ineach of the figures given below, AB || CD. Find the value of x in each case.

Solution 10

Through O draw OF||CD.

Now since OF || CD and OD is transversal.

CDO + FOD = 180o

[sum of consecutive interior angles is 180o]

25o + FOD = 180o

FOD = 180o - 25o = 155o

As OF || CD and AB || CD [Given]

Thus, OF || AB and OB is a transversal.

So,ABO + FOB = 180o    [sum of consecutive interior angles is 180o]

55o + FOB = 180o

FOB = 180o - 55o = 125o

Now, xo = FOB + FOD = 125o + 155o = 280o.

Hence, x = 280.

Question 11

Ineach of the figures given below, AB || CD. Find the value of x in each case.

Solution 11

Through E, draw EF || CD.

Now since EF || CD and EC is transversal.

FEC + ECD = 180o

[sum of consecutive interior angles is 180o]

FEC + 124o = 180o

FEC = 180o - 124o = 56o

Since EF || CD and AB ||CD

So, EF || AB and AE is a trasveral.

So,BAE + FEA = 180o

[sum of consecutive interior angles is 180o]

116o + FEA = 180o

FEA = 180o - 116o = 64o

Thus,xo = FEA + FEC

= 64o + 56o = 120o.

Hence, x = 120.

Question 12

In the given figure, AB || CD. Find the value of x.

Solution 12

Through C draw FG || AE

Now, since CG || BE and CE is a transversal.

So, GCE = CEA = 20o            [Alternate angles]

DCG = 130o - GCE

= 130o - 20o = 110o

Also, we have AB || CD and FG is a transversal.

So, BFC = DCG = 110o          [Corresponding angles]

As, FG || AE, AF is a transversal.

BFG = FAE                           [Corresponding angles]

xo = FAE = 110o.

Hence, x = 110

Question 13

In the given figure, AB || PQ. Find the values of x and y.

Solution 13

Since AB || PQ and EF is a transversal.

So, CEB = EFQ                 [Corresponding angles]

EFQ = 75o

EFG + GFQ = 75o

25o + yo = 75o

y = 75 - 25 = 50

Also, BEF + EFQ = 180o   [sum of consecutive interior angles is 180o]      

  BEF = 180o - EFQ

           = 180o - 75o

  BEF = 105o

FEG + GEB = BEF = 105o

FEG = 105o - GEB = 105o - 20o = 85o

In EFG we have,

xo + 25o + FEG = 180o

   

Hence, x = 70.


Question 14

In the given figure, AB || CD. Find the value of x.

Solution 14

Since AB || CD and AC is a transversal.

So, BAC + ACD = 180o   [sum of consecutive interior angles is 180o]

ACD = 180o - BAC

= 180o - 75o = 105o

ECF = ACD                     [Vertically opposite angles]

ECF = 105o

Now in CEF,


ECF + CEF + EFC =180o

105o + xo + 30o = 180o

x = 180 - 30 - 105 = 45

Hence, x = 45.


Question 15

In the given figure, AB || CD. Find the value of x.

Solution 15

Since AB || CD and PQ a transversal.

So, PEF = EGH [Corresponding angles]

EGH = 85o

EGH and QGH form a linear pair.

So, EGH + QGH = 180o

QGH = 180o - 85o = 95o

Similarly, GHQ + 115o = 180o

GHQ = 180o - 115o = 65o

In GHQ, we have,


xo + 65o + 95o = 180o

x = 180 - 65 - 95 = 180 - 160

x = 20


Question 16

In the given figure, AB || CD. Find the values of x, y and z.

Solution 16

Since AB || CD and BC is a transversal.

So, ABC = BCD

x = 35

Also, AB || CD and AD is a transversal.

So, BAD = ADC

z = 75

In ABO, we have,


xo + 75o + yo = 180o

35 + 75 + y = 180

y = 180 - 110 = 70

x = 35, y = 70 and z = 75.


Question 17

In the given figure, AB || CD. Prove that p + q - r = 180.

Solution 17

Through F, draw KH || AB || CD

Now, KF || CD and FG is a transversal.

KFG = FGD = ro (i)

[alternate angles]

Again AE || KF, and EF is a transversal.

So,AEF + KFE = 180o

KFE = 180o - po (ii)

Adding (i) and (ii) we get,

KFG + KFE = 180 - p + r

EFG = 180 - p + r

q = 180 - p + r

i.e.,p + q - r = 180

Question 18

In the given figure, AB || CD and EF || GH. Find the values of x, y, z and t.

Solution 18

PRQ = xo = 60o            [vertically opposite angles]

Since EF || GH, and RQ is a transversal.

So, x = y                   [Alternate angles]

y = 60

AB || CD and PR is a transversal.

So,         [Alternate angles]

     [since ]

x + QRD = 110o

QRD = 110o - 60o = 50o

In QRS, we have,

QRD + to + yo = 180o

50 + t + 60 = 180

t = 180 - 110 = 70

Since, AB || CD and GH is a transversal

So, zo = to = 70o [Alternate angles]

x = 60 , y = 60, z = 70 and t = 70


Question 19

In the given figure, AB CD and a transversal t cuts them at E and F respectively. If EG and FG are the bisectors of BEF and EFD respectively, prove that EGF = 90°.

  

Solution 19

AB CD and a transversal t cuts them at E and F respectively.

BEF + DFE = 180° (interior angles)

GEF + GFE = 90° ….(i)

Now, in ΔGEF, by angle sum property

GEF + GFE + EGF = 180° 

90° + EGF = 180° ….[From (i)]

EGF = 90° 

Question 20

In the given figure, AB CD and a transversal t cuts them at E and F respectively. If EP and FQ are the bisectors of AEF and EFD respectively, prove that EP FQ.

  

Solution 20

Since AB CD and t is a transversal, we have

AEF = EFD (alternate angles)

PEF = EFQ

But, these are alternate interior angles formed when the transversal EF cuts EP and FQ.

EP FQ

Question 21

In the given figure, BA ED and BC EF. Show that ABC = DEF.

 

  

 

Solution 21

Construction: Produce DE to meet BC at Z.

  

Now, AB DZ and BC is the transversal.

ABC = DZC (corresponding angles) ….(i)

Also, EF BC and DZ is the transversal.

DZC = DEF (corresponding angles) ….(ii)

From (i) and (ii), we have

ABC = DEF 

Question 22

In the given figure, BA ED and BC EF. Show that ABC + DEF = 180°.

 

  

 

Solution 22

Construction: Produce ED to meet BC at Z.

  

Now, AB EZ and BC is the transversal.

ABZ + EZB = 180° (interior angles)

ABC + EZB = 180° ….(i)

Also, EF BC and EZ is the transversal.

BZE = ZEF (alternate angles)

BZE = DEF ….(ii)

From (i) and (ii), we have

ABC + DEF = 180° 

Question 23

In the given figure, m and n are two plane mirrors perpendicular to each other. Show that the incident ray CA is parallel to the reflected ray BD.

 

  

Solution 23

Let the normal to mirrors m and n intersect at P.

Now, OB m, OC n and m n.

OB OC

APB = 90° 

2 + 3 = 90° (sum of acute angles of a right triangle is 90°)

By the laws of reflection, we have

1 = 2 and 4 = 3 (angle of incidence = angle of reflection)

1 + 4 = 2 + 3 = 90° 

1 + 2 + 3 + 4 = 180° 

CAB + ABD = 180° 

But, CAB and ABD are consecutive interior angles formed, when the transversal AB cuts CA and BD.

CA BD 

Question 24

In the figure given below, state which lines are parallel and why?

  

Solution 24

In the given figure,

BAC = ACD = 110° 

But, these are alternate angles when transversal AC cuts AB and CD.

Hence, AB CD. 

Question 25

Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.

Solution 25

  

Let the two parallel lines be m and n.

Let p m.

1 = 90° 

Let q n.

2 = 90° 

Now, m n and p is a transversal.

1 = 3 (corresponding angles)

3 = 90° 

3 = 2 (each 90°)

But, these are corresponding angles, when transversal n cuts lines p and q.

p q. 

Hence, two lines which are perpendicular to two parallel lines, are parallel to each other. 

CBSE Class 9 Maths Homework Help

Clear all your doubts instantly at our “Ask Doubt” section. Get expert help and guidance at your comfort. To know the syllabus in detail, click here.

 

NCERT Textbooks are the rich stimulant for those students who want to score better in the CBSE examinations. By solving our papers the students have achieved a better and higher result. One of the primary objectives of creating ncert solution for class 9  is that the student gets access to an easy solution; which acts as a strong catalyst in improving the marks. We usually focus that the students don’t find any difficulty in understanding the concepts and can memorize them easily.