R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 7 - Lines And Angles

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Chapter 7 - Lines And Angles Exercise 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) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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)

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

Let A = 130° 

In ΔABC, by angle sum property,

B + C + A = 180° 

B + C + 130° = 180° 

B + C = 50° 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Question 5

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

 

(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

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

(a) 40° 

(b) 50° 

(c) 60° 

(d) 70° 

Solution 7

Correct option: (c)

Through B draw YBZ OA CD.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Question 12

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 12

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 13

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 17

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 18

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 20

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Question 21

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

 

(a) 70° 

(b) 60° 

(c) 50° 

(d) 40° 

Solution 21

Correct option: (c)

Let AOC = x° 

Draw YOZ CD AB.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

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°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 22

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Question 23

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

  1. 65°
  2. 55°
  3. 45°
  4. 75°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 23

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 24

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Question 25

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

  1. 50°
  2. 60°
  3. 40°
  4. 35°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 25

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Question 26

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

  1. 50°
  2. 60°
  3. 70°
  4. 50°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 26

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Question 27

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

  1. 20°
  2. 25°
  3. 30°
  4. 35°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 27

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Question 28

In the adjoining figure y =?

  1. 36°
  2. 54°
  3. 63°
  4. 72°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 28

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Chapter 7 - Lines And Angles Exercise 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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesAOB 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(ii)

Find the complement of each of the following angles.

16o

Solution 2(ii)

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

Question 2(iv)

Find the complement of each of the following angles.

46o 30

Solution 2(iv)

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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

Question 2(i)

Find the complement of each of the following angle:

55° 

Solution 2(i)

Complement of 55° = 90° - 55° = 35°  

Question 2(iii)

Find the complement of each of the following angle:

90° 

Solution 2(iii)

Complement of 90° = 90° - 90° = 0° 

Question 3(iv)

Find the supplement of each of the following angles.

75o 36'

Solution 3(iv)

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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

Question 3(i)

Find the supplement of each of the following angle:

42° 

Solution 3(i)

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

Question 3(ii)

Find the supplement of each of the following angle:

90° 

Solution 3(ii)

Supplement of 90° = 180° - 90° = 90° 

Question 3(iii)

Find the supplement of each of the following angle:

124° 

Solution 3(iii)

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

Question 4

Find the measure of an angle which is

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

Solution 4

(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 5

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

Solution 5

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 6

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

Solution 6

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 7

Find the angle which is four times its complement.

Solution 7

Let the required angle be xo

Then, its complement = 90o - xo

The required angle is 72o.

Question 8

Find the angle which is five times its supplement.

Solution 8

Let the required angle be xo

Then, its supplement is 180o - xo

The required angle is 150o.

Question 9

Find the angle whose supplement is four times its complement.

Solution 9

Let the required angle be xo

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

That is we have,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles The required angle is 60o.


Question 10

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

Solution 10

Let the required angle be xo

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

The required angle is 45o.

Question 11

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

Solution 11

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

Then R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 5x = 4(90 - x)

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 5x = 360 - 4x

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 5x + 4x = 360

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 9x = 360

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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


Question 12

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

Solution 12

(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 Exercise Ex. 7B

Question 1

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 1

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Since R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBOC and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCOA form a linear pair of angles, we have

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBOC + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCOA = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles xo + 62o = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles x = 180 - 62

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles x = 118o

Question 2

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

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesAOC, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCOD and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBOD.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 3

Since R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBOD and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesDOA from a linear pair of angles.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBOD + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesDOA = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBOD + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesDOC + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCOA = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles xo + (2x - 19)o + (3x + 7)o = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 6x - 12 = 180

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 6x = 180 + 12 = 192

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles x = 32

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesAOC = (3x + 7)o = (3 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 32 + 7)o = 103o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCOD = (2x - 19)o = (2 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 32 - 19)o = 45o

and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBOD = 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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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  R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

And R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Anglesz = 180o - R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Anglesx - R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Anglesy

= 180o - 60o - 48o

= 180o - 108o = 72o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 7

Since R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCOE and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesDOF are vertically opposite angles, we have,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCOE = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesDOF

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Anglesz = 50o

Also R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBOD and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCOA are vertically opposite angles.

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBOD = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCOA

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Anglest = 90o

As R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCOA and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesAOD form a linear pair,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCOA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesAOD = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCOA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesAOF + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFOD = 180o [R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Anglest = 90o]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles t + x + 50o = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 90o + xo + 50o = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles x + 140 = 180

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles x = 180 - 140 = 40

Since R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEOB and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesAOF are vertically opposite angles

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEOB = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesAOF

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And 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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

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.

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

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 Exercise 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.

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And 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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And 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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And 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?

 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 4

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

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles3x - 2x = 10 + 20

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Anglesx = 30

Question 5

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

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

*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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 6

Since AB || CD and BC is a transversal.

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBCD = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesABC = xo     [Alternate angles]

As BC || ED and CD is a transversal.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBCD + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEDC = 180o    

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles  R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBCD + 75o =180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBCD = 180o - 75o = 105o 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesABC = 105o                 [since R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBCD = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesABC]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles xo = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesABC = 105o

Hence, x = 105.


Question 7

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 7

Since AB || CD and BC is a transversal.

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesABC = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBCD                [atternate interior angles]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles70o = xo + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesECD(i)

Now, CD || EF and CE is transversal.

So,R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesECD + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCEF = 180o    [sum of consecutive interior angles is 180o]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesECD + 130o = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesECD = 180o - 130o = 50o

Putting R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesECD = 50o in (i) we get,

70o = xo + 50o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Anglesx = 70 - 50 = 20

Question 8

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

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(i)

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 9(i)

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

So,R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesGED = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEDC = 65o[Alternate interior angles]

Since EG || CD and AB || CD,

EG||AB and EB is transversal.

So,R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBEG = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesABE = 35o[Alternate interior angles]

So,R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesDEB = xo

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBEG + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesGED = 35o + 65o = 100o.

Hence, x = 100.

Question 9(ii)

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 9(ii)

Through O draw OF||CD.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Now since OF || CD and OD is transversal.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCDO + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFOD = 180o

[sum of consecutive interior angles is 180o]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles25o + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFOD = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFOD = 180o - 25o = 155o

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

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

So,R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesABO + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFOB = 180o    [sum of consecutive interior angles is 180o]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles55o + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFOB = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFOB = 180o - 55o = 125o

Now, xo = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFOB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFOD = 125o + 155o = 280o.

Hence, x = 280.

Question 9(iii)

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 9(iii)

Through E, draw EF || CD.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Now since EF || CD and EC is transversal.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFEC + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesECD = 180o

[sum of consecutive interior angles is 180o]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFEC + 124o = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFEC = 180o - 124o = 56o

Since EF || CD and AB ||CD

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

So,R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBAE + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFEA = 180o

[sum of consecutive interior angles is 180o]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles116o + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFEA = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFEA = 180o - 116o = 64o

Thus,xo = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFEA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFEC

= 64o + 56o = 120o.

Hence, x = 120.

Question 10

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 10

Through C draw FG || AE

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesGCE = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCEA = 20o            [Alternate angles]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesDCG = 130o - R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesGCE

= 130o - 20o = 110o

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

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBFC = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesDCG = 110o          [Corresponding angles]

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBFG = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFAE                           [Corresponding angles]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles xo = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFAE = 110o.

Hence, x = 110

Question 11

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 11

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Since AB || PQ and EF is a transversal.

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCEB = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEFQ                 [Corresponding angles]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEFQ = 75o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEFG + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesGFQ = 75o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 25o + yo = 75o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles y = 75 - 25 = 50

Also, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBEF + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEFQ = 180o   [sum of consecutive interior angles is 180o]      

 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles BEF = 180o - R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEFQ

           = 180o - 75o

  R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBEF = 105o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFEG + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesGEB = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBEF = 105o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFEG = 105o - R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesGEB = 105o - 20o = 85o

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEFG we have,

xo + 25o + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFEG = 180o

    R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Hence, x = 70.


Question 12

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 12

Since AB || CD and AC is a transversal.

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBAC + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesACD = 180o   [sum of consecutive interior angles is 180o]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesACD = 180o - R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBAC

= 180o - 75o = 105o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesECF = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesACD                     [Vertically opposite angles]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesECF = 105o

Now in R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCEF,


R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesECF + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesCEF + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEFC =180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 105o + xo + 30o = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles x = 180 - 30 - 105 = 45

Hence, x = 45.


Question 13

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 13

Since AB || CD and PQ a transversal.

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesPEF = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEGH [Corresponding angles]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEGH = 85o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEGH and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesQGH form a linear pair.

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEGH + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesQGH = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesQGH = 180o - 85o = 95o

Similarly, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesGHQ + 115o = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesGHQ = 180o - 115o = 65o

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesGHQ, we have,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

xo + 65o + 95o = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles x = 180 - 65 - 95 = 180 - 160

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles x = 20


Question 14

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 14

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Since AB || CD and BC is a transversal.

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesABC = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBCD

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles x = 35

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

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesBAD = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesADC

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles z = 75

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesABO, we have,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles xo + 75o + yo = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 35 + 75 + y = 180

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles y = 180 - 110 = 70

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles x = 35, y = 70 and z = 75.


Question 16

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 16

Through F, draw KH || AB || CD

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesKFG = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesFGD = ro (i)

[alternate angles]

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

So,R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesAEF + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesKFE = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesKFE = 180o - po (ii)

Adding (i) and (ii) we get,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesKFG + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesKFE = 180 - p + r

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesEFG = 180 - p + r

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Anglesq = 180 - p + r

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

Question 17

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

Solution 17

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesPRQ = xo = 60o            [vertically opposite angles]

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

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Anglesx = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Anglesy                   [Alternate angles]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles y = 60

AB || CD and PR is a transversal.

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles         [Alternate angles]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles     [since R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Anglesx + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesQRD = 110o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesQRD = 110o - 60o = 50o

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesQRS, we have,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And AnglesQRD + to + yo = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 50 + t + 60 = 180

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles t = 180 - 110 = 70

Since, AB || CD and GH is a transversal

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles x = 60 , y = 60, z = 70 and t = 70


Question 18

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°.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

Solution 18

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

BEF + DFE = 180° (interior angles)

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

GEF + GFE = 90° ….(i)

Now, in ΔGEF, by angle sum property

GEF + GFE + EGF = 180° 

90° + EGF = 180° ….[From (i)]

EGF = 90° 

Question 19

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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

Solution 19

Since AB CD and t is a transversal, we have

AEF = EFD (alternate angles)

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles

PEF = EFQ

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

EP FQ

Question 20

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

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

 

Solution 20

Construction: Produce DE to meet BC at Z.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

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 21

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

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

 

Solution 21

Construction: Produce ED to meet BC at Z.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

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 22

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.

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

Solution 22

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 23

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

Solution 23

In the given figure,

BAC = ACD = 110° 

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

Hence, AB CD. 

Question 24

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

Solution 24

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Lines And Angles 

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.