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 triangleSolution 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) 155Solution 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) 25Solution 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 180^o but less than 360^o 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 = 80^o 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

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 anglesSolution 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 90^o but less than 180^o, is called an obtuse angle.

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

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

(vi) Supplementary angles: Two angles are said to be supplementary, if the sum of their measures is 180^o.
Question 2(ii)

Find the complement of each of the following angles.

16^oSolution 2(ii)

Complement of 16^o = 90 – 16^o = 74^oQuestion 2(iv)

Find the complement of each of the following angles.

46^o 30Solution 2(iv)

Complement of 46^o 30′ = 90^o – 46^o 30′ = 43^o 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.

75^o 36’Solution 3(iv)

Supplement of 75^o 36′ = 180^o – 75^o 36′ = 104^o 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 x^o

Then, its complement = 90^o – x^o

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

(ii) Let the required angle be x^o

Then, its supplement = 180^o – x^o

 The measure of an angle which is equal to its supplement is 90^o.Question 5

Find the measure of an angle which is 36^o more than its complement.Solution 5

Let the required angle be x^o

Then its complement is 90^o – x^o

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

Then, its complement = 90^o – x^o

 The required angle is 72^o.Question 8

Find the angle which is five times its supplement.Solution 8

Let the required angle be x^o

Then, its supplement is 180^o – x^o

 The required angle is 150^o.Question 9

Find the angle whose supplement is four times its complement.Solution 9

Let the required angle be x^o

Then, its complement is 90^o – x^o and its supplement is 180^o – x^o_{That is we have,}

 The required angle is 60^o.
Question 10

Find the angle whose complement is one-third of its supplement.Solution 10

Let the required angle be x^o

Then, its complement is 90^o – x^o and its supplement is 180^o – x^o

 The required angle is 45^o.Question 11

Two complementary angles are in the ratio 4: 5. Find the angles.Solution 11

Let the two required angles be x^o and 90^o – x^o.

Then 

 5x = 4(90 – x)

 5x = 360 – 4x

 5x + 4x = 360

 9x = 360

Thus, the required angles are 40^o and 90^o – x^o = 90 ^o – 40^o = 50^o.
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° 

Exercise 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 = 180^o

 x^o + 62^o = 180^o

 x = 180 – 62

 x = 118^oQuestion 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 = 180^o

 BOD + DOC + COA = 180^o

 x^o + (2x – 19)^o + (3x + 7)^o = 180^o

 6x – 12 = 180

 6x = 180 + 12 = 192

 x = 32

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

 COD = (2x – 19)^o = (2  32 – 19)^o = 45^o

and BOD = x^o = 32^o
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 = 180^o

[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 180^o, 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 180^o, then, measure of 

And z = 180^o – x – y

= 180^o – 60^o – 48^o

= 180^o – 108^o = 72^o

 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 = 180^o

 (4x – 36)^o + (3x + 20)^o = 180^o

 4x – 36 + 3x + 20 = 180

 7x – 16 = 180^o

 7x = 180 + 16 = 196

 The value of x = 28.Question 6

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

Solution 6

Since AOC and AOD form a linear pair.

AOC + AOD = 180^o

 50^o + AOD = 180^o

 AOD = 180^o – 50^o = 130^o

AOD and BOC are vertically opposite angles.

 AOD = BOC

 BOC = 130^o

BOD and AOC are vertically opposite angles.

BOD = AOC

 BOD = 50^oQuestion 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 = 50^o

Also BOD and COA are vertically opposite angles.

So, BOD = COA

 t = 90^o

As COA and AOD form a linear pair,

COA + AOD = 180^o

 COA + AOF + FOD = 180^o [t = 90^o]

 t + x + 50^o = 180^o

 90^o + x^o + 50^o = 180^o

 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 = 90Question 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 = 180^o

 COE + EOA + AOD = 180^o

 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 = 2x^o = 2  18^o = 36^o

COE = 5x^o = 5  18^o = 90^o

and, EOA = BOF = 3x^o = 3  18^o = 54^oQuestion 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 = 180^o

 9x = 180^o

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

and 4x = 4  20^o = 80^oQuestion 10

If two straight lines intersect each other in such a way that one of the angles formed measures 90^o, show that each of the remaining angles measures 90^o.Solution 10

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

Now, AOC = BOD [Vertically opposite angles]

 BOD = 90^o

Also, as AOC and AOD form a linear pair.

 90^o + AOD = 180^o

 AOD = 180^o – 90^o = 90^o

Since, BOC = AOD [Verticallty opposite angles]

 BOC = 90^o

Thus, each of the remaining angles is 90^o.Question 11

Two lines AB and CD intersect at a point O such that BOC +AOD = 280^o, 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 = 280^o [Given]

 AOD + AOD = 280^o

 2AOD = 280^o

 AOD = 

 BOC = AOD = 140^o

As, AOC and AOD form a linear pair.

So, AOC + AOD = 180^o

 AOC + 140^o = 180^o

 AOC = 180^o – 140^o = 40^o

Since, AOC and BOD are vertically opposite angles.

AOC = BOD

 BOD = 40^o

 BOC = 140^o, AOC = 40^o , AOD = 140^o and BOD = 40^o.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 = 90^o

Proof: Since ACD and BCD forms a linear pair.

ACD + BCD = 180^o

ACE + ECD + DCF + FCB = 180^o

 ECD + ECD + DCF + DCF = 180^o

because ACE = ECD

and DCF = FCB

 2(ECD) + 2 (CDF) = 180^o

 2(ECD + DCF) = 180^o

 ECD + DCF = 

 ECF = 90^o (Proved)

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.

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 = 30Question 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 = x^o     [Alternate angles]

As BC || ED and CD is a transversal.

BCD + EDC = 180^o    

  BCD + 75^o =180^o

 BCD = 180^o – 75^o = 105^o 

 ABC = 105^o                 [since BCD = ABC]

 x^o = ABC = 105^o

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]

70^o = x^o + ECD(i)

Now, CD || EF and CE is transversal.

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

ECD + 130^o = 180^o

ECD = 180^o – 130^o = 50^o

Putting ECD = 50^o in (i) we get,

70^o = x^o + 50^o

x = 70 – 50 = 20Question 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(i)

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

Solution 9(i)

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

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

Since EG || CD and AB || CD,

EG||AB and EB is transversal.

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

So,DEB = x^o

BEG + GED = 35^o + 65^o = 100^o.

Hence, x = 100.Question 9(ii)

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

Solution 9(ii)

Through O draw OF||CD.

Now since OF || CD and OD is transversal.

CDO + FOD = 180^o

[sum of consecutive interior angles is 180^o]

25^o + FOD = 180^o

FOD = 180^o – 25^o = 155^o

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

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

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

55^o + FOB = 180^o

FOB = 180^o – 55^o = 125^o

Now, x^o = FOB + FOD = 125^o + 155^o = 280^o.

Hence, x = 280.Question 9(iii)

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

Solution 9(iii)

Through E, draw EF || CD.

Now since EF || CD and EC is transversal.

FEC + ECD = 180^o

[sum of consecutive interior angles is 180^o]

FEC + 124^o = 180^o

FEC = 180^o – 124^o = 56^o

Since EF || CD and AB ||CD

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

So,BAE + FEA = 180^o

[sum of consecutive interior angles is 180^o]

116^o + FEA = 180^o

FEA = 180^o – 116^o = 64^o

Thus,x^o = FEA + FEC

= 64^o + 56^o = 120^o.

Hence, x = 120.Question 10

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

Solution 10

Through C draw FG || AE

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

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

 DCG = 130^o – GCE

= 130^o – 20^o = 110^o

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

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

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

BFG = FAE                           [Corresponding angles]

 x^o = FAE = 110^o.

Hence, x = 110Question 11

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

Solution 11

Since AB || PQ and EF is a transversal.

So, CEB = EFQ                 [Corresponding angles]

 EFQ = 75^o

 EFG + GFQ = 75^o

 25^o + y^o = 75^o

 y = 75 – 25 = 50

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

           = 180^o – 75^o

  BEF = 105^o

 FEG + GEB = BEF = 105^o

 FEG = 105^o – GEB = 105^o – 20^o = 85^o

In EFG we have,

x^o + 25^o + FEG = 180^o

Hence, x = 70.
Question 12

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

Solution 12

Since AB || CD and AC is a transversal.

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

 ACD = 180^o – BAC

= 180^o – 75^o = 105^o

ECF = ACD                     [Vertically opposite angles]

 ECF = 105^o

Now in CEF,

ECF + CEF + EFC =180^o 105^o + x^o + 30^o = 180^o

 x = 180 – 30 – 105 = 45

_{Hence, x = 45.}
Question 13

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

Solution 13

Since AB || CD and PQ a transversal.

So, PEF = EGH [Corresponding angles]

 EGH = 85^o

EGH and QGH form a linear pair.

So, EGH + QGH = 180^o

 QGH = 180^o – 85^o = 95^o

Similarly, GHQ + 115^o = 180^o

 GHQ = 180^o – 115^o = 65^o

In GHQ, we have,

x^o + 65^o + 95^o = 180^o

 x = 180 – 65 – 95 = 180 – 160

 x = 20
Question 14

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

Solution 14

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,

 x^o + 75^o + y^o = 180^o

 35 + 75 + y = 180

 y = 180 – 110 = 70

 x = 35, y = 70 and z = 75.
Question 16

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

Solution 16

Through F, draw KH || AB || CD

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

KFG = FGD = r^o (i)

[alternate angles]

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

So,AEF + KFE = 180^o

KFE = 180^o – p^o (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 = 180Question 17

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

Solution 17

PRQ = x^o = 60^o            [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 = 110^o

 QRD = 110^o – 60^o = 50^o

In QRS, we have,

QRD + t^o + y^o = 180^o

 50 + t + 60 = 180

 t = 180 – 110 = 70

Since, AB || CD and GH is a transversal

So, z^o = t^o = 70^o [Alternate 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°.

Solution 18

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 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.

Solution 19

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 ∥ FQQuestion 20

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

Solution 20

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 21

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

Solution 21

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 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.

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?

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

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.