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

## Chapter 7 - Lines And Angles Exercise MCQ

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

Correct option: (d)

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

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)

Correct option: (b)

Let each interior opposite angle be x.

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

⇒ 2x = 110°

⇒ x = 55°

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

- Acute angled
- Obtuse angled
- Right angled
- an isosceles triangle

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

Correct option: (d)

Let ∠A = 130°

In ΔABC, by angle sum property,

∠B + ∠C + ∠A = 180°

⇒ ∠B + ∠C + 130° = 180°

⇒ ∠B + ∠C = 50°

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

(a) 12

(b) 15

(c) 20

(d) 25

Correct option: (b)

AOB is a straight line.

⇒ ∠AOB = 180°

⇒ 60° + 5x° + 3x° = 180°

⇒ 60° + 8x° = 180°

⇒ 8x° = 120°

⇒ x = 15°

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°

Correct option: (c)

By angle sum property,

2x + 3x + 4x = 180°

⇒ 9x = 180°

⇒ x = 20°

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

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

(a) 40°

(b) 50°

(c) 60°

(d) 70°

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°

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

- An acute angle
- An obtuse angle
- A right angle
- A reflex angle

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.

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

- An acute angle
- An obtuse angle
- A straight angle
- A reflex angle

Correct option: (d)

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

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

- 25°
- 35°
- 65°
- 75°

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

- 72°
^{o} - 54°
- 63°
- 36°

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

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

- 96°
- 86°
- 76°
- 106°

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

- 40°
- 60°
- 80°
- 50°

Which of the following statements is false?

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

Correct option: (a)

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

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

- 15°
- 30°
- 75°
- 150°

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

- 60°
- 80°
- 48°
- 72°

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

- 30°
- 40°
- 45°
- 60°

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

- 65°
- 115°
- 110°
- 125°

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 =?

- 72°
- 18°
- 36°
- 54°

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

(a) 70°

(b) 60°

(c) 50°

(d) 40°

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°

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

- 130°
- 150°
- 80°
- 100°

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

- 65°
- 55°
- 45°
- 75°

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

- 108°
- 126°
- 162°
- 63°

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

- 50°
- 60°
- 40°
- 35°

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

- 50°
- 60°
- 70°
- 50°

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

- 20°
- 25°
- 30°
- 35°

In the adjoining figure y =?

- 36°
- 54°
- 63°
- 72°

## Chapter 7 - Lines And Angles Exercise Ex. 7A

Define the following terms:

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

(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}.

Find the complement of each of the following angles.

16^{o}

Complement of 16^{o} = 90 - 16^{o} = 74^{o}

Find the complement of each of the following angles.

46^{o} 30

Complement of 46^{o} 30' = 90^{o} - 46^{o} 30' = 43^{o} 30'

Find the complement of each of the following angle:

55°

Complement of 55° = 90° - 55° = 35°

Find the complement of each of the following angle:

90°

Complement of 90° = 90° - 90° = 0°

Find the supplement of each of the following angles.

75^{o} 36'

Supplement of 75^{o} 36' = 180^{o} - 75^{o} 36' = 104^{o} 24'

Find the supplement of each of the following angle:

42°

Supplement of 42° = 180° - 42° = 138°

Find the supplement of each of the following angle:

90°

Supplement of 90° = 180° - 90° = 90°

Find the supplement of each of the following angle:

124°

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

Find the measure of an angle which is

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

(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}.

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

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}.

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

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°.

Find the angle which is four times its complement.

Let
the required angle be x^{o}

Then,
its complement = 90^{o} - x^{o}

_{}

_{} The required angle is 72^{o}.

Find the angle which is five times its supplement.

Let
the required angle be x^{o}

Then,
its supplement is 180^{o} - x^{o}

_{}

_{} The required angle is 150^{o}.

Find the angle whose supplement is four times its complement.

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}.

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

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}.

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

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}.

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

(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

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

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^{o}

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

∠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°

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

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}

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

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.

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

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.

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

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^{o}

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.

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 = 90

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.

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^{o}

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

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^{o}

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}.

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}.

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.

Since,
_{}AOD and _{}BOC are vertically opposite angles.

_{}AOD
= _{}BOC

Now,
_{}AOD + _{}BOC = 280^{o} [Given]

_{} _{}AOD + _{}AOD = 280^{o}

_{} 2_{}AOD = 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}.

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

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°

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.

∠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)

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.

∠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°

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.

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.

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

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)

## Chapter 7 - Lines And Angles Exercise Ex. 7C

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.

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)

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.

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)

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.

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)

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

_{}

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

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

*Question modified, back answer incorrect.

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°

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

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.

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

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 = 20

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

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°

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

_{}

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.

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

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.

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

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.

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

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 = 110

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

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.

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

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.}

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

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

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

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.

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

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 = 180

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

_{}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

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°.

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°

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.

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

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

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

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

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°

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.

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

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

In the given figure,

∠BAC = ∠ACD = 110°

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

Hence, AB ∥ CD.

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

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.

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