These Class 7 Maths Extra Questions Chapter 5 Lines and Angles will help students prepare well for the exams.

## Class 7 Maths Chapter 5 Extra Questions Lines and Angles

### Class 7 Maths Lines and Angles Extra Questions – 2 Marks

Question 1.

Two angles are equal and complement to each other. Find the equal angle.

Solution:

Let the equal angle be x°

If two angles are complementary, then their sum is 90°

x° + x° = 90°

2x° = 90°

x° = \(\frac{90^{\circ}}{2}\) ⇒ x° = 45°

∴ Measure of epual angle = 45°

Question 2.

Two angles are equal and supplement to each other. Find the equal angle.

Solution:

Let the equal angle = x°

If two angles are supplementary, then their sum is 180°

x° + x° = 180°

2x° = 180°

x° = \(\frac{180^{\circ}}{2}\) ⇒ x° = 90°

∴ Measure of equal angle = 90°

Question 3.

Two supplementary angles are differ by 38°. Find the angles.

Solution:

Let the first angle = x°

Second angle = (x° + 38°)

If two angles are supplementary, then the sum of the angles is 180°

x° + x° + 38° = 180°

2x° = 180° – 38°

2x° = 142°

x = \(\frac{142^{\circ}}{2}\) ⇒ x = 71°

Other angle = 180° – 71° = 109°

∴ The measure of two angles are 71° and 109°

Question 4.

In the figure find a.

Solution:

In the figure,

∠AOB, ∠BOC, form a linear pair

∴ ∠AOB + ∠BOC = 180°

7a° + 2a° = 180°

9a° = 180° ⇒ a = \(\frac{180^{\circ}}{9}\)

a = 20°

Question 5.

In the figure, find ∠x and ∠y.

Solution:

In the figure ∠x° = 60° (Vertically opposite angles)

y° + 60° = 180° (Linear pair)

y = 180° – 60°

y = 120°

Question 6.

Write down each pair of adjacent angles in the figure.

Solution:

Adjacent angles :

∠BOE, ∠EOD;

∠EOD, ∠DOC; ∠DOC, ∠COA

Question 7.

Find the complement of

i) 73°,

ii) 21°

Solution:

i) Complement of 73° = 90° – 73° = 17°

ii) Complement of 21° = 90° – 21° = 69°

Question 8.

Find the supplementary angle of

i) 70°

ii) 110°

Solution:

i) Supplementary angle of 70° = 180° – 70° = 110°

ii) Supplementary angle of 110° = 180° – 110° = 70°

Question 9.

Name all the pairs of adjacent angles.

Solution:

Adjacent angles :

∠SRP, ∠PRQ; ∠TPQ, ∠RPQ;

∠RQP, ∠RQU

Question 10.

In the below figure what can you say about

i)∠a + ∠b

ii)∠a + ∠b + ∠c + ∠d

Solution:

In the figure

i) ∠a + ∠b = 180° (Linear pair)

ii) ∠c + ∠d = 180° (Linear pair)

∠a + ∠b + ∠c + ∠d = 180° + 180° = 360°

Question 11.

In the below figure a : b = 2 : 3, then find a and b.

Solution:

∠a and ∠b form a linear pair

∠a +∠b = 180°

Let ∠a = 2x°

∠b = 3x°

2x° + 3x° = 180°

5x° = 180°

x° = \(\frac{180^{\circ}}{5}\) ⇒ x° = 36°

∠a = 2 × 36° = 72°

∠b = 3 × 36° = 108°

Question 12.

One of the angles of a linear pair is obtuse. What will be the other angle ?

Solution:

Let the obtuse angle be 100°

Other angle = x°

x° + 100° = 180° (Linear pair)

x° = 180° – 100°

x° = 80°

∴ If one of the angles of a linear pair is obtuse then other angle is acute.

Question 13.

Find x is the following figure.

Solution:

In the figure

3x° = 120° (Vertically opposite angles)

x° = \(\frac{120^{\circ}}{3}\)

⇒ x° = 40°

Question 14.

In the figure find ‘a’.

Solution:

In the figure

80° + 2a = 180°

2a° = 180° – 80°

2a = 100° ⇒ a = \(\frac{100^{\circ}}{2}\)

∴ a = 50°

Question 15.

Draw i) a pair of intersecting lines

ii) a pair of parallel lines

Solution:

Question 16.

In the figure l || m, find the x and y if n is the transversal.

Solution:

l || m, n is the transversal

x° = 80° (Corresponding angles)

x° + y° = 180° (Linear pair)

80° + y = 180°

y = 180° – 80°

∴ y = 100°

Question 17.

If a transversal intersects two coplanar lines then,

i) how many angles are formed?

ii) how many pairs of corresponding angles are formed?

Solution:

i) 8 angles are formed.

ii) 4 pairs of corresponding angles are formed

they are ∠a, ∠e; ∠b, ∠f;

∠d, ∠h; ∠c, ∠g

Question 18.

In the below figure, find

i) ∠1 + ∠4

ii) ∠2 + ∠3

Solution:

i) ∠1 + ∠4 = 180° (Linear Pair)

ii) ∠2 + ∠3 = 180° (Linear Pair)

Question 19.

In the figure s || t, l is the transversal find x.

Solution:

Let us consider the given figure is the following way

a + 70° = 180° (Linear Pair)

a = 180° – 70°

a = 110°

a = x = 110° (Corresponding angles)

x = 110°

Question 20.

In the figure PQR is a straight line. Find x°.

Solution:

From the figure

2x + 15° + 3x + 35° = 180° (Linear Pair)

5x + 50° = 180°

5x° = 180° – 50°

5x = 130° ⇒ x = \(\frac{130^{\circ}}{5}\)

∴ x = 26°

### Lines and Angles Extra Questions Class 7 – 3 Marks

Question 1.

If the sum of an angle and one third of its supplementary angle is 90°. Find the measure of the angle.

Solution:

Let the angle required = x°

Supplementary angle of x° = 180° – x°

One third of 180° – x°

= \(\frac{180^{\circ}-\mathrm{x}^{\circ}}{3}\) = 60 – \(\frac{x^{\circ}}{3}\)

According to the problem

x° + 60 – \(\frac{x^{\circ}}{3}\) = 90°

x° – \(\frac{x^{\circ}}{3}\) = 90° – 60°

\(\frac{3 x^{\circ}-x^{\circ}}{3}\) = 30

2x = 3 × 30

x = \(\frac{3 \times 30}{2}\)

x = 3 × 15

x° = 45°

∴ The measure of the angle is 45°.

Question 2.

In the figure MNR is a straight line. Find x°.

Solution:

From the figure

∠BNA = 90°

MNR is a straight line

2x° + 90° + 3x° = 180°

5x° + 90° = 180°

5x° = 180° – 90°

5x = 90° ⇒ x = \(\frac{90^{\circ}}{5}\)

∴ x° = 18°

Question 3.

In the figure p || q and r is the transversal, then find ‘x’.

Solution:

In the figure p || q

r is the transversal

∴ 3x + 34° = 5x – 14° (Alterante interior angles)

34° + 14° = 5x° – 3x°

48 = 2x° ⇒ x° = \(\frac{48^{\circ}}{2}\) ⇒ x° = 24°

Question 4.

In the figure l || m, r is the transversal, then find x, y.

Solution:

l || m, r is the transversal

70° + x° = 180°

(Interior angles on the same side of the transversal are supplemetary)

∴ x = 180° – 70°

x = 110°

x = y = 110° (Vertically opposite angles)

y = 110°

Question 5.

In the figure, find a & b.

Solution:

In the figure

7a – 2a + 3a = 180° (Linear Pair)

10a – 2a = 180° ⇒ 8a = 180°

a = \(\frac{180^{\circ}}{8}\) = \(\frac{90^{\circ}}{4}\) = \(\frac{45^{\circ}}{2}\) = 22\(\frac{1}{2}^{\circ}\)

a = \(22 \frac{1}{2}^{\circ}\)

3a + b = 180°

\(3\left(22 \frac{1}{2}\right)\) + b = 180°

67\(\frac { 1 }{ 2 }\) + b = 180°

b = 180 – 67\(\frac { 1 }{ 2 }\)

∴ b = 112\(\frac{1}{2} \circ\)

Question 6.

In the figure PQR is a triangle, l || QR, then find a and b.

Solution:

The figure can be taken in the following way.

∠QPS = 50° (l || OR)

(Alternate interior angles)

∴ ∠QPS + ∠a + 60° = 180°

50° + ∠a + 60° = 180°

∠a = 180° – 110°

∠a = 70°

∠b = 60° (Alternate interior angles)

Question 7.

In the figure ∠a = 60°, find other angles.

Solution:

In the figure ∠a = 60°

∠a = ∠c = 60° (Vertically opposite angles)

∠b + ∠a = 180° (Linear Pair)

∠b + 60° = 180°

∠b = 180° – 60°

∠b = 120°

∠b = ∠d = 120° (Vertically opposite angles)

∠d = 120°

Question 8.

Find x in the figure.

Solution:

From the figure

x + 25° + x + x + 35° + 180° (Sum of all linear angles)

3x° + 60° = 180°

3x = 180° – 60°

3x = 120°

⇒ x = \(\frac{120^{\circ}}{3} \)

x = 40°

Question 9.

In the figure is m || n, state reason.

Solution:

In the figure, m and n are two lines

60° = 60°

Corresponding angles are equal

∴ The lines are parallel

∴ m || n

Question 10.

In the figure PQ || RS, ∠PQR = 60°, ∠QPR = 50°, find x, y and z.

Solution:

Given PQ || RS

∠QPR = y = 50° (Alternate interior angles)

y = 50°

In △PQR, 50° + 60° + x = 180° (Sum of the angles in a triangle)

110° + x = 180°

x = 180° – 110°

x° = 70°

x° + y° + z° = 180° (Sum of linear angles)

70° + 50° + z = 180°

120° + z = 180°

z = 180° – 120°

z° = 60°

Question 11.

In the figure find ‘a’ and ‘y’. OC ⊥ AB.

Solution:

a° + 40° = 90°

a° = 90° – 40°

∴ a = 50°

y +90° = 180°

y = 180° – 90°

∴ y = 90°

Question 12.

In the figure MN || JK, find K.

Solution:

MN || JK, t is the transversal

Sum of the interior angles on the same side of the transversal is 180°

7K – 8 + 2K + 17 = 180°

9K + 9 = 180°

9K = 180° – 9°

9K = 171 ⇒ K = \(\frac{171}{9}\)

K = 19°

Question 13.

Find ‘y’ in the figure.

Solution:

Sum of the angles at a point is 360°

8y – 41° + 3y° + 4y – 5 + 3y + 10° = 360°

18y – 36 = 360

18y = 360 + 36

18y = 396

y = \(\frac{396}{18}\)

∴ y = 22°

Question 14.

Write any three conditions to say. “If a transversal intersects two coplanar lines, then they are parallel.”

Solution:

If a transversal intersects two coplanar lines, then to say the lines are parallel if

i) Corresponding angles are equal.

(or)

ii) Sum of interior angles on the same side of the transversal is supplementary.

(or)

iii) Alternate interior angles are equal.

Question 15.

In the figure l || m, p is the transversal. Find ∠x.

Solution:

Method-1 :

Let us draw the given figure in the following way

∠a + 70° = 180° (Linear Pair)

∠a = 180° – 70°

∴ ∠a = 110°

∠a =∠x = 110° (Corresponding angles)

∠x = 110°

Method-2 :

If l || m, p is the transversal, then ‘Sum of exterior angles on the same side of the transversal is 180°.

∴ 70° + ∠x° = 180°

∠x = 180° – 70°

x = 110°

### Extra Questions of Lines and Angles Class 7 – 5 Marks

Question 1.

In the figure p || q, r is the transversal if ∠a = 60°, find all other angles.

Solution:

Given p || q, r is the transversal

∠a = ∠e = 60° (Corresponding angles)

∴ ∠e = 60°

∠b + 60° = 180°

∠b = 180° – 60°

∴ ∠b = 120°

∴ ∠b = ∠d = 120° (Vertically opposite angles)

also, ∠a = ∠c = 60° (Vertically opposite angles)

∴ ∠c = 60°

∠b = ∠f = 120° (Corresponding angles)

∴ ∠f = 120°

∠e = ∠g = 60° (Vertically opposite angles)

∴ ∠g = 60°

∠f = ∠h = 120° (Vertically opposite angles)

∴ ∠h = 120°

Question 2.

In the figure find p and q.

Solution:

Given l || ON

If MN is the transversal

p = 50° (Alternate interior angles)

also, OM is the transversal

q = 60° (Alternate interior angles)

Question 3.

In the figure AB || CD find a, b and c.

Solution:

In the figure

125° + a° = 180° (Linear Pair)

∠a° = 180° – 125°

∠a° = 55°

AB || CD and AD is the transversal

∠c + ∠a = 180°

∠c + 55° = 180°

∠c = 180° – 55°

∠c = 125°

∠a + ∠b = 180°

(AB || CD, BC is the transversal)

55° + ∠b = 180°

∠b = 180° – 55°

∠b = 125°

Question 4.

In the figure find ∠m, AB || PQ, PQ || EF.

Solution:

Let us draw the figure in the following way

Let 120° + n = 180° (Sum of interior angles on the same side of the transversal is 180°)

n = 180° – 120°

n = 60°

AB || PQ, AP is the transversal

∴ 80 = m + n (Alternate interior angles are equal)

80° = m + 60°

m = 80° – 60°

m = 20°

Question 5.

In the figure PQRS is a parallelogram, find the values of a, b and c.

Solution:

60° + c = 180° (Interior angles on the same side of transversal are supplementary)

c = 180° – 60°

c = 120°

60 + a° = 180° (Interior angles on the same side of transversal are supplementary)

a° = 180° – 60°

a° = 120°

a + b = 180° (Interior angles on the same side of transversal are supplementary)

120° + b = 180°

b = 180° – 120°

b = 60°

Question 6.

In the figure find a, b, c and d. if p || q and l and m are transversals.

Solution:

In the figure

∠a = 70° (Vertically opposite angles)

∠a = ∠b = 70°

∠b = 70° (Corresponding angles)

∠b = ∠c = 70° (Corresponding angles)

∴ ∠c = 70°

∠a = ∠d = 70°

∠d = 70° (Corresponding angles)

Question 7.

In the figure AB || CD. Find∠b.

Solution:

Let us draw the figure in the following way

Here l || AB and l || CD.

EF and GF are the transversals

∴ x = 60° (Alternate interior angles)

y = 70° (Alternater interior angles)

b = 60° + 70°

b = 130°

Question 8.

In the figure find y, if AB || CD.

Solution:

Let us draw the figure in the following way.

Let l || CD and l || AB

Let y =∠(1) + ∠(2)

∠1 = 70° (Alternate interior angles)

∠2 = 40° (Alternate interior angles)

∴ y = 70° + 40°

y = 110°

Question 9.

Find y in the following figure

Solution:

y + 90 + 50 + 130 = 360

y + 270 = 360

y = 360 – 270

y = 90°

Solution:

y = 360° – 30° (Reflect angle)

y = 330°

Question 10.

In the figure find x, y, z, if l || m.

Solution:

l || m

y = 30° (Alternate interior angle)

z = 130° (Alternate interior angle)

x + y + z = 180°

x + 30° + 130° = 180°

x° + 160° = 180°

x° = 180° – 160°

x = 20°

Question 11.

In the below figure, if x : y : z = 2 : 3 : 4, find the values of ∠x, ∠y and ∠z.

Solution:

Let x = 2k; y = 3k; z = 4k

∠x + ∠y + ∠z = 180° (Sum of linear angles)

2k + 3k + 4k = 180°

9k = 180°

k = \(\frac{180^{\circ}}{9}\)

∴ k = 20°

x° = 2 × 20° = 40°

y° = 3 × 20° = 60°

z° = 4 × 20° = 80°

Question 12.

In the figure PQ, RS, UT are parallel lines

i) If c =57° and a = \(\frac { c }{ 3 }\) find ‘d’.

Solution:

c = 57°; a = \(\frac { c }{ 3 }\) = \(\frac { 57 }{ 3 }\) = 19°

PQ || UT

∴ ∠a + ∠b = ∠c (Alternate interior angle)

∠a + ∠b = 57°

∠b = 57° – 19° = 38°

∴ ∠b = 38°

PQ || RS

∠b + ∠d = 180° (Co-interior angles)

38° + ∠d = 180°

∠d = 180° – 38° = 142°

∴ ∠d = 142°

ii) If c = 75° and a = \(\frac { 2 }{ 5 }\) c find b .

Solution:

∠c = 75° and ∠a = \(\frac { 2 }{ 5 }\)∠c

∠a = \(\frac { 2 }{ 5 }\) × 75° = 30°

PQ || UT

∠a + ∠b = ∠c

30° + ∠b = 75°

∠b = 75° – 30° = 45°

∴ ∠b = 45°

Question 13.

In the figure PQ and ST intersect at ‘O’. If ∠POR = 90° and x : y = 3 : 2, then find the value of z.

Solution:

∠POQ is a straight line

∴ ∠POQ = 180°

x : y = 3 : 2

Let x = 3k; y = 2k

∴ 90 + x + y = 180°

x + y = 180° – 90°

2k + 3k = 90°

5k = 90° ⇒ k = \(\frac { 90 }{ 5 }\) ⇒ k = 18°

∴ x = 3 × 18 = 54°

y = 2 × 18 = 36°

z + y = 180°

z + 36° = 180°

z = 180° – 36° ⇒ z = 144°

Question 14.

In the figure PQ || ST. Find teh value of x + 2y.

Solution:

130 + y = 180° (Linear Pair)

y = 180° – 130°

y = 50°

x = 80° (Linear Pair)

x + 2y =80° + 2(50°)

= 80° + 100°

=180°

Question 15.

In the figure, find a and b, if PA || BC || DT and A B || DC.

Solution:

Let us draw the diagram in the following way

a = 50° (Alternate interior angles)

a + x = 180° (Sum of interior angles on the same side of transversal is 180°)

50 + x = 180°

x = 180° – 50°

x = 130°

x = b (Alternate interior angles)

∴ x = 130°

Hence a = 50° and b = 130°