Lines and Angles Class 7 Extra Questions with Answers

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°

Lines and Angles Class 7 Extra Questions with Answers

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.
Lines and Angles Class 7 Extra Questions with Answers Img 1
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.
Lines and Angles Class 7 Extra Questions with Answers Img 2
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.
Lines and Angles Class 7 Extra Questions with Answers Img 3
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.
Lines and Angles Class 7 Extra Questions with Answers Img 4
Solution:
Adjacent angles :
∠SRP, ∠PRQ; ∠TPQ, ∠RPQ;
∠RQP, ∠RQU

Question 10.
In the below figure what can you say about
Lines and Angles Class 7 Extra Questions with Answers Img 5
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.
Lines and Angles Class 7 Extra Questions with Answers Img 6
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.
Lines and Angles Class 7 Extra Questions with Answers Img 7
Solution:
In the figure
3x° = 120° (Vertically opposite angles)
x° = \(\frac{120^{\circ}}{3}\)
⇒ x° = 40°

Question 14.
In the figure find ‘a’.
Lines and Angles Class 7 Extra Questions with Answers Img 8
Solution:
In the figure
80° + 2a = 180°
2a° = 180° – 80°
2a = 100° ⇒ a = \(\frac{100^{\circ}}{2}\)
∴ a = 50°

Lines and Angles Class 7 Extra Questions with Answers

Question 15.
Draw i) a pair of intersecting lines
ii) a pair of parallel lines
Solution:
Lines and Angles Class 7 Extra Questions with Answers Img 9

Question 16.
In the figure l || m, find the x and y if n is the transversal.
Lines and Angles Class 7 Extra Questions with Answers Img 10
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:
Lines and Angles Class 7 Extra Questions with Answers Img 11
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
Lines and Angles Class 7 Extra Questions with Answers Img 12
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.
Lines and Angles Class 7 Extra Questions with Answers Img 13
Solution:
Let us consider the given figure is the following way
Lines and Angles Class 7 Extra Questions with Answers Img 14
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°.
Lines and Angles Class 7 Extra Questions with Answers Img 15
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°.
Lines and Angles Class 7 Extra Questions with Answers Img 16
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’.
Lines and Angles Class 7 Extra Questions with Answers Img 17
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.
Lines and Angles Class 7 Extra Questions with Answers Img 18
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.
Lines and Angles Class 7 Extra Questions with Answers Img 19
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.
Lines and Angles Class 7 Extra Questions with Answers Img 20
Solution:
The figure can be taken in the following way.
Lines and Angles Class 7 Extra Questions with Answers Img 21
∠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.
Lines and Angles Class 7 Extra Questions with Answers Img 22
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°

Lines and Angles Class 7 Extra Questions with Answers

Question 8.
Find x in the figure.
Lines and Angles Class 7 Extra Questions with Answers Img 23
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.
Lines and Angles Class 7 Extra Questions with Answers Img 24
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.
Lines and Angles Class 7 Extra Questions with Answers Img 25
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.
Lines and Angles Class 7 Extra Questions with Answers Img 26
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.
Lines and Angles Class 7 Extra Questions with Answers Img 27
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.
Lines and Angles Class 7 Extra Questions with Answers Img 28
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.
Lines and Angles Class 7 Extra Questions with Answers Img 29
Solution:
Method-1 :
Let us draw the given figure in the following way
Lines and Angles Class 7 Extra Questions with Answers Img 31
∠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.
Lines and Angles Class 7 Extra Questions with Answers Img 32
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.
Lines and Angles Class 7 Extra Questions with Answers Img 33
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.
Lines and Angles Class 7 Extra Questions with Answers Img 34
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°

Lines and Angles Class 7 Extra Questions with Answers

Question 4.
In the figure find ∠m, AB || PQ, PQ || EF.
Lines and Angles Class 7 Extra Questions with Answers Img 35
Solution:
Let us draw the figure in the following way
Lines and Angles Class 7 Extra Questions with Answers Img 36
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.
Lines and Angles Class 7 Extra Questions with Answers Img 37
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.
Lines and Angles Class 7 Extra Questions with Answers Img 38
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.
Lines and Angles Class 7 Extra Questions with Answers Img 39
Solution:
Let us draw the figure in the following way
Lines and Angles Class 7 Extra Questions with Answers Img 40
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.
Lines and Angles Class 7 Extra Questions with Answers Img 41
Solution:
Let us draw the figure in the following way.
Let l || CD and l || AB
Lines and Angles Class 7 Extra Questions with Answers Img 42
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
Lines and Angles Class 7 Extra Questions with Answers Img 44
Solution:
y + 90 + 50 + 130 = 360
y + 270 = 360
y = 360 – 270
y = 90°

Lines and Angles Class 7 Extra Questions with Answers Img 45
Solution:
y = 360° – 30° (Reflect angle)
y = 330°

Question 10.
In the figure find x, y, z, if l || m.
Lines and Angles Class 7 Extra Questions with Answers Img 46
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.
Lines and Angles Class 7 Extra Questions with Answers Img 47
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
Lines and Angles Class 7 Extra Questions with Answers Img 48
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.
Lines and Angles Class 7 Extra Questions with Answers Img 49
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°

Lines and Angles Class 7 Extra Questions with Answers

Question 14.
In the figure PQ || ST. Find teh value of x + 2y.
Lines and Angles Class 7 Extra Questions with Answers Img 50
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.
Lines and Angles Class 7 Extra Questions with Answers Img 51
Solution:
Let us draw the diagram in the following way
Lines and Angles Class 7 Extra Questions with Answers Img 52
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°

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