Chapter 5: Lines and Angles

Q11Written answer

In Fig. 5.12, lines l and m intersect each other at a point. Which of the following is false?

(a) ∠a = ∠b (b) ∠d = ∠c

(c) ∠a + ∠d = 180^o (d) ∠a = ∠d

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Solution:-

(d) ∠a = ∠d

∠a ≠ ∠d

∠a = ∠b [because vertically opposite angles]

∠d = ∠c [because vertically opposite angles]

∠a + ∠d = 180^o [Linear pair of angles]

Q12Written answer

If angle P and angle Q are supplementary and the measure of angle P is 60^o, then the measure of angle Q is

(a) 120^o (b) 60^o (c) 30^o (d) 20^o

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Solution:-

(a) 120^o

When the sum of the measures of two angles is 180^o, then the angles are called supplementary angles.

P + Q = 180^o

60^o + Q = 180^o

Q = 180^o – 60^o

Q = 120^o

Q13Written answer

In Fig. 5.13, POR is a line. The value of a is

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 6

(a) 40^o (b) 45^o (c) 55^o (d) 60^o

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Solution:-

(a) 40^o

We know that, when the sum of the measures of two angles is 180^o, then the angles are called supplementary angles.

(3a + 5)^o + (2a – 25)^o = 180^o

3a + 5 + 2a – 25 = 180^o

5a – 20 = 180^o

5a = 180^o + 20

5a = 200

a = 200/5

a = 40^o

Q14Written answer

In Fig. 5.14, POQ is a line. If x = 30°, then ∠ QOR is

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 7

(a) 90^o (b) 30^o (c) 150^o (d) 60^o

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Solution:-

(a) 90^o

Sum of all angles about a straight line given in the figure are equal to 180^o.

Then, 30^o + 2y + 3y = 180^o

30^o + 5y = 180^o

5y = 180^o – 30^o

5y = 150^o

y = 150/5

y = 30^o

So, 2y = 2 × 30 = 60^o

3y = 3 × 30 = 90^o

Therefore, ∠ QOR = 90^o

Q15Written answer

The measure of an angle which is four times its supplement is

(a) 36^o (b) 144^o (c) 16^o (d) 64^o

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Solution:-

(b) 144^o

We know that, when the sum of the measures of two angles is 180^o, then the angles are called supplementary angles.

Let us assume the angle be x.

Then, its supplement angle = (180^o – x)

As per the condition given in the question, x = 4 (180^o – x)

x = 720^o – 4x

x + 4x = 720^o

5x = 720^o

x = 720^o/5

x =144^o

Q16Written answer

In Fig. 5.15, the value of y is

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 8

(a) 30^o (b) 15^o (c) 20^o (d) 22.5^o

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Solution:-

(c) 20^o

The sum of all angles about a straight line given in the figure is equal to 180^o.

Then, 6y + y + 2y = 180^o

9y = 180^o

y = 180/9

y = 20^o

So, value of y is 20^o.

Q17Written answer

In Fig. 5.16, PA || BC || DT and AB || DC. Then, the values of a and b are respectively.

(a) 60^o, 120^o (b) 50^o,130^o (c) 70^o,110^o (d) 80^o,100^o

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Solution:-

(b) 50^o,130^o

We know that, ∠PAB = ∠ABC = 50^o … [because interior alternate angles]

Given, AB || DC so consider it as parallelogram,

In a parallelogram, adjacent angles of the parallelogram are supplementary.

So, ∠ABC + ∠BCD = 180^o

50^o + ∠BCD = 180^o

∠BCD = 180^o – 50^o

∠BCD = 130^o

∠BCD = ∠CDT = 130^o … [because interior alternate angles]

Therefore, a = 50^o and b = 130^o

Q18Written answer

The difference of two complementary angles is 30^o. Then, the angles are

(a) 60^o, 30^o (b) 70^o, 40^o (c) 20^o, 50^o (d) 105^o, 75^o

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Solution:-

(a) 60^o, 30^o

When the sum of the measures of two angles is 90^o, then the angles are called complementary angles.

So, 60^o + 30^o = 90^o

As per the condition in the question, 60^o – 30^o = 30^o

Q19Written answer

In Fig. 5.17, PQ || SR and SP || RQ. Then, angles a and b are respectively

(a) 20^o, 50^o (b) 50^o, 20^o (c) 30^o, 50^o (d) 45^o, 35^o

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Solution:-

(a) 20^o, 50^o

∠QRP = ∠RPS = 50^o … [because interior alternate angles]

∠SRP = ∠RPQ = 20^o … [because interior alternate angles]

Therefore, angle a = 20^o and angle b = 50^o

Q20Written answer

In Fig. 5.18, a and b are

(a) alternate exterior angles

(b) corresponding angles

(c) alternate interior angles

(d) vertically opposite angles

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Solution:-

(c) alternate interior angles

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