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NCERT Exemplar · Class 7 · Maths

Chapter 5: Lines and Angles

Practice each question, then open the answer or worked solution.

Questions

56 questions
Q1Written answer

The angles between North and West and South and East are

(a) complementary (b) supplementary

(c) both are acute (d) both are obtuse

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Worked solution

Solution:-

(b) supplementary

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 1

The angle between North and West is 90o, the angle between South and East is 90o, as shown in the figure above. So, 90o + 90o = 180o.

Then, the angles between North and West and South and East are supplementary.

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

Q2Written answer

Angles between South and West and South and East are

(a) vertically opposite angles (b) complementary angles

(c) making a linear pair (d) adjacent but not supplementary

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Solution:-

(c) making a linear pair

A linear pair is a pair of adjacent angles whose non-common sides are opposite rays.

Q3Written answer

In Fig. 5.9, PQ is a mirror, AB is the incident ray and BC is the reflected ray. If ∠ ABC = 46o, then ∠ ABP is equal to

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 2

(a) 44o (b) 67o (c) 13o (d) 62o

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Worked solution

Solution:-

(b) 67o

As we know that, the angle formed by the incident ray and angle formed by the reflected ray is equal.

From the given figure,

PQ is a straight line,

So, ∠ABP + ∠ABC + ∠CBQ = 180o

Let us assume the ∠ABP = ∠CBQ = x

Then,

x + 46o + x = 180o

2x + 46o = 180o

2x = 180o – 46o

2x = 134o

x = 134o/2

x = 67o

Therefore, the ∠ABP = ∠CBQ = 67o

Q4Written answer

If the complement of an angle is 79o, then the angle will be of

(a) 1o (b) 11o (c) 79o (d) 101o

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Worked solution

Solution:-

(b) 11o

When the sum of the measures of two angles is 90°, the angles are called complementary angles. Each of them is called complement of the other.

The given complement of an angle is 79o

Let the measure of the angle be xo.

Then,

x + 79o = 90o

x = 90o – 79o

x = 11o

Hence, the measure of the angle is 11o.

Q5Written answer

Angles which are both supplementary and vertically opposite are

(a) 95o, 85o (b) 90o, 90o (c) 100o, 80o (d) 45o, 45o

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Solution:-

(b) 90o, 90o

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

Q6Written answer

The angle which makes a linear pair with an angle of 61o is of

(a) 29o (b) 61o (c) 122o (d) 119o

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Solution:-

(d) 119o

A linear pair is a pair of adjacent angles whose non-common sides are opposite rays.

We know that, measure of sum of adjacent angles is equal to 180o.

Let the measure of other angle be xo.

Then,

x + 61o = 180o

x = 180o – 61o

x = 119o

Q7Written answer

The angles x and 90o – x are

(a) supplementary (b) complementary

(c) vertically opposite (d) making a linear pair

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Solution:-

(b) complementary

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

x + 90o – x = 90o

90o = 90o

LHS = RHS

Q8Written answer

The angles x – 10o and 190o – x are

(a) interior angles on the same side of the transversal

(b) making a linear pair

(c) complementary

(d) supplementary

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Solution:-

(d) supplementary

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

x – 10o + 190o – x = 180o

190o – 10 = 180o

180o = 180o

LHS = RHS

Q9Written answer

In Fig. 5.10, the value of x is

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 3

(a) 110o (b) 46o (c) 64o (d) 150o

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Solution:-

(d) 150o

The sum of all angles about a point given in the figure is equal to 360o.

Then, 100o + 46o + 64o + x = 360o

210o + x = 360o

x = 360o – 210o

x = 150o

Q10Written answer

In Fig. 5.11, if AB || CD, ∠ APQ = 50o and ∠PRD = 130o, then ∠ QPR is

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 4

(a) 130o (b) 50o (c) 80o (d) 30o

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Solution:-

(c) 80o

We know that, ∠APR = ∠PRD … [because interior alternate angles]

∠APQ + ∠QPR = 130o

50o + ∠QPR = 130o

∠QPR = 130o – 50o

∠QPR = 80o