CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles InText Questions and Answers are provided by experts in order to help students secure good marks in exams.

## Class 7 Maths NCERT Solutions Chapter 5 Lines and Angles InText Questions

Try These (Page 93)

Question 1.

You already know how to identify different lines, line segments, and angles in a given shape. Can you Identify the different line segments and angles formed in the following figures? Can you also identify whether the angles made are acute or obtuse or right?

Solution:

Yes, we can identify the different line segments and angles formed in the given figures. Let us mark the angles in the given figures as under.

(i)

In figure (i): Line segments are AB, BC, CD, DL, DK, DE, LK, LE, KE, EF, FG, GA, AH, HC, HE, MN, NP, PO, OM, KI, IJ, JL.

Angles formed are marked as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,15, 16, 17, 18, 19, 20, 21, 22, 23.

Acute angles are 1, 3, 17

Obtuse angles are 2, 4, 16, 18

Right angles are 8, 9, 10, 11, 12, 5, 7, 6, 15, 23, 13, 14, 19, 20, 21, 22

(ii)

In figure (ii): Line segments are AB, BC, CD, DA, AL, BM, CN, DO.

Angles formed are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

Angles 9 to 12 are ∠DCN, ∠CDO, ∠ADO, and ∠DAL, respectively.

All the angles are right angles though some angles are seen to obtuse.

(iii)

In figure (iii): Line segments are AB, AE, EC, and CD.

Angles formed are 1, 2, 3, and 4.

Acute angles are 1, 2, and 3.

The obtuse angle is 4.

(iv)

In figure (iv) : Line segments are AB, BC, CA, DE, EF, FD, AE, EC, CF, FB, BD and DA.

Angles formed are 1 to 23.

Acute angles are 1 to 17, 19, 20.

Obtuse angles are 18, 21, 22, 23.

Try These (Page 94)

Question 1.

List ten figures around you and identify the acute, obtuse, and right angles found in them.

Solution:

Try These (Page 95)

Question 1.

Which pairs of following angles are complementary?

Solution:

(i)

Since in this pair, the sum of two angles = 70° + 20° = 90° So, this pair of angles is complementary.

(ii)

Since in this pair, the sum of two angles = 75° + 25° = 100° ≠ 90°. So, this pair of angles is not complementary.

(iii)

Since in this pair, the sum of two angles = 48° + 52° = 100° ≠ 90°. So, this pair of angles is not complementary.

(iv)

Since in this pair, the sum of two angles = 35° + 55°= 90°. So, this pair of angles are complementary.

Question 2.

What is the measure of the complement of each one of the following angles?

Solution:

We know that the sum of the measures of an angle and its complement is 90°. Therefore,

(i) 45°

The complement of an angle of measure 45° is the angle of (90°- 45°) = 45°.

(ii) 65°

The complement of an angle of measure 65° is the angle of (90 °- 65°) = 25°.

(iii) 41°

The complement of an angle of measure 41° is the angle of (90°- 41°) = 49°.

(iv) 54°

The complement of an angle of measure 54° is the angle of (90°- 54°) = 36°.

Question 3.

The difference in the measures of two complementary angles is 12°. Find the measures of the angles.

Solution:

Let one angle be x°. Then, the other angle is (x + 12)°. Now, x° and (x + 12)° are complementary angles.

x + (x +12) = 90

=> 2x +12 = 90

=> 2x =90-12

=> 2x = 78

=> x = 39

∴ Thus, the measures of two angles are 39° and 39° + 12°, i.e., 51°.

Try These (Page 96)

Question 1.

Find the pairs of supplementary angles in the following figures:

Solution:

(i)

In the (i) pair, the sum of the angles = 110°+ 50°= 160 ≠ 180°. So, they are not supplementary.

(ii)

In the (ii) pair, sum of the angles = 105° + 65°= 170° ≠ 180°. So, they are not supplementary.

(iii)

In the (iii) pair, sum of the angles = 130° + 50°= 180°. So, they are supplementary.

(iv)

In the (iv) pair, sum of the angles = 45° + 45°= 90 ≠ 180°. So, they are not supplementary.

Question 2.

What will be the measure of the supplement of each one of the following angles?

Solution:

We know that the sum of the measures of an angle and its supplement is 180°. Therefore,

(i) 100°

The supplement of an angle of 100° is the angle of (180° – 100°), i.e., 80°.

(ii) 90°

The supplement of an angle of 90° is the angle of (180° – 90°), i.e., 90°.

(iii) 55°

The supplement of an angle of 55° is the angle of (180° – 55°), i.e., 125°.

(iv) 125°

The supplement of an angle of 125° is the angle of (180° – 125°), i.e., 55°.

Question 3.

Among two supplementary angles, the measure of the larger angle is 44° more than the measure of the smaller. Find their measures.

Solution:

Let the smaller angle be x°. Then, the larger angle is (x + 44)°. Now, x° and (x + 44)° are supplementary angles.

∴ x + (x + 44) =180

=> 2x + 44 =180

=> 2x =180-44

=> 2x =136

=> x = 68

Thus, the measures of two angles are 68° and (68°+ 44°), i.e., 112°.

Try These (Page 97 – 98)

Question 1.

Are the angles marked 1 and 2 adjacent? If they are not adjacent, say, ‘why’?

Solution:

(i)

Yes, the angles marked 1 and 2 are adjacent.

(ii)

Yes, the. angles marked 1 and 2 are adjacent.

(iii)

No, as the angles marked 1 and 2 do not have a common vertex.

(iv)

No, because the other arms of the angles marked 1 and 2 are not on the opposite of the common arm.

(v)

Yes, the angles marked 1 and 2 are adjacent.

Question 2.

In the given figure, are the following adjacent angles?

(a) ∠AOB and ∠BOC

(b) ∠BOD and ∠BOC

Justify your answer.

Solution:

(a) In the figure, ∠AOB and ∠BOC have a common vertex. Also, they have a common arm OB and their other OA and OC lie on the opposite sides of the common arm OB. Therefore, ∠AOB and ∠BOC are adjacent angles,

(b) ∠BOD and ∠BOC are not adjacent angles, because their other arms OD and OC are not on the opposite of the common arm OB.

Try These (Page 99)

Question 1.

Check which of the following pairs of angles form a linear pair:

Solution:

(i)

In the (i) pair, the sum of the angles = 140° + 40°=180°. So, this pair can form a linear pair.

(ii)

In the (ii) pair, the sum of the angles = 60° + 60° = 120 ≠ 180°. So, this pair cannot form a linear pair.

(iii)

In the (iii) pair, the sum of the angles = 90° + 40° = 130 ≠ 180°. So, this pair cannot form a linear pair.

(iv)

In the (iv) pair, the sum of the angles = 65° + 115° = 180°. So, this pair can form a linear pair.

Try These (Page 100)

Question 1.

One pair of vertically opposite angles is ∠1 and ∠3 in the given figure. Can you name the other pair of vertically opposite angles?

Solution:

The other pair of vertically opposite angles are ∠2 and ∠4.

Try These (Page 101)

Question 1.

Can you prove that ∠2 = ∠4 (Try it !)

Solution:

Let l and m two lines which intersect at O, making angles ∠1, ∠2, ∠3, and ∠4.

We want to prove ∠2 = ∠4

Now, ∠2 = ∠180° – ∠1

[∴ ∠1, ∠2 are a linear pair and ∠1 + ∠2 = 180°]

Also, ∠4 = 180°- ∠1

[∴ ∠1, ∠4 are a linear pair and ∠1 + ∠4 = 180°]

∴ ∠2 = ∠4

Try These (Page 101)

Question 1.

In the given figure, if ∠1 = 30°, find ∠2 and ∠3.

Solution:

Since the lines intersect at a point, therefore, ∠1 and ∠3.

Question 2.

Give an example for vertically opposite angles in your surroundings.

Solution:

A grill of the type shows vertically opposite angles.

Try These (Page 104)

Question 1.

Find examples from your surroundings where lines intersect at right angles.

Solution:

Some examples from our surroundings where the lines intersect at right angles are :

(i) Edges of a blackboard

(ii) Edges of a paper sheet

(iii) Leg and top of the table, etc.

Question 2.

Find the measures of the angles made by the intersecting lines at the vertices of an equilateral triangle.

Solution:

Let ABC be an equilateral triangle. We want to find its angles.

Since all the angles of an equilateral triangle are equal.

Question 3.

Draw any rectangle and find the measures of angles at the four vertices made by the intersecting lines.

Solution:

Let ABCD be the rectangle and we want to find its angles, i. e., ABCD is a parallelogram with ∠A = 90°

Question 4.

If two lines intersect, do they always intersect at right angles?

Solution:

No, two intersecting lines do not always intersect at right angles.

Try These (Page 105)

Question 1.

In the adjoining figure, the line p is not a transversal, although it cuts two lines l and m. Can you say, ‘why’?

Line p is not a transversal to the lines l and m because it does not intersect the lines at different points. In fact, it intersects the two lines at the same point.

Try These (Page 105)

Question 1.

Suppose two lines are given. How many transversals can you draw for these lines?

Solution:

An infinite number of transversals can be drawn for two given lines.

Question 2.

If a line is a transversal to three lines, how many points of intersections are there?

Solution:

If three lines have a transversal, then they may have three or more points of intersections as shown below :

Question 3.

Try to identify a few transversals in your surroundings.

Solution:

Examples of transversals from our surroundings are:

A line is drawn to intersect lines of the ruled sheet of paper, an iron staircase, a towel stand, grills of the window, etc.

The complement of an angle of measures 45° + x°, where x > 0 is the angle of [90°- (45° + x°)]

= 90°- 45°- x° = 45° – x°.

Clearly, 45° + x° > 45° – x°

Hence, the complement of an angle > 45° is less than 45°.

Try These (Page 106)

Question 1.

Name the pairs of angles in each figure:

Solution:

∠1 and ∠2 are corresponding angles.

∠3 and ∠4 are alternate interior angles

∠5 and ∠6 form a pair of interior angles.

∠7 and ∠8 are corresponding angles.

∠9 and ∠10 are alternate interior angles.

∠11 and ∠12 form a linear pair of angles.

Try These (Page 109)

Question 1.

(i)

Solution:

Since l || m and t is a transversal ∴ ∠x = 60° [Alternate angles]

(ii)

Solution:

Since, a || b and c is transversal ∴ ∠y = 55° [Alternate angles]

(iii)

Solution:

Since L and l7 are any two nonparallel lines and t is a transversal

(iv)

Solution:

Since l || m and t is a transversal ∴ 60° + z = 180°

[Sum of the interior angles on the same side of the transversal is 180°] => z = 180° – 60° =120°

(v)

Solution:

Since, l || m and t is a transversal ∠x = 120p [Corresponding angles]

(vi)

Solution:

Since, p || q and l is a transversal a + 60° = 180°

[Sum of the interior ∠s on the same side of a transversal is 180 °] => a =180°- 60° = 120°.

Since l || m and q is a transversal

∠a = ∠1 [pair of corresponding angles]

(v)

Try These (Page 110)

Question 1.

(i)

Solution:

Yes, because l || m, so the alternate angles are equal.

(ii)

Solution:

Yes, because l || m, so the corresponding angles are equal.

∴ ∠1 +130° = 180° [by linear pair]

=> ∠1 = 180° – 130° = 50°

(iii)

Solution:

Given, l || m, and t is a transversal.

∴ x + 70° = 180°

=> x = 180° – 70

=> x = 110°

[∵ sum of the interior angle on the same side of a transversal is 180°.]