CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties InText Questions and Answers are provided by experts in order to help students secure good marks in exams.

## Class 7 Maths NCERT Solutions Chapter 6 The Triangle and its Properties InText Questions

Try These (Page 113)

Question 1.

Can you name the angle opposite to the side AB?

Solution:

Yes, the angle opposite the side AB is C.

Try These (Page 113)

Question 1.

Write the six elements (i.e., the 3 sides and the 3 angles) of ΔABC.

Solution:

The three sides AB, BC, CA, and three angles ∠A, ∠B, ∠C of A ABC are together the six elements of the ΔABC.

Question 2

(i) Side opposite to the vertex Q of ΔPQR.

Solution:

The side opposite to the vertex Q of ΔPQR is RP.

(ii) Angle opposite to the side LM of ΔLMN.

Solution:

The angle opposite to the side LM of ΔLMN is ∠N.

(iii) Vertex opposite to the side RT of ΔRST.

Solution:

The vertex opposite to the side RT of ΔRST is S.

Question 3.

Look at the figure and classify each of the triangles according to its

(a) Sides (b) Angles

Solution:

(a) Classification of triangles by considering the lengths of their sides is as under:

Scalene triangle : (ii)

Isosceles triangles : (i), (iii), (v) and (vi)

Equilateral triangle : (iv)

(b) Classification of triangles by considering the measures of their angles is as under:

Acute-angled triangles : (i) and (iv)

Right-angled triangles : (ii) and (vi)

Obtuse-angled triangles : (iii) and (v)

Try These (Page 118)

Question 1.

An exterior angle of a triangle is of measure 70° and one of its interior opposite angles is of measure 25°. Find the measure of the other interior opposite angle.

Solution:

Let ABC be a triangle whose side BC is produced to form an exterior angle ∠ACD such that ∠ACD = 70°.

Let ∠B = 25°. By exterior angle theorem, we have

∠ACD = ∠B + ∠A

=> 70° = 25° + ∠A

=> ∠A = 70°- 25°

= 45°

Hence, the measure of the other interior opposite angle is 45°.

Question 2.

The two interior opposite angles of an exterior angle of a triangle are 60° and 80°. Find the measure of the exterior angle.

Solution:

Let ABC be a triangle whose side BC is produced to form an exterior angle ∠ACD.

Let ∠A = 80° and ∠B = 60°

By exterior angle theorem, we have

∠ACD = ∠A + ∠B

=> ∠ACD = 80° + 60° = 140°

Hence, the measure of the exterior angle is 140°.

Question 3.

Is something wrong with this diagram? Comment.

Solution:

In this figure, the exterior angle is not equal to the sum of the two interior opposite angles, because of 50° ≠ 50°+ 50°.

So, the data given is incorrect.

Try These (Page 122)

Question 1.

Two angles of a triangle are 30° and 80°. Find the third angle.

Solution:

Let ABC be a triangle such that ∠B = 30° and ∠C = 80°.

Then, we have to find the measure of the third angle A.

Now, ∠B + ∠C = 30° + 80°= 110°

By the angle sum property of a triangle, we have

∠A + ∠B + ∠C = 180°

=> ∠A +110° = 180° [ ∠B+ ∠C = 110°)

=> – ∠A – 180°- 110°

=> ∠A = 70°

Question 2.

One of the angles of a triangle is 80° and the other two angles are equal. Find the measure of each of the equal angles.

Solution:

Let ABC be a triangle such that ∠A = 80° and ∠B = ∠C.

By the angle sum property of a triangle, we have

Hence, the measure of each of the remaining two angles is 50°.

Question 3.

The three angles of a triangle are in the ratio 1:2:1. Find all the angles of the triangle. Classify the triangle in two different ways.

Solution:

Let the angles of the triangle be x, 2x, and x.

Hence, the angles of the triangle are 45°, 90°, and 45°.

We observe that one angle of a given triangle is 90° and each of the two other angles is of measure 45°.

Therefore, the sides opposite of two equal angles is also equal.

Thus, we may classify the triangle in two different ways as follows:

(i) On the basis of the angles triangle is a right angle triangle.

(ii) On the basis of the sides triangle is an isosceles triangle.

Try These (Page 123 – 124)

Question 1.

Find the angle x in each figure :

Solution:

(i)

Since, ΔABC is isosceles with AB = AC

∴ ∠B =∠C

=> x = 40°

(ii)

Since, ΔABC is isosceles with AB = AC

∠B =∠C

=> 45° = ∠C

=> ∠C = 45°

Also, by the angle sum property, we have

∠A + ∠B + ∠C =180°

=> x + 45° + 45° =180°

=> x + 90° =180°

=> x = 180°- 90°

=> x = 90°

(iii)

Since, ΔABC is isosceles with AB = AC

∴ ∠B =∠C

=> x = 50°

(iv)

Since, ΔABC is isosceles with AB = AC

∠B = ∠C => x = ∠C

By the angle sum property, we have

(v)

Since, ΔABC is isosceles with AB = AC

∠B = ∠C => x = ∠C

By the angle sum property, we have

(vi)

Since, ΔABC is isosceles with AB = AC

∴ ∠B =∠C

=> x = ∠C

By the angle sum property, we have

(vii)

Since, ΔABC is isosceles with AB = AC

∴ ∠B = ∠C

=> x = ∠C

Since, in a triangle an exterior angle and the interior adjacent angle form a linear pair, therefore,

∠C + 120° = 180°

=> x = 180°-120°= 60°

(viii)

Since, ΔABC is isosceles with AB = AC

∴ ∠B =∠C

=> ∠B = x

Also, in a triangle, an exterior angle and the interior adjacent angle form a linear pair. Therefore,

110°+∠A = 180°

=> ∠A = 180° – 110°= 70°

By the angle sum property, we have

(ix)

Since, ΔABC is isosceles with AB = AC

∴ ∠B = ∠C

=> ∠B = x

Also, ∠B = 30°

∴ x = 30°

Question 2.

Find angles x and y in each figure.

Solution:

(i)

Since, in a triangle an exterior angle and the interior adjacent angle form a linear pair, therefore,

∠C + 120° = 180°

=> ∠C = 180° – 120° = 60°

Since, AABC is isosceles with AB = AC

∠B = ∠C

=> ∠B = 60° i.e., y = 60°

By the angle sum property, we have

=> x + 60° + 60° = 180°

=> x + 120° = 180°

=> x =180° -120°

=> x =180° – 120° = 60°

Hence, x = 60° and y = 60°.

(ii)

Since, AABC is isosceles with AB = AC

∠B = ∠C

=> ∠B = x

Also, ∠B + ∠C = 90°

=> x + x = 90° [∴ ∠B = x]

=> 2x = 90°

=> x = 45°

But, ZB + y = 180°

=> x + y = 180°

=> y =180° – 45°

=> y = 135°

Hence, x = 45° and y = 135°.

(iii)

Since AABC is isosceles with AB – AC

∠B = ∠C

i.e., ∠B = ∠C = x

and ∠A = 92°

[∴ Vertically opposite angles] By the angle sum property, we have

∠A + ∠B + ∠C = 180°

=> 92°+ x + x = 180°

=> 2x = 180°- 92° = 88°

=> x = \(\left(\frac{88}{2}\right)^{\circ}\) = 44°

ZC + y =180° [Linear pair]

=> y =180°- 44°

= 136°

Hence, x = 44° and y = 136°.

Try These (Page 125)

Question 1.

Draw any three triangles, say ΔABC, ΔPQR, and ΔXYZ in your notebook [Fig. (i), (ii) and (iii)]:

Use your ruler to find the lengths of their sides and then tabulate your results as follows:

Solution:

Clearly, from the above table, we can conclude that Some of the lengths of any two sides of a triangle are greater than the length of the third side. We also find that the difference between the length of any two sides of a triangle is smaller than the length of the third side.

Try These (Page 129 – 130)

Question 1.

Find the unknown length x in the following figures :

Solution:

(i)

ΔABC is right-angled at B.

By Pythagoras theorem, we have

(ii)

ΔABC is right-angles at B.

By Pythagoras theorem, we have

(iii)

ΔABC is right-angles at B.

By Pythagoras theorem, we have

(iv)

ΔABC is right-angles atB.

By Pythagoras theorem, we have

(v)

Using Pythagoras theorem in right-angled triangles ALB and ALC, we have

(vi)

Using Pythagoras theorem in right-angled ΔALB, we have