CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4 Textbook 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 Ex 6.4

**Ex 6.4 Class 7 Maths Question 1.**

Is it possible to have a triangle with the following sides?

**(i)** 2 cm, 3 cm, 5 cm

**(ii)** 3 cm, 6 cm, 7 cm

**(iii)** 6 cm, 3 cm, 2 cm

**Solution:
**

**(i)**2 cm, 3 cm, S cm We have 2 + 3 = 5

⇒ Sum of the lengths of two sides = Length of the third side

This is impossible since the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

**(ii)** 3 cm, 6 cm, 7 cm

We see that 3 + 6 > 7

6 + 7 > 3

7 + 3 > 6

Therefore, it is possible to have a triangle with side lengths 3 cm, 6 cm, 7 cm.

**(iii)** 6 cm, 3 cm, 2 cm

We see that 6 + 3 = 9 > 2

3 + 2 = 5 \(\ngtr \) 6

2 + 6 = 8 > 3

Therefore, it is not possible to have a triangle with side lengths 6 cm, 3 cm, 2 cm.

**Ex 6.4 Class 7 Maths Question 2.**

Take any point O in the interior of a triangle PQR. Is

**(i)** OP + OQ > PQ?

**(ii)** OQ + OR > QR ?

**(iii)** OR + OP > RP ?

**Solution:**

**(i)** Yes, OP + OQ > PQ because on joining OP and OQ, we get an ∆OPQ and in a triangle, the sum of the lengths of any two sides is always greater than the third side.

**(ii)** Yes, OQ + OR > QR, because on joining OQ and

OR, we get an ∆OQR and in a triangle, the sum of the length of any two sides is always greater than the third side.

**(iii)** Yes, OR + OP > RP, because on joining OR and OP, we get an ∆OPR and in a triangle, the sum of the lengths of any two sides is always greater than the third side.

**Ex 6.4 Class 7 Maths Question 3.**

AM is median of a triangle ABC. Is AB + BC + CA > 2AM?

(Consider the sides of triangles ∆ABM and ∆AMC.)

**Solution:**

In ∆ ABM, AB + BM > AM …(1)

The Sum of the lengths of any two sides of a triangle is greater than the length of the third side

In ∆ ACM,

CA + CM > AM …(2)

The Sum of the lengths of any two sides of a triangle is greater than the length of the third side

Sum (1) and (2),

(AB + BM) + (CA + CM) > AM + AM

⇒ AB + (BM + CM) + CA > 2 AM

⇒ AB + BC + CA > 2 AM.

**Ex 6.4 Class 7 Maths Question 4.**

ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?

**Solution:
**

**Ex 6.4 Class 7 Maths Question 5.**

ABCD is a quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)?

**Solution:**

Let ABCD be a quadrilateral and its diagonals AC and BD intersect at O. Using triangle inequality property, we have

In ∆OAB

OA + OB > AB ……(1)

**Ex 6.4 Class 7 Maths Question 6.**

The lengths of the two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

**Solution:**

Let x cm be the length of the third side.

Thus, 12 + 15 > x, x + 12 > 15 and x + 15 > 12

⇒ 27 > x, x > 3 and x > -3

The numbers between 3 and 27 satisfy these.

∴ The length of the third side could be any length between 3 cm and 27 cm.