In ∆PQR,
a. PQ - QR > PR
b. PQ + QR < PR
c. PQ - QR< PR
d. PQ + PR< QR

Solution:

Given, PQR is a triangle.

We have to choose the correct option.

We know that the sum of the lengths of any two sides of a triangle is always greater than the length of the third side.

So, PR + QR > PQ

On rearranging,

PR > PQ - QR

PQ - QR < PR

Therefore, option c is true.

✦ Try This: In ∆ XYZ,
a. XY + YZ > XZ
b. XY + YZ < XZ
c. XY + XZ < YZ
d. XZ + YZ < XY

Given, XYZ is a triangle.

We have to choose the correct option.

We know that the sum of the lengths of any two sides of a triangle is always greater than the length of the third side.

Let the two sides be XY and YZ

Let the third side be XZ

Therefore, XY + YZ > XZ

☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 6

NCERT Exemplar Class 7 Maths Chapter 6 Problem 9

In ∆PQR, a. PQ - QR > PR, b. PQ + QR < PR, c. PQ - QR< PR, d. PQ + PR< QR

Summary:

In ∆PQR, PQ - QR < PR

☛ Related Questions:

• In ∆ ABC, a. AB + BC > AC, b. AB + BC < AC, c. AB + AC < BC, d. AC + BC < AB
• In Fig. 6.8, PB = PD. The value of x is: a. 85°, b. 90°, c. 25°, d. 35°
• Lengths of sides of a triangle are 3 cm, 4 cm and 5 cm. The triangle is: a. Obtuse angled triangle, . . . .