# In the given triangles of Fig. 9.39, perimeter of ∆ABC = perimeter of ∆PQR. Find the area of ∆ABC.

**Solution:**

Given, __perimeter__ of ∆ABC = perimeter of ∆PQR

We have to find the area of ∆ABC.

Perimeter of ∆ABC = AB + BC + AC

= AB + 5 + 13

= AB + 18

Perimeter of ∆PQR = PQ + QR + PR

= 6 + 10 + 14

= 16 + 14

= 30 cm

Given, AB + 18 = 30

AB = 30 - 18

AB = 12 cm.

__Area of triangle__ = 1/2 × base × height

Area of ∆ABC = 1/2 × BC × AB

= 1/2 × 5 × 12

= 6(5)

= 30 cm²

Therefore, the area of ∆ABC = 30 cm²

**✦ Try This:** A circular flower bed is surrounded by a path 5 m wide. The diameter of the flower bed is 77m. What is the area of this path? (π = 3.14)

**☛ Also Check: **NCERT Solutions for Class 7 Maths Chapter 11

**NCERT Exemplar Class 7 Maths Chapter 9 Problem 85**

## In the given triangles of Fig. 9.39, perimeter of ∆ABC = perimeter of ∆PQR. Find the area of ∆ABC.

**Summary:**

In the given triangles of Fig. 9.39, perimeter of ∆ABC = perimeter of ∆PQR. The area of ∆ABC is 30 cm²

**☛ Related Questions:**

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