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
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