Question

Find the area of a triangle whose sides are 9 cm, 12 cm and 15 cm.

Answer 1

Given : Sides of a triangle ∆ are 9 cm, 12 cm, and 15 cm.

Let the sides of the triangle are a = 9 cm, b = 12 cm & c = 15 cm.

Semi Perimeter of the ∆,s = (a + b + c) /2

s = (9 + 12 + 15) / 2

s = 36/2

s = 18 cm

Semi Perimeter of the ∆ = 18 cm

Using Heron’s formula :  

Area of the wall , A = √s (s - a) (s - b) (s - c)

A = √18(18 - 9) (18 - 12) (18 - 15)

A = √18 × 9 × 6 × 3

A = √(2 × 9) × (9) × (2 × 3) × (3)

A = √(2 × 2) × (9 × 9) × (3 × 3)  

A = 2 × 9 × 3

A = 54 cm²

Hence, the  area of a triangle is 54 cm².

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Answer 2

Formula to be used:

Heron's Formula

Heron's Formula: $\tt{\sqrt{s(s-a)(s-b)(s-c)}}$

Where:

s = semi perimeter

a, b, c = sides of triangle

a, b, c = 9 cm, 12 cm, 15 cm respectively

s = (a + b + c)/2 = (9 + 12 + 15)/2 = 36/2 = 18

Area of triangle = $\tt{\sqrt{s(s-a)(s-b)(s-c)}}$

=> $\tt{\sqrt{18(18-9)(18-12)(18-15)}}$

=> $\tt{\sqrt{18(9)(6)(3)}}$

=> 9 × 2 × 3

=> 54 sq cm