Question

an isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. find the area of the triangle. ​

Answer 1

First,let the third side be x.

It is given that the length of the equal sides us 12 cm and it's perimeter is 30 cm.

• So,

$30=12+12+x$

$⇒ 30 = 24 + x$

$⇒24 + x = 30$

$⇒ x= 30 - 24$

$⇒ x = 6$

So,the length of the third side is 6 cm.

Thus,the semi perimeter of the isosceles triangle (s) = 30/2 cm =15 cm

By using Heron's Formula,

Area of the triangle,

$= \sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{15(15 - 12)(15 - 12)(15 - 6)} \: {cm}^{2}$

$= \sqrt{15 \times 3 \times 3 \times 9} \: {cm}^{2}$

$= 9 \sqrt{15} \: {cm}^{2}$

Answer 2

Petimeter = 30 cm

Equal sides = 12 cm

Let the third side be x,

So, the third side :-

30 = 12 + 12 + x

=> x = 30 - 24

=> x = 6cm

Third side = 6cm

Using Heron's Formula,

Area of Δ = √s(s-a)(s-b)(s-c)

and, s = ( a + b + c ) ÷ 2

So, s = 30 ÷ 2 = 15

Therefore,

Area of ∆ = √15(15-12)(15-12)(15-6)

=> √15 × 3 × 3 × 9

=> 3 × 3√15

=> 9√15 cm²

Hence, Area of the triangle is 9√15 cm².