Question

find the area in sq. cm of an isosceles triangle whose base is 16 cm and each equal side is 9 CM​

Answer 1

Let a,b,c be the sides

Perimeter=16+9+9=34cm

S(Semi perimeter)=34/2=17cm

Using heron's formula

Area=

$\sqrt{s(s - a)(s - b)(s - c)} \\ = \sqrt{17(17 - 16)(17 - 9)(17 - 9)} \\ = \sqrt{17 \times 1 \times 8 \times 8} \\ = 32.98 {cm}^{2}$

The area is approx 33cm²