Question

Each of the equal sides of an isosceles triangle is 2 cm more than its height and the base of the triangle is 12 cm. Find the area of the triangle.

Answer 1

Base = 12 cm,

Let the height be h cm

then,

equal sides = (h+2) cm

Area of the triangle = $\frac{1}{2}\times b\times h$

$=\frac{1}{2}\times 12\times h$

$=6hc{m}^2...\left(i\right)$

when equal sides are considered,

semi-perimeter = $\frac{(h+2)+(h+2)+12}{2}$

$=\frac{2h+16}{2}=(h+8)cm$

By heron's formula,

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

$=\sqrt{(h+8)(h+8-h-2)(h+8-h-2)(h+8-2)}$

$=\sqrt{(h+8)\left(6\right)\left(6\right)(h-4)}$

$=\sqrt{(h+8)(h-4)\left(36\right)}$ __ (ii)

Equating (i) and (ii), h = 8 cm

Area = 6h

$=6\times 8$ $=48c{m}^2$