Question

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

Answer 1

Let the height of the triangle be h cm.
Each of the equal sides measures $a=\left(h+2\right)\mathrm{cm}$ and b = 12 cm (base).

Now,
Area of the triangle = Area of the isosceles triangle
$\frac{1}{2}\times \mathrm{base}\times \mathrm{height}=\frac{1}{4}\times b\sqrt{4a^2-b^2}$

$$\begin{gathered} \Rightarrow \frac{1}{2}\times 12\times h=\frac{1}{4}\times 12\times \sqrt{4(h+2{)}^2-144} \\ \Rightarrow 6h=3\sqrt{4h^2+16h+16-144} \\ \Rightarrow 2h=\sqrt{4h^2+16h+16-144} \\ \mathrm{On}\mathrm{squaring}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ \Rightarrow 4h^2=4h^2+16h+16-144 \\ \Rightarrow 16h-128=0 \\ \Rightarrow h=8 \end{gathered}$$

Area of the triangle = $\frac{1}{2}\times b\times h$
$$\begin{gathered} =\frac{1}{2}\times 12\times 8 \\ =48{\mathrm{cm}}^2 \end{gathered}$$