Question

In an isosceles triangle $ABC$, in which $AB=AC=6cm$, is inscribed in a circle of radius $9cm$. Find the area of the triangle.

Answer 1

Let $AP=x$

So, $OP=9-x$

In $△APC$

$\Rightarrow A{P}^2=36-P{C}^2$

$\Rightarrow AP=\sqrt{36-P{C}^2}$

Similarly, $P{C}^2=36-x^2$

Now, In $△OPC$

$\Rightarrow O{C}^2=O{P}^2+P{C}^2$

$\Rightarrow 9^2=(9-x{)}^2+36-x^2$

$\Rightarrow 81=81+x^2-18x+36-x^2$

$\Rightarrow 18x=36$

$\Rightarrow x=2cm$

$\Rightarrow PC=\sqrt{36-4=\sqrt{32}=4\sqrt{2}}$

$\Rightarrow BC=8\sqrt{2}$

Therefore, $Area=\frac{1}{2}\times b\times h$

$\Rightarrow Area=\frac{1}{2}\times 8\sqrt{2}\times 2=8\sqrt{2}c{m}^2$