Question

If an isoceles triangle $ABC$ in which $AB=AC=6cm$ is inscribed in a circle of radius $9cm$, find the area of triangle.

Answer 1

Let $O$ be the centre of the circle and let $P$ be the mid-point of $BC$. Then, $OP\perp BC$.

Since $△ABC$ is isosceles and $P$ is the mid-point of $BC$. Therefore, $AP\perp BC$ as median from the vertex in an isosceles triangle is perpendicular to the base.

Let $AP=x$ and $PB=CP=y$.

Applying Pythagoras theorem in $△$s $APB$ and $OPB$, we have

$A{B}^2=B{P}^2+A{P}^2$ and $O{B}^2=O{P}^2+B{P}^2$

$\Rightarrow 36=y^2+x^2$ . . . $\left(i\right)$ and, $81=(9-x{)}^2+y^2$ . . . $\left(ii\right)$

$\Rightarrow 81-36=(9-x{)}^2+y^2-y^2+x^2$ [Subtracting $\left(i\right)$ from $\left(ii\right)$]

$\Rightarrow 45=81-18x$

$\Rightarrow x=2cm$

Putting $x=2$ in $\left(i\right)$, we get

$36=y^2+4\Rightarrow y^2=32\Rightarrow y=4\sqrt{2}cm$

$\therefore BC=2BP=2y=8\sqrt{2}cm$

Hence, Area of $△ABC=\frac{1}{2}(BC\times AP)$

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

$=8\sqrt{2}c{m}^2$