Question

A circle is inscribed in a triangle whose sides are $40,40$ and $48cm$ respectively. A smaller circle touching two equal sides of the triangle and to the first circles, then the area of smaller circle is

Answer 1

In $△ABC$

$AB=AC=40cm$ and $BC=48cm$

As we know that, $AD$ will be median, perpendicular bisector, angle bisector and altitude of the isosceles $△ABC$.

$\therefore BD=\frac{BC}{2}=\frac{48}{2}=24cm$

Now, in $△ABD$

Using pythagoras theorem,

${AD}^2={AB}^2-{BD}^2$

${AD}^2={40}^2-{24}^2$

$\Rightarrow AD=\sqrt{1024}=32cm$

As we know that centroid divides the median in the ration $2:1$.

$\therefore AO=2OD$

$AD=AO+OD=2OD+OD$

$\Rightarrow OD=\frac{1}{3}AD$

$\therefore OD=\frac{1}{3}\times 32=\frac{32}{3}cm$

$AO=2OD=2\times \frac{32}{3}=\frac{64}{3}cm$

Now, in $△APE$ and $△AOF$

$\angle PAF=\angle OAF\left[\text{Common angle}\right]$

$\angle AEP=\angle AFO=90℃[\because \text{At the point of contact, tangent is}\perp \text{to the radius}]$

By AA similarity

$△APE\sim △AOF$

As we know that corresponding sides of similar tringle are proportional.

$\therefore \frac{PE}{OF}=\frac{AP}{AO}$

$\Rightarrow \frac{r}{\left(\frac{32}{3}\right)}=\frac{AP}{\left(\frac{64}{3}\right)}$

$\Rightarrow AP=2r$

As $AO=AP+PO$

$AP=AO-PO$

$\Rightarrow 2r=\frac{64}{3}-(r+\frac{32}{3})$

$\Rightarrow 2r=\frac{64}{3}-r-\frac{32}{3}$

$\Rightarrow 3r=\frac{32}{3}$

$\Rightarrow r=\frac{32}{9}cm\approx 3.56cm$

As we know that are of circle $=\pi \times {\left(\text{radius}\right)}^2$

$\therefore$ Area of smaller circle $=\pi r^2=\pi \times {\left(3.56\right)}^2=12.67\pi {cm}^2$