Math Series #10

In $∆ABC$ $BC=BA=6 cm$ and $AC=4 cm$

$AM$ and $AO$ are two tangents from Point $A$ ,so $AO=AM$ (see figure)

$∆AOB$ is a right angled triangle . So $AO^2+BO^2=AB^2 \implies BO=\sqrt{AB^2-AO^2} = \sqrt{6^2-2^2}=4\sqrt{2} cm$

Now tangents are perpendicular to radius of the circle . So $∆MPB$ is a right angled triangle .Here $BP=4\sqrt{2}-r$ ( $r$ =Radius of the circle)

So , $MB^2+MP^2=PB^2\implies 4^2+r^2 =(4\sqrt{2}-r)^2 \implies r=\sqrt{2} cm$

Area of the circle $\boxed{π(\sqrt{2})^2= 2π≈6.28 cm^2}$

let $A_T=\text{area of the triangle}$ , $R_C=\text{radius of the incircle}$ , $A_C=\text{area of the incircle}$ and $P_T=\text{perimeter of the triangle}$

Compute the area of the triangle using the Heron's Formula :

$s=\dfrac{a+b+c}{2}=\dfrac{6+6+4}{2}=8$

$A_T=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{8(8-6)(8-6)(8-4)}=8\sqrt{2}$

Compute the radius of the incircle using a formula:

$R_C=\dfrac{2*A_T}{P_T}=\dfrac{2*8\sqrt{2}}{6+6+4}=\sqrt{2}$

Compute the area of incircle:

$A_C=\pi * R^2=3.14 \left(\sqrt{2}\right)^2=\boxed{6.28}$