Question

In the given figure, $\Delta ABC$ is a right-angled triangle with $\angle B={90}^{\circ }$ and $C_1$ is the incircle of $\Delta ABC$ with radius $3.$ A circle $C_2$ of radius $2$ touches the sides $AC,BC$ and the circle $C_1.$ Then the value of $23AB$ is

• A. $12(12+\sqrt{6})$
• B. $12(6+\sqrt{6})$
• C. $12(12+\sqrt{12})$
• D. $12(6+\sqrt{12})$

Answer 1

The correct option is A $12(12+\sqrt{6})$


From the above figure,
$BP=BT=3,PQ=SR=\sqrt{24}=2\sqrt{6}$
Let the center of $C_1$ be $O_1$ and the center of $C_2$ be $O_2$.
$\Delta CQ{O}_2\sim \Delta CP{O}_1\Rightarrow \frac{CQ}{2}=\frac{CP}{3}\Rightarrow \frac{CQ}{2}=\frac{2\sqrt{6}+CQ}{3}\Rightarrow CQ=4\sqrt{6}=CR$

Let $AS=AT=x$
$A{B}^2+B{C}^2=A{C}^2\Rightarrow (x+3{)}^2+(3+6\sqrt{6}{)}^2=(x+6\sqrt{6}{)}^2\Rightarrow x=\frac{75+12\sqrt{6}}{23}\Rightarrow AB=\frac{144+12\sqrt{6}}{23}=\frac{12(12+\sqrt{6})}{23}$