Question

In the adjoining figure, $ABC$ is a triangle in which $\angle B={90}^{\circ }$ and its incircle $C_1$ has radius $3$. A circle $C_2$ of radius $1$ touches sides $AC,BC$ and the circle $C_1$. Then length $AB$ is equal to

• A. $3+6\sqrt{3}$
• B. $10+3\sqrt{2}$
• C. $10+2\sqrt{3}$
• D. $9+3\sqrt{3}$

Answer 1

The correct option is D $9+3\sqrt{3}$

$M{N}^2=O_1{O}_2^2-O_1{P}^2$
$MN=\sqrt{4^2-2^2}$
$MN=2\sqrt{3}=O_{2}P$
$△O_1{O}_{2}P\sim △O_{1}CM$

$\Rightarrow \frac{O_{1}P}{O_{2}P}=\frac{O_{1}M}{CM}\Rightarrow \frac{2}{2\sqrt{3}}=\frac{3}{CM}\Rightarrow CM=3\sqrt{3}=CE$

Now $A{B}^2+B{C}^2=A{C}^2$

$\Rightarrow (x+3{)}^2+(3+3\sqrt{3}{)}^2=(x+3\sqrt{3}{)}^2$

$\Rightarrow x^2+6x+9+9+27+18\sqrt{3}=x^2+6\sqrt{3}+27$

$\Rightarrow x=\frac{18(\sqrt{3}+1)}{6(\sqrt{3}-1)}=6+3\sqrt{3}\Rightarrow AB=3+6+3\sqrt{3}=9+3\sqrt{3}$.