Question

The angles of a right-angled triangle are in $AP$. Then the ratio of the inradius and the perimeter is?

• A. $(2-\sqrt{3}):2\sqrt{3}$
• B. $1:8\sqrt{3}(2+\sqrt{3}):$
• C. $(2+\sqrt{3}):4\sqrt{3}$
• D. none of these

Answer 1

The correct option is A $(2-\sqrt{3}):2\sqrt{3}$

A,B,C must be ${30}^{\circ },{60}^{\circ },{90}^{\circ }$ to be in A.P.

$\frac{\sin 60°}{a}=\frac{\sin 90°}{b}=\frac{\sin 30°}{c}=K$ (Sine rule)

$a=\frac{\sqrt{3}}{2k}$ , $b=\frac{1}{k}$, $c=\frac{1}{2k}$

$s=\frac{a+b+c}{2}\Rightarrow \frac{\frac{\sqrt{3}}{2k}+\frac{1}{k}+\frac{1}{2k}}{2}\Rightarrow \frac{\frac{\sqrt{3}+2+1}{2k}}{2}$

$s=\frac{3+\sqrt{3}}{4k}$

Now,

Inradius, $r=(s-a)$$\tan \left(\frac{A}{2}\right)$

$r=(\frac{3+\sqrt{3}}{4k}-\frac{\sqrt{3}}{2k})\tan \left(30°\right)$

$r=\frac{3+\sqrt{3}-2\sqrt{3}}{4k}\times \frac{1}{\sqrt{3}}$

$r=\frac{3-\sqrt{3}}{4k}\times \frac{1}{\sqrt{3}}$

$\frac{r}{s}\Rightarrow \frac{3-\sqrt{3}}{4k\times \sqrt{3}}\times \frac{4k}{3+\sqrt{3}}\Rightarrow \frac{(3-\sqrt{3})(3-\sqrt{3})}{\sqrt{3(3+\sqrt{3})}(3-\sqrt{3})}$

$\frac{r}{s}=\frac{{\left(3\right)}^2+{\left(\sqrt{3}\right)}^2-2\times 3\times \sqrt{3}}{\sqrt{3}({\left(3\right)}^2-{\left(\sqrt{3}\right)}^2)}\Rightarrow \frac{9+3-6\sqrt{3}}{\sqrt{3}\times 6}$

$\frac{r}{s}=\frac{2-\sqrt{3}}{\sqrt{3}}$

Hence, $(2-\sqrt{3}):2\sqrt{3}$ is the correct answer.