First slide

Properties of triangles

Question

The angle of a right angled triangle are $A . P .$ The ratio of the in-radius and the perimeter is

Moderate

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

Solution

Let ABC be the right angled triangle whose angles
are in $A . P .$ Then,
Now, $A + B + C = 180^{\circ} \Rightarrow 3 B = 180^{\circ} \Rightarrow B = 60^{\circ}$
So, let the angles be $A = 30^{\circ}, B = 60^{\circ}$ and $C = 90^{\circ}$
$\begin{array}{l} ∴ \frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{c}{\sin 90^{\circ}} = 2 R \\ \Rightarrow a = R, b = \sqrt{3} R and c = 2 R \end{array}$
Also,
$\begin{array}{l} Δ = \frac{1}{2} a b \sin 90^{\circ} = \frac{1}{2} a b = \frac{\sqrt{3}}{2} R^{2} \\ ∴ \frac{r}{s} = \frac{Δ}{s^{2}} \\ \Rightarrow \frac{r}{s} = \frac{\sqrt{3} R^{2}}{{(\frac{2}{\sqrt{3} R + 2 R)^{2}} - \frac{\sqrt{3}}{2} \times \frac{4}{(2})}^{2}} = \frac{2 \sqrt{3}}{{\sqrt{3} + \frac{3}{\sqrt{3} + 3})}^{2}} \\ \Rightarrow \frac{r}{s} = \frac{2 \sqrt{3} (\sqrt{3} - 3)^{2}}{(9 - 3)^{2}} \frac{6 \sqrt{3} (\sqrt{3} - 1)^{2}}{36} = \frac{2}{\sqrt{3} (4 - 2 \sqrt{3})} \\ \Rightarrow \frac{r}{s} = \frac{2 - \sqrt{3}}{\sqrt{3} - \Rightarrow \frac{r}{2 s}} = \frac{2 - \sqrt{3}}{2 \sqrt{3}} \end{array}$

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