In $\triangle A B C$, if $A, B, C$ are in arithmetic progression, $\Delta=\frac{\sqrt{3}}{2}$ and $r_1…

In $\triangle A B C$, if $A, B, C$ are in arithmetic progression, $\Delta=\frac{\sqrt{3}}{2}$ and $r_1 r_2=r_3 r$, then $R=$
  1. $\sqrt{3}$
  2. 2
  3. 1
  4. $\sqrt{2}$

Solution

$\begin{aligned} & A, B, C \text { are in AP, } r_1 r_2=r_3 r \\ & \Rightarrow \frac{\Delta}{(s-a)} \cdot \frac{\Delta}{(s-b)}=\frac{\Delta}{(s-c)} \cdot \frac{\Delta}{s} \\ & \Rightarrow \frac{(s-a)(s-b)}{s(s-c)}=1 \Rightarrow \tan ^2 \frac{C}{2}=1 \\ & C=90^{\circ}, A+B+C=180^{\circ} \\ & \Rightarrow 3 B=180^{\circ} \Rightarrow B=60^{\circ} \Rightarrow A=30^{\circ} \\ & C=2 R, B C=2 R \sin 30^{\circ} \\ & a=R, b=A C=2 R \sin 60^{\circ}=\sqrt{3} R \\ & \Delta=\frac{a b}{2}=\frac{\sqrt{3} R^2}{2}=\frac{\sqrt{3}}{2} \Rightarrow R^2=1 \Rightarrow R=1\end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

Practice more Properties of Triangles questions on Aicharya