Q68 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
If $A B C$ is an isosceles triangle with base $B C$, then $r r_1=$
- A$R^2 \cos ^2 A$
- B$\frac{a^2}{2}$
- C$\frac{r}{R}$
- Correct answerD$R^2 \sin ^2 A$
Solution
$r=\frac{\Delta}{s}, r_1=\frac{\Delta}{s-a}$
$\begin{aligned} & r r_1=\frac{\Delta^2}{s(s-a)} \\ & =\frac{s(s-a)(s-b)(s-c)}{s(s-a)}\end{aligned}$ $=(s-b)(s-c)=(s-b)^2 \quad(\because \mathrm{~b}=\mathrm{c})$ $=\frac{(2 s-2 b)^2}{4}$ $=\frac{(a+b+c-2 b)^2}{4}=\frac{a^2}{4} \quad(\because b=c)$ $=R^2 \sin ^2 A$
$\begin{aligned} & r r_1=\frac{\Delta^2}{s(s-a)} \\ & =\frac{s(s-a)(s-b)(s-c)}{s(s-a)}\end{aligned}$ $=(s-b)(s-c)=(s-b)^2 \quad(\because \mathrm{~b}=\mathrm{c})$ $=\frac{(2 s-2 b)^2}{4}$ $=\frac{(a+b+c-2 b)^2}{4}=\frac{a^2}{4} \quad(\because b=c)$ $=R^2 \sin ^2 A$
Hence, the correct answer is option: 'D'.