Question

Let $ABC$ be an isosceles triangle with $BC$ as its base. Then, $r{r}_1=$

• A. $a^2$
• B. $\frac{a^2}{2}$
• C. $R^2{\sin }^{2}A$
• D. $R^2{\sin }^{2}2B$

Answer 1

Answer: (C)

We have, $ABC$ be an isosceles triangle, $BC$ as its base.
$\therefore \angle B=\angle C$

We know that,
$r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
$r_1=4R\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$
$\therefore r{r}_1=16R^2{\sin }^2\frac{A}{2}\sin \frac{B}{2}\cos \frac{B}{2}\sin \frac{C}{2}\cos \frac{C}{2}$
$=4R^2{\sin }^2\frac{A}{2}\cdot \sin B\sin C$
$=4R^2{\sin }^2\frac{A}{2}\cdot {\sin }^{2}B[\because \angle B=\angle C]$
$=4R^2{\sin }^2\frac{A}{2}{\sin }^2\left(\frac{\pi }{2}-\frac{A}{2}\right)$
$=4R^2{\sin }^2\frac{A}{2}{\cos }^2\frac{A}{2}$
$=R^2{\left(2\sin \frac{A}{2}\cos \frac{A}{2}\right)}^2=R^2{\sin }^{2}A$