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. $R^{2} \sin ^{2} 2 B$

Answer 1

Answer: (C)

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

We know that,
$r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
$r_{1} =4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
$\therefore \, r r_{1} =16 R^{2} \sin ^{2} \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2} $
$=4 R^{2} \sin ^{2} \frac{A}{2} \cdot \sin B \sin C $
$=4 R^{2} \sin ^{2} \frac{A}{2} \cdot \sin ^{2} B \, [\because \angle B=\angle C] $
$=4 R^{2} \sin ^{2} \frac{A}{2} \sin ^{2}\left(\frac{\pi}{2}-\frac{A}{2}\right)$
$=4 R^{2} \sin ^{2} \frac{A}{2} \cos ^{2} \frac{A}{2}$
$=R^{2}\left(2 \sin \frac{A}{2} \cos \frac{A}{2}\right)^{2}=R^{2} \sin ^{2}\, A$