Let $A B C$ be an isosceles triangle with $B C$ as its base. Then, $r_1=$

• A

  $a^2$
• B

  $\frac{a^2}{2}$
• Correct answer
  C

  $R^2 \sin ^2 A$
• D

  $R^2 \sin ^2 2 B$

Solution

We have, $A B C$ be an isosceles triangle, $B C$ as its base. $$ \therefore \quad \angle B=\angle C $$
We know that, $$ r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} $$ $$ \begin{aligned} r_1 & =4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \\ \therefore \quad r_1 & =16 R^2 \sin ^2 \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2} \\ & =4 R^2 \sin ^2 \frac{A}{2} \cdot \sin B \sin C \\ & =4 R^2 \sin ^2 \frac{A}{2} \cdot \sin ^2 B \\ & =4 R^2 \sin ^2 \frac{A}{2} \sin ^2\left(\frac{\pi}{2}-\frac{A}{2}\right) \quad[\because \angle B=\angle C] \\ & =4 R^2 \sin ^2 \frac{A}{2} \cos ^2 \frac{A}{2} \\ & =R^2\left(2 \sin \frac{A}{2} \cos \frac{A}{2}\right)^2=R^2 \sin ^2 A \end{aligned} $$

Hence, the correct answer is option: 'C'.