Question

Let $ABC$ be an isosceles triangle with base $BC$. If $r$ is the radius of the circle inscribed in triangle $ABC$ and $r_1$ is the radius of the circle escribed opposite to the angle A,

then the product $r_1\times r$ can be equal to?

• A. $R^{2}si{n}^{2}A$
• B. $R^{2}si{n}^{2}2B$
• C. $\frac{1}{2}{a}^2$
• D. $\frac{a^2}{4}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $R^{2}si{n}^{2}A$
B $R^{2}si{n}^{2}2B$
D $\frac{a^2}{4}$

Since, ABC is an isosceles triangle with base $BC$.
Therefore, $B=C$ and $A-\pi -2B$
$r_{1}r=\left(4R\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}\right)\left(4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\right)$
$\Rightarrow r_{1}r=4R^2{{\sin }^2\frac{A}{2}\left(2\sin \frac{B}{2}\cos \frac{B}{2}\right)}^2=4R^2{\sin }^2\frac{A}{2}{\sin }^{2}B$
$\Rightarrow r_{1}r=4R^2{\sin }^2\left(\frac{\pi -2B}{2}\right){\sin }^{2}B=R^2{\left(2\sin B\cos B\right)}^2$
$\Rightarrow r_{1}r=R^2{\sin }^{2}2B=R^2{\sin }^2(\pi -A)=R^2{\sin }^{2}A$
$\Rightarrow r_{1}r=\frac{a^2}{4}$

Ans: A,C,D