Question

If a triangle ABC with side $a=12$ units is inscribed in a circle of radius 10 units, then in-radius of triangle $ABC$ can be

• A. 4 units
• B. 8 units
• C. 5 units
• D. 2 units

Answer 1

Answer: (A), (D)

The correct options are
A 4 units
D 2 units

Use the relation between 'r' and 'R' for any triangle.
$r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=2R\sin \frac{A}{2}(\cos \frac{B-C}{2}-\cos \frac{B+C}{2})=2R\sin \frac{A}{2}(\cos \frac{B-C}{2}-\sin \frac{A}{2})....\left(1\right)$
Given that $a=12$ and $R=10$. So we get $\sin A=\frac{a}{2R}=\frac{3}{5}$
$\Rightarrow \cos A=\frac{4}{5}$
$\Rightarrow \sin \frac{A}{2}=\sqrt{\frac{1-\cos A}{2}}=\frac{1}{\sqrt{10}}....\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, $r$ will be maximum when $\cos \frac{B-C}{2}$ takes maximum value, which is $1$.
So, we get maximum value of $r=2\times 10\times \frac{1}{\sqrt{10}}(1-\frac{1}{\sqrt{10}})=2(\sqrt{10}-1)$
Since $3<\sqrt{10}<3.2$, we get maximum value of $r$ is such that $4<r_{max}<4.4$
So any possible value of $r$ has to be less than $r_{max}$.
$4$ and $2$ are possible.