Question

In a triangle ABC, $I$ be the incentre and $I_1,I_2$ and $I_3$
be the excetres. If 5 and 2 be the radius of circum circle and incircle
respectively. Then $I{I}_1.I{I}_2.I{I}_3=$

• A. 900
• B. 600
• C. 400
• D. 800

Answer 1

The correct option is D 800

there's some standard result that says:

$I{I}_1=4Rsin\frac{A}{2}$

$I{I}_2=4Rsin\frac{B}{2}$

$I{I}_3=4Rsin\frac{C}{2}$

now $I{I}_1.I{I}_2.I{I}_3$ $=$ $\left(4R{)}^3\right[sin\frac{A}{2}.sin\frac{B}{2}.sin\frac{C}{2}]$ -----equ($1$)

we know that $r=4R.sin\frac{A}{2}.sin\frac{B}{2}.sin\frac{C}{2}$ ------equ($2$)

using equ($1$) and equ($2$)

we get $I{I}_1.I{I}_2.I{I}_3=16R^{2}r$

now given $R=5$ and $r=2$

$I{I}_1.I{I}_2.I{I}_3=16\times (5{)}^2\times 2$

$I{I}_1.I{I}_2.I{I}_3=800$