Question

In a triangle $ABC,$ the ratio of area of the in-circle is to the area of the triangle itself is

• A. $\pi :\cot \frac{A}{2}.\cot \frac{B}{2}.\cot \frac{C}{2}$
• B. $2\pi :\cot \frac{A}{2}.\cot \frac{B}{2}.\cot \frac{C}{2}$
• C. $3\pi :\cot \frac{A}{2}.\cot \frac{B}{2}.\cot \frac{C}{2}$
• D. None of these

Answer 1

The correct option is A $\pi :\cot \frac{A}{2}.\cot \frac{B}{2}.\cot \frac{C}{2}$

$\frac{areaofcircle}{areaoftriangle}=\frac{\pi r^2}{\Delta }=\frac{\pi }{\Delta }.\frac{{\Delta }^2}{s^2}=\pi .\frac{\Delta }{s^2}$ ....(1) $(\because r=\frac{\Delta }{s})$

Now $\cot \frac{A}{2}.\cot \frac{B}{2}.\cot \frac{C}{2}={[\frac{s(s-a)}{(s-b)(s-c)}.\frac{s(s-b)}{(s-a)(s-c)}.\frac{s(s-c)}{(s-b)(s-a)}]}^{1/2}$

$={\left[\frac{s^3}{(s-a)(s-b)(s-c)}\right]}^{1/2}={\left[\frac{s^4}{s(s-a)(s-b)(s-c)}\right]}^{1/2}=\frac{s^2}{\Delta }$

Now $\frac{\pi }{\cot \frac{A}{2}.\cot \frac{B}{2}.\cot \frac{C}{2}}=\frac{\pi }{s^2/\Delta }=\frac{\pi \Delta }{s^2}$ ...(2)

From (1) and (2), we get required ratios as

$\pi :\cot \frac{A}{2}.\cot \frac{B}{2}.\cot \frac{C}{2}$