Question

In $\Delta ABC$, ratio of area of incircle to area of $\Delta ABC=$

• A. $\pi$ : Tan $\frac{A}{2}$ Tan $\frac{B}{2}$ Tan $\frac{C}{2}$
• B. $\pi$ : Cot $\frac{A}{2}$ Cot $\frac{B}{2}$ Cot $\frac{C}{2}$
• C. 1:1
• D. $\pi :r$

Answer 1

Answer: (B)

$\pi$ : Cot $\frac{A}{2}$ Cot $\frac{B}{2}$ Cot $\frac{C}{2}$

$r=(s-a)\tan A/2=(s-b)\tan B/2=(s-c)\tan C/2=\frac{△}{S}\Rightarrow \frac{r^3}{(s-a)(s-b)(s-c)}=\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}\Rightarrow \frac{r^3.s}{s(s-a)(s-b)(s-c)}=\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}\Rightarrow \frac{r^3.s}{{△}^2}=\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}\Rightarrow r^2=△\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}$

$\Rightarrow \frac{\pi r^2}{△}=\pi \times \tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}\Rightarrow \frac{\pi r^2}{△}=\pi :\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$