Question

If $A,A_1,A_2,A_3$ are the areas of the inscribed and escribed circles of a $\Delta ABC,$ then which of the following relations hold true:

• A. $\sqrt{A_1}+\sqrt{A_2}+\sqrt{A_3}=\sqrt{\pi }(r_1+r_2+r_3)$
• B. $\frac{1}{\sqrt{A_1}}+\frac{1}{\sqrt{A_2}}+\frac{1}{\sqrt{A_3}}=\frac{1}{\sqrt{A}}$
• C. $\frac{1}{\sqrt{A_1}}+\frac{1}{\sqrt{A_2}}+\frac{1}{\sqrt{A_3}}=\frac{s^2}{\sqrt{\pi }{r}_1{r}_2{r}_3}$
• D. $\sqrt{A_1}+\sqrt{A_2}+\sqrt{A_3}=\sqrt{\pi }(4R+r)$

Answer 1

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

$\sqrt{A_1}+\sqrt{A_2}+\sqrt{A_3}=\sqrt{\pi }(4R+r)$

We have $A=\pi r^2,A_1=\pi r_1^2,A_2=\pi r_2^2$ and $A_3=\pi r_3^2$
$\therefore \sqrt{A_1}+\sqrt{A_2}+\sqrt{A_3}=\sqrt{\pi }(r_1+r_2+r_3)$
Also, $\frac{1}{\sqrt{A_1}}+\frac{1}{\sqrt{A_2}}+\frac{1}{\sqrt{A_3}}=\frac{1}{\sqrt{\pi }}(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3})=\frac{1}{\sqrt{\pi }}\left(\frac{1}{r}\right)=\frac{1}{\sqrt{\pi r^2}}=\frac{1}{\sqrt{A}}(\because \text{We know}\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{r})\Rightarrow \frac{1}{\sqrt{A_1}}+\frac{1}{\sqrt{A_2}}+\frac{1}{\sqrt{A_3}}=\frac{1}{\sqrt{A}}\cdots \left(i\right)$
Now, $\frac{s^2}{\sqrt{\pi }{r}_1{r}_2{r}_3}=\frac{s^2}{\sqrt{\pi }\frac{{\Delta }^3}{(s-a)(s-b)(s-c)}}=\frac{s^2(s-a)(s-b)(s-c)}{{\Delta }^3\sqrt{\pi }}=\frac{s}{\Delta \sqrt{\pi }}=\frac{1}{r\sqrt{\pi }}=\frac{1}{\sqrt{\pi r^2}}=\frac{1}{\sqrt{A}}$
Hence from equation $\left(i\right)$, we have $\frac{s^2}{\sqrt{\pi }{r}_1{r}_2{r}_3}=\frac{1}{\sqrt{A}}=\frac{1}{\sqrt{A_1}}+\frac{1}{\sqrt{A_2}}+\frac{1}{\sqrt{A_3}}$
Now, $\sqrt{A_1}+\sqrt{A_2}+\sqrt{A_3}=\sqrt{\pi }(r_1+r_2+r_3)\text{We know,}{r}_1+r_2+r_3=(4R+r)$
Hence, $\sqrt{A_1}+\sqrt{A_2}+\sqrt{A_3}=\sqrt{\pi }(4R+r)$