Question

If $A_1,A_2,A_3$ be the area of the incircle and exercise, then $\frac{1}{{\sqrt{A}}_1}+\frac{1}{{\sqrt{A}}_2}+\frac{1}{{\sqrt{A}}_3}$ is equal to

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is A

Area of circle=$\pi \times [radius{]}^2$
$\therefore A=\pi r^2,A_1=\pi r_1^2,A_2=\pi r_2^2,A_3=\pi r_3^2$
$\therefore \frac{1}{{\sqrt{A}}_1}+\frac{1}{{\sqrt{A}}_2}+\frac{1}{{\sqrt{A}}_3}=\frac{1}{r_1\sqrt{\pi }}+\frac{1}{r_2\sqrt{\pi }}+\frac{1}{r_3\sqrt{\pi }}$
=$\frac{1}{\sqrt{\pi }}[\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}]=\frac{1}{\sqrt{\pi }}[\frac{s-a}{△}+\frac{s-b}{△}+\frac{s-c}{△}]$
=$\frac{1}{\sqrt{\pi }}[\frac{3s-(a+b+c)}{△}]=\frac{1}{\sqrt{\pi }}.\frac{3s-2s}{△}$
=$\frac{1}{\sqrt{\pi }}.\frac{S}{△}=\frac{1}{r\sqrt{\pi }}=\frac{1}{\sqrt{\pi r^2}}=\frac{1}{\sqrt{A}}$