Question

The radii $r_1,r_2,r_3$ of escribed circle of a triangle ABC are in H.P. If its area is $24c{m}^2$ and its perimeter is $24cm$, find the length of its sides.

• A. True
• B. False

Answer 1

The correct option is A True

Let $a,b,c$ be the length of the sides of the triangle, $s$ be the semi perimeter and $\Delta$ be the area.
$r_1=\frac{\Delta }{(s-a)}=\frac{24}{(12-a)}{r}_2=\frac{\Delta }{(s-b)}=\frac{\Delta }{(12-b)}{r}_3=\frac{\Delta }{s-c}=\frac{24}{(12-c)}$
Since $r_1,r_2,r_3$ are in H.P.
$\Rightarrow \frac{1}{r_1},\frac{1}{r_2},\frac{1}{r_3}$ are in A.P.
$\Rightarrow \frac{2}{r_1}=\frac{1}{r_1}+\frac{1}{r_3}$ or $2\times \frac{12-b}{24}=\frac{12-a}{24}+\frac{12-c}{24}$
Or $a+c=2b$ .....(1)
Also, $a+b+c=24$ ....(2)
From (1) and (2), we get $b=8cm$ and $a+c=16cm$
Now $\Delta \sqrt{s(s-a)(s-b)(s-c)}$
$\Rightarrow 24=\sqrt{12(12-a)(12-b)(12-c)}\Rightarrow 24\times 24=12(12-a)\times 4\times (12-16+a)\Rightarrow 12=(12-a)(a-4)\Rightarrow a^2-16a+60=0\Rightarrow a=6or10$.
When $a=6,c=10$ and when $a=10,c=6$
Hence sides are $6,8,10$ or $10,8,6$