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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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The radii $r_1,r_2,r_3$ of described circle of a $\Delta$le ABC are in harmonic progression.If its area is 24sq .cm,and its perimeter is 24,find the lengths of its sides

$\begin{array}{1 1}(a)\;a=6,b=8,c=10&(b)\;a=7,b=9,c=11\\(c)\;a=0,b=2,c=3&(d)\;None\;of\;these\end{array}$

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Step 1:
Let $s$ be the semi-perimeter of $\Delta$ABC then $s=12cm$
Given :
$\Delta =24$sq.cm
Let $a,b,c$ be the lengths of sides of $\Delta ABC$
$\Rightarrow \large\frac{24}{12-a}$
$\Rightarrow \large\frac{24}{12-b}$
$r_3= \large\frac{\Delta}{s-c}$
$\Rightarrow \large\frac{24}{12-c}$
Step 2:
$r_1,r_2,r_3$ are in H.P
Step 3:
From (1) and (2) we have
From (1) we have
$\Rightarrow c=16-a$
Step 4:
Now area $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$
$24\times 24=12(12-a)(12-b)(12-c)$
$a=6$ or $a=10$
When $a=6\Rightarrow c=10$
When $a=10\Rightarrow c=6$
$\therefore a=6cm,b=8cm,c=10cm$
answered Oct 21, 2013 by sreemathi.v
edited Mar 3, 2014 by meenakshi.p

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