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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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1 Answer

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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 $\Delta ABC$
$r_1=\large\frac{\Delta}{s-a}$
$\Rightarrow \large\frac{24}{12-a}$
$r_2=\large\frac{\Delta}{s-b}$
$\Rightarrow \large\frac{24}{12-b}$
$r_3= \large\frac{\Delta}{s-c}$
$\Rightarrow \large\frac{24}{12-c}$
Step 2:
$r_1,r_2,r_3$ are in H.P
$\large\frac{1}{r_2}-\frac{1}{r_1}=\frac{1}{r_3}-\frac{1}{r_2}$
$\large\frac{12-b}{24}-\frac{12-a}{24}=\frac{12-c}{24}-\frac{12-b}{24}$
$a-b=b-c$-----(1)
$2b=a+c$
$a+b+c=24$-----(2)
Step 3:
From (1) and (2) we have
$3b=24$
$b=8cm$
From (1) we have
$a+c=2b$
$\qquad\;=16$
$\Rightarrow c=16-a$
Step 4:
Now area $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$
$24\times 24=12(12-a)(12-b)(12-c)$
$48=(12-a)(12-8)(12-16+a)$
$48=4(12-a)(0-4)$
$12=-a^2+160-48$
$a^2-16a+60=0$
$a^2-10a-6a+60=0$
$(a-6)(a-10)=0$
$a=6$ or $a=10$
$c=16-a$
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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