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Q.
The sides of a right angled triangle are in A.P. If the area of the triangle is $24 \,sq$. units, then find the length of its smallest side.
Sequences and Series
Solution:
Let the sides of the triangle be $a-d, a, a+d$.
Then, area $=24$
$ \Rightarrow \frac{1}{2} a(a-d)=24 $
$ \Rightarrow a(a-d)=48 \,\,\,...(i)$
$ a^{2}+(a-d)^{2}=(a+d)^{2}$
$\ldots$ [Pythagoras theorem]
$\Rightarrow 2 a^{2}-2 a d+d^{2}=a^{2}+2 a d+d^{2}$
$\Rightarrow a^{2}=4 a d$
$\Rightarrow a=4 d\,\,\,\, ...(ii)$
Solving (i) and (ii), we get
$a=8, d=2 $
$\Rightarrow a-d=8-2=6$
