Question

The sides of a right angled triangle are in A.P. If the area of the triangle is $24sq$. units, then find the length of its smallest side.

Answer 1

Answer: 6

Let the sides of the triangle be $a-d,a,a+d$.
Then, area $=24$
$\Rightarrow \frac{1}{2}a(a-d)=24$
$\Rightarrow a(a-d)=48...(i)$
$a^2+(a-d{)}^2=(a+d{)}^2$
$\dots$ [Pythagoras theorem]

$\Rightarrow 2a^2-2ad+d^2=a^2+2ad+d^2$
$\Rightarrow a^2=4ad$
$\Rightarrow a=4d...(ii)$
Solving (i) and (ii), we get
$a=8,d=2$
$\Rightarrow a-d=8-2=6$