2022/038) The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24, find the length of smallest sides

Let the sides be a -b , a , a+b

as the tringle is right angled so $(a-b)^2 + a^2 = (a+b)^2$

or $a^2-2ab + b^2 + a ^2 = a^2 + 2ab + b^2$

or   $a^2= 4ab$

or $a= 4b$

as the triangls is right angled so aea = $\frac{1}{2} a(a-b) =  \frac{1}{2} 4b(4b-b) = 6b^2 = 24$ or b= 2

the smallest sde = 3b = 6.