The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24, then what is the length of its smallest side?

$\text{Given: sides of a right triangle are in A.P.}$

$\text{Let the sides be } a-d,\ a,\ a+d$

$\text{Area} = 24$

$\text{To find: the smallest side } (a-d)$

$\text{A.P.: equal gap } d \text{ between consecutive sides}$ $\text{Right angle} \Rightarrow \text{hypotenuse} = a+d\ (\text{largest})$ $\text{The two legs: } a-d\ (\text{smallest}),\ a$ $\text{Plan: } 1.\ \text{Pythagoras} \Rightarrow \text{relation between } a, d$ $2.\ \text{Area} \Rightarrow d$ $3.\ \text{smallest side} = a-d$ $(a-d)^2 + a^2 = (a+d)^2$ $a^2 - 2ad + d^2 + a^2 = a^2 + 2ad + d^2$ $2a^2 - 2ad + d^2 = a^2 + 2ad + d^2$ $a^2 - 4ad = 0$ $a(a - 4d) = 0$ $a \neq 0 \;\Rightarrow\; a = 4d$ $a-d = 3d,\quad a = 4d,\quad a+d = 5d$ $\Rightarrow\ \text{sides are } 3d,\ 4d,\ 5d$ $\text{Area} = \dfrac{1}{2}\,(a-d)(a)$ $\text{Area} = \dfrac{1}{2}(3d)(4d)$ $= 6d^2$ $6d^2 = 24$ $d^2 = 4$ $d = 2$ $\text{smallest side} = a-d = 3d = 3(2) = 6$ $6^2 + 8^2 = 100 = 10^2\ \checkmark$ $\text{smallest side} = \boxed{6}$