The perimeter of a rectangle is 68 cm. If the area of the rectangle is 240 $$cm^2$$, then what is the length of each of its diagonals?

Let's assume the length and breadth of a rectangle are 'L' and 'B'.

The perimeter of a rectangle is 68 cm.

2(L+B) = 68

L+B = 34    Eq.(i)

If the area of the rectangle is 240 $$cm^2$$.

$$L \times B$$ = 240    Eq.(ii)

In Eq.(i) made a square on both of sides.

$$\left(L+B\right)^2=34^2$$

$$L^2+B^2\ +2LB=1156$$

Put Eq.(ii) in the above equation.

$$L^2+B^2\ +2\times240=1156$$

$$L^2+B^2\ +480=1156$$

$$L^2+B^2\ = 1156-480$$

$$L^2+B^2\ = 676$$    Eq.(iii)

Length of each of its diagonals = $$\sqrt{\ L^2+B^2}$$

Put Eq.(iii) in the above equation.

= $$\sqrt{676}$$

= 26 cm