If the sum of the diagonals of a rhombus is 'L' and the perimeter is '4P', find the area of the rhombus.

the perimeter of rhombus is '4P'.

the perimeter of rhombus = $$4\times side$$ = 4P

So side = P

Let's assume the diagonals are 'y' and 'z'.

As we know the sum of the squares of the sides equals the sum of the squares of the diagonal.

$$y^2+z^2\ =\ 4P^2$$    Eq.(i)

If the sum of the diagonals of a rhombus is 'L'.

y+z = L

If we apply square on both sides.

$$\left(y+z\right)^2=L^2$$

$$y^2+z^2+2yz=L^2$$

$$2yz=L^2-(y^2+z^2)$$

Put Eq.(i) in the above equation.

$$2yz=L^2-4P^2$$

$$yz=\frac{\left(L^2-4P^2\right)}{2}$$    Eq.(ii)

Area of the rhombus = $$\frac{1}{2}\times\ product\ of\ diagonals$$

= $$\frac{1}{2}yz$$

Put Eq.(ii) in the above equation.

= $$\frac{1}{2}\times \frac{\left(L^2-4P^2\right)}{2}$$

= $$\frac{\left(L^2-4P^2\right)}{4}$$