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

$\frac{1}{4}(L^2-4P^2)$

We know that,

(d₁/2)² + (d₂/2)² =  (side)² 

Area  = 1/2 × d₁× d₂ 

a² + b² = ( a + b)² - 2ab

Where, d₁ and d₂ are diagonals of the rhombus respectively

And, a and b are 1st and 2nd term respectively

We have,

Sum of the diagonals of a rhombus = L

Perimeter = 4P

According to the question,

Perimeter = 4P 

= 4× Side = 4 P 

= Side = P 

Now, sum of the diagonals = L

And, let us assume one diagonal be d₁ and another diagonal be d₂ 

= (d₁/2)² + ( d₂/2)² = P²

= d₁² + d₂² = 4 P² 

= ( d₁ + d₂)² - 2d₁d₂ = 4P²             [a² + b² = ( a + b)² - 2ab ]

= L² - 2d₁d₂  = 4P²                              [As, d₁ + d₂ = L]

= 2d₁d₂ = L² - 4 P²

= d₁d₂ = ( L² - 4P² )/2 

Area of the Rhombus = \(\frac{1}{2}\) × d₁d₂  = \(\frac{1}{2}\) × (L² - 4P²)/2  = \(\frac{1}{4}\)  ( L² - 4P²)