SSC MTS 2022 PYQ — Question 73

The correct answer is option 4 i.e.  $\frac{1}{4}(L^2-4P^2)$

Sum of the diagonals of the rhombus = L

Perimeter of the rhombus = 4P

So,

The side of the rhombus = P

Now,

Let one diagnol be L₁ and second diagnol be L₂ - 

Therefore,

$\left ( \frac{L_{1}^{2}}{4} \right )+\left ( \frac{L_2^{2}}{4} \right )= P^{2}$

 ⇒ $L_{1}^{2}+L_{2}^{2}=4P^{2}$

Now,

$L=L_{1}+L_{2}$

⇒ $L^{2}=L_{1}^{2}+L_{2}^{2}+2L_{1}L_{2}$

So,

 ⇒ $L^{2}=4P^{2}+2L_{1}L_{2}$

 ⇒ $L^{2}-4P^{2}=2L_{1}L_{2}$

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

Now,

Area of the rhombus = $\frac{1}{2}L_{1}L_{2}$ = $\frac{1}{2}\times \frac{1}{2}\left ( L^{2}-4P^{2} \right )$ = $\frac{1}{4}\left ( L^{2}-4P^{2} \right)$