Correct Answer: $\frac{1}{4}\left(\mathrm{~L}^2-4 \mathrm{P}^2\right)$

Solution : Given: The sum of the diagonals of a rhombus is $L$.
Perimeter = $4P$
⇒ Sides = $\frac{4P}{4}$ = $P$
Now, the sum of the diagonals = $L$
Let one of the diagonal be $d_1$ and another diagonal be $d_2$.
⇒ $(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2 = P^2 $
⇒ $d_1^2+d_2^2=4P^2$
⇒ $(d_1+d_2)^2-2d_1d_2=4P^2$ as $[a^2 + b^2 = ( a + b)^2 - 2ab]$
⇒ $L^2-2d_1d_2=4P^2$ as $[d_1+d_2=L]$
⇒ $2d_1d_2=L^2-4P^2$
⇒ $d_1d_2=\frac{L^2-4P^2}{2}$
$\therefore$ Area of the Rhombus
= $\frac{1}{2} × d_1× d_2 $
= $\frac{1}{2} × \frac{L^2-4P^2}{2}$
= $\frac{1}{4} (L^2-4P^2)$
Hence, the correct answer is $\frac{1}{4} (L^2-4P^2)$.