Correct Answer: $\frac{m^{2} - p^{2}}{4}$ sq. units

Solution : Given that, the perimeter of a rhombus is $2p$ units, and the sum of the lengths of the diagonals is $m$ units.
Let $d_1$ and $d_2$ be the diagonal of the rhombus.
The perimeter of a rhombus $=2\sqrt{d_1^2+d_2^2}$
⇒ $2p=2\sqrt{d_1^2+d_2^2}$
On squaring both sides,
$p^2=d_1^2+d_2^2$ ___(i)
We have,
$d_1^2+d_2^2=m$
Squaring both sides,
$(d_1+d_2)^2=m^2$
⇒ $d_1^2+d_2^2+2d_1d_2=m^2$
From equation (i),
$p^2+2d_1d_2=m^2$
⇒ $d_1d_2=\frac{m^2-p^2}{2}$
The area of the rhombus $=\frac{1}{2}d_1d_2$
$=\frac{1}{2} \times \frac{m^2-p^2}{2}$
$=\frac{m^{2} – p^{2}}{4}$
Hence, the correct answer is $\frac{m^{2} – p^{2}}{4}$ sq. units.