If the perimeter of a rhombus is 40 cm and one of its diagonals is 16 cm, what is the area (in cm2) of the rhombus?

$\mathrm{AB}=\frac{40}{4}=10 \mathrm{~cm} .$

$d_{1}=\mathrm{AC}=20 \mathrm{~A}=16 \mathrm{~cm} .$

$\therefore \mathrm{OA}=8 \mathrm{~cm} .$

$\angle \mathrm{AOB}=90^{\circ}$

$\therefore$ In $\triangle \mathrm{OAB}$,

$\mathrm{OB}=\sqrt{\mathrm{AB}^{2}-\mathrm{OA}^{2}}=\sqrt{10^{2}-8^{2}}=\sqrt{100-64}$

$=\sqrt{36}=6 \mathrm{~cm} .$

$\therefore d_{2}=\mathrm{BD}=12 \mathrm{~cm} .$

$\therefore \quad$ Area of rhombus $\mathrm{ABCD}=\frac{1}{2} d_{1} d_{2}$

$=\frac{1}{2} \times 16 \times 12=96$ sq. $\mathrm{cm} .$