Choose the correct answer of the following question:

WE have,

$\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}=20 \mathrm{~cm}$ and $\mathrm{BD}=24 \mathrm{~cm}$

Also, $\mathrm{BO}=\frac{\mathrm{BD}}{2}=\frac{24}{2}=12 \mathrm{~cm}$

In $\Delta \mathrm{AOB}$,

Using Pythagoras theorem

$\mathrm{AO}^{2}=\mathrm{AB}^{2}-\mathrm{BO}^{2}$

$=20^{2}-12^{2}$

$=400-144$

$\Rightarrow \mathrm{AO}^{2}=256$

$\Rightarrow \mathrm{AO}=\sqrt{256}$

$\Rightarrow \mathrm{AO}=16 \mathrm{~cm}$

$\Rightarrow \mathrm{AC}=2 \mathrm{AO}=2 \times 16=32 \mathrm{~cm}$

Now, the area of the rhombus $\mathrm{ABCD}=\frac{1}{2} \times \mathrm{AC} \times \mathrm{BD}$

$=\frac{1}{2} \times 32 \times 24$

$=384 \mathrm{~cm}^{2}$

Hence, the correct answer is option (d).