**Question:**

Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.

**Solution:**

Given:

Side of the rhombus $=20 \mathrm{~cm}$

Length of a diagonal $=24 \mathrm{~cm}$

We know : If $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$ are the lengths of the diagonals of the rhombus, then side of the rhombus $=\frac{1}{2} \sqrt{\mathrm{d}_{1}^{2}+\mathrm{d}_{2}^{2}}$

So, using the given data to find the length of the other diagonal of the rhombus:

$20=\frac{1}{2} \sqrt{24^{2}+\mathrm{d}_{2}^{2}}$

$40=\sqrt{24^{2}+\mathrm{d}_{2}^{2}}$

Squaring both sides to get rid of the square root sign:

$40^{2}=24^{2}+\mathrm{d}_{2}^{2}$

$\mathrm{~d}_{2}^{2}=1600-576=1024$

$\mathrm{~d}_{2}=\sqrt{1024}=32 \mathrm{~cm}$

$\therefore$ Area of the rhombus $=\frac{1}{2}(24 \times 32)=384 \mathrm{~cm}^{2}$