Question : In a right-angled triangle $\Delta PQR, PR$ is the hypotenuse of length 20 cm, $\angle PRQ = 30^{\circ}$, the area of the triangle is:
Option 1: $50\sqrt{3}\text{ cm}^{2}$
Option 2: $100\sqrt{3}\text{ cm}^{2}$
Option 3: $25\sqrt{3}\text{ cm}^{2}$
Option 4: $\frac{100}{\sqrt{3}}\text{ cm}^{2}$
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Correct Answer: $50\sqrt{3}\text{ cm}^{2}$
Solution :
$\sin 30^\circ=\frac{PQ}{PR}$
⇒ $\frac{1}{2}=\frac{PQ}{20}$
⇒ $PQ=\frac{20}{2}=10$ cm
$\cos 30^\circ=\frac{QR}{PR}$
⇒ $\frac{\sqrt3}{2}=\frac{QR}{20}$
⇒ $QR=\frac{20\sqrt3}{2}=10\sqrt3$ cm
Area of $\triangle PQR=\frac{1}{2}\times PQ \times QR$
$=\frac{1}{2}\times 10 \times10\sqrt3$
$=50\sqrt3\text{ cm}^2$
Hence, the correct answer is $50\sqrt3\text{ cm}^2$.
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