Question : In the given figure, O is the centre of the circle, OQ is perpendicular to RS, and $\angle SRT=30^{\circ}$. If $RS=10\sqrt{2}$ units, what is the value of PR2?
Option 1: $200\left ( 1+\sqrt{3} \right )$ sq. units
Option 2: $300\left ( 2+\sqrt{3} \right )$ sq. units
Option 3: $200\left ( 2+\sqrt{3} \right )$ sq. units
Option 4: $100\left ( 3+2\sqrt{3} \right )$ sq. units
Correct Answer: $200\left ( 2+\sqrt{3} \right )$ sq. units
Solution :
Here RQ = QS = $5\sqrt{2}$
$\angle$SRT = 30°
In $\triangle$ROQ,
$\angle$ORQ = 90° – 30° = 60°
⇒ OR $=\frac{5\sqrt{2}}{\cos 60º} = 10\sqrt{2} = OP$
OQ $=5\sqrt{2} \times \tan 60º = 5\sqrt{6}$
PQ = OP + OQ = $10\sqrt{2} + 5\sqrt{6}$
In $\triangle$PQR,
PR
2
= RQ
2
+ PQ
2
= $(5\sqrt{2})^2 + (10\sqrt{2} + 5\sqrt{6})^2$
= $50 + 200 + 150 + 100\sqrt{12}$
= $400 + 200\sqrt{3} = 200(2 + \sqrt{3})$ sq. units
Hence, the correct answer is $200\left ( 2+\sqrt{3} \right )$ sq. units.
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