In the figure given below, O is the center of the circle. OQ is perpendicular to RS and ∠SRT = 30°. If RS = 10√2, then what is the value of PR2?
Explanation:
RQ = QS = 1/2 × 10√2 = 5√2
∴ ∠ORQ = 90 - 30 = 60°
∆ORQ is a 30°-60°-90° triangle
∴ Hypotenuse will be twice the side opposite to 30° ⇒ OR = 10√2 = radius ⇒ OQ = √3/2 × OR = 5√6
In ∆PRQ, PR2 = PQ2 + RQ2 ⇒ PR2 = (10√2 + 5√6)2 + (5√2)2 ⇒ PR2 = 200 + 150 + 100√12 + 50 ⇒ PR2 = 400 + 200√3 = 200(2 + √3)
Hence, option (c).
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