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In the circle given below,let OA=1 unit, OB=13 unit and PQ\perp OB. Then, the area of the triangle PQB (in square units ) is :
 
Option: 1 26\sqrt{2}
Option: 2 26\sqrt{3}
Option: 3 24\sqrt{2}  
Option: 4 24\sqrt{3}

Answers (1)

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To solve this concept, we will use two concepts that we have learnt in class 9 or 10.

The perpendicular from the centre of a circle to a chord bisects the chord.

If two chords of a circle intersect each other, then the products of the lengths of their segments are equal.

PQ and OB are the chords of the circle

PQ\perp OB

Given OA = 1 unit, OB = 13 unit, so, AB = 12 unit

\text{Let PA = AQ }=\lambda

\\\mathrm{OA\cdot OB =PA\cdot AQ}\\\Rightarrow 1\times 12=\lambda\times \lambda\\\Rightarrow \lambda=2\sqrt3\\

\\\text { Area } \Delta \mathrm{PQB}=\frac{1}{2} \times 2 \lambda \times \mathrm{AB} \\ \Delta=\frac{1}{2} \cdot 4 \sqrt{3} \times 12 \\ =24 \sqrt{3}

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himanshu.meshram

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