# 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 $24\sqrt{3}$

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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