In the circle given below,

Question:

In the circle given below, let $\mathrm{OA}=1$ unit, $\mathrm{OB}=13$ unit and $\mathrm{PQ} \perp \mathrm{OB}$. Then, the area of the triangle PQB (in square units) is

  1. $24 \sqrt{2}$

  2. $24 \sqrt{3}$

  3. $26 \sqrt{3}$

  4. $26 \sqrt{2}$


Correct Option: , 2

Solution:

$\mathrm{PA}=\mathrm{AQ}=\lambda$

$\mathrm{OA} \cdot \mathrm{AB}$

$=\mathrm{AP} \cdot \mathrm{AQ}$.3

$\Rightarrow 1.12=\lambda . \lambda$

$\Rightarrow \lambda=2 \sqrt{3}$

Area $\triangle \mathrm{PQB}=\frac{1}{2} \times 2 \lambda \times \mathrm{AB}$

$\Delta=\frac{1}{2} \cdot 4 \sqrt{3} \times 12$

$=24 \sqrt{3}$

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