Question:
In the circle given below, let $O A=1$ unit, $O B=13$ unit and $P Q \perp O B$.
Then, the area of the triangle PQB (in square units) is:
Correct Option: , 3
Solution:
$\mathrm{OC}=\frac{13}{2}=6.5$
$\mathrm{AC}=\mathrm{CO}-\mathrm{AO}$
$=6.5-1$
$=5.5$
In $\Delta \mathrm{PAC}$
$\mathrm{PA}=\sqrt{6.5^{2}-5.5^{2}}$
$P A=\sqrt{12}$
$\Rightarrow P_{Q}=2 P A=2 \sqrt{12}$
Now, area of $\Delta \mathrm{PQB}=\frac{1}{2} \times \mathrm{PQ} \times \mathrm{AB}$
$=\frac{1}{2} \times 2 \sqrt{12} \times 12$
$=12 \sqrt{12}$
$=24 \sqrt{3}$
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