Question

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 :

• A. $26\sqrt{3}$
• B. $24\sqrt{2}$
• C. $24\sqrt{3}$
• D. $26\sqrt{2}$

Answer 1

The correct option is C $24\sqrt{3}$


$OC=\frac{13}{2}=6.5$
$AC=CO-AO$
$=6.5-1=5.5$

In $\Delta PAC$
$PA=\sqrt{{6.5}^2-{5.5}^2}$
$\Rightarrow PA=\sqrt{12}$
$\Rightarrow PQ=2PA=2\sqrt{12}$

Now, area of $\Delta PQB=\frac{1}{2}\times PQ\times AB$
$=\frac{1}{2}\times 2\sqrt{12}\times 12$
$=12\sqrt{12}$
$=24\sqrt{3}$ sq. units