Question

In the circle given below, let $OA=1$ unit, $OB=13$ unit and $PQ$ perpendicular to $OB$. 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

$24\sqrt{3}$

Explanation For The Correct Option:

Finding the area of the triangle $PQB$:

Given that $PQ\perp OB$

We know that, perpendicular bisector of a chord passes through the center and they are equidistant.

Let, $PA=AQ=\lambda$

$$\begin{gathered} PQ=PA+AQ \\ =\lambda +\lambda \\ PQ=2\lambda \dots \dots \left(i\right) \end{gathered}$$

According to the property of circles,

$$\begin{gathered} \Rightarrow OA.AB=AP.AQ \\ \Rightarrow 1\times 12=\lambda .\lambda [\text{given}OA=1\&OB=13] \\ \Rightarrow \lambda =2\sqrt{3} \end{gathered}$$

Area of triangle $PQB$ $=\frac{1}{2}\times \left(\text{base}\right)\times \left(\text{height}\right)$

$$\begin{gathered} \Rightarrow \frac{1}{2}\times PQ\times AB=\frac{1}{2}\times 2\lambda \times 12[\text{from equation}(i\left)\right] \\ =\frac{1}{2}\times 4\sqrt{3}\times 12[\lambda =2\sqrt{3}] \\ =24\sqrt{3} \end{gathered}$$

Hence, option C is correct .