In the circle given below, let OA = 1 unit, OB = 13 unit and PQ ⊥ OB . Then, the area of the triangle PQB (in square units) is :

Question

Quantrex Academy · Accepted Answer

Correct answer: A. 24 3

Since x-axis is normal to the circle at $O$ , and normal to the circle passes through its centre. So, $OB$ is a diameter.

Now, perpendicular from centre of circle to chord bisects the chord, so $PA = AQ = \lambda ( Let )$

Also, product of segments of chords in a circle is equal i.e. $OA \cdot AB = AP \cdot AQ$

$⇒ 1 \cdot 12 = \lambda \cdot \lambda$

$⇒ \lambda = 2\sqrt{3}$ units

In the circle given below, let OA = 1 unit, OB = 13 unit and PQ ⊥ OB . Then, the area of the triangle PQB (in square

Now, area of $\delta PQB = \dfrac{1}{2} \times PQ \times AB = \dfrac{1}{2} \times 2\lambda \times AB$

i.e. $\Delta = \dfrac{1}{2} \cdot 4\sqrt{3} \times 12$ sq. units

$= 24\sqrt{3}$ sq. units

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