Question

From a point P, tangents PQ and PR are drawn to a circle with centre O and radius 6 cm. If OP = 10 cm, then area of quadrilateral PQOR is :

• A. 48 \(\text{cm}^2\)
• B. 24 \(\text{cm}^2\)
• C. 96 \(\text{cm}^2\)
• D. 72 \(\text{cm}^2\)

Hint:

Remember the standard Pythagorean triplet \((6, 8, 10)\) to instantly find the tangent length of 8 cm without full manual calculation during the exam!

Answer 1

The Correct Option is A

Step 1: Find the tangent length.
Since the radius is perpendicular to the tangent, $\Delta OQP$ is right-angled, so by Pythagoras:
\[ PQ = \sqrt{OP^2 - OQ^2} = \sqrt{10^2 - 6^2} = \sqrt{64} = 8 \text{ cm} \]
Step 2: Treat PQOR as one combined shape instead of doubling a triangle's area.
The quadrilateral $PQOR$ is made of two congruent right triangles sharing the same two legs $OQ$ and $PQ$, so its area equals the product of these two legs directly.
Step 3: Compute the area.
\[ \text{Area}(PQOR) = OQ \times PQ = 6 \times 8 \]
\[ \boxed{48 \text{ cm}^2} \]