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: Understanding the Question:
Tangents \(PQ\) and \(PR\) are drawn from an external point \(P\) to a circle centered at \(O\). The radius of the circle is \(6 \text{ cm}\) and the distance \(OP\) is \(10 \text{ cm}\). We need to calculate the area of the quadrilateral \(PQOR\).

Step 2: Key Formula or Approach:
- The radius is perpendicular to the tangent at the point of contact: \(\angle OQP = \angle ORP = 90^\circ\).
- The right-angled triangles \(\Delta OQP\) and \(\Delta ORP\) are congruent.
- Area of quadrilateral \(PQOR\) = \(2 \times \text{Area of } \Delta OQP\).
- Use Pythagoras theorem to find the length of tangent \(PQ\):
\[ PQ = \sqrt{OP^2 - OQ^2} \]

Step 3: Detailed Explanation:

• In right-angled triangle \(\Delta OQP\):
Using Pythagoras theorem:
\[ OP^2 = OQ^2 + PQ^2 \]
Substitute the given values (\(OP = 10\) and \(OQ = 6\)):
\[ 10^2 = 6^2 + PQ^2 \]
\[ 100 = 36 + PQ^2 \]
\[ PQ^2 = 64 \implies PQ = 8 \text{ cm} \]

• Calculate the area of right-angled triangle \(\Delta OQP\):
\[ \text{Area}(\Delta OQP) = \frac{1}{2} \times \text{base} \times \text{height} \]
\[ \text{Area}(\Delta OQP) = \frac{1}{2} \times OQ \times PQ \]
\[ \text{Area}(\Delta OQP) = \frac{1}{2} \times 6 \times 8 = 24 \text{ cm}^2 \]

• Find the total area of quadrilateral \(PQOR\):
\[ \text{Area}(PQOR) = 2 \times \text{Area}(\Delta OQP) \]
\[ \text{Area}(PQOR) = 2 \times 24 = 48 \text{ cm}^2 \]


Step 4: Final Answer:
The area of the quadrilateral \(PQOR\) is 48 \(\text{cm}^2\).