4. From a point \(P\) at a distance 13 cm from the centre \(O\) of a circle of radius 5 cm, tangents \(PQ\) and \(PR\) are drawn. The area of quadrilateral \(PQOR\) is
\(60\,\text{cm}^2\)
\(65\,\text{cm}^2\)
\(30\,\text{cm}^2\)
\(32.5\,\text{cm}^2\)
Step 1: We are given:
Step 2: In right triangle \(OQP\), \(OQ\) is the radius (5 cm), and \(OP\) is the hypotenuse (13 cm). The tangent \(PQ\) is perpendicular to the radius at point \(Q\). So, using Pythagoras theorem:
\[ PQ = \sqrt{OP^2 - OQ^2} = \sqrt{13^2 - 5^2} \]
\[ PQ = \sqrt{169 - 25} = \sqrt{144} = 12\,\text{cm} \]
Step 3: Quadrilateral \(PQOR\) is made of two congruent right triangles: \(\triangle OPQ\) and \(\triangle OPR\).
Step 4: Area of one right triangle:
\[ \text{Area}(\triangle OPQ) = \tfrac{1}{2} \times OQ \times PQ \]
\[ = \tfrac{1}{2} \times 5 \times 12 = 30\,\text{cm}^2 \]
Step 5: Since both triangles have the same area, total area is:
\[ \text{Area}(PQOR) = 30 + 30 = 60\,\text{cm}^2 \]
Final Answer: The area of quadrilateral \(PQOR\) is \(60\,\text{cm}^2\).