O is the centre of a circle of radius 5 cm. From a point P at a distance of 13 cm from O, two tangents PQ and PR are drawn to the circle. What is the area of the quadrilateral PQOR in square centimetres?

Introduction / Context:
This problem deals with tangents drawn from an external point to a circle and asks for the area of a quadrilateral formed by the point, the two tangent points and the centre of the circle. It tests your understanding of properties of tangents, right triangles involving radii and tangents, and basic area calculation using triangles. Such questions are standard in geometry sections of competitive exams.

Given Data / Assumptions:

- Circle has centre O and radius 5 cm.
- A point P lies outside the circle at a distance OP = 13 cm.
- Tangents PQ and PR are drawn from P to the circle, touching it at Q and R respectively.
- Quadrilateral PQOR is formed by joining P, Q, O and R in sequence.
- We need the area of quadrilateral PQOR in square centimetres.

Concept / Approach:
A radius drawn to the point of tangency is perpendicular to the tangent. Therefore, OQ is perpendicular to PQ, and OR is perpendicular to PR. This means triangles OPQ and OPR are right triangles. Also, PQ equals PR since tangents from an external point to a circle are equal in length. The quadrilateral PQOR can be decomposed into two congruent right triangles OPQ and OPR. Finding the area of one triangle and doubling it gives the area of the quadrilateral.

Step-by-Step Solution:
Step 1: In triangle OPQ, OP is the hypotenuse, OQ is the radius and PQ is the tangent.Step 2: Given OP = 13 cm and OQ = 5 cm. Angle at Q is 90 degrees because the radius is perpendicular to the tangent.Step 3: By Pythagoras theorem, PQ^2 = OP^2 - OQ^2.Step 4: Compute PQ^2 = 13^2 - 5^2 = 169 - 25 = 144, so PQ = 12 cm.Step 5: Area of triangle OPQ = (1 / 2) * OQ * PQ = (1 / 2) * 5 * 12 = 30 sq. cm.Step 6: Triangle OPR is congruent to triangle OPQ, so its area is also 30 sq. cm.Step 7: Area of quadrilateral PQOR = area of triangle OPQ + area of triangle OPR = 30 + 30 = 60 sq. cm.

Verification / Alternative check:
You can visualise PQOR as a kite with equal sides OP and OP on each side, and radii OQ and OR. Because OP is fixed and both tangents have equal length, the shape is symmetric about line OP. Computation of each right triangle area is straightforward, and the sum gives the quadrilateral area. There is no other part to add or subtract, so 60 sq. cm is consistent and unique.

Why Other Options Are Wrong:
30 sq. cm would be the area of only one of the triangles, not the entire quadrilateral. 32.5 sq. cm and 65 sq. cm arise from misusing wrong values or mixing radius and tangent lengths. 90 sq. cm would require each triangle to have area 45 sq. cm, which is not compatible with the given side lengths. Only 60 sq. cm correctly accounts for both congruent right triangles formed by the tangents and radii.

Common Pitfalls:
Some students forget that the radius to the point of tangency is perpendicular to the tangent, so they do not treat OPQ as a right triangle. Others mistakenly use OP as a leg and PQ as the hypotenuse. Another common error is to compute the area of only one triangle and overlook the need to double it for the quadrilateral. Always carefully identify right angles and count all constituent regions in composite figures.

Final Answer:
The area of the quadrilateral PQOR is 60 sq. cm.