From an external point P, two tangents PA and PB are drawn to a circle with centre O, touching the circle at A and B. If ∠AOB = 120° and the length of AP is 6 cm, what is the area (in sq cm) of triangle APB?

Introduction / Context:
This problem combines properties of tangents to a circle with angles formed at the centre and at an external point. A key idea is that tangents from an external point to a circle are equal in length, and the quadrilateral formed by the radii and tangent segments has right angles at the points of tangency. The relation between the central angle and the angle at the external point then allows direct computation of the area of triangle APB.

Given Data / Assumptions:

• Circle with centre O.
• From external point P, tangents PA and PB are drawn to the circle, touching at A and B respectively.
• ∠AOB, the central angle subtending chord AB, is 120°.
• Length of tangent AP is 6 cm, so PB is also 6 cm since tangents from P are equal.
• We must find the area of triangle APB.

Concept / Approach:
Any radius drawn to a point of tangency is perpendicular to the tangent. Hence, in quadrilateral AOPB, angles at A and B are right angles. The sum of internal angles of a quadrilateral is 360°. Using this and the given central angle ∠AOB, we can determine the angle at P, ∠APB. Once we know ∠APB and the equal sides AP and BP, we can use the formula for the area of a triangle given two sides and the included angle: Area = (1 / 2) * a * b * sin(C).

Step-by-Step Solution:
Step 1: In quadrilateral AOPB, ∠OAB = 90° and ∠OBP = 90° because radii OA and OB are perpendicular to tangents at A and B. Step 2: Let ∠APB be denoted by θ. The four angles of the quadrilateral are 90°, 90°, 120° (∠AOB) and θ (∠APB). Step 3: Sum of interior angles of a quadrilateral is 360°, so 90° + 90° + 120° + θ = 360°. Step 4: Simplify: 300° + θ = 360°, so θ = 60°. Hence ∠APB = 60°. Step 5: Triangle APB has sides AP = 6 cm, BP = 6 cm and included angle ∠APB = 60°. Step 6: Area of triangle APB = (1 / 2) * AP * BP * sin(60°) = (1 / 2) * 6 * 6 * (√3 / 2) = 18 * (√3 / 2) = 9√3 sq cm.

Verification / Alternative check:
We can also note that triangle APB is isosceles with AP = BP. An isosceles triangle with sides 6, 6 and included angle 60° is actually equilateral with all sides equal to 6, because the angle opposite the base is 60°. For an equilateral triangle with side 6, the area is (√3 / 4) * side^2 = (√3 / 4) * 36 = 9√3 sq cm. This matches the result obtained via the sine formula, confirming our answer.

Why Other Options Are Wrong:
An area of 6√3 or 8√3 would correspond to smaller effective side lengths or incorrect included angle. The value 9 (without √3) comes from omitting the sine factor and is a common mistake when applying the area formula. The extra option 12√3 would require a larger side or angle than is given by the geometry. Only 9√3 sq cm is consistent with both the triangle and quadrilateral angle relationships.

Common Pitfalls:
Learners might forget that the angle at P in a quadrilateral formed by two tangents and two radii is supplementary to the central angle, leading to wrong angle values. Another frequent mistake is not using the fact that AP and BP are equal, which simplifies the triangle classification. Additionally, some students incorrectly use the formula (1 / 2) * a * b without the sine of the included angle. Always check that all angle relationships in the figure are correctly used before computing the triangle area.

Final Answer:
The area of triangle APB is 9√3 sq cm.