In triangle PQR, PS is the internal bisector of angle ∠QPR and PT is the internal bisector of angle ∠QPS. If ∠QPT = 30°, PT = 9 cm and TR = 15 cm, what is the area of triangle PTR in square centimetres?

Introduction / Context:
This question involves a triangle with nested angle bisectors and asks for the area of a smaller triangle formed using one side and one of the bisector segments. The data include one angle, one side of the small triangle and another side of that triangle. By using angle bisector properties and some angle chasing, we can show that triangle PTR is right angled, which allows us to compute its area using a straightforward formula involving the product of perpendicular sides.

Given Data / Assumptions:
- PQR is a triangle. - PS is the internal angle bisector of ∠QPR. - PT is the internal angle bisector of ∠QPS. - ∠QPT = 30°. - PT = 9 cm. - TR = 15 cm. - We must find the area of triangle PTR.

Concept / Approach:
Because PT bisects angle ∠QPS, angle ∠QPT equals angle ∠SPT. Since ∠QPT is given as 30°, this directly gives ∠SPT = 30°. Also, PS bisects angle ∠QPR, so angle ∠QPS equals angle ∠SPR. By combining this information, we deduce that angle ∠QPR is 120°. Then angle ∠TPR, which is the angle between PT and PR, is the sum of angles between PT and PS and between PS and PR, that is 30° + 60° = 90°. Thus triangle PTR is right angled at P, with legs PT and PR and hypotenuse TR. Knowing PT and TR allows us to find PR using the Pythagoras relation and then area as one half of the product of the perpendicular sides PT and PR.

Step-by-Step Solution:
Step 1: PT bisects angle ∠QPS, so ∠QPT = ∠SPT = 30°. Step 2: Therefore ∠QPS = ∠QPT + ∠SPT = 30° + 30° = 60°. Step 3: PS bisects angle ∠QPR, so ∠QPS = ∠SPR = 60°. Step 4: Hence ∠QPR = ∠QPS + ∠SPR = 60° + 60° = 120°. Step 5: Angle between PT and PR at P is ∠TPR = ∠TPS + ∠SPR = 30° + 60° = 90°. Step 6: Thus triangle PTR is right angled at P with PT and PR as perpendicular legs and TR as the hypotenuse. Step 7: Use Pythagoras theorem in triangle PTR: PR^2 + PT^2 = TR^2. Step 8: Substitute PT = 9 and TR = 15 to get PR^2 + 9^2 = 15^2, so PR^2 + 81 = 225. Step 9: Compute PR^2 = 225 − 81 = 144, which gives PR = 12 cm. Step 10: Area of triangle PTR = (1 / 2) * PT * PR = (1 / 2) * 9 * 12 = 54 square centimetres.

Verification / Alternative check:
We can verify the right angle more directly by summing the angles around P in the small configuration. From the construction, angle between PQ and PT is 30°, between PT and PS another 30°, and between PS and PR 60°. Total angle between PQ and PR is 120°, consistent with the earlier result for ∠QPR. The angle between PT and PR is then exactly 90°, confirming that triangle PTR is right angled at P. Substituting PT = 9 and TR = 15 into Pythagoras is a standard 9 12 15 right triangle triple, whose ratio matches the familiar 3 4 5 triple scaled by 3. This further confirms PR = 12 and the area value.

Why Other Options Are Wrong:
Option 36 would be the area if one of the perpendicular sides were only 8 cm instead of the correct 12 cm. Option 72 would arise if we incorrectly treated TR as one of the perpendicular legs in the area formula. Option 216 corresponds roughly to four times the correct area and would require much larger side lengths than those given.

Common Pitfalls:
A common mistake is to misinterpret which angles are being bisected and to miscalculate ∠QPR. Another frequent error is to assume the given 30° angle itself is the right angle in triangle PTR, which is not true. Some learners also forget that the area formula using sine requires the included angle and mistakenly use the hypotenuse with an incorrect angle. Recognising the 3 4 5 type ratio hidden in 9 12 15 helps confirm the shape is a right triangle.

Final Answer:
The area of triangle PTR is 54 square centimetres.