Parallel segment in a triangle and area of the lower trapezium:\nPoints P on AB and Q on AC satisfy PQ ∥ BC. If AP:PB = 2:3 and area(ΔAPQ) = 8 sq cm, find the area of trapezium PQCB.

Introduction / Context:
A segment parallel to the base of a triangle creates a smaller similar triangle at the top. Areas scale with the square of the linear ratio. We use the given small-triangle area to recover the full triangle’s area, then subtract.

Given Data / Assumptions:

• P on AB, Q on AC, PQ ∥ BC.
• AP:PB = 2:3 → AP/AB = 2/5.
• Area(ΔAPQ) = 8 sq cm.

Concept / Approach:
ΔAPQ ~ ΔABC with linear ratio 2/5. Therefore, area(APQ) : area(ABC) = (2/5)^2 = 4/25.

Step-by-Step Solution:

Verification / Alternative check:
Proportional reasoning aligns with standard results from the Basic Proportionality Theorem / similarity of triangles.

Why Other Options Are Wrong:
50 is the whole triangle; 18 and 14 are too small; 32 does not match the derived difference.

Common Pitfalls:
Using linear ratio directly on area (forgetting the square), or misreading AP:PB as AP:AB.

Final Answer:
42 sq cm