In triangle ABC, points P and Q lie on sides AB and AC respectively such that segment PQ is parallel to side BC. If AP : PB = 2 : 3 and the area of triangle APQ is 8 sq. cm, what is the area of trapezium PQCB?

Introduction / Context:
This problem involves similar triangles and area ratios within a triangle when a line segment is drawn parallel to one of its sides. The line PQ is parallel to BC, so triangle APQ is similar to triangle ABC. Using the ratio of corresponding sides, we can relate the areas of the smaller and the larger triangle and then compute the area of the remaining trapezium PQCB. Such questions check understanding of similarity, proportionality, and area relationships.

Given Data / Assumptions:

• Triangle ABC is any triangle.
• P lies on AB and Q lies on AC such that PQ ∥ BC.
• AP : PB = 2 : 3, so AB is divided internally in that ratio.
• The area of triangle APQ is 8 sq. cm.
• We are asked to find the area of trapezium PQCB, the region between PQ and BC.

Concept / Approach:
Since PQ is parallel to BC, triangle APQ is similar to triangle ABC by angle angle similarity. The ratio of similarity between corresponding sides is AP / AB. Given AP : PB = 2 : 3, the total AB is 2 + 3 = 5 parts, so AP / AB = 2 / 5. The ratio of areas of similar triangles is the square of the ratio of corresponding sides. Once we find the area of triangle ABC, we subtract the area of triangle APQ to obtain the area of trapezium PQCB.

Step-by-Step Solution:
AP : PB = 2 : 3 implies AB is 5 equal parts.So AP / AB = 2 / 5.Triangles APQ and ABC are similar, so area(APQ) / area(ABC) = (AP / AB)^2 = (2 / 5)^2 = 4 / 25.Given area(APQ) = 8 sq. cm, we get 8 / area(ABC) = 4 / 25, so area(ABC) = 8 * 25 / 4 = 50 sq. cm.Area of trapezium PQCB = area(ABC) − area(APQ) = 50 − 8 = 42 sq. cm.

Verification / Alternative check:
We can think of triangle ABC as being cut by a line parallel to BC at some fraction down from A. The scaling factor for lengths from A down to PQ is 2 / 5, and the scaling factor for lengths from PQ down to BC is 3 / 5. The area of the lower part, trapezium PQCB, must then be the remaining fraction of the total area, which is 1 − 4 / 25 = 21 / 25 of area(ABC). Since area(ABC) = 50, this gives 21 / 25 * 50 = 42, matching the earlier calculation.

Why Other Options Are Wrong:
Option 50 sq. cm is the total area of triangle ABC, not the area of the trapezium. Option 18 sq. cm is too small and might come from subtracting incorrectly or misusing ratios. Option 14 sq. cm does not reflect the correct fraction of the total area. Option 32 sq. cm is another miscalculated residual area that does not follow from the correct area proportionality for similar triangles.

Common Pitfalls:
Many students forget to square the side ratio when comparing areas and instead use 2 / 5 directly. Others misinterpret AP : PB as AP : AB or confuse which parts of the triangle correspond. Some also mistakenly treat trapezium PQCB as a standalone shape without using similarity, which makes calculations much more complex and error prone.

Final Answer:
The area of trapezium PQCB is 42 sq cm.