In triangle ABC, points P and Q lie on sides AB and AC respectively such that segment PQ is parallel to side BC. If the ratio AP : PB is 1 : 4 and the area of triangle APQ is 4 square centimetres, what is the area of trapezium PQCB?

Introduction / Context:
This problem uses similar triangles and area ratios. It is a standard type of question in geometry and aptitude tests where a line parallel to a side of a triangle divides it into a smaller similar triangle and a trapezium, and the areas are related by the square of the similarity ratio.

Given Data / Assumptions:
- Triangle ABC is given.
- P lies on AB and Q lies on AC.
- Segment PQ is parallel to side BC.
- AP : PB = 1 : 4, so P divides AB in this ratio.
- Area of triangle APQ is 4 sq cm.
- We must find the area of trapezium PQCB, which is the region between PQ and BC inside triangle ABC.

Concept / Approach:
When a line is drawn parallel to one side of a triangle, it cuts the other two sides proportionally and forms a smaller triangle similar to the original. Here, triangle APQ is similar to triangle ABC. For similar triangles, the ratio of their areas equals the square of the ratio of corresponding sides. Once we find the ratio of similarity, we can compute the area of the larger triangle and then subtract the area of the smaller one to get the area of the trapezium.

Step-by-Step Solution:
Step 1: From AP : PB = 1 : 4, let AP = 1k and PB = 4k for some k, so AB = AP + PB = 5k. Step 2: In similar triangles APQ and ABC, the scale factor of similarity is AP / AB = 1k / 5k = 1 / 5. Step 3: The ratio of areas of similar triangles equals the square of the ratio of corresponding sides. Therefore, Area(APQ) / Area(ABC) = (1 / 5)^2 = 1 / 25. Step 4: Given Area(APQ) = 4 sq cm, then 4 / Area(ABC) = 1 / 25, so Area(ABC) = 4 * 25 = 100 sq cm. Step 5: The trapezium PQCB occupies the rest of the triangle beyond triangle APQ. Hence Area(trapezium PQCB) = Area(ABC) − Area(APQ) = 100 − 4 = 96 sq cm.

Verification / Alternative Check:
A quick check is to imagine actual side lengths scaled such that AB is 5 units and AP is 1 unit. The small triangle then has linear measure 1/5 of the full triangle, so its area is 1/25 of the full area. This fully agrees with the algebraic method. The numbers remain consistent and yield the same area for the trapezium, 96 sq cm.

Why Other Options Are Wrong:
Option a, 60 sq cm, would imply that triangle APQ had area 40 sq cm, which contradicts the given 4 sq cm and the similarity ratio.
Option b, 16 sq cm, is smaller than the area of triangle APQ, which is impossible because the trapezium covers more region than the small triangle.
Option d, 21 sq cm, does not match any reasonable proportion arising from the 1:5 linear ratio and leads to a noninteger total area when combined with triangle APQ.

Common Pitfalls:
Students sometimes forget to square the side ratio when relating areas and instead use Area ratio = 1 / 5, which would give the wrong total area. Another common mistake is to misinterpret AP : PB as AP : AB, leading to an incorrect similarity factor. Carefully identifying corresponding sides and remembering that area scales with the square of linear dimensions is the key to solving this type of question.

Final Answer:
The area of trapezium PQCB is 96 sq cm.