Points P and Q lie on sides AB and AC of triangle ABC respectively such that segment PQ is parallel to side BC. If the ratio AP : PB is 2 : 5 and the area of triangle APQ is 4 square centimetres, what is the area of trapezium PQCB?

Introduction / Context:
This geometry problem involves a triangle with a line segment drawn parallel to one side, creating a smaller similar triangle and a trapezium. The question requires using properties of similar triangles and area ratios to find the area of the trapezium formed between the parallel line and the base of the triangle. It tests understanding of similarity, proportional segments, and area scaling.

Given Data / Assumptions:

• Triangle ABC is given.
• Point P lies on side AB and point Q lies on side AC.
• PQ is parallel to BC, so triangle APQ is similar to triangle ABC.
• AP : PB = 2 : 5, so AB is divided in that ratio.
• Area of triangle APQ is 4 square centimetres.
• We seek the area of trapezium PQCB, which is region ABC minus region APQ.

Concept / Approach:
Since PQ is parallel to BC, triangle APQ is similar to triangle ABC. The similarity ratio between the smaller triangle and the larger triangle is AP / AB. From AP : PB = 2 : 5, AB is divided into 2 + 5 = 7 equal parts and AP is 2 of those parts, so AP / AB = 2 / 7. For similar triangles, the ratio of areas is the square of the similarity ratio. Therefore, area(APQ) : area(ABC) = (2 / 7)^2. With the area of APQ known, we can compute the area of ABC and then subtract to get the area of pqcb.

Step-by-Step Solution:
Step 1: From AP : PB = 2 : 5, AB = 2 + 5 = 7 parts and AP = 2 parts.Step 2: So AP / AB = 2 / 7.Step 3: Because the triangles APQ and ABC are similar, area(APQ) / area(ABC) = (AP / AB)^2 = (2 / 7)^2 = 4 / 49.Step 4: Let area(ABC) = K. Then 4 / 49 = area(APQ) / K.Step 5: Given area(APQ) = 4 square centimetres, we have 4 / 49 = 4 / K, so K = 49.Step 6: Area of triangle ABC is therefore 49 square centimetres.Step 7: The area of trapezium PQCB is area(ABC) minus area(APQ) = 49 - 4 = 45 square centimetres.

Verification / Alternative check:
We can verify the area ratio by imagining scaling all sides of triangle APQ by a factor of 7 / 2 to obtain triangle ABC.Since area scales with the square of the linear scale factor, the area of ABC should be (7 / 2)^2 times the area of APQ, that is (49 / 4) * 4 = 49 square centimetres.Subtracting 4 from 49 again yields 45 square centimetres, matching the previous calculation.

Why Other Options Are Wrong:
An area of 49 square centimetres would be the area of the whole triangle ABC, not the trapezium.Values like 25 or 21 square centimetres underestimate the size of trapezium PQCB, as they ignore correct use of the similarity ratio.A value like 36 square centimetres corresponds to a different incorrect difference in areas and does not arise from the given ratio 2 : 5.

Common Pitfalls:
A common mistake is to treat AP : AB as 2 : 5 instead of 2 : 7, forgetting that AB equals AP plus PB.Another frequent error is to use a direct ratio of 2 : 7 for areas rather than squaring the ratio, which leads to wrong results.Some may try to treat the trapezium as a separate shape without using the powerful tool of similar triangles and area ratios.

Final Answer:
The area of trapezium PQCB is 45 sq cm.