Parallel segment in a triangle and area of the lower trapezium: Points 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.
-
A50 sq cm
-
B18 sq cm
-
C42 sq cm
-
D14 sq cm
-
E32 sq cm
Answer
Correct Answer: 42 sq cm
Explanation
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:
area(ABC) = area(APQ) * (25/4) = 8 * (25/4) = 50 sq cm.Area(trapezium PQCB) = area(ABC) − area(APQ) = 50 − 8 = 42 sq cm.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