Question

In a trapezium \(ABCD AB || CD\), the diagonals AC and BD intersect at ‘P’. If \(AB: CD = 2:1\), then area of \(△CPD: \)area of \(△APB = \)

• A. 1:4
• B. 2:1
• C. 1:2
• D. 4:1

Answer 1

The Correct Option is A

In a trapezium, the diagonals \( AC \) and \( BD \) divide each other in the same ratio as the parallel sides.
Given \( AB : CD = 2 : 1 \), the diagonals \( AC \) and \( BD \) will divide each other in the same ratio.
Hence, the ratio of the areas of the triangles \( \triangle CPD \) and \( \triangle APB \) is also \( 1 : 4 \), because the area of a triangle is proportional to the base when the height is the same.

The correct answer is option (A): \(1:4\)