In trapezium ABCD, sides AB and CD are parallel, and ∠ADC = 90°. The lengths of the parallel sides are AB = 15 cm and CD = 40 cm, and the diagonal AC has length 41 cm. What is the area (in cm²) of trapezium ABCD?

Introduction / Context:
This problem deals with a trapezium (trapezoid) in which one non-parallel side and one parallel side meet at a right angle. We are given the lengths of the two parallel sides and one diagonal. The goal is to find the area of the trapezium. To solve this, we interpret the trapezium as a figure that contains right triangles and a rectangle-like structure, and we use the Pythagoras theorem to find the height, which is the perpendicular distance between the parallel sides. Once the height is known, we can use the standard area formula for a trapezium.

Given Data / Assumptions:

• AB and CD are parallel sides of trapezium ABCD.
• ∠ADC = 90°, so AD is perpendicular to CD.
• AB = 15 cm.
• CD = 40 cm.
• Diagonal AC = 41 cm.
• We must find the area of ABCD in square centimetres.

Concept / Approach:
Place the trapezium on a coordinate plane for clarity. Because ∠ADC = 90°, we can treat AD as a vertical segment (height) and CD as horizontal. Then AB will be parallel to CD at the top. By using the coordinates of A, D, and C, and the length of diagonal AC, we apply the Pythagoras theorem to find the height AD. Once the height is known, we use the area formula for a trapezium: Area = (1 / 2) * (sum of parallel sides) * height.

Step-by-Step Solution:
Let D be at (0, 0) and place CD along the x-axis, so C = (40, 0).Since ∠ADC = 90°, take AD along the y-axis. Let A = (0, h), where h is the height.AB is parallel to CD and has length 15 cm, so B has coordinates (15, h).We know the diagonal AC has length 41 cm.Coordinates: A = (0, h) and C = (40, 0).Use the distance formula: AC^2 = (40 − 0)^2 + (0 − h)^2 = 40^2 + h^2.Given AC = 41 cm, so 41^2 = 1600 + h^2.Compute 41^2 = 1681, so 1681 = 1600 + h^2.Thus h^2 = 1681 − 1600 = 81.So h = 9 cm (height of the trapezium).Area of trapezium = (1 / 2) * (AB + CD) * height = (1 / 2) * (15 + 40) * 9.Compute: (15 + 40) = 55, so area = (1 / 2) * 55 * 9 = 27.5 * 9 = 247.5 cm².

Verification / Alternative check:
It is worth checking the internal right triangle dimensions. With height h = 9 cm, AD = 9 and CD = 40, triangle ADC is a right triangle with legs 9 and 40, giving AC as √(9^2 + 40^2) = √(81 + 1600) = √1681 = 41, which matches the given diagonal. This confirms that the coordinate placement and height calculation are consistent with the problem statement and supports the correctness of the final area value.

Why Other Options Are Wrong:
An area of 245 cm² or 240 cm² would result from using an incorrect height approximation or miscomputing the Pythagoras step. An area of 250 cm² would come from slightly miscalculating (55 * 9) or rounding. The option 225 cm² is too low and does not match the correct height. Only 247.5 cm² exactly matches the precise area obtained from the correct height and trapezium area formula.

Common Pitfalls:
One frequent error is to forget that the diagonal AC connects points (0, h) and (40, 0) and so misapply the distance formula. Another mistake is to assume AD equals 41 or to treat the trapezium as a rectangle. Some candidates also forget to take the half factor in the area formula for a trapezium or incorrectly add the side lengths. Carefully setting up the coordinate geometry and applying Pythagoras leads to a reliable solution.

Final Answer:
The area of trapezium ABCD is 247.5 cm².