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In a trapezium ABCD, AB is parallel to DC, BC is perpendicular to DC and \(∠BAD=45°\). If DC=5cm , BC=4 cm, the area of the trapezium in sq cm is ? [This Question was asked as TITA]

Updated On: Jan 15, 2026

  • \(25 \ cm^2\)

  • \(26 \ cm^2\)

  • \(28\  cm^2\)

  • \(30\ cm^2\)

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The Correct Option is C

Solution and Explanation

Given:
\( \angle BAD = 45^\circ \), \( DC = 5\,\text{cm} \), \( BC = 4\,\text{cm} \), and \( BC \perp DC \).
Given that AB is parallel to DC, the objective is to determine the area of trapezium ABCD.

Step 1: The condition \( BC \perp DC \) indicates that BC serves as the height of the trapezium.
Therefore, the height is \( h = BC = 4\,\text{cm} \).

Step 2: Triangle \( \triangle ABD \) is established as right-angled at A, with \( \angle BAD = 45^\circ \).
The length of \( AD \) can be calculated using the trigonometric relation: \[ \tan \angle BAD = \frac{AD}{BC} \] Substituting the given values: \[ \tan 45^\circ = \frac{AD}{4} \Rightarrow 1 = \frac{AD}{4} \Rightarrow AD = 4\,\text{cm} \]

Step 3: Since AB, AD, and DC are collinear and AB is parallel to DC, we consider the formation of the trapezium. The side AB's length is determined by the sum of AD and DC: \[ AB = AD + DC = 4 + 5 = 9\,\text{cm} \]

Step 4: The formula for the area of a trapezium is applied: \[ \text{Area} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height} \] The parallel sides are identified as \( AB = 9\,\text{cm} \) and \( DC = 5\,\text{cm} \), with a height of \( 4\,\text{cm} \).
The calculation proceeds as follows: \[ \text{Area} = \frac{1}{2} \times (9 + 5) \times 4 = \frac{1}{2} \times 14 \times 4 = 28\,\text{cm}^2 \]

Final Answer: The computed area of trapezium ABCD is: \[ \boxed{28\,\text{cm}^2} \]

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