Question:
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
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
Updated On: Jan 15, 2026
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Solution and Explanation
We are given a composite figure consisting of a rectangle and an isosceles triangle.
Step 1: Drop a perpendicular $DE$ from point $D$ onto line $AB$.
This divides the figure into two parts:
- A rectangle $BCED$ with dimensions $4 \times 5$ (length = 5 cm, breadth = 4 cm)
- An isosceles triangle $AED$ with base $AE = 4$ cm and height $DE = 4$ cm
Step 2: Calculate the area of the rectangle:
$\text{Area of rectangle} = \text{length} \times \text{breadth} = 5 \times 4 = 20 \;\text{cm}^2$
Step 3: Calculate the area of the triangle $AED$:
$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8 \;\text{cm}^2$
Step 4: Add both areas to get the total area:
$\text{Total area} = 20 + 8 = 28 \;\text{cm}^2$
∴ Required Area = $28 \;\text{cm}^2$
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