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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

Updated On: Jan 15, 2026
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Solution and Explanation

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A composite figure is presented, comprising a rectangle and an isosceles triangle.

Step 1: Construct perpendicular $DE$ from point $D$ to line $AB$.

This segmentation yields:

  • Rectangle $BCED$ with dimensions $4 \times 5$ (length = 5 cm, breadth = 4 cm).
  • Isosceles triangle $AED$ with base $AE = 4$ cm and height $DE = 4$ cm.

Step 2: Compute the area of the rectangle.

$\text{Area of rectangle} = \text{length} \times \text{breadth} = 5 \times 4 = 20 \;\text{cm}^2$

Step 3: Compute the area of 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: Sum the computed areas for the total area.

$\text{Total area} = 20 + 8 = 28 \;\text{cm}^2$

∴ Required Area = $28 \;\text{cm}^2$

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