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

Answer 1

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 = $28 \;\text{cm}^2$