Question

In the following figure, a piece of cardboard, in the shape of a trapezium ABCD, and AB || DC and ∠BCD = 90°, quarter circle BFEC is removed. Given AB = BC = 3.5 cm and DE = 2 cm. Calculate the area of the remaining piece of the cardboard `("Take"  π = 22/7)`.

Answer 1

Given:

AB = BC = 3.5 cm

DE = 2 cm

AB || DC

∠BCD = 90°

A quarter circle BFEC is removed.

`π = 22/7`

Step 1: Understand the figure

Since ∠BCD = 90° and BC = 3.5 cm the quarter circle has:

Radius = BC = 3.5 cm

Also, DC = DE + EC

But EC = radius = 3.5 cm

So, DC = 2 + 3.5 = 5.5 cm

Step 2: Area of trapezium ABCD

Formula:

Area = `1/2 (AB + DC) xx "height"`

Here, AB = 3.5, DC = 5.5

Height = BC = 3.5

Area = `1/2 (3.5 + 5.5) xx 3.5`

= `1/2 (9) xx 3.5`

= 4.5 × 3.5

= 15.75 cm²

Step 3: Area of quarter circle removed

Area = `1/2 πr^2`

= `1/4 xx 22/7 xx (3.5)^2`

= `1/4 xx 22/7 xx 12.25`

= `1/4 xx 38.5`

= 9.625 cm²

Step 4: Remaining area

= 15.75 – 9.625

= 6.125 cm²