In the figure (ii) given below, a piece of cardboard, in the shape of a trapezium ABCD and AB || CD 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.

From figure,

Radius of quadrant = BC = 3.5 cm.

EC = BC = 3.5 cm (As both equal to radius of quadrant BFEC)

By formula,

Area of trapezium = $\dfrac{1}{2}$ × (Sum of || sides) × distance between them

= $\dfrac{1}{2}$ × (AB + DC) × BC

= $\dfrac{1}{2}$ × (AB + EC + DE) × BC

= $\dfrac{1}{2}$ × (3.5 + 3.5 + 2) × 3.5

= $\dfrac{1}{2}$ × 9 × 3.5

= 4.5 × 3.5

= 15.75 cm².

So, the area of quadrant BCEF = $\dfrac{πr^2}{4}$

$= \dfrac{\dfrac{22}{7} \times (3.5)^2}{4} \\[1em] = \dfrac{\dfrac{22}{7} \times 12.25}{4} \\[1em] = \dfrac{22 \times 12.25}{4 \times 7} \\[1em] = \dfrac{269.5}{28} \\[1em] = 9.625 \text{ cm}^2$

Area of shaded portion = Area of trapezium - Area of quadrant

= 15.75 – 9.625 = 6.125 cm².

Hence, area of shaded portion = 6.125 cm².