In the given figure, ABCD is a piece of cardboard in the shape of trapezium in which AB || DC, ∠ABC = 90°. From this piece, quarter circle BEFC is removed. Given DC = BC = 4.2 cm and AE = 2 cm. Calculate the area of the remaining piece of the cardboard.

Given,

AB ∥ DC

∠ABC = 90°

DC = 4.2 cm

BC = 4.2 cm

AE = 2 cm

Quarter circle BEFC is removed.

Since it is a quarter circle with centre at B:

So, radius = BC = 4.2 cm

From figure,

AB = AE + EB

AB = 2 + 4.2 = 6.2 cm.

Calculating the area of trapezium ABCD,

$\text{Area of trapezium ABCD} = \dfrac{1}{2} \times \text{(sum of parallel sides)} \times \text{height} \\[1em] = \dfrac{1}{2} \times (6.2 + 4.2) \times 4.2 \\[1em] = \dfrac{1}{2} \times (10.4) \times 4.2 \\[1em] = 5.2 \times 4.2 \\[1em] = 21.84 \text{ cm}^2.$

Calculating the area of quarter circle,

$\text{Area of quarter circle} = \dfrac{1}{4}πr^2 \\[1em] = \dfrac{1}{4} \times \dfrac{22}{7} \times (4.2)^2 \\[1em] = \dfrac{11}{2 × 7} \times 17.64 \\[1em] = 11 \times 1.26 \\[1em] = 13.86 \text{ cm}^2.$

From figure,

Area of remaining piece of cardboard = Area of trapezium - Area of quarter circle

= 21.84 - 13.86

= 7.98 cm².

Hence, area of remaining piece of cardboard = 7.98 cm².