In the figure (i) given below, ABCD is a rectangle. AB = 14 cm, BC = 7 cm. From the rectangle, a quarter circle BFEC and a semicircle DGE are removed. Calculate the area of the remaining piece of the rectangle.

Area of rectangle = AB × BC = 14 × 7 = 98 cm².

Since, ABCD is a rectangle.

∴ CD = AB = 14 cm.

BFEC is a quadrant of radius, BC = r₁ = 7 cm.

∴ CE = 7 cm.

From figure,

DE = CD - CE = 14 - 7 = 7 cm.

From figure,

DE is the diameter of semi-circle DGE.

So, radius (r) = $\dfrac{DE}{2} = \dfrac{7}{2}$ = 3.5 cm.

Area of semi-circle = $\dfrac{πr^2}{2}$

$= \dfrac{\dfrac{22}{7} \times (3.5)^2}{2} \\[1em] = \dfrac{22 \times 12.25}{14} \\[1em] = \dfrac{269.5}{14} \\[1em] = 19.25 \text{ cm}^2.$

Area of quadrant BFEC = $\dfrac{πr_1^2}{4}$

$= \dfrac{\dfrac{22}{7} \times 7^2}{4} \\[1em] = \dfrac{22 \times 7}{4} \\[1em] = \dfrac{154}{4} = 38.5 \text{ cm}^2.$

Area of remaining piece of rectangle = Area of rectangle - Area of semicircle DGE - Area of quadrant BFEC

= 98 - 19.25 - 38.5

= 40.25 cm².

Hence, area of remaining piece of rectangle = 40.25 cm².