In the figure (i) given below, a piece of cardboard in the shape of a quadrant of a circle of radius 7 cm is bounded by the perpendicular radii OX and OY. Points A and B lie on OX and OY respectively such that OA = 3 cm and OB = 4 cm. The triangular part OAB is removed. Calculate the area and the perimeter of the remaining piece.

Area of quadrant = $\dfrac{πr^2}{4}$

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

Area of △OAB = $\dfrac{1}{2}$ × base × height

$= \dfrac{1}{2} \times OA \times OB \\[1em] = \dfrac{1}{2} \times 3 \times 4 \\[1em] = 6 \text{ cm}^2.$

Area of shaded region = Area of quadrant - Area of △OAB

= 38.5 - 6 = 32.5 cm².

From figure,

BY = OY - OB = 7 - 4 = 3 cm.

AX = OX - OA = 7 - 3 = 4 cm.

In right angle triangle OAB,

⇒ AB² = OA² + OB²

⇒ AB² = 3² + 4²

⇒ AB² = 9 + 16

⇒ AB² = 25

⇒ AB = $\sqrt{25}$ = 5 cm.

Perimeter of shaded region = AB + BY + AX + Circumference of quadrant

= 5 + 3 + 4 + $\dfrac{2πr}{4}$

= 12 + $\dfrac{2 \times \dfrac{22}{7} \times 7}{4}$

= 12 + 11

= 23 cm.

Hence, area of shaded region = 32.5 cm² and perimeter of remaining piece = 23 cm.