The shape of the top of a table in a restaurant is that of a segment of a circle with centre O and ∠BOD = 90°. BO = OD = 60 cm. Find :

(i) the area of the top of the table;

(ii) the perimeter of the table.

Since,

OB = OD = 60 cm

Thus, radius of circle (r) = 60 cm

∠BOD = 90°.

(i) Area of circle = πr²

= 3.14 × (60)²

= 3.14 × 3600 = 11304 cm².

Area of sector of 90° = $\dfrac{90}{360}$ × πr²

= $\dfrac{1}{4}$ × 11304 = 2826 cm².

From figure,

Area of top of the table = Area of circle - Area of sector 90°

Area of top of the table = 11304 - 2826

= 8478 cm^2.

Hence, area of top of the table = 8478 cm².

(ii) Circumference of circle = 2πr

= 2 × 3.14 × 60 = 376.8 cm.

Length of 270° arc = $\dfrac{270}{360}$ × 376.8

= $\dfrac{3}{4}$ × 376.8

= 3 × 94.2 = 282.6 cm.

From figure,

Perimeter of table top = Circumference of major arc (of 270°) + OB + OD

= 282.6 + 60 + 60

= 282.6 + 120

= 402.6 cm.

Hence, perimeter of the table = 402.6 cm.