Question

From a rectangular solid of metal 42 cm by 30 cm by 20 cm, a conical cavity of diameter 14 cm and depth 24 cm is drilled out. Find:

(i) the surface area of remaining solid,

(ii) the volume of remaining solid

(iii) the weight of the material drilled out if it weighs 7 gm per cm³.

Answer 1

(i) Total surface area of cuboid = 2(ℓb + bh + ℓh)

= 2 (42 × 30 + 30 × 20 + 20 × 42)

= 2 (1260 + 600 + 840)

= 2 × 2700

= 5400 cm²

Diameter of the cone = 14 cm

Radius of the cone = 14/2 = 7cm

Surface area of remaining part = 5400 + 550 – 154 = 5796 cm2

(ii) Dimensions of rectangular solids = (42 × 30 × 20) cm

volume = (42 × 30 × 20) = 25200 cm³

Radius of conical cavity (r) =7 cm

height (h) = 24 cm

Volume of remaining solid = (25200 - 1232) = 23968 cm³

(iii) Weight of material drilled out

=1232 × 7 g = 8624g = 8.624 kg