Question

A metallic bucket, open at the top, of height 24 cm is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are 7 cm and 14 cm respectively. Find : (i) the volume of water which can completely fill the bucket. (ii) the area of the metal sheet used to make the bucket. [Use  = 22/7] are the answers: i) 8624 cm 3 ii) 1650 cm 2

Answer 1

Solution:-

Let the slant height of the frustum of the cone be 'l' and the radius of the upper end be 'R' and the radius of the lower end be 'r' respectively.
Therefore,
l² = √(R - r)² + h²
l² = √(14 - 7)² + 24²
l² = √7² + 24²
l² = √49 + 576
l² = √625
l = 25 cm
So, slant height is 25 cm

Volume = 1/3πh(R² + r² + R*r)
⇒ 1/3*22/7*24*(14² + 7² + 14*7)
⇒ 1/3*22/7*24*(196 + 49 + 98)
⇒ 1/3*22/7*24*343
⇒ 181104/21
⇒ 8624 cm³

Curved surface area = πl(R + r)
⇒ 22/7*25*(14 + 7)
⇒ 22/7*25*21
⇒ 11550/21
⇒ 1650 cm²

Area of the base (lower end) of the bucket = πr²
⇒ 22/7*7*7
⇒ 154 cm²

Area of the metal sheet used to make the bucket = Curved surface area + Area of the base (lower end)
= 1650 + 154 
= 1804 cm²

Answer.

Answer 2

Pih÷3(R^2+r^2+Rr)

= 22÷7×24÷3(14^2+7^2+14×7)

= 8624cm^3

Pil(R+r)+pir^2

22÷7×25(14+7)+22÷7×7×7

=1804 cm^3