Question

A bucket is in the form of a frustum of a cone with a capacity of 12317.6 cm³ of water. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of the metal sheet used in its making. (Use π = 3.14).

Answer 1

Given: Capacity of bucket, \[V = 12317.6\ \mathrm{cm^3}\]

Radius of top end, \[r_1 = 20\ \mathrm{cm}\]

Radius of bottom end, \[r_2 = 12\ \mathrm{cm}\] \[\pi = 3.14\]

To find: (i) Height of the bucket, [h]

(ii) Area of metal sheet used, [A]

Formula: Volume: \[V = \dfrac{1}{3}\pi h\left(r_1^2 + r_2^2 + r_1 r_2\right)\]

Slant height: \[l = \sqrt{h^2 + (r_1 - r_2)^2}\]

Area of metal sheet: \[A = \pi(r_1 + r_2)l + \pi r_2^2\]

Calculation:

1. Height of the bucket:

\[12317.6 = \dfrac{1}{3} \times 3.14 \times h \times \left(20^2 + 12^2 + 20 \times 12\right)\] 

\[12317.6 = \dfrac{1}{3} \times 3.14 \times h \times (400 + 144 + 240)\]

\[12317.6 = \dfrac{1}{3} \times 3.14 \times h \times 784\]

\[12317.6 = \dfrac{2461.76}{3} \times h\]

\[h = \dfrac{12317.6 \times 3}{2461.76} = 15\ \mathrm{cm}\]

2. Slant height of the bucket:

\[l = \sqrt{15^2 + (20 - 12)^2}\]

\[= \sqrt{225 + 8^2}\]

\[= \sqrt{225 + 64}\] 

\[= \sqrt{289}\]

\[= 17\ \mathrm{cm}\]

3. Area of the metal sheet used:

\[A = \pi(r_1 + r_2)l + \pi r_2^2\] 

\[A = 3.14 \times (20 + 12) \times 17 + 3.14 \times 12^2\] 

\[A = 3.14 \times [32 \times 17 + 144]\] 

\[A = 3.14 \times [544 + 144] = 3.14 \times 688 = 2160.32\ \mathrm{cm^2}\]

Answer: Height of bucket = \[15\ \mathrm{cm}\]

Area of metal sheet = \[2160.32\ \mathrm{cm^2}\]