Question

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. Also, find the surface area of the toy. (Take 𝜋 = 3⋅14)

Answer 1

Given: Height of the cone \[h = 2\ \mathrm{cm}\] 

Diameter of the base \[d = 4\ \mathrm{cm} \implies r = \dfrac{4}{2} = 2\ \mathrm{cm}\] 

Radius of the hemisphere \[r = 2\ \mathrm{cm}\]

Value of \[\pi = 3.14\]

To find:

1. Volume of the toy
2. Surface area of the toy

Formula: \[V = \text{Volume of cone} + \text{Volume of hemisphere} = \dfrac{1}{3}\pi r^{2}h + \dfrac{2}{3}\pi r^{3}\] 

\[\text{Slant height of cone } l = \sqrt{r^{2} + h^{2}}\] 

\[\text{Surface area of toy } S = \text{CSA of cone} + \text{CSA of hemisphere} = \pi r l + 2\pi r^{2} = \pi r(l + 2r)\]

Substitution & Calculation:

1. Volume of the toy:

\[V = \dfrac{1}{3}\pi (2)^{2}(2) + \dfrac{2}{3}\pi (2)^{3}\] 

\[V = \dfrac{8\pi}{3} + \dfrac{16\pi}{3} = \dfrac{24\pi}{3} = 8\pi\ \mathrm{cm^{3}}\] 

Substitute \[\pi = 3.14\]: \[V = 8 \times 3.14 = 25.12\ \mathrm{cm^{3}}\]

2. Surface area of the toy: Slant height (l):

\[l = \sqrt{2^{2} + 2^{2}} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2 \times 1.414 = 2.828\ \mathrm{cm}\] 

Surface area: \[S = \pi r(l + 2r) = 3.14 \times 2 \times (2.828 + 2 \times 2)\] 

\[S = 6.28 \times (2.828 + 4) = 6.28 \times 6.828 \approx 42.88\ \mathrm{cm^{2}}\]

Answer: \[\text{Volume of the toy} = 25.12\ \mathrm{cm^{3}}\]

\[\text{Surface area of the toy} = 42.88\ \mathrm{cm^{2}}\]