[tex] \huge \mathrm {➢ \: \: \: \: Answer }[/tex]
The above figure consists of three different shapes, that is :
Let's solve for surface area of whole figure :
1. Curved surface area of hemisphere : -
[tex] \large \boxed{ \boxed{2\pi {r }^{2} }}[/tex]
[tex]➢ \: \: 2 \times 3.14 \times 4 \times 4[/tex]
[tex]➢ \: \: 100.48 \: m {}^{2} [/tex]
2. lateral surface area of Cylinder : -
[tex] \large \boxed {\boxed{2\pi rh}}[/tex]
[tex]➢ \: \: 2 \times 3.14 \times 4 \times 10[/tex]
[tex]➢ \: \: 251.2 \: { m}^{2} [/tex]
3. curved surface area of cone :
[tex]slant \: height = \sqrt{4 {}^{2} + 3 {}^{2} } [/tex]
(by Pythagoras theorem)
[tex] \large \boxed { \boxed{c.s.a = \pi rl}}[/tex]
[tex]➢ \: \: 3.14 \times 4 \times 5[/tex]
[tex]62.8 \: m {}^{2} [/tex]
Surface area of the figure :
[tex]➢ \: \: 100.48 + 251.2 + 62.8[/tex]
[tex]➢ \: \: 414.48 \: \: m {}^{2} [/tex]
[tex]➢ \: \: 414.5 \: m {}^{2} \: \: \: (approx)[/tex]
Now, let's solve for volume :
Volume of given figure will be combined volume of the all three shapes,
1. Volume of hemisphere :
[tex] \boxed{ \boxed{\dfrac{2}{3} \pi {r}^{3} }}[/tex]
[tex]➢ \: \: \dfrac{2}{3} \times 3.14 \times 4 \times 4 \times 4[/tex]
[tex]➢ \: \: 133.97 \: \: { m}^{3} [/tex]
2. Volume of Cylinder :
[tex] \large\boxed{ \boxed{ \pi{r}^{2}h }}[/tex]
[tex]➢ \: \: 3.14 \times 4 \times 4 \times 10[/tex]
[tex]➢ \: \: 502.4 \: m {}^{3} [/tex]
3. Volume of cone :
[tex] \large \boxed {\boxed{ \frac{1}{3} \pi {r}^{2}h' }}[/tex]
[tex]➢ \: \: \dfrac{1}{ 3} \times 3.14 \times 4 \times 4 \times 3[/tex]
[tex]➢ \: \: 50.24 \: {m}^{3} [/tex]
The total volume of the given figure :
[tex]➢ \: \: 133.97 + 502.4 + 50.24[/tex]
[tex]➢ \: \: 686.61 \: {m}^{3} [/tex]
or
[tex]➢ \: \: 686.6 \: \: m {}^{3} \: \: \: (approx)[/tex]
[tex] \mathrm{✌TeeNForeveR✌}[/tex]
