Question

Find the surface area and volume of the composite figure. Use 3.14 for π and round to the nearest tenth if needed.

Please show step by step, I'm really struggling with this.

Answer 1

`\huge \mathrm {➢ \: \: \: \: Answer }`

The above figure consists of three different shapes, that is :

• A hemisphere
• A Cylinder
• A Cone

Let's solve for surface area of whole figure :

1. Curved surface area of hemisphere : -

• radius (r) = 4 m

`\large \boxed{ \boxed{2\pi {r }^(2) }}`

`➢ \: \: 2 * 3.14 * 4 * 4`

`➢ \: \: 100.48 \: m {}^(2)`

2. lateral surface area of Cylinder : -

• radius (r) = 4 m
• height (h) = 10 m

`\large \boxed {\boxed{2\pi rh}}`

`➢ \: \: 2 * 3.14 * 4 * 10`

`➢ \: \: 251.2 \: { m}^(2)`

3. curved surface area of cone :

• radius (r) = 4 m
• height of cone (h') = 3 m.

`slant \: height = \sqrt{4 {}^(2) + 3 {}^(2) }`

(by Pythagoras theorem)

`\large \boxed { \boxed{c.s.a = \pi rl}}`

`➢ \: \: 3.14 * 4 * 5`

`62.8 \: m {}^(2)`

Surface area of the figure :

`➢ \: \: 100.48 + 251.2 + 62.8`

`➢ \: \: 414.48 \: \: m {}^(2)`

`➢ \: \: 414.5 \: m {}^(2) \: \: \: (approx)`

Now, let's solve for volume :

Volume of given figure will be combined volume of the all three shapes,

1. Volume of hemisphere :

`\boxed{ \boxed{(2)/(3) \pi {r}^(3) }}`

`➢ \: \: (2)/(3) * 3.14 * 4 * 4 * 4`

`➢ \: \: 133.97 \: \: { m}^(3)`

2. Volume of Cylinder :

`\large\boxed{ \boxed{ \pi{r}^(2)h }}`

`➢ \: \: 3.14 * 4 * 4 * 10`

`➢ \: \: 502.4 \: m {}^(3)`

3. Volume of cone :

• height of cone (h') = 3 m

`\large \boxed {\boxed{ (1)/(3) \pi {r}^(2)h' }}`

`➢ \: \: (1)/( 3) * 3.14 * 4 * 4 * 3`

`➢ \: \: 50.24 \: {m}^(3)`

The total volume of the given figure :

`➢ \: \: 133.97 + 502.4 + 50.24`

`➢ \: \: 686.61 \: {m}^(3)`

or

`➢ \: \: 686.6 \: \: m {}^(3) \: \: \: (approx)`

`\mathrm{✌TeeNForeveR✌}`