Explanation -:
In this question we are provided with the height and radius. We are asked to calculate the volume of a composite solid
First we will find the volume of a cone
We know,
![\orange{\star \: \small\boxed{ \sf{ Volume_((cone)) = (1)/(3)πr²h}}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/dp4lwg5scvzfyr5ev5ul7z1oje1g8pa5k8.png)
Where,
- r stand for radius
- h stand for height
- Assuming π as 3.14
Substituting the values we get
![\small\bf Volume_((cone)) = (1)/(3) * 3.14×4 × 4×5](https://img.qammunity.org/2023/formulas/mathematics/high-school/u60xpledrk0tzyg8r0hksllyo0qyht4dpc.png)
![\rightarrow \small\rm{ Volume_((cone)) = (1)/(3)×251.2}](https://img.qammunity.org/2023/formulas/mathematics/high-school/4ta1c2vlk38jngpdez7odxna5r92og429f.png)
![\small\sf{ Volume_((cone)) = 83.73 \:cubic \: units }](https://img.qammunity.org/2023/formulas/mathematics/high-school/q4s3twz3ng4f34d01o6q92144f7z1b940v.png)
Now we will calculate the volume of a hemisphere
We know,
![\red{\star \: \small \boxed{\sf{ Volume_((hemisphere)) = (2)/(3)πr³}}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/8l8i6u4bi5zvumed2kwmjmk8ut6eubgbtf.png)
Substituting the values we get
![\small\bf{ Volume_((hemisphere)) = (2)/(3) * 3.14 × 4 × 4 × 4}](https://img.qammunity.org/2023/formulas/mathematics/high-school/s8937b3r8ac9yfuh8xdjyz54usuldbiake.png)
![\rightarrow\small\rm{ Volume_((hemisphere)) = (2)/(3) * 200.96}](https://img.qammunity.org/2023/formulas/mathematics/high-school/wzg12168ahmi7cyhplhsrf26xjer4qjbde.png)
![\rightarrow\small\rm{Volume_((hemisphere)) = 2 * 66.98}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ch4rsn9pq7evzl95c7pkcdvtbg15xrbp88.png)
![\small\sf{ Volume_((hemisphere)) =133.97}](https://img.qammunity.org/2023/formulas/mathematics/high-school/20utnd12nkb864i60r1ig8a9km2cxv8646.png)
Now we will calculate the volume
Volume = 83.73 + 133.97 = 217.70 cubic units