Answer:
Explanation:
Volume of a triangular pyramid =
![(1)/(3)(\text{Area of the triangular base})(\text{Height})](https://img.qammunity.org/2022/formulas/mathematics/high-school/pluo6e4007nt3q7bqo6gjxenvevp4ob5wt.png)
1). Volume of the pyramid =
![(1)/(3)(72)(15)](https://img.qammunity.org/2022/formulas/mathematics/high-school/e5ag767slvp0wdb36w8iql45rl77hwcesm.png)
= 360 cm³
2). Volume of the pyramid =
![(1)/(3)(\text{Area of the triangular base})(\text{Height})](https://img.qammunity.org/2022/formulas/mathematics/high-school/pluo6e4007nt3q7bqo6gjxenvevp4ob5wt.png)
Area of the triangular base =
![(1)/(2)(\text{Base})(\text{height})](https://img.qammunity.org/2022/formulas/mathematics/high-school/u9qe3kamrh9yusn4u8qw87bnyf9be4if4c.png)
=
= 20 cm²
Therefore, volume of the pyramid =
![(1)/(3)(20)(12)](https://img.qammunity.org/2022/formulas/mathematics/high-school/l62be564t74h0t7q7wosz3hgf9gvfrtvfv.png)
= 80 cm³
5). Volume of the cone =
![(1)/(3)\pi r^(2)h](https://img.qammunity.org/2022/formulas/mathematics/high-school/g2o1qnk94h22ha6zh90h67y42yei3uoash.png)
Here, r = radius of the circular base
h = Height of the cone
Volume of the cone =
![(1)/(3)\pi (2)^2(4)](https://img.qammunity.org/2022/formulas/mathematics/high-school/d0ljrft6es83sayz3l41lgb8fzjg6dgrox.png)
=
![(16\pi )/(3)](https://img.qammunity.org/2022/formulas/mathematics/high-school/rxk4wsqwjv8vxqyqzmxmrhs2hla2l8uter.png)
= 16.76 cm³