Q9. a. The diagram below shows a triangular piece of land LMN in which LM = 23.7m, MN = 16.6m and angle LMN 106.

The radius is half of the diameter.

$$r_1 = \frac{6 \text{ cm}}{2} = 3 \text{ cm}$$ $$r_2 = \frac{18 \text{ cm}}{2} = 9 \text{ cm}$$

The height of the frustum is $h = 10 \text{ cm}$.

The formula for the volume of a frustum of a cone is:

$$V = \frac{1}{3} \pi h (r_1^2 + r_1 r_2 + r_2^2)$$

where $h$ is the height, $r_1$ is the radius of the top base, and $r_2$ is the radius of the bottom base.

$$V = \frac{1}{3} \times 3.142 \times 10 \text{ cm} \times (9 \text{ cm}^2 + 27 \text{ cm}^2 + 81 \text{ cm}^2)$$ $$V = \frac{1}{3} \times 3.142 \times 10 \text{ cm} \times (117 \text{ cm}^2)$$

$$V = \frac{1}{3} \times 3.142 \times 1170 \text{ cm}^3$$ $$V = 3.142 \times 390 \text{ cm}^3$$ $$V = 1225.38 \text{ cm}^3$$

The volume of the frustum is $\text{1225.38 cm}^3$.