1. **State the problem:** We have a frustum formed by removing a smaller cone from a larger cone. The larger cone has height 45 cm, the smaller cone removed has height 9 cm and base radius 4 cm. We need to find: a) The length $x$, which is the radius of the frustum's bottom base. b) The total surface area of the frustum in terms of $\pi$. 2. **Find the radius $x$ of the larger cone's base:** Since the cones are similar, their corresponding dimensions are proportional. Let $R$ be the radius of the larger cone's base (which is $x$). The ratio of heights equals the ratio of radii: $$\frac{R}{45} = \frac{4}{9}$$ Solve for $R$: $$R = 45 \times \frac{4}{9} = 5 \times 4 = 20$$ So, $x = 20$ cm. 3. **Calculate the slant heights:** - Slant height of the larger cone $l = \sqrt{45^2 + 20^2} = \sqrt{2025 + 400} = \sqrt{2425}$ cm. - Slant height of the smaller cone $l_s = \sqrt{9^2 + 4^2} = \sqrt{81 + 16} = \sqrt{97}$ cm. - Slant height of the frustum $l_f = l - l_s = \sqrt{2425} - \sqrt{97}$ cm. 4. **Calculate the total surface area of the frustum:** The total surface area $A$ includes: - The area of the top circle (smaller base): $\pi \times 4^2 = 16\pi$ - The area of the bottom circle (larger base): $\pi \times 20^2 = 400\pi$ - The lateral surface area of the frustum: $$\pi (R + r) l_f = \pi (20 + 4)(\sqrt{2425} - \sqrt{97}) = 24\pi (\sqrt{2425} - \sqrt{97})$$ So, $$A = 16\pi + 400\pi + 24\pi (\sqrt{2425} - \sqrt{97}) = 416\pi + 24\pi (\sqrt{2425} - \sqrt{97})$$ **Final answers:** - a) $x = 20$ cm - b) Total surface area $= 416\pi + 24\pi (\sqrt{2425} - \sqrt{97})$ cm$^2$