A frustum of a cone has a total surface area of 450 m². The top has a diameter of 2 m and the bottom has a diameter of 5 m. What is the curved surface area of the frustum? (Use pi = 3.14)

First, draw a frustum of a cone according to the given conditions. Assume the top cone's radius is r, the bottom cone's radius is R, the frustum height is h, and the slant height is l.

Given, the diameter of the top of the cone is 2 m.

Therefore, r = 2/1 = 1 m

 The diameter of the bottom of the cone is 5 m.

Therefore, R = 5/2 = 2.5 m

Apply the formula for the total surface area of a frustum,

$\mathrm{T}.\mathrm{S}.\mathrm{A}=\mathrm{C}.\mathrm{S}.\mathrm{A}+\mathrm{\pi }\left({\mathrm{r}}^2+{\mathrm{R}}^2\right)$

$\Rightarrow 450=\mathrm{C}.\mathrm{S}.\mathrm{A} + 3.14(1^2+2.5^2)$

$\Rightarrow \mathrm{C}.\mathrm{S}.\mathrm{A}=450 - 3.14\times 7.25$

$\Rightarrow \mathrm{C}.\mathrm{S}.\mathrm{A}=450 - 22.765$

$\Rightarrow \mathrm{C}.\mathrm{S}.\mathrm{A}=427.235 {\mathrm{m}}^2$

Therefore, the curved surface area of the frustum of the cone is 427.235 m².