A frustum of a frustum cone has a total surface area of $450{ \mathrm{m}}^2$. The top cone has a diameter of $2 \mathrm{m}$ and the bottom cone has a diameter of $5 \mathrm{m}$.What is the frustum cone's curved surface area?

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. Using the relationships radius=diameter2, calculate the values of r and R. Apply the formula for total surface area of a frustum,$T.S.A=\pi \left(r+R\right)l+\pi \left(r^2+R^2\right),$where $C.S.A=\pi \left(R+r\right)l.$

Total surface area $=\pi (\mathrm{R}+\mathrm{r})\mathrm{s}+\pi {\mathrm{R}}^2+\pi {\mathrm{r}}^2$

Curved surface area $=\pi (\mathrm{R}+\mathrm{r})\mathrm{s}$

Diameter = radius $\times 2$

Radius, $\mathrm{R}=2.5 \mathrm{m},\mathrm{r}=1 \mathrm{m}$

Therefore,  $450=$Curved surface area $+\pi \left(2.5^2+1^2\right)$

$450=$ Curved surface area $+3\times (6.25+1)$

$450-21.75=$ Curved surface area

Curved surface area$=428.25{ \mathrm{m}}^2$