Question

A deep-cone clarifier tank is being built. The radius is 5 feet, the height of the cylinder is 8 feet and the height of the cone is 6 feet. Assuming there is no sheet metal overlap, how much sheet metal is needed to build this tank including the top? Must show work

Answer 1

Is needed
`452.32\ ft^(2)` of sheet metal to build this tank

Explanation:

we know that

To find how much sheet metal is needed to build this tank including the top, calculate the lateral area of the cone plus the lateral area of the cylinder plus the area of the top of the cylinder

`A=\pi rl+2\pi rh1+\pi r^(2)`

we have

`r=5\ ft`

`h1=8\ ft` ----> height of the cylinder

`h2=6\ ft` ----> height of the cone

Find the slant height of the cone l

Applying Pythagoras Theorem

`l^(2)=r^(2)+h2^(2)`

substitute

`l^(2)=5^(2)+6^(2)`

`l^(2)=61`

`l=√(61)\ ft`

assume

`\pi =3.14`

`A=(3.14)(5)(√(61))+2(3.14)(5)(8)+(3.14)(5)^(2)`

`A=122.62+251.2+78.5=452.32\ ft^(2)`