Question

A piece of wood is designed in the shape of a cone.

A cylindrical hole of radius 4cm and depth 2cm has been drilled into the base.

The surface area of the piece of wood has to be painted, except for the area inside the hole.

Workout the surface area to be painted.

Give your answer in terms of pi.

Answer 1

The surface area of the piece of wood to be painted is:

`A = \pi \cdot (R^(2)-16\,cm^(2)) + 2\pi \cdot R \cdot \sqrt{R^(2)+h^(2)}`

Explanation:

The surface area to be painted is equal to the surface area of the cone minus the cross area of the hole. That is:

`A = \pi \cdot R^(2) + 2\pi\cdot R \cdot \sqrt{R^(2)+h^(2)} - \pi \cdot r^(2)`

`A = \pi \cdot (R^(2)-r^(2)) + 2\pi \cdot R \cdot \sqrt{R^(2)+h^(2)}`

Where:

`R` - Radius at the bottom of the cylinder, measured in centimeters.

`h` - Height of the cylinder, measured in centimeters.

`r` - Radius of the cylindrical hole, measured in centimeters.

If
`r = 4\,cm`, the surface area of the piece of wood to be painted is:

`A = \pi \cdot (R^(2)-16\,cm^(2)) + 2\pi \cdot R \cdot \sqrt{R^(2)+h^(2)}`