Question

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

Answer 1

Answer:

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

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

Step-by-step 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:

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