Problem 945

The total surface area of the cone includes the circular base and the lateral surface.

For the given unpainted cone, the base has area $\text{r 2 = (3 cm ) 2 = 9 cm 2}$ and the lateral surface has area $\text{rs = (3 cm )(5 cm ) = 15 cm 2}$.

Thus, the total surface area of the unpainted cone is $\text{9 cm 2 + 15 cm 2 = 24 cm 2}$.

Since the height and base are perpendicular, the lengths $s$, $h$ and $r$ form a right-angled triangle with hypotenuse $s$.

For this cone, by the Pythagorean Theorem, $\text{h 2 = s 2 - r 2 = (5 cm ) 2 - (3 cm ) 2 = 16 cm 2}$ and so $\text{h = 4 cm}$.

When the unpainted cone is placed in the container of paint so that the paint rises to a depth of 2 cm, the base of the cone (area $\text{9 cm 2}$) is covered in paint.

Also, the bottom portion of the lateral surface is covered in paint.

[[IMAGE0]]

Hide/Reveal Alternative Format for the Cone and Triangle

Two diagrams are side by side. Left diagram shows a cone where the lower portion of the cone is shaded so that the upper unshaded portion looks like a smaller cone within the original one. Right diagram shows a right-angled triangle whose height looks to be the same as the cone's height, and whose base looks to be half the base of the cone. The upper portion of the right-angled triangle is a smaller right-angled triangle whose height is labelled 2 centimetres. Also, the height of the lower remainder of the original triangle is labelled 2 centimetres.

The unpainted portion of the cone is itself a cone with height 2.

When we take a vertical cross-section of the cone through its top vertex and a diameter of the base, the triangle formed above the paint is similar to the original triangle and has half its dimensions. The triangles are similar because both are right-angled and they share an equal angle at the top vertex. The ratio is $2:1$ since their heights are $\text{4 cm}$ and $\text{2 cm}$.

Therefore, the unpainted cone has radius 1.5 cm (half of the original radius of 3 cm) and slant height 2.5 cm (half of the original slant height of 5 cm).

Thus, the unpainted lateral surface area is the lateral surface area of a cone with radius 1.5 cm and slant height 2.5 cm, and so has area $\text{(1.5 cm )(2.5 cm ) = 3.75 cm 2}$.

This means that the painted lateral surface area is $\text{15 cm 2 - 3.75 cm 2 = 11.25 cm 2}$. (This is in fact three-quarters of the total surface area. Can you explain why this is true in a different way?)

Thus, the fraction of the total surface area of the cone that is painted is $$\text{9 cm 2 + 11.25 cm 2 24 cm 2 = 20.25 24 = 81 96 = 27 32}$$ Since $\dfrac{27}{32}$ is in lowest terms, then $p + q = 27 + 32 = 59$.