Frustums

The volume of a cone is \dfrac{1}{3}\pi r^2 h. To find the volume of the frustum we need the volume of the whole cone and the volume of the missing portion. To calculate the latter, we will need the radius of the top of the frustum by using similarity.

 

The whole cone and the missing portion are similar. To find the scale factor between the dimensions of these two shapes, we must divide the height of one by the height of the other. The height of the missing portion is 50-30=20\text{cm}, so the scale factor is 50\div20 = 2.5. Therefore, the radius of the top face of the frustum is 10\div 2.5 = 4\text{cm}. So, we get:

 

\text{Volume of original cone}=\dfrac{1}{3}\pi\times 10^2\times 50=\dfrac{5000}{3}\pi

 

\text{Volume of missing portion}=\dfrac{1}{3}\pi\times 4^2\times 20=\dfrac{320}{3}\pi

 

Finally, we get:

 

\text{Volume of frustum} = \dfrac{5000}{3}\pi - \dfrac{320}{3}\pi = 4,900 \text{cm}^3

To work out the surface area of this frustum we need the area of both the top square and the bottom square as well as the 4 trapezia that form the sides of the frustum. Since the apex is directly above the centre of the base, all these trapezia will be identical.

 

Working out the area of the squares is easy. The formula for the area of a trapezium is

 

\dfrac{1}{2}(a + b)h

 

where a and b are the two parallel sides. In this case, we know the two parallel sides are lengths 4m and 8m, but we don’t know the perpendicular height of each trapezium face. To do this, we’re going to need to use 3D trigonometry.

 

By constructing a right-angled triangle between the apex, the centre of the base, and the centre of one of the base’s sides, we can find the height of one of the triangular sides of the whole pyramid.

 

 

We know that the base of this triangle will be half the base of the whole pyramid, so 4m, and we know the height is 7.5m. So, using Pythagoras, we get:

 

c^2 = 4^2 + 7.5^2 \text{ therefore } c = \sqrt{72.25} = 8.5\text{m}

 

The whole pyramid and the missing portion are similar shapes. Given that the base of one has width 8m and the base of the other has 4m, the scale factor is clearly 2. Therefore, now we know that the perpendicular height of one of the triangular faces of the whole pyramid is 8.5m, we can calculate the perpendicular height of one of the trapezoidal faces of the frustum:

 

8.5 \div 2 = 4.25\text{m}

 

Therefore, we get:

\text{Area of trapezium} = \dfrac{1}{2}(4 + 8)\times 4.25 = 25.5\text{ m}^2

 

To get the whole area, we must add 4 of these areas to the areas of the top face and the bottom face. We get:

 

\text{Total area } = (4 \times 25.5) + 4^2 + 8^2 = 182\text{ m}^2

The whole cone and the missing portion are similar shapes. Given that the base of one has a radius of 7 cm and the base of the other has a radius of 3 cm, the scale factor is, 3 \div 7 = \dfrac{3}{7}

 

Therefore the height of the smaller cone is:

 10 \times \dfrac{3}{7} =\dfrac{30}{7} \text{cm}

 

Thus, subtracting the height of the smaller cone from that of the original we are left with:

 

\text{Height of frustum}=10 -\dfrac{30}{7} \approx 5.71 \text{cm}

The whole cone and the missing portion are similar shapes. Given that the base of one has a radius of 12 cm and the base of the other has a radius of 3 cm, the scale factor is, 12 \div 3 = 4

 

Therefore the slanted height of the smaller cone is:

 

15 \div 4 = 3.75 \text{cm}

 

Subtracting the slanted height of the smaller cone from the complete cone we are left the the slanted height of the frustum:

 

\text{slanted height of frustum}= 15-3.75 = 11.25\text{cm}

First we must find the volume of the complete triangular pyramid using the formula,

\text{volume}=\dfrac{1}{3}\times \text{base area}\times \text{height}.

 

the area of a triangle can be calculated using \dfrac{1}{2}\text{ab}\sin{(c)}, which in the case of equilateral triangle where all three sides are the same and each interior angle is 60 degrees, this becomes,

 

\begin{aligned} \dfrac{1}{2}\text{ab}\sin{(c)} &=  \dfrac{1}{2}\text{a}^2\sin{(60)} \\ &=  \dfrac{1}{2}\times a^2 \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{4} a^2\end{aligned}

 

In this instance a = 11 \text{cm}, thus

 

\text{base area}=\dfrac{\sqrt{3}}{4}\times 11^2

 

The height of the original pyramid is 20 \text{cm} so,

 

\text{volume of complete pyramid}=\dfrac{1}{3}\times\dfrac{\sqrt{3}}{4}\times 11^2 \times 20 =349.3\text{cm}^3

 

The volume of the smaller pyramid that is removed from the top can be calculated in much the same way:

 

\text{volume of top pyramid}=\dfrac{1}{3}\times \dfrac{\sqrt{3}}{4}\times 7^2 \times 14 = 99 \text{cm}^3

 

Subtracting the volume of the smaller pyramid from the complete pyramid we are left the the volume of the frustum:

 

\text{volume of frustum}= 349.3 - 99 = 350.3\text{cm}^3