Frustum of a Cone: Definition, Properties & Examples

Frustum is a Latin word that means piece cut off. Its plural is frusta. When we cut a cone by plane into two parts, the part of a cone obtained is referred to as the frustum of a cone. The upper part of the cone remains the same in shape, but the bottom part makes a frustum. To get this part of the right circular cone, the cone must be divided horizontally, parallel to the base. The area and volume of both these pieces differ distinctly. A glass tumbler resembles a cone with its pointed part removed. It can be said that a tumbler is a resemblance to the frustum of a cone.

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Cone: Definition

A cone is a typical three-dimensional geometric figure with a flat surface, and a curved surface pointed towards the top. The pointed end of the cone is known as the vertex, whereas the flat surface is called the base.

A cone is a shape formed by using a set of lines that connect a common point, called the vertex, to all the points of a circular base. The distance from the vertex of the cone to the bottom is the height \(h\) of the cone. The circular base has a measured value of radius \(r\). The length of the cone from apex to any point on the circumference of the base is the slant height \(l\).

Surface Area and Volume of a Cone

Curved Surface Area of a Cone

The curved surface area of a cone is \(\pi rl\).
Where, \(l = \sqrt {\left({{h^2} + {r^2}} \right)} \)

Total Surface Area of a Cone

The surface area of the cone is the sum of lateral surface area and the area of the circular base.
Therefore, the total surface area of cone \( = \pi rl + \pi {r^2}\)
\( \Rightarrow \pi r(l + r)\)
So, the total surface area of cone \(=\pi r(l+r)\)

Volume of a Cone

The volume \(V\) of a cone with a radius \((r)\), height \((h)\) is given by \(V=\frac{1}{3} \pi r^{2} h\)

Frustum of a Cone

If a plane parallel to its base cuts off a right circular cone, then the portion of the cone between the cutting plane and the base of the cone is called a frustum of the cone.

A frustum of a right circular cone has two unequal flat circular bases and a curved surface.
Let us now define some other terms like height, slant height, etc., related to a frustum of a cone.

Height: The height \((h)\) or thickness of the frustum is the perpendicular distance between its two circular bases.
Slant Height: The slant height \((l)\) of a frustum of a right circular cone is the length of the line segment joining the extremities of two parallel radii, drawn in the same direction of the two circular bases.
Radii: As frustum is having two circular bases, there will be two radii in a frustum.

Examples of Frustum of a Cone

There are various real-life examples for a frustum of a cone. Some well-known examples are the shade of a table lamp, bucket, glass tumbler, etc.

Surface Area of Frustum of a Cone

The curved surface area of a frustum of a cone \(=\pi\left(r_{1}+r_{2}\right) l\)
where, \(l=\sqrt{h^{2}+\left(r_{1}-r_{2}\right)^{2}}\)
The total surface area of a frustum of a cone \(=\pi l\left(r_{1}+r_{2}\right)+\pi r_{1}^{2}+\pi r_{2}^{2}\)
where, \(l=\sqrt{h^{2}+\left(r_{1}-r_{2}\right)^{2}}\)
These formulas can be derived using the idea of the similarity of triangles.

Volume of Frustum of a Cone

Let \(h\) be the height, \(l\) be the slant height and \(r_{1}\) and \(r_{2}\) be the radii of the bases \(\left(r_{1}>r_{2}\right)\) of the frustum of a cone. Then we can directly find the volume and is given by,
The volume of a frustum of the cone \(=\frac{1}{3} \pi h\left(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right)\)

Solved Examples – Frustum of a Cone

Question 1: If the radii of the circular ends of a conical bucket which is \(45\,{\text{cm}}\) high are \(28\,{\text{cm}}\) and \(7\,{\text{cm}}\). Find the capacity of the bucket. (Use \(\pi = \frac{{22}}{7}\))
Ans: The bucket forms a frustum of cone such that the radii of its circular ends are \(r_{1}=28 \mathrm{~cm}, r_{2}=7 \mathrm{~cm}\) and height \(h=45 \mathrm{~cm}\). Let \(V\) be the capacity or the volume of the bucket.
Then, the volume of the frustum \(\Rightarrow V=\frac{1}{3} \pi h\left(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right)\)
\(\Rightarrow V=\frac{1}{3} \times \frac{22}{7} \times 45\left[28^{2}+7^{2}+(28 \times 7)\right]\)
\(\Rightarrow V=22 \times 15 \times[(28 \times 4)+7+28]\)
\(\Rightarrow V=330 \times 147 \mathrm{~cm}^{3}\)
\(\Rightarrow V=48510 \mathrm{~cm}^{3}\)
Therefore, the capacity of the bucket is \(48510 \mathrm{~cm}^{3}\).

Q.2. An open metal bucket is in the form of a frustum cone of height \(21 \mathrm{~cm}\) with radii of its lower and upper ends as \(10 \mathrm{~cm}\) and \(20 \mathrm{~cm}\), respectively. Find the cost of milk which can fill the bucket at \(₹ 30\) per litre.
Ans: Let \(r_{1}\) and \(r_{2}\) be the radii of the top and base of the bucket, respectively. Let \(h\) be the height.
Given: \(r_{1}=20 \mathrm{~cm}, r_{2}=10 \mathrm{~cm}\) and \(h=21 \mathrm{~cm}\)
The capacity of the bucket \(=\) The volume of a frustum of the cone
We know that the volume of the frustum \(\Rightarrow V=\frac{1}{3} \pi h\left(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right)\)
\(\Rightarrow V=\frac{1}{3} \times \frac{22}{7} \times 21\left[20^{2}+10^{2}+(20 \times 10)\right] \mathrm{cm}^{3}\)
\(\Rightarrow V=22[500+200] \mathrm{cm}^{3}\)
\(\Rightarrow V=22 \times 700 \mathrm{~cm}^{3}\)
\(\Rightarrow V=15400 \mathrm{~cm}^{3}\)
Therefore, the volume of milk that the bucket can hold \(=\frac{15400}{1000}\) litres \(=15.4\) litres

Q.3. Hanumappa and his wife Gangamma are busy making jaggery out of sugarcane juice. They have processed sugarcane juice to make molasses, which is poured into moulds in the shape of a frustum of a cone having the diameters of its two circular faces as \(30 \mathrm{~cm}\) and \(35 \mathrm{~cm}\) and the vertical height of the mould is \(14 \mathrm{~cm} .\) If each molasses has a mass of about \(1.2 \mathrm{~g}\), find the mass of the molasses that can be poured into each mould. Take \(\pi=\frac{22}{7}\),
Ans: Let \(r_{1}\) and \(r_{2}\) be the radii of the larger and smaller base of the moulds, respectively. Let \(h\) be the height.
Given:
\(d_{1}=35 \mathrm{~cm} \Rightarrow r_{1}=\frac{35}{2}, d_{2}=30 \mathrm{~cm} \Rightarrow r_{2}=\frac{30}{2} \mathrm{~cm}\) and \(h=14 \mathrm{~cm}\)
Since mould is in the shape of a frustum of a cone, quantity (volume) of molasses that can be poured into it \(=\frac{1}{3} \pi h\left(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right)\)
\(\Rightarrow V=\frac{1}{3} \times \frac{22}{7} \times 14\left[\left(\frac{35}{2}\right)^{2}+\left(\frac{30}{2}\right)^{2}+\left(\frac{35}{2} \times \frac{30}{2}\right)\right] \mathrm{cm}^{3}\)
\(\Rightarrow V=\frac{44}{3}[306.25+225+262.5] \mathrm{cm}^{3}\)
\(\Rightarrow V=11641.7 \mathrm{~cm}^{3}\)
It is given that \(1 \mathrm{~cm}^{3}\) of molasses has a mass of \(1.2 \mathrm{~g}\). So, the mass of molasses that can be poured into each mould \(=(11641.7 \times 1.2) g=13970.04 g=13.97 \mathrm{~kg} \approx 14 \mathrm{~kg}\)

Q.4. A bucket is in the form of a cone with a capacity of \(12308.8 \mathrm{~cm}^{3}\) of water. The radii of the top and bottom circular ends are \(20 \mathrm{~cm}\) and \(12 \mathrm{~cm}\), respectively. Find the height of the bucket and the area of the metal sheet used in its making. Use \(\pi=3.14\)

Ans: Given: \(r_{1}=20 \mathrm{~cm}, r_{2}=12 \mathrm{~cm}\) and the volume \(V=12308.8 \mathrm{~cm}^{3}\). Let \(h\) be the height of the bucket.
Here, the volume of the bucket \(=\) the volume of the frustum of a cone.
We know that the volume of the frustum \(\Rightarrow V=\frac{1}{3} \pi h\left(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right)\)
\(\Rightarrow \frac{1}{3} \pi h\left(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right)=12308.8\)
\(\Rightarrow \frac{1}{3} \times 3.14 \times h\left[(20)^{2}+(12)^{2}+(20 \times 12)\right]=12308.8\)
\(\Rightarrow 784 h=\frac{12308.8 \times 3}{3.14}\)
\(\Rightarrow h=\frac{12308.8 \times 3}{3.14 \times 784}\)
\(\Rightarrow h=15\)
Therefore, the height is \(15 \mathrm{~cm}\)
We know that the slant height of bucket \(l=\sqrt{h^{2}+\left(r_{1}-r_{2}\right)^{2}}\)
\(\Rightarrow l=\sqrt{(15)^{2}+(20-12)^{2}} \mathrm{~cm}\)
\(\Rightarrow l=\sqrt{(15)^{2}+(8)^{2}} \mathrm{~cm}=\sqrt{289} \mathrm{~cm}=17 \mathrm{~cm}\)
Area of the metal sheet used \(=\) curved surface area \(+\) area of the bottom
\(=\left[\pi l\left(r_{1}+r_{2}\right)+\pi r^{2}\right]\)
\(=[3.14 \times 17 \times(20+12)+3.14 \times 12 \times 12] \mathrm{cm}^{2}\)
\(=3.14(544+144) \mathrm{cm}^{2}\)
\(=(3.14 \times 688) \mathrm{cm}^{2}\)
\(=2160.32 \mathrm{~cm}^{2}\)
Therefore, the area of metal sheet used is \(2160.32 \mathrm{~cm}^{2}\)

Q.5. The slant height of a frustum of a cone is \(4 \mathrm{~cm}\), and the perimeters (circumference) of its circular ends are \(18 \mathrm{~cm}\) and \(6 \mathrm{~cm}\). Find the curved surface area of the frustum.
Ans: Given: \(l=4 \mathrm{~cm}\)
Circumference of a circular end \(=18 \mathrm{~cm}\)
\(\Rightarrow 2 \pi r_{1}=18 \mathrm{~cm}\)
\(\Rightarrow \pi r_{1}=9 \mathrm{~cm}\)
Circumference of another circular end \(=6 \mathrm{~cm}\)
\(\Rightarrow 2 \pi r_{2}=6 \mathrm{~cm}\)
\(\Rightarrow \pi r_{2}=3 \mathrm{~cm}\)
We know that curved surface area \(=\pi l\left(r_{1}+r_{2}\right)\)
\(=4 \times(9+3) \mathrm{cm}^{2}\)
\(=48 \mathrm{~cm}^{2}\)
Therefore, the curved surface area of the frustum is \(48 \mathrm{~cm}^{2}\).

Summary

In the above article, we have learned the definition of a cone, volume and surface area of a cone. Also, we have studied the frustum of a cone, shape of a frustum of a cone. We learned formulas to find the volume of the frustum of a cone and surface area of the frustum of a cone and solved some examples using the same.

Frequently Asked Questions (FAQ) – Frustum of a Cone

Frequently asked questions related to frustum of a cone is listed as follows:

Q.1. Is a frustum a prism?
Ans: A frustum is a solid portion, between two parallel planes cutting it. If all the edges are identical, then a frustum becomes a uniform prism. But the frustum of a cone is not a prism as a prism cannot have curved faces.

Q.2. What is TSA of a frustum of a cone?
Ans: The total surface area of a frustum of a cone is \(\pi l\left(r_{1}+r_{2}\right)+\pi r_{1}^{2}+\pi r_{2}^{2}\)
where, \(l=\sqrt{h^{2}+\left(r_{1}-r_{2}\right)^{2}}\).

Q.3. What is the volume of a frustum of a cone?
Ans: The volume of a frustum of a cone is given by \(V=\frac{1}{3} \pi h\left(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right)\).

Q.4. Define the frustum of a cone.
Ans: If a plane parallel to its base cuts off a right circular cone, then the portion of the cone between the cutting plane and the base of the cone is called a frustum of the cone.

Q.5. Write the formula to find the curved surface area of a frustum of a cone.
Ans: The curved surface area of a frustum of a cone \(=\pi\left(r_{1}+r_{2}\right) l\)
where, \(l=\sqrt{h^{2}+\left(r_{1}-r_{2}\right)^{2}}\).

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