Area and Volume of Frustum – SS1 Mathematics Lesson Note (Third Term)

A frustum is the part of a cone or pyramid that remains when a plane parallel to the base cuts off the top.

Behavioural objectives

By the end of the lesson, students should be able to:

1. find the volume of a frustum of a cone or pyramid by subtracting the smaller solid from the complete solid, using similar triangles;
2. calculate the volume of a composite solid made from a cube and a square-based pyramid;
3. show that the volume of metal in a cylindrical tube is V = πlt(2R – t), and use the result;
4. prove that the exterior angle of a triangle equals the sum of the two opposite interior angles;
5. use a ratio to find the angles of a triangle.

What is a frustum?

A frustum is the solid left when a cone or pyramid is cut by a plane parallel to its base and the top part is removed.

• Frustum of a cone: has two circular faces, a large base and a smaller top.
• Frustum of a pyramid: has two similar polygon faces, such as two squares, one larger than the other.
• Small cone or pyramid: the top part that is cut off. It is similar to the complete solid.
• Height of frustum: the perpendicular distance between the two parallel faces.
• Slant height: the length along the sloping side of a frustum of a cone, from the top edge to the base edge.

How do you find the volume of a frustum?

The volume of a frustum is the volume of the complete cone or pyramid minus the volume of the small cone or pyramid cut off.

1. Draw the complete solid and mark the small solid cut off the top.
2. Let the small solid have height h and the complete solid have height H.
3. Use similar triangles to link the radii or base lengths: r/R = h/H.
4. Find the volume of the complete solid.
5. Find the volume of the small solid.
6. Subtract: volume of frustum = big solid – small solid.

Useful facts

• Volume of a cone: V = 1/3 πr²h.
• Volume of a pyramid: V = 1/3 × base area × height.
• Similar solids: if lengths are in the ratio 1 : k, volumes are in the ratio 1 : k³.
• Frustum of a cone (formula): V = 1/3 π(H – h)(R² + Rr + r²).
• Curved surface area of a frustum of a cone: π(R + r)l, where l is the slant height.
• Total surface area of a frustum of a cone: curved surface area + πR² + πr².

Example: frustum of a cone

The volume of a right circular cone is 5 litres. Calculate the volumes of the two parts into which the cone is divided by a plane parallel to the base, one-third of the way down from the vertex to the base. Give your answers to the nearest ml.

Solution

1. Let the small cone have height h and radius r. Let the big cone have height H and radius R.
2. The cut is one-third of the way down, so h = 1 part and H = 3 parts. Thus H = 3h.
3. By similar triangles, r/R = h/H = 1/3. Thus R = 3r.
4. Volume of big cone = 5 litres = 5 × 1000 ml = 5000 ml.
5. So 1/3 πR²H = 5000, which gives πR²H = 15000 …….. (1)
6. Use h = H/3 and r = R/3. Volume of small cone = 1/3 × π × (R/3)² × (H/3) = πR²H/81.
7. From (1), volume of small cone = 15000/81 = 185.19 ml (2 decimal places).
8. Volume of frustum = 5000 – 185.19 = 4814.81 ml = 4815 ml (nearest ml).

• Small cone: 185 ml.
• Frustum: 4815 ml.
• Quick check: the lengths are in the ratio 1 : 3, so the volumes are in the ratio 1 : 27. Then 5000 ÷ 27 = 185.19 ml.

How do you find the volume of a composite solid?

A composite solid is a solid made by joining basic solids, and its volume is the sum of the volumes of its parts.

Example 1: cube and pyramid

The figure below shows a composite solid consisting of a cube of edge 28 cm and a square-based pyramid of height 16 cm. Calculate the volume of the solid.

1. Volume of solid = volume of pyramid + volume of cube.
2. = 1/3 b²h + l³
3. = 1/3 × 28 × 28 × 16 + 28³ cm³
4. = 12544/3 + 21952 cm³
5. = (12544 + 65856)/3 cm³
6. = 78400/3 cm³
7. = 26133 1/3 cm³
8. = 26133 cm³ (to the nearest whole number).

Example 2: volume of metal in a tube

The outer radius of a cylindrical metal tube is R and t is the thickness of the metal.

• (a) Show that the volume V of metal in a length, l units, of the tube is given by V = πlt(2R – t).
• (b) Hence calculate V when R = 7.5, t = 1 and l = 20.

Solution to (a)

The volume of metal is the volume of the outside cylinder minus the volume of the inside cylinder.

1. Let R = outer radius, r = inner radius, t = thickness and l = length.
2. Volume of metal = πR²l – πr²l …….. (1)
3. R = t + r, so r = R – t …….. (2)
4. Substitute (2) in (1): V = πR²l – π(R – t)²l
5. = πR²l – π(R² – 2Rt + t²)l
6. = πR²l – πR²l + 2πRtl – πt²l
7. = 2πRtl – πt²l
8. = πlt(2R – t)

Solution to (b)

1. Use π = 22/7. V = πlt(2R – t)
2. = 22/7 × 20 × 1 × (2 × 7.5 – 1)
3. = 22/7 × 20 × 14
4. = 44 × 20
5. = 880 cubic units.

What are supplementary angles?

Supplementary angles are two angles that add up to 180°.

• Angles on a straight line: they add up to 180°.
• Adjacent angles x and y on a line: x + y = 180° (supplementary angles).

What is the exterior angle theorem of a triangle?

The exterior angle of a triangle is equal to the sum of the two opposite interior angles.

• Exterior angle: the angle formed outside a triangle when one side is extended.
• Opposite interior angles: the two interior angles that are not next to the exterior angle.

Proof

• Given: any triangle ABC, with BC produced to D.
• To prove: ∠ACD = x₁ + y₁, where x₁ = ∠BAC and y₁ = ∠ABC.
• Construction: draw CE parallel to BA.

1. Let ∠ACE = x and ∠ECD = y.
2. x₁ = x (alternate angles, BA // CE).
3. y₁ = y (corresponding angles, BA // CE).
4. But ∠ACD = x + y.
5. Therefore ∠ACD = x₁ + y₁.

Example: angles in a ratio

The ratio of the angles of a triangle is 3:4:5. Find the smallest and the largest angles.

Solution

1. Total parts = 3 + 4 + 5 = 12.
2. The angles of a triangle add up to 180°.
3. 1st angle = 3/12 × 180° = 45°.
4. 2nd angle = 4/12 × 180° = 60°.
5. 3rd angle = 5/12 × 180° = 75°.
6. Check: 45° + 60° + 75° = 180°.

The smallest angle is 45° and the largest angle is 75°.

Evaluation

1. What is a frustum?
2. Explain how to find the volume of a frustum of a cone or a pyramid.
3. A small cone is cut from the top of a big cone. Its height is one-third of the height of the big cone. What fraction of the volume of the big cone is the volume of the small cone?
4. Write down the formula for the volume of metal in a cylindrical tube of outer radius R, thickness t and length l.
5. State the exterior angle theorem of a triangle.

Assignment

1. A right pyramid on a base 10 m square is 15 m high.
   
   (a) Find the volume of the pyramid.
   
   (b) If the top 6 m of the pyramid are removed, what is the volume of the remaining frustum?