10.07 Volume

A prism is defined as a solid geometric figure whose two end faces are exactly the same shape, straight and parallel.

If the solid is a prism, its volume can be worked out by using the following rule: \\\text{Volume}=\text{Area of base}\times \text{Height}, where the base is one of the two faces that are identical and parallel to one another and the height is perpendicular to the base.

In this shape, the base is the L-shaped section. It is possible to calculate the volume of an irregular prism by first calculating the area of the base and multiplying this by the height.

\text{Volume of an irregular prism } = \text{Area of the base (L shape) } \times \text{Height of the prism }

Volume of common prisms:

\text{Volume of rectangular prism } = \text{Length }\times \text{Width }\times \text{Height }

\text{Volume of triangular prism }=\text{Area of the triangle }\times \text{Height of the prism }

\text{Volume of trapezium based prism }=\text{Area of the trapezium }\times \text{Height of the prism}

You are about to go camping for the first time and need to buy a tent. You visit the local store that sells camping gear and one tent immediately attracts your attention but it appears to be a little too small for your liking. When you ask one of the salesmen for a tent with a larger volume he points you to a tent that has a similar length but double the height and that he claims is the largest tent in store. When you ask another salesman he points you to a tent that has a similar height to your original choice but double the length and that he claims is the largest tent in store. Who is right? Can they both be right?

If you think back to the formula for the volume of a triangular prism, \\ \dfrac{1}{2}\times \text{base }\times \text{height }\times \text{length of prism } if the prism is laid out like a tent, you will notice that when the height doubles, so too will the volume; and when the length doubles, so too will the volume. So in the case of the salesmen, both of them could be right because both of the tents they point out have the same volume.

Examples

Example 1

Find the volume of the cube shown.

Worked Solution

Create a strategy

Use the volume of a cube formula.

Apply the idea

\displaystyle V	\displaystyle =	\displaystyle s^3	Use the volume formula
	\displaystyle =	\displaystyle (12)^3	Substitute s=12
	\displaystyle =	\displaystyle 1728\text{ cm}^3	Evaluate

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Example 2

Find the volume of the triangular prism shown.

Worked Solution

Create a strategy

Find the area of the triangle, then use the volume of a triangular prism formula.

Apply the idea

\displaystyle A	\displaystyle =	\displaystyle \dfrac{1}{2} bh	Use the area of a triangle formula
	\displaystyle =	\displaystyle \dfrac{1}{2} \times 4 \times 2	Substitute b=4 and h=2
	\displaystyle =	\displaystyle 4 \text{ cm}^2	Evaluate
\displaystyle V	\displaystyle =	\displaystyle Ah	Use the volume of a triangular prism formula
	\displaystyle =	\displaystyle 4 \times 8	Substitute A=4 and h=8
	\displaystyle =	\displaystyle 32 \text{ cm}^3	Evaluate

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Example 3

Find the volume of the prism by finding the base area first.

Worked Solution

Create a strategy

Find the area of the trapezium, then multiply it to the height of the prism.

Apply the idea

\displaystyle A	\displaystyle =	\displaystyle \dfrac{1}{2}\times \left(a+b\right)\times h	Use the area of a trapezium formula
	\displaystyle =	\displaystyle \dfrac{1}{2}\times \left(10+13\right)\times 7	Substitute a=10,\,b=13 and h=7
	\displaystyle =	\displaystyle \dfrac{1}{2}\times 23 \times 7	Evaluate the sum
	\displaystyle =	\displaystyle 80.5\text{ cm}^2	Evaluate

\displaystyle V	\displaystyle =	\displaystyle Ah	Use the volume of a ttrapezoidal prism formula
	\displaystyle =	\displaystyle 80.5 \times 5	Substitute A=80.5 and h=5
	\displaystyle =	\displaystyle 402.5\text{ cm}^2	Evaluate

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A cylinder is a type of prism where the base shape is a circle.

\text{Volume of cylinder }=\pi r^2h

Examples

Example 4

Find the volume of the cylinder shown.

Round your answer to two decimal places.

Worked Solution

Create a strategy

Use the volume of a cylinder formula.

Apply the idea

\displaystyle V	\displaystyle =	\displaystyle \pi r^2 h	Use the formula
	\displaystyle =	\displaystyle \pi \times 3^2 \times 13	Substitute r=3 and h=13
	\displaystyle =	\displaystyle 367.57 \text{ cm}^3	Evaluate and round

Example 5

A 13\text{ cm} concrete cylindrical pipe has an outer radius of 6\text{ cm} and an inner radius of 4\text{ cm} as shown. Find the volume of concrete required to make the pipe, correct to two decimal places.

Worked Solution

Create a strategy

Subtract the volume of the inner cylinder from the volume of the outer cylinder.

Apply the idea

Volume of the outer cylinder:

\displaystyle V	\displaystyle =	\displaystyle \pi r^{2} h	Use the formula
	\displaystyle =	\displaystyle \pi \times 6^{2} \times 13	Substitute r=6 and h=13
	\displaystyle =	\displaystyle 1470.26\text{ cm}^3	Evaluate

Volume of the inner cylinder:

\displaystyle V	\displaystyle =	\displaystyle \pi r^{2} h	Use the formula
	\displaystyle =	\displaystyle \pi \times 4^{2} \times 13	Substitute r=4 and h=13
	\displaystyle =	\displaystyle 653.45\text{ cm}^3	Evaluate

The volume of concrete required to make the pipe:

\displaystyle V	\displaystyle =	\displaystyle 1470.26 - 653.45	Subtract the volume of the cylinders
	\displaystyle =	\displaystyle 816.81\text{ cm}^3	Evaluate

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A pyramid is formed when the vertices of a polygon are projected up to a common point (called a vertex). A right pyramid is formed when the apex is perpendicular to the midpoint of the base.

\text{Volume of Pyramid} = \dfrac{1}{3}\times \text{Area of base} \times \text{Height}

Examples

Example 6

Find the volume of the square pyramid shown.

Worked Solution

Create a strategy

Use the formula for the volume of a pyramid given by V=\dfrac{1}{3} A h, where the area of a square is given by A=s^2.

Apply the idea

We are given s=9 and h=3.

\displaystyle V	\displaystyle =	\displaystyle \dfrac{1}{3} s^2 h	Write the formula
	\displaystyle =	\displaystyle \frac{1}{3}\times 9^2\times 3	Subsitute the values
	\displaystyle =	\displaystyle 81 \text{ cm}^3	Evaluate

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Example 7

A small square pyramid of height 6 \text{ cm} was removed from the top of a large square pyramid of height 12 \text{ cm} forming the solid shown. Find the exact volume of the solid.

Worked Solution

Create a strategy

Subtract the volume of the smaller pyramid from the volume of the larger pyramid.

Apply the idea

Volume of the bigger pyramid:

\displaystyle V	\displaystyle =	\displaystyle \dfrac{1}{3}Ah	Use the formula
	\displaystyle =	\displaystyle \dfrac{1}{3}\times 4^2 \times (6 + 6)	Substitute A=4^2=16 and h=6+6=12
	\displaystyle =	\displaystyle 64\text{ cm}^3	Evaluate

Volume of the smaller pyramid:

\displaystyle V	\displaystyle =	\displaystyle \dfrac{1}{3}Ah	Use the formula
	\displaystyle =	\displaystyle \dfrac{1}{3}\times 2^2 \times 6	Substitute A=2^2=4 and h=6
	\displaystyle =	\displaystyle 8\text{ cm}^3	Evaluate

Exact volume of the solid:

\displaystyle V	\displaystyle =	\displaystyle 64-8	Subtract the volume of the pyramids
	\displaystyle =	\displaystyle 56\text{ cm}^3	Evaluate

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The volume of a sphere with radius r can be calculated using the following formula.

\text{Volume of sphere }=\dfrac{4}{3}\pi r^3

Examples

Example 8

Find the volume of the sphere shown. Round your answer to two decimal places.

Worked Solution

Create a strategy

Use the formula for the volume of the sphere: V = \dfrac{4}{3} \pi r^3

Apply the idea

\displaystyle V	\displaystyle =	\displaystyle \dfrac{4}{3} \times \pi \times 4^3	Substitute r = 4
	\displaystyle \approx	\displaystyle 268.08 \text{ cm}^3	Evaluate and round

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Example 9

A sphere has a radius of r\text{ cm} and a volume of \dfrac{512 \pi}{3}\text{ cm}^3. Find the radius of the sphere to two decimal places. Round your answer to two decimal places and write as an equation.

Worked Solution

Create a strategy

Use the formula for the volume of the sphere: V = \dfrac{4}{3} \pi r^3

Apply the idea

\displaystyle \dfrac{512 \pi}{3}	\displaystyle =	\displaystyle \dfrac{4}{3} \times \pi \times r^3	Substitute V=\dfrac{512 \pi}{3}
\displaystyle \dfrac{512}{3}	\displaystyle =	\displaystyle \dfrac{4}{3}\times r^3	Divide both sides by \pi
\displaystyle r^3	\displaystyle =	\displaystyle \dfrac{512}{3} \times \dfrac{3}{4}	Divide both sides of the equation by \dfrac{4}{3}
\displaystyle r^3	\displaystyle =	\displaystyle \dfrac{512}{4}	Cancel out the common factors
\displaystyle r	\displaystyle =	\displaystyle \sqrt[3]{\frac{512}{4}}	Take the cube root of both sides
\displaystyle r	\displaystyle =	\displaystyle 5.04\text{ cm}	Evaluate

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