Lesson Explainer: Volumes of Pyramids Mathematics

In this explainer, we will learn how to find the volume of a pyramid and solve problems including real-life situations.

Definition: Pyramids

Pyramids are three-dimensional geometric shapes where the base is a polygon and all other sides are triangles that meet at the apex, or vertex.

A right pyramid is a pyramid whose apex lies above the centroid of the base.

A regular pyramid is a right pyramid whose base is a regular polygon: all the sides of the base are of equal length and all the pyramid’s lateral edges are of equal length.

The perpendicular height of a pyramid is the distance from the apex to the base.

The slant height of a pyramid is the distance measured along a lateral face from the apex to the base edge. In other words, it is the height of the triangle comprising a lateral face.

We can note the distinction between the slant height and perpendicular heights in the diagram below. The perpendicular height is the height of the pyramid measured at a right angle from the base.

Now that we have learned what a pyramid is, let us look at its volume.

Imagine that we can fill a pyramid completely with, say, water. If we poured this water into a prism of the same base and height as the pyramid, we would observe that the level of water is exactly at one-third of the height of the prism.

This is a general rule for any pyramid.

Definition: The Volume of a Pyramid

The volume of a pyramid is one-third of the volume of the prism of the same base and perpendicular height: 𝑉=13(π΄Γ—β„Ž).pyramidbase

In some problems, we may be given the area of the base, or we may be expected to calculate the area of a familiar two-dimensional shape at the base using its given dimensions.

We will now see how we can apply this formula to find the volume of a pyramid. In the first example, we will find the volume of a triangular pyramid.

Example 1: Finding the Volume of a Based Pyramid with a Right Triangular Base

Determine, to the nearest hundredth, the volume of the shown pyramid.

Answer

We can recall that the volume of a pyramid is one-third of the volume of the prism of the same base and perpendicular height: 𝑉=13(π΄Γ—β„Ž).pyramidbase

We are not given the area of the triangular base of the pyramid, but we can calculate it using the given dimensions. The area of a triangle is given by 𝐴=Γ—2.trianglebaseperpendicularheight

Note that the height of the triangle is the height of the base triangle and is not the same as the perpendicular height of the pyramid.

We can take the base length of this triangle as 6 m, and, as the triangle is a right triangle we can easily determine that the height is 4.7 m. Substituting these into the area formula gives 𝐴=6Γ—4.72=14.1.trianglem

As the triangle is the base of the pyramid, we can substitute 𝐴=14.1basem and β„Ž=9m into the volume formula to give 𝑉=13(14.1Γ—9)=42.3.pyramidm

Note that since we have multiplied an area with square units by a length to give a volume, then the volume is given in cubic units.

We can give our answer to the nearest hundredth as 42.30 m3.

In the next example, we will see how we can apply the volume formula to find an unknown height in a real-world context.

Example 2: Solving Real-World Problems Using the Volume of a Pyramid

The Louvre Pyramid in Paris has a square base whose sides are 112 feet long. Given that its volume is 296β€Žβ€‰β€Ž875 cubic feet, find the height of the pyramid to the nearest foot.

Answer

In order to find the volume of this pyramid, we can use the formula 𝑉=13(π΄Γ—β„Ž),pyramidbase where 𝐴base is the area of the base of the pyramid and β„Ž is the height.

We will need to determine the area of the square at the base. Recall that the area of a square is found by calculating the square of the length of one side. Thus, we have 𝐴=112=12544.baseft

We can then substitute the given volume of the pyramid, 𝑉=296875pyramid, and 𝐴=12544base into the volume formula and simplify. This gives us 296875=13(12544Γ—β„Ž)296875Γ—3=12544β„Ž890625=12544β„Ž89062512544=β„Ž71.00…=β„Ž.

We can round this value to the nearest foot, giving the answer that the height of the Louvre Pyramid is 71 ft.

In the next example, we will see how we can use a given volume of a pyramid and its height to find the perimeter of its base.

Example 3: Finding the Base Perimeter of a Squared-Based Pyramid given Its Volume and Height

Given that a square pyramid has a volume of 372 cm3 and a height of 31 cm, determine the perimeter of its base.

Answer

We have been given information about the volume and height of this pyramid. We can use the formula that connects these variables, along with the area of the base of the pyramid, to then find the perimeter of its base. The volume of a pyramid is given by 𝑉=13(π΄Γ—β„Ž),pyramidbase where 𝐴base is the area of the base of the pyramid and β„Ž is the height.

Substituting 𝑉=372pyramidcm and β„Ž=31cm, we have 372=13(𝐴×31)372Γ—3=31×𝐴111631=𝐴36=𝐴.basebasebasebasecm

Thus, the area of the base of this pyramid is 36 cm2. As we are given that the base is a square, we can calculate the length of the side by recalling that any square of length 𝑙 has an area given by 𝐴=𝑙.square

Therefore, we have 36=π‘™βˆš36=𝑙6=𝑙.

As the length, 𝑙, of the square must be positive, we only need to consider the positive value of the square root. Thus, the length of the square on the base of the pyramid is 6 cm.

The perimeter of a shape is the distance around the outside edge, and as a square has 4 sides of equal length, we calculate perimetercm=6Γ—4=24.

We can then give the answer that the perimeter of the base of this pyramid is 24 cm.

In the following example, we will see how we can use the slant height and the length of the lateral edge of a pyramid to calculate its volume.

Example 4: Finding the Volume of a Pyramid given Its Slant Height and Lateral Edge Length

Find the volume of the following regular pyramid approximating the result to the nearest hundredth.

Answer

In this figure, we have the lengths of the slant height and the lateral edge. In a regular pyramid, the lateral faces are congruent isosceles triangles. Therefore, the lateral edge will be the length of one of the congruent lengths in the isosceles triangles.

In order to work out the volume of a pyramid, we can use the formula 𝑉=13(π΄Γ—β„Ž),pyramidbase where 𝐴base is the area of the base of the pyramid and β„Ž is the height.

Before we can use this formula, we will need to work out the area of the base and the perpendicular height using the given lengths.

Consider the following triangular section of the pyramid.

In a regular pyramid, the apex of the pyramid lies above the centroid of the base. The height, β„Ž, of this triangle is also the height of the pyramid. The unknown base length of this triangle can be defined as π‘₯ cm. However, as we are only given the length of one of the sides of this triangular section, we cannot work out the value of β„Ž.

We can, however, calculate the value of π‘₯ by considering the right triangle that is half of one of the isosceles triangles that form the lateral faces.

The base length of this triangle would be congruent to the base of the previous triangle, π‘₯ cm. The perpendicular height is the same as the slant height of the pyramid, 15 cm, and the hypotenuse of the triangle is 17 cm. We can apply the Pythagorean theorem, which states that, for every right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. Thus, we substitute the values of the lengths, 15 cm and 17 cm, and simplify to find π‘₯. This gives 17=15+π‘₯289=225+π‘₯289βˆ’225=π‘₯64=π‘₯√64=π‘₯.

As π‘₯ is a length, we need only consider the positive value of the square root; hence, π‘₯=8.cm

We can now use this value of π‘₯ to calculate the value of β„Ž in the first triangle. Applying the Pythagorean theorem once more, we have 15=π‘₯+β„Ž15=8+β„Ž225=64+β„Ž225βˆ’64=β„Ž161=β„Žβˆš161=β„Ž.

As 161 is not a square number, we can keep the value of β„Ž=√161 in this radical form.

Thus far, we have calculated the perpendicular height of the pyramid as √161 cm. We still require the area of the base in order to find the volume of the pyramid. In fact, we are close to having the length of one of the sides of the base. As we are given that this is a regular pyramid, its base must be a regular polygon. This must mean we have a square base. The value of π‘₯=8 that we calculated is half of the length of one of the square’s sides. Therefore, the length of the square at the base is 8Γ—2=16cm.

Recall that, to find the area of a square, we square its length. Hence, the area of the square at the base of the pyramid is calculated by 𝐴=16=256.basecm

Finally, we can substitute these values, 𝐴=256basecm and β„Ž=√161cm, into the formula to find the volume of a pyramid, 𝑉=13(π΄Γ—β„Ž)pyramidbase. This gives us 𝑉=13ο€»256Γ—βˆš161=1082.7586….pyramidcm

Approximating to the nearest hundredth, we can give the answer that the volume of the pyramid is 1β€Žβ€‰β€Ž082.76 cm3.

So far, we have seen examples of pyramids that have triangular or quadrilateral bases, for which their volume is easier to find using geometric techniques. However, it can be more complex to find the volume of pyramids with bases of 5 sides or more if we are not given the area of the base. But, if we have a regular 𝑛-sided polygon at the base, we can use the formula below to find its area.

Definition: Area of a Regular Polygon

The area of a regular 𝑛-sided polygon of side length π‘₯ is given by 𝐴=𝑛π‘₯4ο€Ό180π‘›οˆ.polygon∘cot

We will now see an example of how we can apply this area formula to find the volume of a regular pyramid.

Example 5: Finding the Volume of a Pentagonal Pyramid

A regular pentagonal pyramid has base side length 41 cm and height 71 cm. Compute, to one decimal place, the volume of the pyramid.

Answer

We can sketch the pyramid with the given dimensions as shown below.

We recall that, to find the volume of a pyramid, we use the formula 𝑉=13(π΄Γ—β„Ž),pyramidbase where 𝐴base is the area of the base and β„Ž is the height.

We are not given the area of the base, but we can calculate this area using the information that this is a regular pentagonal pyramid, with side length 41 cm.

The area of a regular 𝑛-sided polygon of side length π‘₯ is given by 𝐴=𝑛π‘₯4ο€Ό180π‘›οˆ.polygon∘cot

As the pentagon has 5 sides, we substitute 𝑛=5 and the given length, π‘₯=41cm. Hence, we calculate 𝐴=5ο€Ή414ο€Ό1805=8405436.pentagon∘∘cotcot

Multiplying by the cotangent is equal to dividing by the tangent; thus, we can write this as 𝐴=8405436.pentagontan∘

Using a calculator, we can find the decimal equivalent of this as 𝐴=2892.122….pentagoncm

When we use this value in the next part of the calculation, it is useful to keep this found value as precise as possible, rather than using a rounded value.

We can now use the formula for the volume of the pyramid, 𝑉=13(π΄Γ—β„Ž)pyramidbase, substituting the area of the pentagon as 𝐴base and the height, β„Ž=71cm. This gives 𝑉=13(2892.122…×71)=68446.899….pyramidcm

As required, we round the value to one decimal place. Hence, the volume of the pyramid can be approximated as 68β€Žβ€‰β€Ž446.9 cm3.

We now summarize the key points.

Key Points

  • Pyramids are three-dimensional geometric shapes, where the base is a polygon and all other sides are triangles that meet at the apex.
  • A right pyramid is a pyramid whose apex lies above the centroid of the base.
  • A regular pyramid is a right pyramid whose base is a regular polygon: all the sides of the base are of equal length, and all the pyramid’s lateral edges are of equal length.
  • The volume of a pyramid is one-third of the volume of the prism of the same base and height: 𝑉=13(π΄Γ—β„Ž).pyramidbase
  • In order to find the area of the base of a regular pyramid, we may need to apply the formula to find the area of an 𝑛-sided polygon of length π‘₯, which is given by 𝐴=𝑛π‘₯4ο€Ό180π‘›οˆ.polygon∘cot

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