Lesson Explainer: The Pythagorean Theorem in 3D Mathematics • 8th Grade

In this explainer, we will learn how to use the Pythagorean theorem to solve problems in three dimensions.

We recall that the Pythagorean theorem describes the relationship between the lengths of the three sides of a right triangle, and we should already be familiar with applying this result to problems in two dimensions.

Theorem: The Pythagorean Theorem in Two Dimensions

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the two shorter sides. For the triangle below, π‘Ž+𝑏=𝑐.

The Pythagorean theorem can be applied to three-dimensional problems in a number of ways. One method is to use right triangles contained within the faces of the three-dimensional object or to take two-dimensional β€œslices” through its interior. To identify such triangles, we need to be familiar with the parts and properties of some common three-dimensional objects, such as pyramids and cones.

We recall first two special types of pyramids. First, a right pyramid is a pyramid whose apex is vertically above the center, or in the case of a triangle, the centroid, of its base. Second, a regular pyramid is a right pyramid whose base is a regular polygon.

Key Terms: Parts of a Pyramid

Consider the regular square pyramid in the figure below. We recall the terminology used to describe the lengths and parts of a pyramid. The perpendicular height of a pyramid is the perpendicular distance between the apex of the pyramid and its base, and it is perpendicular to any straight line it intersects in the pyramid’s base. The slant height of a pyramid is the perpendicular distance from one of the sides of the base to the apex of the pyramid. We can note that the slant height will be the same in the four lateral faces.

Key Terms: Parts of a Cone

In the figure below, we recall the terminology used to describe the lengths and parts of a cone. We note that the main lengths in a cone are the slant height, the perpendicular height, and the base radius of the cone. These three lengths form a right triangle.

In the course of this explainer, we will see examples of how we can apply the Pythagorean theorem to right triangles within both pyramids and cones to calculate unknown lengths. In our first example, however, we will consider how to find the area of a diagonal shape contained within a cube, given the cube’s edge length.

Example 1: Finding the Area of a Diagonal Shape in a Cube given Its Dimensions

Given that 𝐴𝐡𝐢𝐷𝐸𝐹𝐺𝐻 is a cube whose edge length is 6√2 cm and 𝑋 is the midpoint of 𝐴𝐡, find the area of the rectangle π·π‘‹π‘ŒπΈ.

Answer

We recall first that the area of a rectangle is found using the formula arealengthwidth=Γ—. For rectangle π·π‘‹π‘ŒπΈ, the area will therefore be equal to π·π‘‹Γ—π‘‹π‘Œ.

The length of π‘‹π‘Œ is 6√2 cm as it is parallel to the edge of the cube and has the same length. To find the length of 𝐷𝑋 consider the square base of the cube. As 𝑋 is the midpoint of 𝐴𝐡, it divides 𝐴𝐡 into two equal parts, each of length 6√22=3√2cm. As all of the interior angles of a square are 90∘, the triangle formed by 𝐴𝑋, 𝐴𝐷, and 𝐷𝑋 is a right triangle at 𝐴.

The hypotenuse of this triangle is 𝐷𝑋 and so, by the Pythagorean theorem, 𝐷𝑋=𝐴𝑋+𝐴𝐷.

Substituting 3√2 for 𝐴𝑋 and 6√2 for 𝐴𝐷 gives 𝐷𝑋=ο€»3√2+ο€»6√2.

We solve this equation by first evaluating each square and then simplifying: 𝐷𝑋=18+72=90.

Taking the square root of each side of the equation gives 𝐷𝑋=√90=3√10.cm

The area of rectangle π·π‘‹π‘ŒπΈ is therefore equal to π·π‘‹Γ—π‘‹π‘Œ=3√10Γ—6√2=18√20=18√4Γ—5=36√5.cm

We now consider two examples relating to pyramids. In the first example, we calculate the height of a pyramid whose base is an equilateral triangle, given the triangle’s side length and the length of the pyramid’s lateral edge.

Example 2: Finding the Height of a Pyramid by Applying the Pythagorean Theorem

𝑀𝐴𝐡𝐢 is a regular pyramid whose base 𝐴𝐡𝐢 is an equilateral triangle whose side length is 32 cm. If the length of its lateral edge is 88 cm, find the height of the pyramid to the nearest hundredth.

Answer

We begin by sketching the pyramid as shown below (not to scale).

We are asked to calculate the height of the pyramid. This is the vertical distance from its vertex 𝑀 to the center of the base of the pyramid, which we will label 𝑋. We draw this height on our diagram and also the line connecting one vertex of the base to point 𝑋.

𝐴𝑀 is the lateral edge of this pyramid, the length of which is given in the question. 𝐴𝑋 is a line segment on the base of the pyramid, and 𝑀𝑋 is the vertical height of the pyramid, which is perpendicular to 𝐴𝑋. Hence, we have a right triangle formed by 𝐴𝑀, 𝐴𝑋, and 𝑀𝑋. By the Pythagorean theorem, 𝐴𝑋+𝑀𝑋=𝐴𝑀.

The length of 𝐴𝑀 is given in the question. In order to calculate the length of 𝑀𝑋, it is first required that we calculate the length of 𝐴𝑋.

Now consider the equilateral triangular base of the pyramid. Point 𝑋 is the centroid of this triangle, the point at which the three medians of the triangle intersect. Point π‘Œ is the midpoint of side 𝐡𝐢.

As triangle 𝐴𝐡𝐢 is equilateral, its three medians are all of equal length. We also know that the centroid of a triangle divides each median in the ratio 2∢1 from the vertex. Hence, 𝐴𝑋=𝐡𝑋=𝐢𝑋 and 𝐴𝑋=23π΄π‘Œ.

Now consider triangle π΄π΅π‘Œ. This is a right triangle at π‘Œ as the median π΄π‘Œ divides the equilateral triangle 𝐴𝐡𝐢 into two congruent right triangles. The length of π΅π‘Œ is half the length of 𝐡𝐢, so it is 16 cm. We can therefore apply the Pythagorean theorem to calculate the length of π΄π‘Œ, which will then enable us to calculate the length of 𝐴𝑋.

Applying the Pythagorean theorem in triangle π΄π΅π‘Œ, we have π΄π‘Œ+π΅π‘Œ=𝐴𝐡.

Substituting π΅π‘Œ=16 and 𝐴𝐡=32 gives π΄π‘Œ+16=32π΄π‘Œ=32βˆ’16=768.

We solve for π΄π‘Œ by square rooting: π΄π‘Œ=√768=16√3.cm

We can then calculate the length of 𝐴𝑋 by recalling that 𝐴𝑋=23π΄π‘Œ: 𝐴𝑋=23Γ—16√3=32√33.cm

Finally, we return to our application of the Pythagorean theorem in triangle 𝐴𝑋𝑀, 𝐴𝑋+𝑀𝑋=𝐴𝑀.

Substituting 𝐴𝑋=32√33 and 𝐴𝑀=88 gives ο€Ώ32√33+𝑀𝑋=88.

Subtracting ο€Ώ32√33ο‹οŠ¨ from each side of the equation and evaluating the squares gives 𝑀𝑋=88βˆ’ο€Ώ32√33=222083.

We solve for 𝑀𝑋 by square rooting: 𝑀𝑋=ο„ž222083=86.038β€¦β‰ˆ86.04.

The height of the pyramid, to the nearest hundredth, is 86.04 cm.

We now consider a problem in which we are given the net of a pyramid and are required to calculate its vertical height and its slant height. We will need to interpret the net carefully to understand which of the pyramid’s dimensions have been given.

Example 3: Finding the Vertical Height and Slant Height of a Square Pyramid given the Dimensions of Its Net

Consider the net of a square pyramid with the shown dimensions.

  1. Find, to the nearest hundredth, the pyramid’s height.
  2. Find, to the nearest hundredth, the pyramid’s slant height.

Answer

Part 1

Let us begin by sketching the pyramid in three dimensions. From the net, we identify that the side length of the square base is 5 cm. The lines of length 4 cm represent the other two sides of the pyramid’s triangular faces, and so we deduce that the length of the pyramid’s lateral edge is 4 cm.

Point 𝑋 on our sketch is the point at which the vertical line drawn from the apex of the pyramid (𝐸) meets the pyramid’s square base. This point is in the center of the base. The vertical height of the pyramid is 𝑋𝐸, which is perpendicular to any line in the pyramid’s base passing through point 𝑋. In particular, it is perpendicular to the line segments joining point 𝑋 to each of the pyramid’s vertices. Hence, the triangle formed by the vertical height 𝑋𝐸, the lateral edge 𝐴𝐸, and 𝐴𝑋 is a right triangle at 𝑋, as shown below.

Hence, by the Pythagorean theorem, 𝑋𝐸+𝐴𝑋=𝐴𝐸.

The length of 𝐴𝐸 is given, but it is necessary that we calculate the length of 𝐴𝑋 before we can calculate the length of 𝑋𝐸. Consider the square base of the pyramid, and recall that 𝑋 lies in the center of this square. We also recall that the center of a square is the intersection point of the two diagonals, and both diagonals bisect each other.

The length of 𝐴𝑋 is half the length of diagonal 𝐴𝐢. This diagonal divides the square base into two congruent right triangles. Hence, by the Pythagorean theorem in triangle 𝐴𝐡𝐢, 𝐴𝐢=𝐴𝐡+𝐡𝐢.

Substituting 5 for each of 𝐴𝐡 and 𝐡𝐢 gives 𝐴𝐢=5+5=25+25=50.

We solve for 𝐴𝐢 by square rooting: 𝐴𝐢=√50=5√2.cm

The length of 𝐴𝑋 is half this value: 𝐴𝑋=12𝐴𝐢=5√22.cm

Returning to our earlier statement of the Pythagorean theorem in triangle 𝐴𝑋𝐸, we substitute 5√22 for 𝐴𝑋 and 4 for 𝐴𝐸 to give the following equation: ο€Ώ5√22+𝑋𝐸=4.

This equation can be solved by first evaluating the squares and then isolating π‘‹πΈοŠ¨: 252+𝑋𝐸=16𝑋𝐸=72.

Finally, we solve for 𝑋𝐸 by finding the square root of each side of this equation: 𝑋𝐸=ο„ž72=1.870β€¦β‰ˆ1.87.

The vertical height of the pyramid, to the nearest hundredth, is 1.87 cm.

Part 2

We now draw in the slant height of the pyramid along one of its faces. We define π‘Œ to be the midpoint of side 𝐡𝐢.

The slant height forms a right triangle with the perpendicular height of the pyramid and the line segment connecting points 𝑋 and π‘Œ, that is, the line segment connecting the center of the base with the midpoint of one of the sides of the base. Hence, by the Pythagorean theorem, π‘ŒπΈ=π‘‹π‘Œ+𝑋𝐸.

In the first part of the question, we determined the value of π‘‹πΈοŠ¨ to be 72. π‘‹π‘Œ is parallel to side 𝐴𝐡 of the pyramid’s base, and as 𝑋 is the center of the base, π‘‹π‘Œ=12𝐴𝐡. Hence, the length of π‘‹π‘Œ is 52 cm. Substituting these values into our statement of the Pythagorean theorem in triangle π‘‹π‘ŒπΈ gives π‘ŒπΈ=ο€Ό52+72.

Evaluating and simplifying gives π‘ŒπΈ=254+72=394.

To solve for π‘ŒπΈ, we find the square root of each side of this equation: π‘ŒπΈ=ο„ž394=3.122β€¦β‰ˆ3.12.cm

The slant height of the pyramid, to the nearest hundredth, is 3.12 cm.

We now consider problems related to cones. In our next example, we will calculate the circumference and base area of a right circular cone by first finding its radius. Recall that, in such a cone, the relationship between the base radius, the perpendicular height, and the slant height can be described using the Pythagorean theorem.

Example 4: Finding the Circumference and Area of the Base of a Right Circular Cone given Its Heights by Applying the Pythagorean Theorem

A right circular cone has height 90 cm and slant height 106 cm. Find the circumference and area of the base in terms of πœ‹.

Answer

We recall first the formulas for calculating the circumference and area of a circle: circumference=2πœ‹π‘Ÿ and area=πœ‹π‘ŸοŠ¨, where π‘Ÿ represents the radius. Hence, we are required to first calculate the radius of the cone’s circular base.

We now sketch the cone. Recall that the height of a right circular cone, often denoted by β„Ž, is the perpendicular distance between the center of the base and the apex of the cone. The slant height, denoted by 𝑙, is the distance from any point on the circumference of the circular base to the apex, along the surface of the cone.

Next, we note that if we draw the radius of the cone, the triangle formed by the radius, the height, and the slant height of the cone is a right triangle.

Hence, by the Pythagorean theorem, π‘Ÿ+90=106.

Subtracting 90 from each side of the equation and simplifying give π‘Ÿ=106βˆ’90=3136.

We solve for π‘Ÿ by taking the positive square root: π‘Ÿ=√3136=56.cm

Substituting π‘Ÿ=56 into the formula for the circumference of a circle gives circumferencecm=2Γ—πœ‹Γ—56=112πœ‹.

The question specifies that our answers should be given in terms of πœ‹, so we are not required to evaluate this.

Substituting the value of π‘ŸοŠ¨ from our penultimate stage of working into the formula for the area of a circle gives areacm=3136πœ‹.

Hence, the circumference of the circular base is 112πœ‹ cm and the area is 3136πœ‹ cm2.

Example 5: Using the Pythagorean Theorem to Calculate the Height of a Cone given Its Net

A piece of paper in the shape of a circular sector having a radius of 72 cm and an angle of 275∘ is folded in a way so that the points 𝐴 and 𝐡 meet to form a circular cone of the greatest possible area. Determine the cone’s height to the nearest hundredth.

Answer

The given figure represents the net of the cone. We will begin by sketching the cone in three dimensions. The slant height of the folded cone will be equivalent to the length of the line segments joining each of points 𝐴 and 𝐡 to the center of the circular sector. This is the length of the radius of the sector, which is given in the question as 72 cm. If the cone is to have the greatest possible area, then the paper must not overlap, and so the circumference of the circle at the base of the cone must be equal to the arc length of the circular sector.

We recall that the base radius, perpendicular height, and slant height of a cone form a right triangle. Hence, by applying the Pythagorean theorem to this right triangle in the cone above, we have π‘Ÿ+β„Ž=72.

To determine the height of the cone, we must first calculate its base radius. We will begin by calculating the arc length of the circular sector, which is equal to the circumference of the circular base. Recall the formula arclength=πœƒ360Γ—2πœ‹π‘…,∘ where πœƒ is the central angle of the sector, measured in degrees, and 𝑅 is the radius. Note that we have used a capital 𝑅 here to distinguish this length from the radius of the cone’s base. We can then substitute πœƒ=275∘ and 𝑅=72 to calculate the arc length: arclengthcm=275360Γ—2πœ‹Γ—72=110πœ‹.∘∘

As the arc length of the circular sector is equal to the circumference of the cone’s circular base, we can calculate the radius of the circle by recalling the formula 𝑐=2πœ‹π‘Ÿ. Hence, 2πœ‹π‘Ÿ=110πœ‹.

Dividing both sides of this equation by 2πœ‹ gives π‘Ÿ=110πœ‹2πœ‹=55.cm

Finally, we substitute this value of π‘Ÿ into our earlier statement of the Pythagorean theorem: π‘Ÿ+β„Ž=7255+β„Ž=72.

Evaluating the squares gives 3025+β„Ž=5184.

To solve for β„Ž, we first subtract 3β€Žβ€‰β€Ž025 from each side of the equation and then find the square root: β„Ž=2159β„Ž=46.465β€¦β‰ˆ46.47.cm

The height of the cone, to the nearest hundredth, is 46.47 cm.

In the previous problems, we have seen how we can take two-dimensional β€œslices” through the interior of a three-dimensional object in order to create right triangles in which we can apply the Pythagorean theorem. An alternative way in which the Pythagorean theorem can be applied to three-dimensional problems is in a three-dimensional extension of the theorem itself. We will demonstrate this for the case of calculating the length of the diagonal of a cuboid.

First, we consider more specifically what is meant by the diagonal of a cuboid. We must be careful to distinguish the interior diagonals of a cuboid from its face diagonals, those that lie within its two-dimensional faces. For example, in the case of the cuboid below, the line segments 𝐡𝐸, 𝐺𝐸, and 𝐢𝐹 are each face diagonals of the cuboid.

In fact, a cuboid has 12 face diagonals in total: 2 in each of its 6 faces.

The diagonals, also sometimes called the space or interior diagonals of a cuboid, are different. These are the line segments that connect opposite vertices of the cuboid and pass through its interior. For example, in the cuboid above, one of the cuboid’s diagonals is 𝐢𝐸. In fact, a cuboid has four diagonals, which for the cuboid above are 𝐴𝐺, 𝐡𝐻, 𝐢𝐸, and 𝐷𝐹. These four line segments are congruent and all intersect at a common point at the center of the cuboid, as shown in the figure below.

We now consider how to calculate the length of the diagonal of a cuboid. Consider a cuboid 𝐴𝐡𝐢𝐷𝐸𝐹𝐺𝐻 with sides of lengths π‘Ž, 𝑏, and 𝑐 units, respectively, as shown below. Suppose we wish to calculate the length of the cuboid’s diagonal, 𝑑.

We can achieve this using two iterations of the Pythagorean theorem. We begin by drawing the base diagonal of the cuboid, which we will label as 𝑒.

By applying the Pythagorean theorem in the right triangle 𝐴𝐡𝐸 on the base of the cuboid, in which 𝑒 is the hypotenuse, we have 𝑒=π‘Ž+𝑏.

To calculate the length of 𝑑, we can apply the Pythagorean theorem again in the right triangle 𝐡𝐢𝐸. Side 𝑑 is the hypotenuse of this triangle and the two shorter sides are 𝑐 and 𝑒, so we have 𝑑=𝑒+𝑐.

Substituting the expression for π‘’οŠ¨ obtained from our first application of the Pythagorean theorem gives 𝑑=π‘Ž+𝑏+𝑐.

This is the three-dimensional extension of the Pythagorean theorem.

Theorem: The Three-Dimensional Extension of the Pythagorean Theorem

In a cuboid with side lengths π‘Ž, 𝑏, and 𝑐 and interior diagonal 𝑑 as shown below, 𝑑=π‘Ž+𝑏+𝑐.

We now demonstrate the application of the Pythagorean theorem in three dimensions to calculate the length of the diagonal of a cuboid given its length, width, and depth.

Example 6: Finding the Length of the Diagonal of a Cuboid Using the Pythagorean Theorem

Find the length of the diagonal of a rectangular prism of sides 3 cm, 4 cm, and 6 cm. Round your answer to the nearest hundredth.

Answer

Although not essential, we begin by sketching the cuboid.

The diagonals of this cuboid are 𝐴𝐺, 𝐡𝐻, 𝐢𝐸, and 𝐷𝐹. These four line segments each connect opposite vertices of the cuboid and pass through its interior. The diagonals of a cuboid are of equal length, so it is entirely arbitrary which we choose to calculate. Let us choose 𝐢𝐸 and add this line segment to our diagram.

The three-dimensional extension of the Pythagorean theorem states that 𝑑=π‘Ž+𝑏+𝑐, where 𝑑 is the diagonal of a cuboid and π‘Ž, 𝑏, and 𝑐 are its three side lengths. Here π‘Ž, 𝑏, and 𝑐 are 3 cm, 4 cm, and 6 cm, and the diagonal is 𝐢𝐸, so we have 𝐢𝐸=3+4+6.

Simplifying gives 𝐢𝐸=9+16+36=61.

We solve for 𝐢𝐸 by finding the square root, ignoring the negative solution as 𝐢𝐸 is a length: 𝐢𝐸=√61=7.810β€¦β‰ˆ7.81.

The length of the diagonal of the rectangular prism, to the nearest hundredth, is 7.81 cm.

In the previous problem, we chose to apply the three-dimensional version of the Pythagorean theorem. It would also have been possible to apply two iterations of the Pythagorean theorem in two dimensions. If we had drawn the base diagonal 𝐡𝐸, then applying the Pythagorean theorem in the right triangle 𝐴𝐡𝐸, in which 𝐡𝐸 is the hypotenuse, would give 𝐡𝐸=𝐴𝐸+𝐴𝐡=6+3=45.

Then, applying the Pythagorean theorem a second time in triangle 𝐡𝐢𝐸, in which 𝐢𝐸 is the hypotenuse, would give 𝐢𝐸=𝐡𝐸+𝐡𝐢=45+4=61.

We see that this is consistent with the penultimate stage of working in our previous method. Both methods of course yield the same result, but applying the three-dimensional extension of the Pythagorean theorem is more efficient.

Let us finish by recapping some key points from this explainer.

Key Points

  • The two-dimensional Pythagorean theorem can be applied to right triangles within the faces of a three-dimensional object, or to two-dimensional slices through its interior, in order to calculate unknown lengths.
  • The perpendicular height of a regular pyramid is perpendicular to any straight line passing through the center of its base and lying on this base, and hence a right triangle can be formed with the lateral edge. The slant height of a regular pyramid is the perpendicular distance from any side of the base to the pyramid’s apex.
  • The base radius (π‘Ÿ), vertical height (β„Ž), and slant height (𝑙) of a cone form a right triangle and are hence related by the Pythagorean theorem as follows: π‘Ÿ+β„Ž=𝑙.
  • The extension of the Pythagorean theorem to three dimensions states that for a cuboid with side lengths π‘Ž, 𝑏, and 𝑐 and a diagonal of length 𝑑, as shown below, π‘Ž+𝑏+𝑐=𝑑.
    This is analogous to applying two iterations of the Pythagorean theorem in two dimensions.

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