 Lesson Video: Applying the Pythagorean Theorem to Pyramids and Cones | Nagwa Lesson Video: Applying the Pythagorean Theorem to Pyramids and Cones | Nagwa

# Lesson Video: Applying the Pythagorean Theorem to Pyramids and Cones Mathematics • 8th Grade

In this video, we will learn how to identify the parts of pyramids and cones and use the Pythagorean theorem to find their dimensions.

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### Video Transcript

In this video, we will learn how to identify the parts of pyramids and cones and use the Pythagorean theorem to find their dimensions.

Recall that the Pythagorean theorem states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. 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 slices 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, for example, the regular square pyramid.

The perpendicular height of the pyramid is the perpendicular distance between the apex of the pyramid and its base. And its line is therefore perpendicular to any straight line contained in the pyramid’s base that intersects it. The lateral edge of the pyramid is the edge connecting its apex to one of its other vertices that make up the 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.

A right circular cone is similar to a pyramid, but its base is circular. 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, as shown in the diagram.

In this video, we will look at examples of how we can apply the Pythagorean theorem to right angles within 3D shapes, including pyramids and cones, to calculate unknown lengths. Let’s look at an example of finding an unknown length within a cube.

Given that 𝐴𝐵𝐶𝐷𝐸𝐹𝐺𝐻 is a cube whose edge length is six root two centimeters and 𝑋 is the midpoint of the line segment 𝐴𝐵, find the area of the rectangle 𝐷𝑋𝑌𝐸.

The area of a rectangle is equal to the product of its side lengths. In this case, the side lengths of the rectangle 𝐷𝑋𝑌𝐸 are 𝐷𝑋 and 𝑋𝑌. We are given by the question that the side length of the cube is six root two centimeters. The line 𝑋𝑌 is parallel to an edge of the cube and extends between two other edges. Therefore, it must also be of length six root two.

To find the length of 𝐷𝑋, consider the top-down view of the base of the cube. The shaded triangle is a right triangle, since the internal angles of the face of a cube, like 𝐷𝐴𝑋, are 90 degrees. The side 𝐴𝐷 is an edge of the cube, so its length is six root two. Since 𝑋 is the midpoint of an edge of the cube, the side 𝐴𝑋 is half the edge length, three root two.

𝐷𝑋 is the hypotenuse of this right triangle. So the length of the side 𝐷𝑋 can therefore be found from the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle, 𝑐, is equal to the sum of the squares of the other two sides, 𝑎 and 𝑏. In this case, therefore, 𝐷𝑋 squared is equal to 𝐴𝐷 squared plus 𝐴𝑋 squared. 𝐴𝐷 is equal to six root two, and 𝐴𝑋 is equal to three root two. So 𝐷𝑋 squared is equal to six root two all squared plus three root two all squared. Six root two all squared is equal to 36 times two, which is 72. And three root two all squared is equal to nine times two, which is 18. And adding these together gives us 90.

Therefore, the length of 𝐷𝑋 is equal to square root of 90, which can be simplified to three root 10. The area of the rectangle 𝐷𝑋𝑌𝐸 is therefore equal to 𝐷𝑋 times 𝑋𝑌, which is three root 10 times six root two. Collecting surds, we get three times six, which is 18, and the square root of 10 times two, which is root 20. And this simplifies to give us the area of 𝐷𝑋𝑌𝐸, 36 root five, and the unit is centimeters squared.

Let’s now look at an example where we calculate the height of a pyramid whose base is an equilateral triangle.

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

Let’s begin by sketching the pyramid. The pyramid has an equilateral triangular base. So the three sides 𝐴𝐵, 𝐵𝐶, and 𝐴𝐶 have the same length of 32 centimeters. It’s also a regular pyramid, so the lateral edges also have the same length of 88 centimeters.

By the symmetry of the sides, the height of the pyramid is given by the length of the line 𝑀𝑋, where 𝑋 is the centroid of the triangular base. This line is vertical. So it is perpendicular to any line contained within the base of the triangle, such as the line 𝐴𝑋. The triangle 𝐴𝑀𝑋 is therefore a right triangle. So by the Pythagorean theorem, the square of the hypotenuse of this right triangle, 𝐴𝑀, is equal to the sum of the squares of the other two sides, 𝐴𝑋 and 𝑀𝑋.

𝑀𝑋 squared is the square of the height of the pyramid, the quantity we wish to find. So we’ll rearrange this equation for 𝑀𝑋. We can do this by subtracting 𝐴𝑋 squared from both sides and taking the square root, giving us 𝑀𝑋 equals the square root of 𝐴𝑀 squared minus 𝐴𝑋 squared. 𝐴𝑀 is the lateral edge of the pyramid, which we are given in the question, 88 centimeters. 𝐴𝑋 is currently unknown, so we will need to find it from the equilateral triangular base.

The base 𝐴𝐵𝐶 is an equilateral triangle with a centroid at the point 𝑋. If we continue the line 𝐴𝑋 to the other side of the triangle, it intercepts the line 𝐵𝐶 at its midpoint, 𝑌. A median passing from the vertex of a triangle through its centroid is divided into a ratio of two to one. So 𝐴𝑋 is twice the length of 𝑋𝑌. This also means that 𝐴𝑋 is two-thirds the length of the complete line 𝐴𝑌.

This subtriangle 𝐴𝐵𝑌 is a right triangle. The hypotenuse 𝐴𝐵 is the side length of the base, 32 centimeters. And the side 𝐵𝑌 is half the side length of the base, 16 centimeters. We can apply the Pythagorean theorem again to this right triangle. The square of the hypotenuse, 𝐴𝐵, is equal to the sum of the squares of the other two sides, 𝐴𝑌 and 𝐵𝑌. We can rearrange for 𝐴𝑌 by subtracting 𝐵𝑌 squared from both sides and taking the square root, giving 𝐴𝑌 equals the square root of 𝐴𝐵 squared minus 𝐵𝑌 squared.

Since 𝐴𝑋 is equal to two-thirds 𝐴𝑌, 𝐴𝑋 is equal to two-thirds times the square root of 𝐴𝐵 squared minus 𝐵𝑌 squared. Taking the square finally gives us 𝐴𝑋 squared, which is equal to four-ninths times 𝐴𝐵 squared minus 𝐴𝑌 squared.

We can now express the height of the pyramid, 𝑀𝑋, entirely in known quantities. 𝑀𝑋 is equal to the square root of 𝐴𝑀 squared minus four-ninths 𝐴𝐵 squared minus 𝐵𝑌 squared. 𝐴𝑀 is the lateral edge of the pyramid, equal to 88 centimeters. 𝐴𝐵 is the side length of the base, equal to 32 centimeters. And 𝐵𝑌 is half the side length of the base, equal to 16 centimeters. Therefore, 𝑀𝑋 is equal to the square root of 88 squared minus four-ninths times 32 squared minus 16 squared. Performing this calculation gives us the height of the pyramid, 86.04 centimeters, to the nearest hundredth.

In the next example, we will use the net of a pyramid to calculate its vertical height and slant height.

Consider the net of the square pyramid with the shown dimensions. Find, to the nearest hundredth, the pyramid’s height. And find, to the nearest hundredth, the pyramid’s slant height.

Let’s begin by sketching the constructed pyramid in 3D. We can label the four base vertices 𝐴, 𝐵, 𝐶, and 𝐷 and the top vertex 𝐸. From the net, the side length of the square base is five centimeters and the lateral edge of the pyramid is four centimeters. We can draw a vertical line from the vertex 𝐸 to the base of the pyramid. The line meets the base of the pyramid at its centroid 𝑋. And its length is the height of the pyramid.

This line is perpendicular to any line contained within the base, such as the line 𝐴𝑋. So it forms a right triangle 𝐴𝑋𝐸. Therefore, by the Pythagorean theorem, the square of the hypotenuse of this triangle, 𝐴𝐸, is equal to the sum of the squares of the other two sides, 𝑋𝐸 and 𝐴𝑋. Rearranging for the height, 𝑋𝐸, by subtracting 𝐴𝑋 squared and taking the square root, we get 𝑋𝐸 equals the square root of 𝐴𝐸 squared minus 𝐴𝑋 squared. 𝐴𝐸 is the slant height of the pyramid, which is four centimeters. 𝐴𝑋 is currently unknown.

To find the length of 𝐴𝑋, consider the top-down view of the base of the pyramid. The center of the base, 𝑋, is the midpoint of the diagonal of the square, 𝐴𝐶. The triangle 𝐴𝐵𝐶 is a right triangle. Therefore, by the Pythagorean theorem, the square of the hypotenuse of this triangle, 𝐴𝐶, is equal to the sum of the squares of the other two sides, 𝐴𝐵 and 𝐵𝐶.

𝐴𝐶 is twice the length of 𝐴𝑋, so the left-hand side could be rewritten as two 𝐴𝑋 all squared. Also, since the base is a square, 𝐴𝐵 is equal to 𝐵𝐶, so the right-hand side is equal to two 𝐴𝐵 squared. The left-hand side has a factor of four, and the right-hand side has a factor of two. Dividing both sides by four, therefore, gives us 𝐴𝑋 squared equals one-half 𝐴𝐵 squared.

Therefore, we can express the height of the pyramid entirely with known quantities. 𝑋𝐸 is equal to square root of 𝐴𝐸 squared minus half 𝐴𝐵 squared. 𝐴𝐸 is the lateral edge of the pyramid, four centimeters, and 𝐴𝐵 is the side length of the base, five centimeters. Therefore, 𝑋𝐸 is equal to the square root of four squared minus one-half times five squared. Performing this calculation gives us the height of the pyramid, 1.87 centimeters to the nearest hundredth.

For the second part of the question, let’s clear some space and redraw the pyramid. The slant height of the pyramid is the length of the line from the top vertex, 𝐸, to the midpoint of one of its base edges, say, 𝐵𝐶. As before, we can draw a vertical line from the vertex 𝐸 to the point 𝑋, the centroid of the base. The angle between this line and the line 𝑋𝑌 is a right angle. So the triangle 𝑋𝑌𝐸 is a right triangle.

We can therefore once again use the Pythagorean theorem to find the slant height, 𝑌𝐸. This is the hypotenuse of the right triangle 𝑋𝑌𝐸. So its square is equal to the sum of the squares of the other two sides, 𝑋𝑌 and 𝑋𝐸. 𝑋𝐸 is already known from the previous part of the question. So we just need to find 𝑋𝑌. 𝑋 is the center of the base, and 𝑋𝑌 is parallel to the edge 𝐴𝐵. Therefore, 𝑋𝑌 is half the length of the edge, which is five over two centimeters.

Taking the square root of the above equation gives us the slant height explicitly. 𝑌𝐸 equals the square root of 𝑋𝑌 squared plus 𝑋𝐸 squared. 𝑋𝑌 squared is equal to five over two all squared, and 𝑋𝐸 is equal to the square root of four squared minus one-half times five squared. So 𝑋𝐸 squared is just four squared minus one-half times five squared. Performing this calculation gives us 𝑌𝐸, the slant height of the pyramid, 3.12 centimeters to the nearest hundredth.

In the final example, we will calculate the circumference and base area of a right circular cone by finding its radius.

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

Let’s start by sketching the cone. The height is 90 centimeters, the slant height is 106 centimeters, and the radius of the base, 𝑟, is unknown. The base is circular. And recall that the circumference of a circle is equal to two 𝜋 times the radius 𝑟. And the area of a circle is given by 𝜋 times the radius 𝑟 squared.

The perpendicular height, the slant height, and the radius form a right triangle. Therefore, we can use the Pythagorean theorem, which states that the square of the hypotenuse of this right triangle, the slant height 𝑠, is equal to the sum of the squares of the other two sides, the radius 𝑟 and the height ℎ.

Rearranging for 𝑟 by subtracting ℎ squared from both sides and taking the square root gives us the radius, 𝑟, equal to the square root of 𝑠 squared minus ℎ squared. The circumference of the circle is equal to two 𝜋 times 𝑟, which is therefore equal to two 𝜋 times the square root of 106 squared minus 90 squared. Performing this calculation gives us the circumference of the base, 112𝜋 centimeters, which we can leave in this form as the question tells us. And the area of the base is equal to 𝜋 times 𝑟 squared, which is 𝜋 times 106 squared minus 90 squared. Performing this calculation gives us the area of the base of the cone, 3136𝜋 centimeters squared, which again we leave in terms of 𝜋.

Let’s finish this video by recapping some key points. The Pythagorean theorem states that the square of the hypotenuse of a right triangle, 𝑐, is equal to the sum of the squares of the other two sides, 𝑎 and 𝑏. This theorem can be applied to three-dimensional shapes, such as cubes, pyramids, and cones, by constructing two-dimensional slices through them to make right triangles, where two of the sides are known so the third may be found from the Pythagorean theorem.

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 𝑟, vertical height ℎ, and slant height 𝑙 of a cone form a right triangle, with the slant as its hypotenuse. Therefore, by the Pythagorean theorem, 𝑟 squared plus ℎ squared is equal to 𝑠 squared.