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 pyramidβ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. Letβs look at an example of finding an unknown length within a cube.

Given that π΄π΅πΆπ·πΈπΉπΊπ» is a cube whose edge length is six root two centimeters and π is the midpoint of the line segment π΄π΅, find the area of the rectangle π·πππΈ.

The area of a rectangle is equal to the product of its side lengths. In this case, the side lengths of the rectangle π·πππΈ are π·π and ππ. We are given by the question that the side length of the cube is six root two centimeters. The line ππ 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 π·π, 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 π·π΄π, are 90 degrees. The side π΄π· is an edge of the cube, so its length is six root two. Since π is the midpoint of an edge of the cube, the side π΄π is half the edge length, three root two.

π·π is the hypotenuse of this right triangle. So the length of the side π·π can therefore be found from the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle, π, is equal to the sum of the squares of the other two sides, π and π. In this case, therefore, π·π squared is equal to π΄π· squared plus π΄π squared. π΄π· is equal to six root two, and π΄π is equal to three root two. So π·π 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 π·π is equal to square root of 90, which can be simplified to three root 10. The area of the rectangle π·πππΈ is therefore equal to π·π times ππ, 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 π·πππΈ, 36 root five, and the unit is centimeters squared.

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

ππ΄π΅πΆ is a regular pyramid whose base π΄π΅πΆ 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.

Letβs begin by sketching the pyramid. The pyramid has an equilateral triangular base. So the three sides π΄π΅, π΅πΆ, and π΄πΆ have the same length of 32 centimeters. Itβ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 ππ, where π 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 π΄π. The triangle π΄ππ is therefore a right triangle. So by the Pythagorean theorem, the square of the hypotenuse of this right triangle, π΄π, is equal to the sum of the squares of the other two sides, π΄π and ππ.

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

The base π΄π΅πΆ is an equilateral triangle with a centroid at the point π. If we continue the line π΄π to the other side of the triangle, it intercepts the line π΅πΆ at its midpoint, π. A median passing from the vertex of a triangle through its centroid is divided into a ratio of two to one. So π΄π is twice the length of ππ. This also means that π΄π is two-thirds the length of the complete line π΄π.

This subtriangle π΄π΅π is a right triangle. The hypotenuse π΄π΅ is the side length of the base, 32 centimeters. And the side π΅π 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, π΄π΅, is equal to the sum of the squares of the other two sides, π΄π and π΅π. We can rearrange for π΄π by subtracting π΅π squared from both sides and taking the square root, giving π΄π equals the square root of π΄π΅ squared minus π΅π squared.

Since π΄π is equal to two-thirds π΄π, π΄π is equal to two-thirds times the square root of π΄π΅ squared minus π΅π squared. Taking the square finally gives us π΄π squared, which is equal to four-ninths times π΄π΅ squared minus π΄π squared.

We can now express the height of the pyramid, ππ, entirely in known quantities. ππ is equal to the square root of π΄π squared minus four-ninths π΄π΅ squared minus π΅π squared. π΄π is the lateral edge of the pyramid, equal to 88 centimeters. π΄π΅ is the side length of the base, equal to 32 centimeters. And π΅π is half the side length of the base, equal to 16 centimeters. Therefore, ππ 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 pyramidβs height. And find, to the nearest hundredth, the pyramidβs slant height.

Letβs begin by sketching the constructed pyramid in 3D. We can label the four base vertices π΄, π΅, πΆ, and π· and the top vertex πΈ. 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 πΈ to the base of the pyramid. The line meets the base of the pyramid at its centroid π. And its length is the height of the pyramid.

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

To find the length of π΄π, consider the top-down view of the base of the pyramid. The center of the base, π, is the midpoint of the diagonal of the square, π΄πΆ. The triangle π΄π΅πΆ is a right triangle. Therefore, by the Pythagorean theorem, the square of the hypotenuse of this triangle, π΄πΆ, is equal to the sum of the squares of the other two sides, π΄π΅ and π΅πΆ.

π΄πΆ is twice the length of π΄π, so the left-hand side could be rewritten as two π΄π all squared. Also, since the base is a square, π΄π΅ is equal to π΅πΆ, so the right-hand side is equal to two π΄π΅ 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 π΄π squared equals one-half π΄π΅ squared.

Therefore, we can express the height of the pyramid entirely with known quantities. ππΈ is equal to square root of π΄πΈ squared minus half π΄π΅ squared. π΄πΈ is the lateral edge of the pyramid, four centimeters, and π΄π΅ is the side length of the base, five centimeters. Therefore, ππΈ 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, letβs clear some space and redraw the pyramid. The slant height of the pyramid is the length of the line from the top vertex, πΈ, to the midpoint of one of its base edges, say, π΅πΆ. As before, we can draw a vertical line from the vertex πΈ to the point π, the centroid of the base. The angle between this line and the line ππ is a right angle. So the triangle πππΈ is a right triangle.

We can therefore once again use the Pythagorean theorem to find the slant height, ππΈ. This is the hypotenuse of the right triangle πππΈ. So its square is equal to the sum of the squares of the other two sides, ππ and ππΈ. ππΈ is already known from the previous part of the question. So we just need to find ππ. π is the center of the base, and ππ is parallel to the edge π΄π΅. Therefore, ππ 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. ππΈ equals the square root of ππ squared plus ππΈ squared. ππ squared is equal to five over two all squared, and ππΈ is equal to the square root of four squared minus one-half times five squared. So ππΈ squared is just four squared minus one-half times five squared. Performing this calculation gives us ππΈ, 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.

Letβs start by sketching the cone. The height is 90 centimeters, the slant height is 106 centimeters, and the radius of the base, π, is unknown. The base is circular. And recall that the circumference of a circle is equal to two π times the radius π. And the area of a circle is given by π times the radius π 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 π , is equal to the sum of the squares of the other two sides, the radius π and the height β.

Rearranging for π by subtracting β squared from both sides and taking the square root gives us the radius, π, equal to the square root of π  squared minus β squared. The circumference of the circle is equal to two π times π, 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 π 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.

Letβs finish this video by recapping some key points. The Pythagorean theorem states that the square of the hypotenuse of a right triangle, π, is equal to the sum of the squares of the other two sides, π and π. 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 pyramidβs apex. The base radius π, vertical height β, and slant height π of a cone form a right triangle, with the slant as its hypotenuse. Therefore, by the Pythagorean theorem, π squared plus β squared is equal to π  squared.