Video Transcript
In this video, we’ll learn how to
use the Pythagorean theorem to solve problems in three dimensions. We should already be familiar with
the Pythagorean theorem in two dimensions, which describes the relationship between
the lengths of the three sides in a right triangle. So let’s begin by recapping this
theorem.
In a right triangle, the square of
the hypotenuse is equal to the sum of the squares of the two shorter sides. So if we take this right triangle
below and label the hypotenuse, that’s the longest side, with the letter 𝑐 and the
two shorter sides with the letters 𝑎 and 𝑏, then by the Pythagorean theorem, we
can say that 𝑎 squared plus 𝑏 squared is equal to 𝑐 squared. And as we’ll see in this video, it
is possible to apply this Pythagorean theorem in two dimensions to shapes in three
dimensions.
And we do this by finding
two-dimensional slices or cross sections of a three-dimensional shape. For example, we could use this
right triangle to find the length of one of the face diagonals. However, we do need to be a little
careful. For example, if we wanted to find
the length of this space diagonal, then we would first need to find the length of
this face diagonal in order to apply the Pythagorean theorem in two dimensions.
In this video, we’ll see how we can
apply this Pythagorean theorem in two dimensions to various three-dimensional
shapes, including cones and pyramids. But let’s begin with an example
where we apply the Pythagorean theorem within a cube.
Given that 𝐴𝐵𝐶𝐷𝐸𝐹𝐺𝐻 is a
cube whose edge length is six root two centimeters and 𝑋 is the midpoint of line
segments 𝐴𝐵, find the area of the rectangle 𝐷𝑋𝑌𝐸.
We are told that this
three-dimensional shape is a cube with an edge length of six root two
centimeters. And that means that the length, the
width, and the height are all six root two centimeters. We observe that we have this
additional line segment 𝑋𝑌, where we are given that 𝑋 is the midpoint of line
segment 𝐴𝐵. The line segment 𝑋𝑌 forms one of
the sides of the rectangle 𝐷𝑋𝑌𝐸, whose area we need to calculate.
We can recall that the area of a
rectangle is equal to the length multiplied by the width. Therefore, to find the area of
𝐷𝑋𝑌𝐸, we can calculate 𝐷𝑋 multiplied by 𝑋𝑌. Now the length of 𝑋𝑌 is
relatively easy to work out. Because we know that the height of
the cube is six root two centimeters, then the length of 𝑋𝑌 is also six root two
centimeters. However, the length of 𝐷𝑋 will
require some more calculations.
Let’s consider a two-dimensional
drawing of the base of the cube, which will form a square. We can also add in the line segment
𝑋𝐷. Given that 𝑋 is the midpoint of
the line segment 𝐴𝐵, then the line segment 𝐴𝑋 has a length of half of six root
two centimeters, which is three root two centimeters. We can now use the Pythagorean
theorem in the right triangle 𝐷𝐴𝑋 to work out the length of the line segment
𝐷𝑋.
Recall that the Pythagorean theorem
states that in every right triangle, the square of the hypotenuse is equal to the
sum of the squares of the two shorter sides. Applying this in triangle 𝐷𝐴𝑋,
we note that 𝐷𝑋 is the hypotenuse. So we have 𝐴𝑋 squared plus 𝐴𝐷
squared is equal to 𝐷𝑋 squared. Substituting in the length values
gives us three root two squared plus six root two squared is equal to 𝐷𝑋
squared. Simplifying this gives 90 is equal
to 𝐷𝑋 squared. We can then find the length of 𝐷𝑋
by taking the square root of both sides. As root 90 can be written as root
nine times root 10, we have that 𝐷𝑋 is equal to three root 10 centimeters.
We now have enough information to
return to the area calculation. The area of 𝐷𝑋𝑌𝐸, which is 𝐷𝑋
multiplied by 𝑋𝑌, can be given as three root 10 multiplied by six root two. This is equal to 18 root 20. We can further simplify this by
considering that root 20 is equal to root four times five. And so the area of 𝐷𝑋𝑌𝐸 is 36
root five. And as this is an area, then the
units will be square units. We can therefore give the answer
that the area of rectangle 𝐷𝑋𝑌𝐸 is 36 root five square centimeters.
In the next example, we’ll see how
we could apply the Pythagorean theorem within a pyramid. But first, let’s recap some of the
key terms that we use when describing lengths in a pyramid.
The top of a pyramid is called the
vertex or apex of the 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 on the pyramid’s
base. Next, the slant height of a pyramid
is the perpendicular distance from one of the sides of the base to the apex of the
pyramid. And finally, a lateral edge
represents the length of one of the sides of the triangle making up the sides of the
pyramid. We must be careful not to confuse
slant height and lateral edge length.
In the next example, we’ll see how
we can apply the Pythagorean theorem to find the height of a pyramid.
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 this problem by
sketching the pyramid. The important information to
show on the diagram is that 𝑀 is the vertex of the pyramid. The base is 𝐴𝐵𝐶. And we know that that’s an
equilateral triangle. Because it’s equilateral, we
know that all of the side lengths of the triangle at the base will be 32
centimeters. As this is a regular pyramid,
then all the lateral edge lengths will be 88 centimeters.
We need to calculate the height
of the pyramid, which is the vertical distance from the vertex 𝑀 to the center
of the base of the pyramid, which we can label with the letter 𝑋. We can connect this center
point to one of the vertices of the triangle at the base. Observe that since we have a
right triangle in 𝑀𝐴𝑋, then we could calculate the height 𝑀𝑋 if we knew the
length of this line segment 𝐴𝑋. So let’s consider a
two-dimensional drawing of the triangle 𝐴𝐵𝐶 at the base of the pyramid and
see how we can calculate the length of 𝐴𝑋.
Point 𝑋 is the centroid of the
triangle, which is the point at which the three medians of the triangle
intersect. There are two important
properties that we’ll need to solve this problem. The first is that in an
equilateral triangle all three medians are of equal length. The second property is that the
centroid of a triangle divides each median in the ratio two to one from the
vertex. From these properties then, we
can say that 𝐴𝑋 is equal to 𝐵𝑋 is equal to 𝐶𝑋. And if we label the midpoint of
𝐵𝐶 with the letter 𝑌, then from the second property, we can say that 𝐴𝑋 is
equal to two-thirds of 𝐴𝑌.
We can now take a closer look
at triangle 𝐴𝐵𝑌. This will in fact have a right
angle at 𝑌 since the median 𝐴𝑌 divides the equilateral triangle 𝐴𝐵𝐶 into
two congruent right triangles. 𝐵𝑌 is half the length of
𝐵𝐶, so it is 16 centimeters. As we know two of the lengths
in this right triangle, we can apply the Pythagorean theorem to find the length
of 𝐴𝑌. As 𝐴𝐵 is the hypotenuse, we
will have that 𝐴𝑌 squared plus 𝐵𝑌 squared is equal to 𝐴𝐵 squared.
Substituting in the length of
𝐵𝑌, which is 16 centimeters, and 𝐴𝐵, which is 32 centimeters, we have 𝐴𝑌
squared plus 16 squared equals 32 squared. By rearranging and simplifying,
we have that 𝐴𝑌 squared is equal to 768. To find the value of 𝐴𝑌, we
take the square root of both sides, leaving us with 16 root three
centimeters. We have therefore worked out
that the length of the entire median 𝐴𝑌 is 16 root three centimeters. But of course, we’re really
just interested in the length of the line segment 𝐴𝑋.
Remember, however, that we have
already noted that 𝐴𝑋 is equal to two-thirds of 𝐴𝑌. So 𝐴𝑋 is equal to two-thirds
multiplied by 16 root three. That’s 32 root three over three
centimeters. But now when we return to the
pyramid, we can see that we have two sides in the triangle 𝑀𝐴𝑋, and we can
work out the third side 𝑀𝑋 using the Pythagorean theorem.
We can write that 𝐴𝑋 squared
plus 𝑀𝑋 squared is equal to 𝐴𝑀 squared. Filling in the lengths, we have
32 root three over three squared plus 𝑀𝑋 squared equals 88 squared. By rearranging and using our
calculators, we can work out that 𝑀𝑋 squared is equal to 22208 over three. We can then find the value of
𝑀𝑋 by taking the square root. And as we need the answer to
the nearest hundredth, then we can convert this to a decimal. As 𝑀𝑋 is the height of the
pyramid, then the answer is that the height of the pyramid to the nearest
hundredth is 86.04 centimeters.
In the next example, we’ll see how
we can apply the Pythagorean theorem within a cone.
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.
We can begin by representing this
circular cone, which has a height of 90 centimeters and a slant height of 106
centimeters. As this is a right cone, then the
height is the perpendicular distance from the apex of the cone to the center of the
base. The slant height is the distance
from any point on the circumference of the circular base to the apex along the
surface of the cone.
Notice that if we draw the radius
of the circle at the center of this cone, then the radius, the perpendicular height,
and the slant height all form a right triangle. As we need to find the
circumference and the area of this circular base, then knowing the radius is very
important. The circumference is equal to two
times 𝜋 times the radius, and the area of a circle is equal to 𝜋𝑟 squared.
Recall that the Pythagorean theorem
tells us that in any right triangle, the square of the hypotenuse is equal to the
sum of the squares of the two shorter sides. By considering this two-dimensional
right triangle created within the cone then, we can say that 𝑟 squared plus 90
squared is equal to 106 squared, where 𝑟 is the radius of the circle. By rearranging and solving, taking
the square root of both sides, we have that 𝑟 is equal to 56 centimeters. The radius of the circle at the
base of this cone is 56 centimeters.
We can therefore determine that the
circumference is equal to two times 𝜋 times 56, which is 112𝜋 centimeters. The area, which is equal to 𝜋
times the radius squared, is therefore equal to 𝜋 times 56 squared. This simplifies to 3136𝜋. And remember, as this is an area,
then we’ll have the units of square centimeters. We can therefore give the two
answers then that the base of this cone, which is a circle, has a circumference of
112𝜋 centimeters and an area of 3136𝜋 square centimeters.
We will now see how the Pythagorean
theorem in two dimensions can be extended into three dimensions.
Let’s say that we have this cuboid
𝐴𝐵𝐶𝐷𝐸𝐹𝐺𝐻 and we want to find the length of the diagonal 𝑑, which is the
line segment 𝐴𝐺. If the dimensions of this cuboid
are given as 𝑎, 𝑏, and 𝑐, then in order to find the length 𝑑, we would first
need to find the length 𝑒, which would be created by the line segment 𝐴𝐹. By creating a right triangle and
applying the two-dimensional Pythagorean theorem, we know that 𝑒 squared is equal
to 𝑎 squared plus 𝑏 squared. And if we consider the right
triangle 𝐴𝐹𝐺, we can say that 𝑑 squared is equal to 𝑒 squared plus 𝑐
squared.
But since we know that 𝑒 squared
is equal to 𝑎 squared plus 𝑏 squared, we can then combine these two equations. 𝑑 squared must be equal to 𝑎
squared plus 𝑏 squared plus 𝑐 squared. And so we have proved the
three-dimensional extension of the Pythagorean theorem. This states that in a cuboid with
side lengths 𝑎, 𝑏, and 𝑐 and an interior diagonal 𝑑, then 𝑎 squared plus 𝑏
squared plus 𝑐 squared is equal to 𝑑 squared.
We have now seen how we can apply
the two-dimensional Pythagorean theorem in a range of three-dimensional shapes and
also the three-dimensional extension of the Pythagorean theorem. So now let’s summarize the key
points.
We saw that 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 a cone, the base radius 𝑟,
vertical height ℎ, and slant height 𝑙 form a right triangle. So by applying the Pythagorean
theorem, we have that 𝑟 squared plus ℎ squared is equal to 𝑙 squared. And finally, we saw the extension
of the Pythagorean theorem to three dimensions, which states that for a cuboid with
side lengths 𝑎, 𝑏, and 𝑐 and a diagonal of length 𝑑, then 𝑎 squared plus 𝑏
squared plus 𝑐 squared is equal to 𝑑 squared.
