### Video Transcript

In this video, we will learn how to
use the Pythagorean theorem to solve problems in three dimensions. Weβll begin by recapping what the
Pythagorean theorem is. The Pythagorean theorem tells us
that, for every right triangle, the area of the square on the hypotenuse, this
longest side in the triangle, is equal to the sum of the squares on the other two
sides. If we represent the hypotenuse with
the letter π and the other two sides as π and π, then we can write the
Pythagorean theorem as π squared equals π squared plus π squared.

When weβre using the Pythagorean
theorem in three dimensions, we could still use this formula. But we need to be careful that we
find a triangle in two dimensions and that itβs a right triangle. For example, if we wanted to
calculate the length of this face diagonal on the cuboid or rectangular prism, then
as we have a prism, we know that this is a right angle. And therefore, we could use the
other two sides, π and π, to calculate the length of the face diagonal. Alternatively, if we wanted to
calculate the length of the diagonal that runs through this cuboid, then this right
triangle would have a length that is the face diagonal of the base of the
cuboid. Before we use the Pythagorean
theorem, letβs first look at a question where we identify the diagonals of a
cuboid.

Identify a pair of points between
which diagonal of the cuboid can be drawn. Option (A) π΄ and πΆ, option (B) πΈ
and π·, option (C) πΉ and πΆ, option (D) πΊ and π·, option (E) π΄ and πΊ.

Here, we have a cuboid, which is
often sometimes seen as a rectangular prism. When weβre asked to draw a diagonal
of this cuboid, what weβre looking for is a line which passes through the interior
of the cuboid, often called a space diagonal. Notice that itβs different to a
face diagonal, which is a diagonal in two dimensions. So letβs say that the line is drawn
between π· and π΅. We have then created a diagonal of
the plane π΄π΅πΆπ·. However, as this plane is in two
dimensions, we have only created a face diagonal. Therefore, this line would not be a
diagonal of the cuboid. In the same way, we could create
the line between π» and πΆ. However, once again, weβve found a
diagonal of the plane, this time πΆπ·π»πΊ. And therefore, the diagonal would
be a face diagonal but not a diagonal of the cuboid.

So what would a diagonal of the
cuboid actually look like? Letβs say we wanted to start at
vertex π» and create a diagonal from there. Drawing a line to vertex πΆ would
give us a face diagonal. So instead, traveling in three
dimensions through the interior of the cuboid would take us to vertex π΅. We can therefore say that π»π΅ is
the diagonal of the cuboid. A diagonal of the cuboid starting
at vertex π· and traveling through the interior of the cuboid would take us to
vertex πΉ.

And so π·πΉ is also a diagonal of
the cuboid. In fact, there are a total of four
space diagonals in a cuboid. Although itβs a little trickier to
see on this diagram, the line joining πΆ and πΈ is a diagonal. And finally, π΄πΊ is our fourth
diagonal. Out of the answer options, (A)
through (E), that we were given, the only one which is a pair of points between
which a diagonal can be drawn is option (E), π΄ and πΊ. The other four options here would
all create face diagonals.

Itβs important to note that when
weβre working with the diagonals of three-dimensional shapes, for example, if weβre
using the Pythagorean theorem, then the face diagonals will be a different length to
the space diagonals. For example, the diagonal π΅π» will
be longer than the face diagonal π΅πΈ.

Letβs now look at some questions
where we apply the Pythagorean theorem in three dimensions.

π΄π΅πΆπ· π΄ prime π΅ prime πΆ prime
π· prime is a cube. Determine the lengths π΄ prime π΅
and π΄πΆ.

As weβre told that this is a cube,
the first thing we can say is that all of the dimensions will be 97 centimeters. Letβs look at the first length
weβre asked to find, π΄ prime π΅. This will be a face diagonal
joining π΄ prime and π΅. As this is a cube, we know that
weβll have right angles in the corner of this face. We can therefore use the
Pythagorean theorem to help us answer this. The Pythagorean theorem tells us
that the square on the hypotenuse is equal to the sum of the squares on the other
two sides. It can be helpful to draw the
triangle in two dimensions so we can work with it a little bit easier. And when weβre drawing the
triangle, itβs particularly useful to label the vertices.

We can see that our base length is
π΄π΅ and the height of our triangle is the line joining π΄ and π΄ prime. We can label our sides with 97
centimeters. And the length, which we wish to
find out, π΄ prime π΅, we can call the letter π₯. When weβre applying the Pythagorean
theorem, we must be careful to establish which length is the hypotenuse. Itβs usually easily found as itβs
opposite the right angle. We can then fill in our values into
the Pythagorean theorem in the form π squared equals π squared plus π
squared. With the hypotenuse as π₯ and the
other two sides as 97 each, weβll have π₯ squared equals 97 squared plus 97
squared.

Evaluating the squares gives us π₯
squared equals 9409 plus 9409, which is 18818. Taking the square root of both
sides gives us that π₯ is equal to the square root of 18818 centimeters. We can then simplify this answer in
two ways. We could go ahead and put this into
our calculator, giving us a decimal value, which we could round. Alternatively, we could simplify
this answer in its square root form. We can make this a little bit
easier if we rewrite this earlier line. If π₯ squared is equal to 9409 plus
9409, then we could also write this as π₯ squared equals two times 9409.

So therefore, when it comes to
taking the square root, weβd be finding the square root of two times 9409. This is equivalent to the square
root of two times the square root of 9409. This may not seem any easier until
we recall that we know the square root of 9409. It is, in fact, 97. So π₯ is then equal to the square
root of two times 97, or more simply 97 root two centimeters. So now, we found the value of
π₯. That means weβve found the value of
π΄ prime π΅, 97 root two centimeters.

Letβs take a look at the second
part of the question, finding the length π΄πΆ. We could apply the same
principle. We have a right triangle in
triangle π΄π΅πΆ. So we could draw out our
two-dimensional triangle and fill in our values of 97 centimeters and 97
centimeters, using π¦ perhaps to designate the length π΄πΆ which we wish to find
out. However, we might then notice that
the triangle weβve drawn before also had the two shorter sides of 97
centimeters. So when we apply the Pythagorean
theorem and start working out our values, we can see that we will get the same
answer for π¦ as we did for π₯. So the length π΄πΆ is also 97 root
two centimeters. In fact, all of the face diagonals
in a cube would be the same length. But here, we can note our answer
that π΄ prime π΅ and π΄πΆ are both 97 root two centimeters.

Letβs have a look at another
question.

Rectangle π΄π΅πΆπ· has π΄π΅ equals
25 and π΅πΆ equals 36. Suppose perpendiculars π΅π» and
π΄π are both of length 27. What is the area of πΆπ·ππ»?

The first thing to do in a question
like this is to sketch a diagram. Weβre told that we have a
rectangle. And coming from this will be two
perpendiculars. Therefore, weβll be working in
three dimensions. So it might be useful to draw our
rectangle like this, ready to add in the perpendiculars. Our first perpendicular starts at
point π΅ and goes up to a point π». Our second perpendicular starts at
π΄ and goes up to a point π. Weβre told that these
perpendiculars are both of length 27. So our perpendiculars should be
roughly the same length. We can then add in the dimensions
to our diagram.

Itβs worth noting that there are a
number of different diagrams we couldβve drawn. What weβre really looking for is
something to help us with our calculations. Here, weβre asked to find the area
of πΆπ·ππ». πΆπ·ππ» would look like this on
our diagram, and it would be a rectangle. In order to find out the area of
this rectangle, πΆπ·ππ», we would need to know the length and the width, which we
could multiply together to find the area. The width here would be equal to
the line π΄π΅, which means it will be 25 units long. What we do need to work out here is
the length, the line ππ·.

As we know that ππ΄ is a
perpendicular, that means that we have a right triangle in ππ΄π·. We can therefore apply the
Pythagorean theorem, which tells us that the square on the hypotenuse is equal to
the sum of the squares on the other two sides. Itβs often helpful to draw out the
triangles that weβre going to use. We know that the length ππ΄ is 27
and the length π΄π· will be the same as the length π΅πΆ, 36. We want to find the length ππ·,
which we can define as any length, but here we can call it π₯. Filling in the values into the
Pythagorean theorem, we have the hypotenuse as π₯ and the two shorter sides of 27
and 36. And it doesnβt matter which way
round we write those. We can evaluate our calculation as
π₯ squared equals 1296 plus 729. So π₯ squared equals 2025. To find π₯, we take the square root
of both sides, so π₯ equals the square root of 2025.

Usually, we keep our answer in this
square root form. But actually, 2025 is a square
number. So π₯ would be 45 units. Now, weβve calculated this length
ππ· as 45. We can go ahead and work out the
area. Multiplying our values 45 and 25,
which give us that the area of πΆπ·ππ» is equal to 1125. We werenβt given any units in the
question, but if we needed to give units here, they would of course be square units
for the area.

In the next question, weβll see how
the Pythagorean theorem can be extended directly into a three-dimensional
theorem.

Find the length of the diagonal of
a cuboid of sides three centimeters, four centimeters, and six centimeters.

We can start this question by
drawing a diagram of the cuboid. Weβre asked to find the length of
the diagonal of the cuboid. And so weβre looking for the line
that runs through the interior of the cuboid. There will, in fact, be a total of
four of these space diagonals. These diagonals would all be the
same length, and weβll see why at the end of this question.

Letβs begin by finding the length
of this orange diagonal. It can be useful to label the
vertices in the cuboid to help with referencing the line segments. We can start by noticing that the
diagonal πΉπΆ will be longer than any of the face diagonals. For example, it will be longer than
the diagonal π΄πΆ. So now, to calculate the length
πΉπΆ, weβll need to create a right triangle within the cuboid. As this is a cuboid, we know that
there will be a right angle at angle π΄. We can then apply the Pythagorean
theorem, which tells us that the square on the hypotenuse is equal to the sum of the
squares on the other two sides.

We can see that the length π΄πΉ
would be three centimeters. However, we donβt know the length
of this face diagonal π΄πΆ. Weβll need to work that out before
we can find the length of our diagonal πΉπΆ. We can create another right
triangle in the triangle π΄π΅πΆ. When weβre working in three
dimensions, itβs often helpful to redraw any two-dimensional triangles to help us
with the calculations. We can see that the length π΄π΅
would be four centimeters as itβs the same as the length πΈπ», which weβve marked as
four centimeters. The length π΅πΆ is six centimeters,
and the length π΄πΆ, which we wish to find out, can be called any letter. But here, letβs use π.

Plugging in the values to the
Pythagorean theorem gives us π squared equals four squared plus six squared. As four squared is 16 and six
squared is 36, weβll have π squared equals 52. To find π, we take the square root
of both sides, giving us that π is equal to the square root of 52 centimeters. As we havenβt finished with this
calculation, we can leave it in the square root form. Now, we found the value of
π΄πΆ. We can continue with this right
triangle πΉπ΄πΆ. We know that the length πΉπ΄ is
three centimeters, the length π΄πΆ is root 52 centimeters, and we can call the
length πΉπΆ anything. But here, letβs use the variable
π‘.

Applying the Pythagorean theorem
here, we have that π‘ squared equals three squared plus root 52 squared. As the square root of 52 squared
gives us 52, weβll have π‘ squared equals nine plus 52, which is 61. To find π‘, we take the square root
of both sides. So we have π‘ equals the square
root of 61 centimeters. Our value π‘ was the length πΉπΆ,
which was a diagonal of the cuboid. So our answer is root 61
centimeters.

However, there is another way we
couldβve approached this problem. We can observe that our answer of
the square root of 52 for our value π was found by taking the square root of four
squared plus six squared. So therefore, when it came to
working out our value of π‘, our calculation for π‘ squared would be three squared
plus four squared plus six squared. So our value π‘, the diagonal of
the cuboid, is found by taking the square root of three squared plus four squared
plus six squared. And these are the dimensions of our
cuboid: three centimeters, four centimeters, and six centimeters.

And that leads us to an extension
or analogue of the Pythagorean theorem in three dimensions. If we wanted to calculate the
length of this diagonal marked π, we would calculate π squared equals π₯ squared
plus π¦ squared plus π§ squared, where π₯, π¦, and π§ are the dimensions of the
cuboid. Applying this method in our cuboid,
weβd have π squared equals four squared plus six squared plus three squared. π squared equals 16 plus 36 plus
nine. So π squared equals 61. And therefore, π equals root 61
centimeters, confirming our earlier answer for the diagonal of the cuboid.

Looking at this extension of the
Pythagorean theorem, we can also see why all of the space diagonals would be the
same length in a cuboid. Each space diagonal needs to take
into account each of the three dimensions π₯, π¦, and π§. Since it doesnβt matter which order
we square and add them in and then finally take the square root, then these will all
give the same answer. We can note our answer for the
diagonal of the cuboid here as root 61 centimeters.

Weβll now summarize the key points
of this video. We saw how we can apply the
Pythagorean theorem in right triangles in three-dimensional objects. We saw that sometimes we need to
apply the Pythagorean theorem twice to find particular lengths. A handy tip for this is to keep the
result of the first calculation in our square root form, which means that the answer
for the second part would be more accurate. And finally, we saw how the
Pythagorean theorem can be extended into three dimensions by using the formula π
squared equals π₯ squared plus π¦ squared plus π§ squared.