Video: The Pythagorean Theorem in 3D

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

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Video Transcript

In this video, we will learn how to use the Pythagorean theorem to solve problems in three dimensions. We’ll 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 𝑐 and the other two sides as π‘Ž and 𝑏, then we can write the Pythagorean theorem as 𝑐 squared equals π‘Ž squared plus 𝑏 squared.

When we’re 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 it’s 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‘Ž and 𝑏, 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, let’s 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) 𝐴 and 𝐢, option (B) 𝐸 and 𝐷, option (C) 𝐹 and 𝐢, option (D) 𝐺 and 𝐷, option (E) 𝐴 and 𝐺.

Here, we have a cuboid, which is often sometimes seen as a rectangular prism. When we’re asked to draw a diagonal of this cuboid, what we’re looking for is a line which passes through the interior of the cuboid, often called a space diagonal. Notice that it’s different to a face diagonal, which is a diagonal in two dimensions. So let’s say that the line is drawn between 𝐷 and 𝐡. We have then created a diagonal of the plane 𝐴𝐡𝐢𝐷. 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 𝐻 and 𝐢. However, once again, we’ve found a diagonal of the plane, this time 𝐢𝐷𝐻𝐺. 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? Let’s say we wanted to start at vertex 𝐻 and create a diagonal from there. Drawing a line to vertex 𝐢 would give us a face diagonal. So instead, traveling in three dimensions through the interior of the cuboid would take us to vertex 𝐡. We can therefore say that 𝐻𝐡 is the diagonal of the cuboid. A diagonal of the cuboid starting at vertex 𝐷 and traveling through the interior of the cuboid would take us to vertex 𝐹.

And so 𝐷𝐹 is also a diagonal of the cuboid. In fact, there are a total of four space diagonals in a cuboid. Although it’s a little trickier to see on this diagram, the line joining 𝐢 and 𝐸 is a diagonal. And finally, 𝐴𝐺 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), 𝐴 and 𝐺. The other four options here would all create face diagonals.

It’s important to note that when we’re working with the diagonals of three-dimensional shapes, for example, if we’re using the Pythagorean theorem, then the face diagonals will be a different length to the space diagonals. For example, the diagonal 𝐡𝐻 will be longer than the face diagonal 𝐡𝐸.

Let’s now look at some questions where we apply the Pythagorean theorem in three dimensions.

𝐴𝐡𝐢𝐷 𝐴 prime 𝐡 prime 𝐢 prime 𝐷 prime is a cube. Determine the lengths 𝐴 prime 𝐡 and 𝐴𝐢.

As we’re told that this is a cube, the first thing we can say is that all of the dimensions will be 97 centimeters. Let’s look at the first length we’re asked to find, 𝐴 prime 𝐡. This will be a face diagonal joining 𝐴 prime and 𝐡. As this is a cube, we know that we’ll 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 we’re drawing the triangle, it’s particularly useful to label the vertices.

We can see that our base length is 𝐴𝐡 and the height of our triangle is the line joining 𝐴 and 𝐴 prime. We can label our sides with 97 centimeters. And the length, which we wish to find out, 𝐴 prime 𝐡, we can call the letter π‘₯. When we’re applying the Pythagorean theorem, we must be careful to establish which length is the hypotenuse. It’s usually easily found as it’s opposite the right angle. We can then fill in our values into the Pythagorean theorem in the form 𝑐 squared equals π‘Ž squared plus 𝑏 squared. With the hypotenuse as π‘₯ and the other two sides as 97 each, we’ll 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, we’d 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 we’ve found the value of 𝐴 prime 𝐡, 97 root two centimeters.

Let’s take a look at the second part of the question, finding the length 𝐴𝐢. We could apply the same principle. We have a right triangle in triangle 𝐴𝐡𝐢. So we could draw out our two-dimensional triangle and fill in our values of 97 centimeters and 97 centimeters, using 𝑦 perhaps to designate the length 𝐴𝐢 which we wish to find out. However, we might then notice that the triangle we’ve 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 𝑦 as we did for π‘₯. So the length 𝐴𝐢 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 𝐴 prime 𝐡 and 𝐴𝐢 are both 97 root two centimeters.

Let’s have a look at another question.

Rectangle 𝐴𝐡𝐢𝐷 has 𝐴𝐡 equals 25 and 𝐡𝐢 equals 36. Suppose perpendiculars 𝐡𝐻 and 𝐴𝑂 are both of length 27. What is the area of 𝐢𝐷𝑂𝐻?

The first thing to do in a question like this is to sketch a diagram. We’re told that we have a rectangle. And coming from this will be two perpendiculars. Therefore, we’ll 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 𝐡 and goes up to a point 𝐻. Our second perpendicular starts at 𝐴 and goes up to a point 𝑂. We’re 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.

It’s worth noting that there are a number of different diagrams we could’ve drawn. What we’re really looking for is something to help us with our calculations. Here, we’re asked to find the area of 𝐢𝐷𝑂𝐻. 𝐢𝐷𝑂𝐻 would look like this on our diagram, and it would be a rectangle. In order to find out the area of this rectangle, 𝐢𝐷𝑂𝐻, 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 𝐴𝐡, which means it will be 25 units long. What we do need to work out here is the length, the line 𝑂𝐷.

As we know that 𝑂𝐴 is a perpendicular, that means that we have a right triangle in 𝑂𝐴𝐷. 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. It’s often helpful to draw out the triangles that we’re going to use. We know that the length 𝑂𝐴 is 27 and the length 𝐴𝐷 will be the same as the length 𝐡𝐢, 36. We want to find the length 𝑂𝐷, 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 doesn’t 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, we’ve calculated this length 𝑂𝐷 as 45. We can go ahead and work out the area. Multiplying our values 45 and 25, which give us that the area of 𝐢𝐷𝑂𝐻 is equal to 1125. We weren’t 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, we’ll 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. We’re asked to find the length of the diagonal of the cuboid. And so we’re 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 we’ll see why at the end of this question.

Let’s 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 𝐹𝐢 will be longer than any of the face diagonals. For example, it will be longer than the diagonal 𝐴𝐢. So now, to calculate the length 𝐹𝐢, we’ll 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 𝐴. 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 𝐴𝐹 would be three centimeters. However, we don’t know the length of this face diagonal 𝐴𝐢. We’ll need to work that out before we can find the length of our diagonal 𝐹𝐢. We can create another right triangle in the triangle 𝐴𝐡𝐢. When we’re working in three dimensions, it’s often helpful to redraw any two-dimensional triangles to help us with the calculations. We can see that the length 𝐴𝐡 would be four centimeters as it’s the same as the length 𝐸𝐻, which we’ve marked as four centimeters. The length 𝐡𝐢 is six centimeters, and the length 𝐴𝐢, which we wish to find out, can be called any letter. But here, let’s use 𝑝.

Plugging in the values to the Pythagorean theorem gives us 𝑝 squared equals four squared plus six squared. As four squared is 16 and six squared is 36, we’ll have 𝑝 squared equals 52. To find 𝑝, we take the square root of both sides, giving us that 𝑝 is equal to the square root of 52 centimeters. As we haven’t finished with this calculation, we can leave it in the square root form. Now, we found the value of 𝐴𝐢. We can continue with this right triangle 𝐹𝐴𝐢. We know that the length 𝐹𝐴 is three centimeters, the length 𝐴𝐢 is root 52 centimeters, and we can call the length 𝐹𝐢 anything. But here, let’s use the variable 𝑑.

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

However, there is another way we could’ve approached this problem. We can observe that our answer of the square root of 52 for our value 𝑝 was found by taking the square root of four squared plus six squared. So therefore, when it came to working out our value of 𝑑, our calculation for 𝑑 squared would be three squared plus four squared plus six squared. So our value 𝑑, 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 𝑐, we would calculate 𝑐 squared equals π‘₯ squared plus 𝑦 squared plus 𝑧 squared, where π‘₯, 𝑦, and 𝑧 are the dimensions of the cuboid. Applying this method in our cuboid, we’d have 𝑐 squared equals four squared plus six squared plus three squared. 𝑐 squared equals 16 plus 36 plus nine. So 𝑐 squared equals 61. And therefore, 𝑐 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‘₯, 𝑦, and 𝑧. Since it doesn’t 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.

We’ll 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 𝑐 squared equals π‘₯ squared plus 𝑦 squared plus 𝑧 squared.

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