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### Course: Basic geometry and measurement>Unit 13

Lesson 3: Pythagorean theorem application

# Pythagorean theorem in 3D

Discover how to find the length of an edge in a 3D shape using the Pythagorean theorem! This fun math lesson explores right pyramids and rectangular prisms, guiding you through solving for unknown lengths by applying the theorem multiple times.

## Want to join the conversation?

• Could you use the 3D pythagorean theorem which is a^2 + b^2 + c^2 = d^2 to figure out the edge of a pyramid given the side lengths of the base and the vertical height.
• Yes, you could use it but i personally like typing `√ a^2 + b^2 + c^2 = d` so it is calculated at once.
• I am colorblind and can not tell what is going on. Is there another video that someone could suggest?
• Let L, W, and H represent the dimensions (length, width, and height) of a rectangular prism, let C represent a diagonal of the bottom face, and let D represent a long diagonal of the prism.

We use the regular (2-dimensional) Pythagorean theorem on two right triangles.

One right triangle has legs L & W and hypotenuse C. This gives L^2+W^2=C^2.

The other right triangle has legs C & H and hypotenuse D. This gives C^2+H^2=D^2.

Substituting the first equation into the second equation gives the 3-dimensional result L^2+W^2+H^2=D^2. Notice how this is just like the 2-dimensional Pythagorean theorem, except that one more square is being added.
• Can you make more another video to help explain the Harder Pythagorean theorem in 3D problems?
• I find the whole video confusing. I would have preferred that he used a cube. When I looked up this video I thought it would thoroughly explain how to solve Pythagoras' theorem questions involving cubes but he seems to be using some sort of pyramid and it is all so confusing. I searched this to clarify this- http://www.bbc.co.uk/schools/gcsebitesize/maths/geometry/pythagoras3drev1.shtml

But I'm still as confused as I was after looking at that website
• why do we need the other dimensions then like the 3?
• I think what he meant was basically the 3 in the problem was useless. And yes, it is useless. Sometimes you're given a problem with information you won't need. You just left that out to the side.
• 2006 was 1 year after I was born! Did you know that?!
• no :0 but now i do thanks
• There is too much going on here for my mush of a brain to understand :(
• fr same
• At I don’t quite understand how Sal knows it’s a 90 degree angle
• It's a 90 degree angle because if you just look at the rectangle based pyramid, the height is straight, which means that any line across the rectangle based would make it into a right angle.
• What I find helpful is breaking the prism into two dimensional shapes. In this particular scenario, the pythagorean theorem is used twice, once to find the base and once to find the actual length we are looking for.

What we are looking for is the lateral edge of the pyramid, which basically amounts to a diagonal "slice" of the pyramid. So that's a 2D right triangle. Now, we know the height (which we'll assign the variable a) of that triangle (1 unit) but we are not given the length of the base of the triangle.

To find the base (which we'll call b), we look at the 2x4 rectangle that forms the base of the pyramid. What we're looking for is the distance from the center of the rectangle (where a originates) to one of the corners of the rectangle, where our lateral edge begins. If you draw it out, a point halfway along the edge of the rectangle, the start of a, and the corner where the lateral edge begins forms a right triangle. To calculate the distance from the start of a to the start of the lateral edge, all we need to do is find the hypotenuse of the right triangle.

So:
A^2 + B^2 = C^2
1^2 + 2^2 = 5
so sqrt(5) is the distance between the start of A and the start of the lateral edge. So the base of our final triangle, b, is sqrt(5).

Great! Now we can plug the base and the height into the equation for the lateral edge.

a^2 + b^2 = c^2
1^2 + sqrt(5)^2 = ?

one squared is one.
when sqrt(5) is squared, the sqrt and the square cancel out, so 5.

1+5 = 6
six is not a perfect square, so we leave our solution as *sqrt(6)*.

Hope this helps! ^-^