Question: 3-D Diagonal

Comment on 3-D Diagonal


I just wanted to confirm, when you meant that the hypotenuse should be corresponding to the Pythagoras triplets, you did not mean even the legs of the right triangle should be corresponding right?
gmat-admin's picture

Hi Jagadish,
When working with right triangles, we should always be on the lookout for Pythagorean triplets. This means examining the legs AND the hypotenuse for Pythagorean triplets or MULTIPLES of Pythagorean triplets.

Hey just wanted to know that at 1:37, how did you conclude that the computed diagnol of 10 inches is perpendicular to the height(7 inches) of the cuboid?
gmat-admin's picture

Great question.

At 1:37 in the video, the diagonal in question is the hypotenuse of a right triangle with legs that are 6 inches and 8 inches long (see 1:21 in the video for more on this)

So, if we let h represent the hypotenuse of this right triangle, we can write: 6² + 8² = h², and when we solve that equation, we see that h = 10

Since that 10-inch diagonal lies on the bottom of the "box" then the side with length 7 must be perpendicular to that 10-inch diagonal.

Does that help?

Hey Brent,
In one of the previous questions, you had used a different approach. The answer then will be √245.
Check this out-
gmat-admin's picture

There's a big difference between the video question above and the linked question:

In the linked question, the ant walks from point A to point B along the OUTSIDE of the cube. In the video question above, we're looking for the DIRECT distance from A to B.

Does that help?


Hi Brent, can you please explain this question. I was unable to understand what the question means. I was trying to solve it by applying formula sqrt(side^2 + side^2 + sides^2).
gmat-admin's picture

Question link:

This question is different from the question in the above video.
In the above video, the distance from points A to B is the shortest path THROUGH the cube. That is, AB is an imaginary straight line passing through the center of the cube. In that case, the formula you provided works perfectly.

In the linked question, we aren't allowed to travel through the cube. Instead, we must travel on the SURFACE of the cube

Incidentally, I created a very similar question here:
You might want to review that question/solution, since it's pretty much identical to the linked question you're referring to. The only difference is that the cube in my question is 1x1x1, whereas the cube in the question you posted is 4x4x4

I hope that helps.

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