Quant questions on the GMAT occasionally ask us to find the length of the longest distance between two vertices in a rectangular solid. To solve these, usually we solve by (1) drawing the figure to visualize it, and (2) carefully applying the Pythagorean Theorem twice.

For example:

*In a cube with a side length of 12 cm, A is the midpoint of an edge that lies on the base and B is the midpoint of a vertical edge. What is the greatest possible distance between A and B rounded to the nearest integer?*

*(A) 10*

*(B) 12*

*(C) 14*

*(D) 16*

*(E) 18*

The greatest distance will occur when B is not on the same face as A. To find the distance of AB, we must find the distance from A to the far right corner of the cube. Let’s use Pythagorean Theorem:

6^{2 }+ 12^{2 }= c^{2}

36 + 144 = c^{2}

√180 = c

6√5 = c

Now we have the two side lengths of the triangle formed with A and B (shown with the dotted lines in the figure). We can use Pythagorean Theorem again to find the distance between A and B:

(6√5)^{2} + 6^{2} = c^{2}

36*5 + 26 = c^{2}

180 + 26 = c^{2}

206 = c^{2}

Since 14^{2} = 196 and 15^{2} = 225, we can estimate that the nearest integer of √206 is 14. The answer is (C).

But rather than have to go through these steps, there’s a much easier shortcut formula we could have applied! Given a prism A x B x C, the longest segment between two vertices of the solid will always equal **√(A ^{2} + B^{2} + C^{2}).** For the above question, since it asked about midpoints, the dimensions would have been 6 x 12 x 6. Without even drawing it:

√(6^{2} + 12^{2} + 6^{2})

√(36 + 144 + 36)

√(216) = 14.69. The answer is (C). Much less work!

There’s nothing wrong with drawing the figure if you need to visualize it, but using this extension of the Pythagorean Theorem is definitely more efficient! Now try one on your own:

*A rectangular solid has dimensions of 3 x 5 x 7. What is the length of the longest segment whose endpoints are vertices of the solid?*

*(A) 8.9*

*(B) 9.1*

*(C) 9.5*

*(D) 9.8*

*(E) 10.1*

–

Here A = 3, B = 5, and C = 7. All we have to do is plug in: 3^{2} + 5^{2} + 7^{2} = 9 + 25 + 49 = 83

The √83 will be a little bit larger than 9, since √81 = 9. The answer is 9.1, or (B).

On the GMAT, saving just a few seconds on one solids question can give you more time for tougher Word Problems later on — look for opportunities to use this condensed formula, and avoiding drawing the figure if you already know the dimensions!

Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!

*Vivian Kerr is a regular contributor to the Veritas Prep blog, providing advice to help students better prepare for the GMAT and the SAT. *

Great Post!

Just a small typo – square root 180 + 36 (not 26). So you should get square root 216 instead of 206. Thanks.