Deluxe Pythagorean Theorem OG 177

[EMAN]

A rectangular box is 10 inches wide, 10 inches long and 5 inches high. What is the greatest possible (straight-line) distance, in inches, between any two points on the box?

(A)15
(B)20
(C)25
(D)10 x SQRT 2
(E)10 x SQRT 3

The OG has a convoluted answer about using the Pythagorean theorem twice. Thank you Manhattan GMAT for showing us the Deluxe Pythagorean Theorem. All you have to do is 10^2 + 10^2 + 5^2 which equals 225. Then just find the square root of that which will be 15. The Deluxe Pythagorean Theorem is simply a^2 + b^2 + c^2 = d^2. If someone knows of cases where you can't do this for the appropriate three dimensional solid, please feel free to correct me.

[glorydefined]

Hey Eman,

Both are same, think of it as a rectangluar box (cuboid), the longest rod what one can fit into it, obviously has to be diagonal. So basically, three sides of box are given and you are asked to find the diagonal of the cuboid which is d= sqrt(a^2 + b^2 + c^2). Jus put the values and you get 15.
OG did not say this simple formula because GMAC does not want to make their problem sound so simple. However what they mean by using the pythogoreous theorem twice is same as finding diagonal for the cuboid..hope u got the picture