mid level geometric problem

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vini vidi vici: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Sun Oct 05, 2008 8:30 am

by vini vidi vici » Thu Oct 16, 2008 3:51 am

A rectangular wooden crate has inside dimensions 3 meters by 4 meters by 12 meters. What is the length, in meters, of the longest, straight, inflexible rod of negligible diameter that can be placed completely within the crate?

A. 12
B. 12.6
C. 13
D. 19
E. 24

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Source: Beat The GMAT — Problem Solving | Thu Oct 16, 2008 3:51 am

stop@800: Legendary Member; Posts: 871; Joined: Wed Aug 13, 2008 7:48 am; Thanked: 48 times

Re: mid level geometric problem

by stop@800 » Thu Oct 16, 2008 3:58 am

vini vidi vici wrote:A rectangular wooden crate has inside dimensions 3 meters by 4 meters by 12 meters. What is the length, in meters, of the longest, straight, inflexible rod of negligible diameter that can be placed completely within the crate?

A. 12
B. 12.6
C. 13
D. 19
E. 24

13
diag of 1 face
= sqrt (3^2 + 4^2)
= 5

largest diag
= sqrt (5^2 + 12^2)
= 13

C

OA??

www.interrogationpoint.com

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Thu Oct 16, 2008 4:05 am

stop@800's solution above is perfect. I'd just add that there's a 3-dimensional version of the Pythagorean Theorem that can be used here. Just as the diagonal d of a rectangle with sides a and b satisfies d^2 = a^2 + b^2, the diagonal d between opposite corners of a rectangular box measuring a by b by c satisfies:

d^2 = a^2 + b^2 + c^2

So

d^2 = 3^2 + 4^2 + 12^2 = 169
d = 13

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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vini vidi vici: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Sun Oct 05, 2008 8:30 am

by vini vidi vici » Thu Oct 16, 2008 10:13 am

thanks guys! noted

i was on the lookout for that formula too. should be very useful

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