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mid level geometric problem

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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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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??
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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
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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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