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by vaivish » Fri Aug 01, 2008 1:49 am
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As the figure above shows, the spiral is formed with 11 right triangles. What is the length of the last triangle's hypotenuse?
A. 11
B. root11
C. root12
D. root13
E. root14

OA is C
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by pepeprepa » Sat Aug 02, 2008 12:43 am
Can someone give us a hand with this one?
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by egybs » Sat Aug 02, 2008 2:01 am
I think there's a piece of info that's cut off. Vaivish is great at leaving out key pieces of info. :?

For this to work, the bottom triangle needs to have both non-hypotenuse sides length 1.

With that you end up with the first hypotenuse being sqrt(2), the second, sqrt(3), etc... until you get to x which would be equal to sqrt(12).


You need to know that first length for this to be possible.

Also, I question the question... the diagram labeled the "last" triangle's hypotenuse as x... why would the question phrase it so ambiguously? Which triangle is the last?
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by Ian Stewart » Sat Aug 02, 2008 2:47 am
Well, it's impossible to answer from that diagram- one of the other sides needs to be labeled. The first triangle in the spiral (the smallest one, at the bottom), should be labeled with a base of 1. Call this triangle 1, and we'll work counterclockwise around the spiral:

hypotenuse of triangle 1 = root(2) by Pythagoras
hypotenuse of triangle 2 = root(3) by Pythagoras
hypotenuse of triangle 3 = root(4) by Pythagoras
...
hypotenuse of triangle n = root(n+1)

So the hypotenuse of the 11th triangle is root(12) = 2root(3).
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