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| Math Central | Quandaries & Queries |
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Question from Aaron: I want to make a table with a puzzle embedded in it. The table top would be a 36" circle and the puzzle is 20"x27" I'm thinking that it's not going to fit, but not sure. Any help would be appreciated. |
Hi Aaron,
It's surprising how many times Pythagoras Theorem is helpful.
Triangle $ABC$ is a right triangle and hence Pythagoras Theorem tells us that
\[|AB|^2 + |BC|^2 = |CA|^2\]
or
\[|CA|^2 = 27^2 + 20^2 = 1129\]
so
\[|CA| = \sqrt{1129} = 33.6 \mbox{ inches.}\]
Thus it will fit but it is tight. Each corner of the puzzle will only be 1.8 inches from the circle.
Penny
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