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- Thread starter Tyrion101
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collinsmark

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If the problem involves right triangles (utilizing the Pythagorean therem, *c*^{2} = *a*^{2} + *b*^{2}) then it ends up pretty easy if the triangles are 3-4-5 triangles (5^{2} = 3^{2} + 4^{2}).

You can also scale the 3, 4 and 5 by any constant and it still works out. For example, let's multiply them all by 123.

*c*^{2} = 369^{2} + 492^{2}.

Solve for*c*.

You can also scale the 3, 4 and 5 by any constant and it still works out. For example, let's multiply them all by 123.

Solve for

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AlephZero

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lisab

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Not very hard for those who can easily solve them forwards and back. But for someone first learning, it's an interesting observation. I wonder how many students are shocked the first time they get an answer that has a digits after the decimal, and realize it's a correct answer!

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