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If the hypotenuse is 41 inches, and one of the legs is 40 inches, what is the length of the other leg in inches?
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very much like the last one
start with \[a^2+40^2=41^2\] then solve for \(a^2\) and then for \(a\)
yeah.. thats the one i was looking at and got mixed up with..
the other leg is 9 inches
or you can go right to \[a=\sqrt{40^2-41^2}\]
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ah that is the one you have before, right?
thanks but i am going the long way. 81?
yes, 81 is \(a^2\)
yeah it is
and so this time \(a=9\)
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