Simple algebra involving pythagorean theorum

NewSlang

New member
Joined
Sep 10, 2006
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6
I'm trying to use an equation to determine the length of the leg of a right triangle, knowing the length of the other two sides. One leg is 90 and the hypotenuse is 150. It seems to me that the correct equation to use to solve this is:

90^2+x^2=150^2

Yet when I try to solve this equation by taking the square root of both sides, I end with "x=60" as my answer. This clearly cannot be true, and through trial and error I found the correct answer to be "x=120". Am I starting with an incorrect equation or am I solving it wrong? Either way, how do I get to the right answer by setting up an equation?
 

galactus

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Sep 28, 2005
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You have it set up right, just do some very minor algebra.

\(\displaystyle 90^{2}+x^{2}=150^{2}\)

\(\displaystyle x^{2}=150^{2}-90^{2}\)

\(\displaystyle x=\sqrt{150^{2}-90^{2}}\)
 

jonboy

Full Member
Joined
Jun 8, 2006
Messages
547
Hello, NewSlang!


NewSlang said:
Solve:\(\displaystyle \L \;90^2\,+\,x^2\,=\,150^2\)

First you always simplify the exponents:\(\displaystyle \L \;8,100\,+\,x^2\,=\,22,500\)


Isolate the \(\displaystyle x^2\):\(\displaystyle \L \;x^2\,=\,14,400\)



.................Sqrt both sides:\(\displaystyle \L \;x\,=\,120\)


Check:\(\displaystyle \L \;90^2\,+\,(120)^2\,=\,150^2\)



.................\(\displaystyle \L \;22,500\,=\,22,500\)
 

NewSlang

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Sep 10, 2006
Messages
6
thanks that makes perfect sense now.
 
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