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Mathematics 18 Online
OpenStudy (anonymous):

Can someone help me here? Solve for x and y. Be prepared to explain what theorem or postulate you used.

OpenStudy (anonymous):

|dw:1425684406448:dw|

OpenStudy (anonymous):

It isnt going straight through sadly, so no its not 45-45

OpenStudy (anonymous):

@LeofisTaylor

OpenStudy (anonymous):

Ok so say we have the third side marked as \(z\): |dw:1425690978072:dw| We can use the Pythagorean theorem and say $$ \begin{array}{ll|ll} x^2 = y^2 + 9^2 &&& z^2 = y^2 + 3^2\\ x^2 = y^2 + 81 &&& z^2 = y^2 + 9 \end{array} \\ \; \\ \begin{array}{cccl} x^2 &+ &z^2 &= (9 + 3)^2 \\ (y^2 + 81) &+ &(y^2 + 9) &= 12^2 \end{array}\\ 2y^2 + 90 = 144 \\ y^2 = \frac{144 - 90}{2} = \frac{54}{2} = 27 \implies y = \sqrt{27} \\ x^2 = y^2 + 81 \implies x^2 = 27 + 81= 108 \implies x = \sqrt{108} $$

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