Let \(XOY\) be a triangle with \(\angle XOY = 90^\circ\). Let \(M\) and \(N\) be the midpoints of legs \(OX\) and \(OY\), respectively. Suppose that \(XN = 19\) and \(YM = 22\). What is \(XY\)?

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## Comments

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TopNewestis XY 26????

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Yes .XY=26 is the correct answer.

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Let \(XM=MO=a\) and \(ON=NY=b\)

By applying PT, we get

\(4a^{2}+b^{2}=19^{2} \rightarrow Eq.1\)

\(a^{2}+4b^{2}=22^{2} \rightarrow Eq.2\)

Eq.1+Eq.2

\(a^{2}+b^{2}=\frac{19^{2}+22^{2}}{5}=13^{2}\)

\( \Rightarrow \sqrt{4a^{2}+4b^{2}}=XY=\sqrt{4× 169}=\boxed{26}\)

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Length of XY is 26

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XY = 26

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XY=26

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26

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26

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