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In △PQR, QS is an altitude. Solve for x and y.
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x=√78; y=6 x=6; y=3√13 x=6; y=2√6 x=2; y=6√2
@hang254 do you know Pythagoras theorem?
yes. a^2+b^2=c^2
Remember the geometric means? \[\frac 4x = \frac x9\]\[\frac9y = \frac{y}{13}\]
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Yes, i do
JUst solve the proportions that I set up.
Ok
x=36, y=10.816
For x: \[x^{2} = 36\]\[x = 6\] For y: \[y^{2} = 117\]\[y = 3\sqrt{13}\]
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Oh, thanks. I just realized that i didnt square the 36.
Alright. No Problem
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