+1946 This is off-topic, but Hello EP! It's been a while!
I sound like an idiot right now, but I had an old account back in the days, and that account had MANY questions answered by EP!
Just sayin hello! Good to see you!
Also, solve this question maybe? Just to refresh your math knowledge...
In triangle $PQR,$ $M$ is the midpoint of $\overline{QR}.$ Find $PM.$
PQ = 6, PR = 9, QR = 15
Thanks! :) (Anyone can answer, not just EP)
~NTS
+130466 Note, NTS, that PQR can't be a triangle because
PQ + PR = QR
But the triangle inequality says that
PQ + PR must be > QR
+1946 Oh, your right! Sorry about that. I was coming up with numbers...sorry.
How about
PQ = 8, PR = 9, QR = 15
Nice catch...should have caught that!
+130466 Here's a similar prob that I answered earlier :
https://web2.0calc.com/questions/geometry_2526
See if you can solve it now....
+1946 Ok, let me give this a shot....
The Law of Cosines state that \(cos PMQ = - cos PMR \)
\(PQ^2 = QM^2 + PM^2 - 2(QM * PM) * (-cos PMR) \\ PR^2 = RM^2 + PM^2 - 2 (RM * PM) * ( cos PMR)\)
Subsituting in the values we already know from the problem. we get
\( 8^2 = 7.5^2 + PM^2 + 2(7.5 * PM)cos(PMR) \\ 9^2 = 7.5^2 + PM^2 - 2(7.5 * PM) cos (PMR)\)
Now, adding these two, we get
\(8^2 + 9^2 = 2 * 7.5^2 + 2PM^2 \)
\(145= 112.5+ 2PM^2 \\ 32.5/ 2 = PM^2 \\ 16.25 = PM^2 \\ PM=\sqrt{16.25}\approx 4.03112887415\)
Is this correct, CPhill? I'm not sure. Can u verify?
Thanks! :)
+1946 Thanks for the guidance, CPhill.
Sorry for the original problem. :)
