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

NotThatSmart Jul 23, 2024

#1**+1 **

Note, NTS, that PQR can't be a triangle because

PQ + PR = QR

But the triangle inequality says that

PQ + PR must be > QR

CPhill Jul 24, 2024

#2**+1 **

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!

NotThatSmart
Jul 24, 2024

#3**0 **

Here's a similar prob that I answered earlier :

https://web2.0calc.com/questions/geometry_2526

See if you can solve it now....

CPhill Jul 24, 2024

#4**+1 **

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! :)

NotThatSmart Jul 24, 2024