+478 In triangle $PQR,$ $M$ is the midpoint of $\overline{QR}.$ Find $PM.$
PQ = 5, PR = 8, QR = 11
+1950 Well, it's been a while since I've answered a question on this website, and what better way to celebrate with a geometry question :)
So, we don't have a right triangle, so we're gonna have to settle with Law of Cosines.
Also, note that we have
\((cos PMQ) = (- cos PMR )\)
Now, let's apply Law of Cosines and add subsitute using equation above.
\(PQ^2 = QM^2 + PM^2 - 2(QM * PM) * (-cos PMR) \\ PR^2 = RM^2 + PM^2 - 2 (RM * PM) * ( cos PMR)\)
Simplify
\(5^2 = 5.5^2 + PM^2 + 2(5.5 * PM)cos(PMR) \\ 8^2 = 5.5^2 + PM^2 - 2(5,5 * PM) cos (PMR) \)
Now, we're stuck...but we can add these two equations to keep progressing.
\(5^2 + 8^2 = 2 * 5.5^2 + 2PM^2 \\ 89 = 60.5 + 2PM^2 \\ 28.5 / 2 = PM^2\\ 14.25 = PM^2 \\ \sqrt {14.25} = PM ≈ 3.77\)
And we're done!
Also, side note, this is gonna be my 1000 answer on this website which is a weird flex but whatever.
I might be gone for another while, but hopefully be back in the summer.
~NTS
