In triangle $PQR,$ $M$ is the midpoint of $\overline{QR}.$ Find $PM.$

PQ = 5, PR = 8, QR = 11

RedDragonl Jul 1, 2024

#1**+1 **

I'm not the best at these questions, but heres my understanding. We have

\(f_Q(x)=\sqrt{5^2-x^2}\\ f_R(x)=\sqrt{8^2-(x-11)^2}\\ 25-x_P^2=64-x_P^2+22x_P-121\\ x_P=\dfrac{121+25-64}{22}\\ x_P=3.\overline{72}\\ y_P=3.333\)

We are using coordinate geometry to solve this problem. Now, since we know these two values, we find that

\(P(3.\overline{72},\ 3.333)\\ M(5.5,\ 0)\\\)

Now, we can use the distance formula to find PM. We have

\( \overline{PM}=\sqrt{(x_M-x_P)^2+y_P^2}=\sqrt{(5.5-3.727)^2+3.333^2}\\\overline{PM}=3.775\)

I don't think this was the clearest explenation, but that's all I could understand.

So 3.775 is our answer.

Thanks! :)

NotThatSmart Jul 1, 2024

#2**+2 **

We can use Apollonius's Theorem*

*Apollonius's theorem states that in any triangle, the square of the length of a median is equal to the sum of the squares of the lengths of the two sides containing the median, minus one-quarter of the square of the third side.

So: PM^2 = (2PQ^2 + 2PR^2 − QR^2)/4

Since PQ = 5, PR = 8, QR = 11

PM^2 = (2*25 + 2*64 - 121)/4

PM^2 = 57/4

PM = \(\frac{\sqrt{57}}{2}\)

So the answer is \(\boxed{\frac{\sqrt{57}}{2}}\)

aboslutelydestroying Jul 1, 2024