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In triangle $PQR,$ $M$ is the midpoint of $\overline{QR}.$ Find $PM.$
PQ = 5, PR = 8, QR = 11

 Jun 13, 2024
 #1
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First, let's set a variable. Let's set x as angle PQR. 

 

From the problem, we can write that 

\(cos(x) = \frac{8^2 + 11² - 5²}{2*8*11}=10/11\)

 

Now, we can sue the law of cosines to write

\( (PM)² = 5.5² + 8² - 2*5.5*8 * cos(x)\\ \)

Now, we can find PM. 

\(PM^2 = 10/11*6.25\\ PM \approx 2.38365647311\)

 

I'm not exactly sure if this is correct. I may have made a mistake. 

 

Thanks! :)

 Jun 13, 2024

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