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# GEOMETRY PLS HELP

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135
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In triangle PQR, PQ=13, QR=14, and PR=15. Let M be the midpoint of QR. Find PM.

Jul 18, 2021

### 3+0 Answers

#1
+26222
+2

In triangle PQR, PQ=13, QR=14, and PR=15.
Let M be the midpoint of QR.
Find PM.

cos Rule:
$$\begin{array}{|rcll|} \hline 15^2&=&13^2+14^2-2*13*14*\cos(Q) \\ 2*13*14*\cos(Q) &=& 13^2+14^2-15^2 \\ 2*13*14*\cos(Q) &=& 140 \quad | \quad : 2 \\ 13*14*\cos(Q) &=& 70 \qquad (1) \\ \hline \end{array}$$

cos Rule:
$$\begin{array}{|rcll|} \hline PM^2 &=& 13^2+\left(\dfrac{14}{2}\right)^2-2*13*\dfrac{14}{2}*\cos(Q) \\\\ PM^2 &=& 13^2+\left(\dfrac{14}{2}\right)^2-13*14*\cos(Q) \\\\ 13*14*\cos(Q) &=& 13^2+\left(\dfrac{14}{2}\right)^2-PM^2 \\\\ 13*14*\cos(Q) &=& 13^2+7^2-PM^2 \\\\ 13*14*\cos(Q) &=& 218-PM^2\qquad (2) \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (1)=(2):& 70 &=& 218-PM^2 \\ & PM^2&=& 218 - 70 \\ & PM^2&=& 148 \\ & PM^2&=& 4*37 \\ & \mathbf{PM}^2 &=&\mathbf{ 2\sqrt{37} } \\ \hline \end{array}$$

Jul 18, 2021
#2
+26222
+2

Sorry, There is a typo!

$$\mathbf{PM} = \mathbf{ 2\sqrt{37} }$$

heureka  Jul 19, 2021
#3
+1588
+3

In triangle PQR, PQ=13, QR=14, and PR=15. Let M be the midpoint of QR. Find PM.

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PM = [sqrt(2 * 132 + 2 * 152 - 142)] / 2

PM = 12.16552506

Jul 19, 2021