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Suppose we have \(PQ = 6\)\(QR = 7\), and \(PR = 9\), as in the picture below:

Find the length of the median from \(R\) to \(\overline{PQ}\).

 Jul 18, 2021
 #1
avatar+63 
0

I won't directly solve this problem, but hint: try using sin on triangle RPM. You don't need theta, just pull up a chart of all sin values and compare the results. After that, use theta to calculate side d.

 Jul 18, 2021
edited by PBJcatalinasandwich  Jul 18, 2021
 #2
avatar+1425 
+2

PR = a = 9          QR = b = 7           PQ = c = 6          RM = d = ?

 

d = 1/2[sqrt(2*a2 + 2*b2 - c2)]

 

d = √224 / 2 ≈ 7.483

 

 Jul 19, 2021
 #3
avatar+26122 
+1

Suppose we have \(PQ=6\), \(QR=7\), and \(PR=9\), as in the picture below:


Find the length of the median from \(R\) to \(\overline{PQ}\).

 

cos Rule:

\(\begin{array}{|rcll|} \hline 9^2&=&(2*3)^2+7^2-2*2*3*7*\cos(Q) \\ 9^2&=&6^2+7^2-2*2*3*7*\cos(Q) \\ 2*2*3*7*\cos(Q) &=& 6^2+7^2-9^2 \\ 2*2*3*7*\cos(Q) &=& 4 \quad | \quad :2 \\ 2*3*7*\cos(Q) &=& 2 \qquad (1) \\ \hline \end{array}\)

 

cos Rule:

\(\begin{array}{|rcll|} \hline d^2&=&3^2+7^2-2*3*7*\cos(Q) \\ 2*3*7*\cos(Q) &=& 3^2+7^2-d^2 \\ 2*3*7*\cos(Q) &=& 58-d^2 \qquad (2) \\ \hline \end{array}\\ \begin{array}{|lrcll|} \hline (1)=(2):& 2 &=& 58-d^2 \\ & d^2&=& 58-2 \\ & d^2&=& 56 \\ & d^2&=& 4*14 \\ & \mathbf{d}&=&\mathbf{ 2\sqrt{14} } \\ \hline \end{array}\)

 

laugh

 Jul 19, 2021
edited by heureka  Jul 19, 2021
edited by heureka  Jul 19, 2021

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