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

May 24, 2020

#1
-3

A

13                 15

B       14                C

Let  B  = (0,0)   and C  = ( 14,0)

So......M  =  (7,0)  ....so   BM  =  7

Using  the Law  of Cosines

AC^2 = AB^2  + BC^2  - 2(AB * BC) cos (ABC)

15^2  =  13^2  + 14^2  - 13 * 14) cos (ABC)

[ 15^2  - 13^2  - 14^2 ] / [  13 * 14]  = cos ABC

-10/13  =  cos ABC

And applying this again we have  that

AM^2 = 7^2 + 13^2 - 7*13*(-10/13) = 288

AM = sqrt(288) = 12*sqrt(2)

May 24, 2020
#2
-3

P

13                 15

Q       14                R

Let  Q  = (0,0)   and R  = ( 14,0)

So......M  =  (7,0)  ....so   QM  =  7

Using  the Law  of Cosines

PR^2 = PQ^2  + QR^2  - 2(PQ * QR) cos (PQR)

15^2  =  13^2  + 14^2  - 13 * 14) cos (PQR)

[ 15^2  - 13^2  - 14^2 ] / [  13 * 14]  = cos PQR

-10/13  =  cos PQR

And applying this again we have  that

PM^2 = 7^2 + 13^2 - 7*13*(-10/13) = 288

PM = sqrt(288) = 12*sqrt(2)

May 24, 2020
#3
+1387
+3

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

Let angle PRQ be  >β<

1)   Let's find β  first:             cos(β) = (15² + 14² - 13²) / 2*15*14       β = 53.13°

2)   Now we can find  PM      (PM)² = 7² + 15² - 2*7*15 * cos(β)

PM = 12.166

May 25, 2020
edited by Dragan  May 26, 2020