+185 In triangle PQR, PQ=13, QR=14, and PR=15. Let M be the midpoint of QR. Find PM.
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)
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)
