Geometry

Question

0

39

1

+613

In triangle $PQR,$ $M$ is the midpoint of $\overline{QR}.$ Find $PM.$

Pythagorearn Apr 23, 2024

CPhill · Answer 1 · 2024-04-23T01:09:11

Law of Cosines

18^2 = 12.5^2 + PM^2 - 2 ( 12.5 * PM) cos (QMP)

23^2 = 12.5^2 + PM^2 - 2(12.5 * PM) (-cos (QMP)

Simplify

(18^2 - 12.5^2 - PM^2 ) / [ -25 PM ] = cos (QMP)

[23^2 - 12.5^2 - PM^2 ] / [ 25 PM = cos (OMP)

Equate cosines and simplify

- [ 167.75 - PM^2 ] = [372.75 - PM^2]

2PM^2 = 540.5

PM^2 = 270.25

PM =sqrt [ 270.25] ≈ 16.44