Medians line AX and line BY of triangle ABC intersect perpendicularly at point O. We know the lengths AX=12 and BC=4sqrt(13). Find the length of the third median, CZ.

beastrohit
Nov 28, 2018

#1**+1 **

O is the centroid of ABC

And since AX = 12.....then OX is 1/3 of this distance = 4

And BX = 2sqrt(13) = sqrt (52)

And since AX and BY meet at right angles at O, then BOX is a right triangle

BX is the hypotenuse of this triangle, and OX is a leg

So....by the Pythagorean Theorem, BO = sqrt ( BX^2 - OX^2 ) = sqrt [52 - 16] = sqrt (36) = 6

So cos angle OBX = BO / BX = 6/[2sqrt(13)] = 3/sqrt (13)

Using the Law of Cosines

CX^2 = BO^2 + BC^2 - 2(BO* BC) cos ( OBX)

CX^2 = 36 + (4sqrt(13) )^2 - 2 (6 * 4sqrt(13)) ( 3 /sqrt (13) )

CX^2 = 36 + 208 - [ 2 * 6 * 4 * 3 ]

CX^2 = 36 + 208 - [ 144]

CX^2 = 100

CX = sqrt (100) = 10

And CZ = (3/2)CX = (3/2) * 10 = 15

CPhill
Nov 29, 2018

#2**+1 **

These questions are from AoPS, a math site. I am doing it, and these are the homework problems. AoPS specifically said not to post any of these questions on the site, so just saying.

CoolStuffYT
Nov 30, 2018