+159 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.
+129907 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
+1252 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.
