We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# Please Help #11

0
250
3
+111

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.

Nov 28, 2018

### 3+0 Answers

#1
+103049
+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

Nov 29, 2018
#2
+794
0

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
#3
+794
0

See here:

https://web2.0calc.com/questions/please-help-10_2#r2

CoolStuffYT  Nov 30, 2018