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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
 #1
avatar+99523 
+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

 

 

cool cool cool

 Nov 29, 2018
 #2
avatar+626 
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
avatar+626 
0

See here:

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

CoolStuffYT  Nov 30, 2018

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