# Geometry Help

• Jan 11th 2006, 07:13 PM
wpcp15
Geometry Help
In Triangle ABC, segmentS CM, BP, and AN are medians

If CM=3X+6 and CE=x+12, find the value of x. Being that e is the centroid.
I know that from the vertex to the centroid is 2/3 of the entire median, but i cant figure out what x equals
I think the value of x equals 18, but im not sure

If EN= X-5 and AE=X+17, find the value of AN.
• Jan 11th 2006, 07:26 PM
ThePerfectHacker
Quote:

Originally Posted by wpcp15
In Triangle ABC, segmentS CM, BP, and AN are medians

If CM=3X+6 and CE=x+12, find the value of x. Being that e is the centroid.
I know that from the vertex to the centroid is 2/3 of the entire median, but i cant figure out what x equals
I think the value of x equals 18, but im not sure

If EN= X-5 and AE=X+17, find the value of AN.

Since, $CM$ is the entire median and $CE$is the centroid then just like you said.
$\frac{2}{3}CM=CE$
But, $CM=3x+6$ and $CE=x+12$
Thus, (substitute)
$\frac{2}{3}(3x+6)=x+12$
Now, open paranthese,
$2x+4=x+12$
Subtract $x$ from both sides,
$x+4=12$
Subtract $8$ from both sides,
$x=8$
Thus, $CM=30$ and $CE=20$.
• Jan 11th 2006, 07:52 PM
ThePerfectHacker
Remember that the ratio from the vertex to the centriod to to the remaining part of the median is 2:1.
Thus,
$\frac{AE}{EN}=\frac{2}{1}$
But, $AE=x+17$ and $EN=x-5$
Thus, (substitute)
$\frac{x+17}{x-5}=\frac{2}{1}$
Cross multiply,
$(x+17)1=2(x-5)$
Open parantheses,
$x+17=2x-10$
Subtract $x$ from both sides,
$17=x-10$
$27=x$
Thus, $AE=44$ and $EN=22$.