# Geometry Help

• Jan 11th 2006, 06:13 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, 06:26 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, $\displaystyle CM$ is the entire median and $\displaystyle CE$is the centroid then just like you said.
$\displaystyle \frac{2}{3}CM=CE$
But, $\displaystyle CM=3x+6$ and $\displaystyle CE=x+12$
Thus, (substitute)
$\displaystyle \frac{2}{3}(3x+6)=x+12$
Now, open paranthese,
$\displaystyle 2x+4=x+12$
Subtract $\displaystyle x$ from both sides,
$\displaystyle x+4=12$
Subtract $\displaystyle 8$ from both sides,
$\displaystyle x=8$
Thus, $\displaystyle CM=30$ and $\displaystyle CE=20$.
• Jan 11th 2006, 06:52 PM
ThePerfectHacker
Remember that the ratio from the vertex to the centriod to to the remaining part of the median is 2:1.
Thus,
$\displaystyle \frac{AE}{EN}=\frac{2}{1}$
But, $\displaystyle AE=x+17$ and $\displaystyle EN=x-5$
Thus, (substitute)
$\displaystyle \frac{x+17}{x-5}=\frac{2}{1}$
Cross multiply,
$\displaystyle (x+17)1=2(x-5)$
Open parantheses,
$\displaystyle x+17=2x-10$
Subtract $\displaystyle x$ from both sides,
$\displaystyle 17=x-10$
$\displaystyle 27=x$
Thus, $\displaystyle AE=44$ and $\displaystyle EN=22$.