+0  
 
0
20
2
avatar+0 

In triangle $STU$, let $M$ be the midpoint of $\overline{ST},$ and let $N$ be on $\overline{TU}$ such that $\overline{SN}$ is an altitude of triangle $STU$.  If $ST = 8$, $SU = 8$, $TU = 8$, and $\overline{SN}$ and $\overline{UM}$ intersect at $X$, then what is $SX$?

 Feb 9, 2024
 #1
avatar+129850 
+1

 

This one is pretty easy

The triangle is equilateral

The altitude of an equilateral triangle  is a median

And since M is the midpoint of ST, then UM is also a median

Their intersection, X, is the centroid of STU

SN = sqrt [ST^2 - TN^2]  = sqrt [ 8^2 - 4^2]  = sqrt [ 48]  = 4sqrt (3)

 

By a property of centroids ....SX = (2/3)(SN)

 

So

 

SX = (2/3) * 4 (sqrt3)  = 8sqrt (3)  / 3  =  8 / sqrt 3   ≈  4.618

 

cool cool cool

 Feb 10, 2024
 #2
avatar+297 
+1

Just as a side note, I'm not criticizing, but most of the time Vxrtate, you answer the question by rationalizing the denominator. In other words, you should put \(\frac{8\sqrt{3}}{3}\)unless it specifically states otherwise.

 Feb 10, 2024

3 Online Users

avatar