Let \(G\) be the centroid of \(\triangle ABC\); that is, the point where the segments joining each vertex to the midpoint of the opposite side all meet. If \(\triangle ABG\) is equilateral with \(AB=2\), then compute the perimeter of \(\triangle ABC\).
+235 First, you graph the triangle. It's going to be a very tall isosceles triangle. Using the fact that the centroid of a triangle splits the median into a 2:1 ratio, and the properties of a 30-60-90 triangle, we can see that the altitude of this triangle is \(3\sqrt 3\). Using the pythagorean theorem, we can see that each of the longer sides is \(\sqrt {82}\). So adding all three sides up, including the one that was already given, we can see that the perimeter of \(2 + 2\sqrt{82}\)
Sorry! I messed up. All the above steps were correct, I just messed up with the Pythagorean theorem. The answer is actually 2 + 4sqrt7
Hope this helps!
Sorry!
+128475 Thanks, Lag!!!
Here is a pic :
The height of ABG = sqrt(3)
And because G is a centroid, then CG = 2sqrt(3)
So...the height of ABC = 3sqrt(3) = sqrt (27)
And using the Pythagorean Theorem
AC^2 = (height of ABC)^2 + [(1/2)AB]^2
AC^2 = ( 27) + (1)
AC^2 = 28
AC = sqrt (28) = 2sqrt(7) = BC
So.....the perimeter of ABC = 2 + 2 (2 sqrt(7) = 2 + 4sqrt(7)
Just as Lag found !!!!
