SOLUTION: I need explanation of this problem: In an equilateral triangle ABC with AB = BC = CA = 4cm, the point D is the midpoint of BC and E is a point between D and C. If AE2 = 13EC2, ca

Algebra ->  Triangles -> SOLUTION: I need explanation of this problem: In an equilateral triangle ABC with AB = BC = CA = 4cm, the point D is the midpoint of BC and E is a point between D and C. If AE2 = 13EC2, ca      Log On


   

Question 865788: I need explanation of this problem:
In an equilateral triangle ABC with AB = BC = CA = 4cm, the point D is the midpoint of BC and E
is a point between D and C. If AE2 = 13EC2, calculate the area of the triangle AEC. ( sqr 3 = 1.732)
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Maybe they want you to apply Law of Cosines.

The angles in an equilateral triangle measure 60%5Eo .
The law of cosines, applied to AEC, with side lengths measured in cm, says
AE%5E2=EC%5E2%2BAC%5E2%2B2%28AC%29%28EC%29cos%28ACE%29 or
13EC%5E2=EC%5E2%2B4%5E2%2B2%2A4%2A%28EC%29cos%2860%5Eo%29
13EC%5E2=EC%5E2%2B16%2B2%2A4%2A0.5%28EC%29
13EC%5E2=EC%5E2%2B16%2B4EC
Let's call EC x for short, and we can write
13x%5E2=x%5E2%2B16%2B4x-->13x%5E2-x%5E2-4x-16=0-->12x%5E2-4x-16=0
Here, I would divide both sides of the equal sign by 4 to get
3x%5E2-x-4=0
and then I would factor to solve
3x%5E2-x-4=0-->%283x-4%29%28x%2B1%29=0-->system%283x-4=0%2C%22or%22%2Cx%2B1=0%29-->system%28x=4%2F3%2C%22or%22%2Cx=-1%29
Since x=EC is a length, it cannot be negative.
So EC=4%2F3cm .
Now we can calculate the area of ACE as
%28AC%29%28EC%29sin%28ACE%29%2F2 or %281%2F2%29%28AC%29%28EC%29sin%28ACE%29
In square centimeters, that area would be
%281%2F2%29%2A4%2A%284%2F3%29%2A%28sqrt%283%29%2F2%29=4sqrt%283%29%2F3= approx.4%2A1.732%2F3= approx.2.309 .
So the area of ACE is highlight%282.309%29cm%5E2