Question 1119151: If DE II (I think this means parallel) BC and triangle ABC is an equilateral triangle with area 4 times the square root of 3, and D bisects AB, what is the area of triangle ADE? (again a picture is given and helpful but I don't know where to upload that)
Answer choices are square root 2, square root 3, 2, 3, and 2 times the square root of 3.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Segment DE joins the midpoints of two sides of the equilateral triangle. Place point F at the midpoint of the third side of the triangle and draw segments DF and EF. It should be clear that the original equilateral triangle is now composed of 4 congruent equilateral triangles, each with area 1/4 of the original triangle.
So the area of each of the smaller triangles, including triangle ADE, is sqrt(3).
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