equilateral triangle AED has been constructed on the side of square ABCD. if the area of a square is 16, what is the area of the triangle AED?

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i believe this problem would have a ratical in it & can you explain how you knew what to do to get to the answer, thanks
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The side of the square is the square root of its area=sqrt(16)=4. The area of the triangle is half the base times the height. So we need the height of the equilateral triangle. The perpendicular from vertex E bisects AD and the angle AED. The equilateral triangle now consists of two back to back right-angled congruent triangles. Angle EAD=60 (same as  the other two angles of triangle AED) so half the angle is 30 degrees. The common side of the two right-angled triangles is the height of the equilateral triangle and has length=sqrt(4^2-2^2), by Pythagoras, because the hypotenuse is length 4 (same as the side of the square) and the shortest side is 2, half of AD. The height is sqrt(12)=sqrt(4*3)=2sqrt(3). The area of the triangle is 1/2*4*2sqrt(3)=4sqrt(3)=6.93 sq units approx.

Another way of finding the area is to take the two right-angled triangles and join them together along their common hypotenuse forming a rectangle of sides 2sqrt(3) and 2. The hypotenuse is a diagonal of the rectangle. The area of the rectangle (2sqrt(3)*2=4sqrt(3)=6.93) must be the same size as the area of the equilateral triangle, because the rectangle and the equilateral triangle are made up of the same two right-angled triangles.

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