-
magical cook
- Master | Next Rank: 500 Posts
- Posts: 484
- Joined: Sun Jul 30, 2006 7:01 pm
- Thanked: 2 times
- Followed by:1 members
My answer would be sqrt(2) / 4.
First, I draw a diagonal line connecting A and D. Since BE = DE, the triangle BED is an isosceles triangle whose height is 1 (the given length of CE) + half of the square ABCD’s diagonal.
The area of that triangle BED is ½ * sqrt(2) * (1 + sqrt(2)/2) which is (sqrt(2) + 1) / 2.
Now, the area of the two triangles BCE and DCE together is (sqrt(2) + 1) / 2 – ½ = sqrt(2) / 2. The area of the triangle BCE alone is half that which should be sqrt(2)/4.
