Remark
Nice little problem. There is an awful lot of geometry in this question; there might be a quicker way of doing it, but this is the only way I could find easily.
Step One
Label the sides AB as 2 in length and AC as 3 in length. Be careful that you understand why they are labeled that way. It is because AC has to be shorter than AB. If you call AB = 1 then AC is 1.5. To make the numbers easier to use, I multiplied both of them by 2 so that AB = 2 and AC = 3
Step Two
Call the square side = s. The trick is to show that triangle BFE is similar to triangle CDE. That is the key to the entire problem.
FE is parallel to AD and by extension FE is parallel to AE.
CB is a transversal across these two parallel lines. Therefore <FEB = <C
In the same way <B = <DEC By AA the two triangles are similar.
The proportion of the triangles can be set up
ED/BF = DC/FE
Givens
ED = s [it's the side of a square]
BF = 2 - s
DC = 3 - s
FE = s
s/(2 - s) = (3 - s) / s Study this awhile. Cross Multiply
s^2 = (2 - s)(3 - s) Remove the brackets.
s^2 = 6 - 2s - 3s + s^2 Subtract s^2 from both sides. Combine on the right.
0 = 6 - 5s Subtract 6 from both sides.
-6 = - 5s Divide by minus 5
-6/-5 = s
s = 1.2
The side of the square is 1.2
Area of the square = s^2 = 1.2^2 = 1.44
Step Three
Find the area of the large triangle.
Area = 1/2 AC * AB
Area = 1/2 * 3 * 2
Area = 3
Step Four
Compare the areas.
Area of square / Area of Triangle = 1.44 / 3 = 0.48
Step five
Change to a fraction
0.48 * 100 / 100 = 48 / 100 Divide top and bottom by 4
12 / 25 is your answer which is D
D <<<<< Answer