For Geometry Lovers

[N:Dure]

Looks easy

[Rahul@gurome]

Triangle ABE will be similar to triangle CDE because angle EAB = 90 = angle EDC, angle AEB = angle CED being vertically opposite angles and obviously angle ABE = angle ECD ( since if two angles of two triangles are same, third angle has to be same) .

So AE/ED = AB/CD.
Let AE = x. So ED = 4-x.
Or x/(4-x) = 3/9 = 1/3.
So x = 1 = AE.
So ED = 4-x = 3.
So area of triangle ACD = ½ *CD * AD = ½ *9*4 = 18.
Area of triangle ECD = ½*CD*ED = ½*9*3 = 13.5.
So area of triangle ACE = 18 - 13.5 = 4.5.
The correct answer is D.

[rkanthilal]

We need the area of triangle AEC. This can be easily calculated if we know the length of AE. Then we can calculated the area of triangle ADC and subtract the area of triangle EDC. That will leave us with the area of triangle AEC. We know the area of triangle ADC from the given information. It is (AD X CD) / 2 or 18. Finding the length of AE is the tricky part. I'm sure there are multiple ways to do this but here is the only way I can think of.

First, draw a line from point B staight down to intersect with line CD (you need to extend line CD to the right). Call this point F. Now you have a big right triangle BFC. Imagine this triangle on a coordinate plane. Put point C at the origin (0,0) and point B would be (12,4). Now calculate the slope (rise/run) of line CB. This is 4/12 or 1/3. This means for every 3 you move horizontally you move one up (or down). The length of AB is 3 so that means point E must be a length of 1 from point A. In other words AE equals 1 and ED equals 3. Now you can calculate the area of triangle EDC, which is 13.5. When you subtract this from the area of triangle ADC which was 18 you get the area of AEC which is 4.5.

Is that the right answer???

[GMATGuruNY]

For a solution to this problem, please see the attached file.