Fawn Nguyen published a great geometry problem on her blog last night:
It was really neat to see the pictures of the work her students were doing. What also struck me is that there are so many different ways to approach this problem.
My own personal style of problem solving has always been heavy on computation. I saw Fawn’s problem as a similar triangle problem:
I was actually having a hard time understanding the approach her students were taking in some of the pictures and asked Fawn about it. It turned out that the approach was not the computationally heavy similar triangle approach, but a much more simple approach with areas:
Sweet!
That solution got me wondering about other geometric solutions and also how my son would solve the problem. Since we are just now finishing up a section in his geometry book about similar triangles, I guessed he would look at similar triangles. He didn’t. He actually drew in the extra piece to make the large rectangle exactly as Fawn’s student’s had done (!), but then he went in a slightly different direction from Fawn’s diagrams above which I wasn’t expecting at all (and sorry for the interruption from the cat):
Finally, I noticed a (slightly) different area solution which is, of course, is a little more computational. I wanted to go through this solution just to get in a little extra algebra practice and also to show my son a different approach to the problem that also uses areas:
I love problems that have lots of different types of solutions. The setup in Fawn’s problem seems so simple – just a rectangle and a triangle – and it is amazing to me that there are so many different kinds of solutions. Fawn’s lesson here is a great reminder to me to focus more time on the more geometric solutions since my natural instincts seem to be to focus on the computation.
Thanks for sharing another great problem, Fawn!
Mike's Math Page 
I did this by computation the first time. But now looking back on it, you could spilt AFDE into two triangles ADE and AFD. The first is a right triangle with sides 10 and 16 so it has area 80. For the second triangle, notice that ABCD is a trapezoid with AC and BD as its diagonals. Thus AFD and BFC have the same area, 24. So the total area is 80 + 24 = 104.