Subject: Re: Areas in a trapezoid
Date: Sat, 5 Apr 1997 13:46:18 -0500
From: Alexander Bogomolny

Dear Roland:

I have a feeling that from the time of your last letter you have solved a lot of geometric problems. Hope you enjoyed yourself too.

Here is the solution, but please check it.

For simplicity, three letters like ABC will denote the area of the triangle ABC, OK?

Let K be the midpoint of AB. Then

  1. FCD+ECD=2*KCD because they have the same base while the height of KCD is the average of heights in FCD and ECD.
  2. FCD+ECD=OCF+OED + 2*OCD = S + 22, S being the sum you are looking for.
  3. Therefore, 2*KCD = S + 22.
  4. Show that 2*KCD = ABCD. This is an exercise.
  5. From 4, S = 34-22=12.

I am curious as to where you take your problems from? Do you like geometry?

Best regards,

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