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The NCTM holds to his promise to post interesting Problems to Ponder in the NCTM's newsletter. Here's one from the July 15 issue.

In the figure below, quadrilateral ABCD is a square, and E is the midpoint of the side AD. How do the areas of regions I, II, III, and IV compare? Another way to think about this is to consider the question, What are the ratios of the areas of the four subdivisions, I : II : III : IV?

This problem also has many interesting extensions for your students. For example:

• What happens if E, instead of being the midpoint of AD, is located at the 1/3 mark on segment AD?

• Suppose that the original figure ABCD is actually a rectangle that is not a square. Does this change affect the ratios of the respective areas?

  And you can probably pose some additional extensions of your own.

The problem has relevance to a construction problem that once captured the NCTM headlines.

Solution →

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Let O be the point of intersection of BD and CE. Then OB = 2·DO so that the altitude in ΔDEO from O is 1/3 the side AB. Since its base DE is half the side its area is 1/3·1/2·1/2 = 1/12 of the are of the square.

In ΔCDO, the altitude from O is 1/3 of BC hence its area is 1·1/3·1/2 = 1/6 the area of the square. Similarly, the area of ΔOBC is 1·2/3·/1 = 1/3 the area of the square.

The portion occupied by the rectangle ABOE equals 1 - 1/12 - 1/6 - 1/3 = 5/12. It follows that the areas are in the following ratios