**From New Scientist #2709, 23rd May 2009** [link]

Joe cut out a cardboard rectangle 12.5 cm by 8 cm. He asked Penny to cut it into two pieces with one straight cut and then, with a second straight cut, to cut one of the pieces into two. Her problem was to work out where to make the cuts so that the three pieces could be arranged to form a square.

Penny found it quite easy once she realised that in each case one of the two cut pieces was a triangle.

What will be the area of the largest of the three pieces?

**Note:** There are actually two ways to solve this puzzle, and they give different answers for the area of the largest piece.

[enigma1546]

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I solved this one by cutting up pieces of paper. I’ve yet to come up with a programmatic approach to this problem.

The first cut removes triangle

A, and the other piece is then cut into two to produce quadrilateralBand triangleC. These three pieces can then be reassembled as a 10×10 square as shown.The area of the pieces is: A = 24, B = 38.5, C = 37.5.

Solution:The area of the largest piece is 38.5 sq cm..

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Here is a second solution, although some of the straight lines in the perimeters of the pieces are irrational, when the pieces are rearranged into a square they are all in their original orientation.

The first cut removes triangle

A, and the remaining piece is cut into pentagonBand triangleC. The three pieces can be assembled into a 10×10 square as shown,The area of the pieces is: A = 40, B = 57.5, C = 2.5.

Solution:The largest piece is 57.5 sq cm..

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