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Solutions for Session 6, Part A
See solutions for Problems: A1 | A2 | A3
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Problem A1 | |
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Draw a square with side lengths of 5 units around the given square so that the vertices of the given square are on the sides of the new square. The area of the new square is 25 square units (since its side length is 5), and its area is larger than the area of the original square by the area of the four right triangles whose legs have lengths of 2 and 3 units. So the area of the original square is 25 - 4 • (3 • 2 / 2) = 13 square units.
It looks like this:  |
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Another method is the following: 
Again, the area of the square is 4 • (2 • 3 / 2) + 1 • 1 = 13 square units. |
<< back to Problem A1
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