Problem 4-3
4-3 A
4-3 A
It appears that no matter how we drag the points of our right triangle to change its sides, the area of two smaller equilateral triangles that are connected to the two legs will always sum up to the area of the largest equilateral triangle that's attached to the hypotenuse.
4-3 B
4-3 B
Like the previous one, this one adds up the same way. So the areas of smaller two squares which are connected to the two legs of our right triangle sum up to be the value of our largest square that is connected to the hypotenuse. So if we call the leg of the right triangle that is connected to SquareA a, the leg connected to SquareB b, and the hypotenuse connected to SquareC c, then we know that the areas of these squares are , , and . So then, knowing that the two smaller ones sum up to SquareC, we get the pythagorean theorem, .
4-3 C
4-3 C
With this problem, we only see our pattern of the two smaller shapes adding to the larger one in one case. When the extra right triangles (so no the original) are moved so they are all 45-45-90 triangles, then the two triangles connected to the legs add up to the larger right triangle. This works for any angles on our original blue triangle (note: point C is also on a hidden semicircle and thus can be moved around to better see all the different angles the right triangle can have). But once we move our red right triangles to be at angles that are not 45 degrees, this conjecture does not hold.