SUMMARIZE
1. Start by observing that the area of the larger square (formed by the outer square) is equal to the sum of the areas of the four smaller squares (formed by the right-angled triangles).
- The area of the larger square is (a + b)^2.
- The area of each smaller square is a^2 and the sum of all four squares is 4a^2.
2. Simplifying the equation, we have:
(a + b)^2 = 4a^2
3. Expand the square on the left side:
a^2 + 2ab + b^2 = 4a^2
4. Move all terms to one side of the equation:
2ab + b^2 = 3a^2
5. Rearrange the terms:
2ab = 3a^2 - b^2
6. Divide both sides of the equation by a:
2b = (3a^2 - b^2)/a
7. Simplify:
2b = 3a - (b^2/a)
8. Recognize that (b^2/a) is equal to c^2, the square of the hypotenuse of each right-angled triangle.
9. Substitute c^2 for (b^2/a):
2b = 3a - c^2
10. Rearrange the terms:
2b + c^2 = 3a
11. Divide both sides of the equation by 3:
(2b + c^2)/3 = a
12. Recognize that (2b + c^2)/3 is equal to a^2, the square of the shorter side of each right-angled triangle.
13. Substitute a^2 for (2b + c^2)/3:
a^2 = (2b + c^2)/3
14. Rearrange the terms:
3a^2 = 2b + c^2
15. Finally, observe that 3a^2 is equal to c^2, the square of the longer side of each right-angled triangle.
From steps 14 and 15, we can conclude that:
c^2 = a^2 + b^2
This is the Pythagorean Theorem, which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.