Question

In the image, the larger square has a side length of \(a + b\), so its area is \((a+b)^2\). This larger square encloses a smaller tilted square with side length \(c\) and four right triangles with legs \(a\) and \(b\). The area of the four triangles is \(4 \times \frac{1}{2}ab = 2ab\). The area of the tilted square is \(c^2\). The total area of the large square is also the sum of the areas of the smaller square and the triangles, which gives: \((a+b)^2 = c^2 + 2ab\). Expanding the left side, \((a+b)^2 = a^2 + 2ab + b^2\). Equating both expressions gives: \(a^2 + 2ab + b^2 = c^2 + 2ab\). Subtract \(2ab\) from both sides: \(a^2 + b^2 = c^2\), proving the Pythagorean Theorem. simplify this

Answer 1

write it as a paragraph

Answer 2

The proof shows how the Pythagorean Theorem can be derived using geometric areas. Here’s a simplified version:

1. Start with a larger square with side length \(a + b\). The area of this square is:
\[
(a + b)^2
\]

2. Inside this larger square, there is a smaller tilted square with side length \(c\) and four right triangles, each with legs \(a\) and \(b\).

3. The area of the four triangles is:
\[
4 \times \frac{1}{2}ab = 2ab
\]

4. The area of the smaller tilted square is:
\[
c^2
\]

5. The total area of the larger square can be expressed as the sum of the area of the tilted square and the triangles:
\[
(a + b)^2 = c^2 + 2ab
\]

6. Expanding the left side gives:
\[
a^2 + 2ab + b^2
\]

7. Setting the two expressions equal gives:
\[
a^2 + 2ab + b^2 = c^2 + 2ab
\]

8. Subtract \(2ab\) from both sides:
\[
a^2 + b^2 = c^2
\]

This proves the Pythagorean Theorem: \(a^2 + b^2 = c^2\).