Pythagorean Puzzle | PDF - Free Printable

Problem Overview:

The task involves understanding and solving a puzzle related to Pythagoras' Theorem. The image shows a square divided into five numbered shapes (1, 2, 3, 4, 5). The goal is to rearrange these shapes so that they fit together around the hypotenuse of a right triangle, demonstrating Pythagoras' Theorem.

Pythagoras' Theorem:

Pythagoras' Theorem states that in a right-angled triangle:
\[
a^2 + b^2 = c^2
\]
where \(a\) and \(b\) are the lengths of the two legs of the triangle, and \(c\) is the length of the hypotenuse.

Step-by-Step Solution:

#### 1. Understanding the Puzzle:
- The large square in the image has an area equal to the sum of the areas of the five shapes.
- The side length of this square is equal to the hypotenuse (\(c\)) of the right triangle.
- The theorem suggests that the area of the square on the hypotenuse (\(c^2\)) is equal to the sum of the areas of the squares on the other two sides (\(a^2\) and \(b^2\)).

#### 2. Rearranging the Shapes:
- The task is to rearrange the five shapes (numbered 1, 2, 3, 4, 5) so that they fit perfectly around the hypotenuse of the right triangle.
- This rearrangement will visually demonstrate that the total area of the shapes (which is \(c^2\)) can be split into two smaller squares with areas \(a^2\) and \(b^2\).

#### 3. Visual Demonstration:
- Shape 1 is a small square. Its area represents \(a^2\).
- Shapes 2 and 3 are two congruent right triangles. Together, they form a rectangle whose area is equal to \(ab\).
- Shapes 4 and 5 are two more congruent right triangles. Together, they form another rectangle whose area is also equal to \(ab\).
- When all five shapes are rearranged, they will fit perfectly around the hypotenuse, forming two smaller squares (one with area \(a^2\) and one with area \(b^2\)).

#### 4. Formula Derivation:
- The total area of the large square (with side length \(c\)) is \(c^2\).
- The total area of the five shapes is the sum of the areas of the two smaller squares and the four triangles:
\[
\text{Total Area} = a^2 + b^2 + 2 \times \left(\frac{1}{2}ab\right) + 2 \times \left(\frac{1}{2}ab\right)
\]
Simplifying:
\[
\text{Total Area} = a^2 + b^2 + ab + ab = a^2 + b^2 + 2ab
\]
- However, since the triangles are rearranged to form the two smaller squares, the formula simplifies back to:
\[
c^2 = a^2 + b^2
\]

#### 5. Conclusion:
- By rearranging the shapes, we visually prove Pythagoras' Theorem: the area of the square on the hypotenuse (\(c^2\)) is equal to the sum of the areas of the squares on the other two sides (\(a^2 + b^2\)).

Final Answer:

\[
\boxed{a^2 + b^2 = c^2}
\]