Fill in the steps in Garfield's proof of Pythagoras' Theorem:

Consider the right-angled triangle ABC, with side lengths a,b,c
:

Triangle ABC with a right angle at ACB.

Construct a congruent triangle, and by rotating and translating it place it such that it creates a straight line of length a+b
. Join the points B and the image of A, which we will call E, to create a Answer
 (name of the shape formed).

Quadrilateral CBED which is created by rotating and translating the triangle ABC.

Area of CBED =

12(a+b)2
12(a+b)
(a+b)2

 ∠BAE=
 Answer
∘

Area of triangle ABC = 

12(a+b)
(ab)2
12ab
Area of triangle BAE = 

12c
12c2
c2
Area of triangle ADE = 

(ab)2
12(a+b)
12ab
Since the area of the trapezium is equal to the sum of the areas of the three triangles:

12(a+b)(a+b)=12ab+12
Answer
+12
Answer
2

Which expands and simplifies to give:

a2+
Answer
2=
Answer
2

Let's go through the steps in Garfield's proof of the Pythagorean Theorem, filling in the blanks:

1. **Consider the right-angled triangle ABC, with side lengths a, b, c:**
   - Triangle ABC is a right triangle with a right angle at ∠ACB.

2. **Construct a congruent triangle, and by rotating and translating it, place it such that it creates a straight line of length \(a + b\). Join the points B and the image of A, which we will call E, to create a **trapezium** (name of the shape formed).**

3. **Quadrilateral CBED, which is created by rotating and translating the triangle ABC.**
   - The quadrilateral CBED is a trapezium because it has one pair of parallel sides (CB and ED).

4. **Area of CBED =**
   - \(\frac{1}{2}(a+b) \times (a+b) = \frac{1}{2}(a+b)^2\)

5. **\(\angle BAE = 90^\circ\)**
   - Since triangle ABC is a right triangle and E is formed by rotating triangle ABC, \(\angle BAE\) is also a right angle.

6. **Area of triangle ABC = \(\frac{1}{2} \times ab = \frac{1}{2}ab\)**

7. **Area of triangle BAE = \(\frac{1}{2} \times c \times c = \frac{1}{2}c^2\)**

8. **Area of triangle ADE = \(\frac{1}{2}ab\)** (same as triangle ABC because it's congruent).

9. **Since the area of the trapezium is equal to the sum of the areas of the three triangles:**
   \[
   \frac{1}{2}(a+b)(a+b) = \frac{1}{2}ab + \frac{1}{2}c^2 + \frac{1}{2}ab
   \]

10. **Which expands and simplifies to give:**
    \[
    a^2 + b^2 = c^2
    \]

This is the final result, demonstrating that the sum of the squares of the legs of a right triangle equals the square of the hypotenuse, which is the Pythagorean Theorem.

---

Do you want more details or have any questions?

Here are 5 related questions to explore further:

1. How does Garfield's proof compare to other proofs of the Pythagorean Theorem?
2. What other methods can be used to prove the Pythagorean Theorem?
3. Can the same method be applied to non-right triangles?
4. How can Garfield's proof be visualized using geometric shapes?
5. What are some real-life applications of the Pythagorean Theorem?

**Tip:** When proving geometric theorems, constructing congruent shapes and using their properties can often lead to elegant and simple proofs.

Explore the steps in Garfield's proof of Pythagoras' Theorem, including the construction of congruent triangles and the calculation of areas. This proof involves fundamental concepts of geometry and the Pythagorean Theorem, demonstrating how the sum of the squares of the legs equals the square of the hypotenuse. Suitable for students in Grades 9-12.

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