Question

c^{2}=4(\frac{1}{2}bh)_{triangle}+(lw)_{square}
c^{2}=4(\frac{1}{2}ab)+(a - b)(a - b)
c^{2}=2ab+a^{2}-2ab + b^{2}
simplify the equation to c^{2}=a^{2}+b^{2}.
b. why does this describe the area of the smaller central square?
c. explain how you obtain this trinomial from the product (a - b)(a - b).

Explanation:

Step1: Expand \((a - b)(a - b)\)

Using the FOIL method (First - Outer - Inner - Last), we have \((a - b)(a - b)=a\times a+a\times(-b)+(-b)\times a+(-b)\times(-b)=a^{2}-ab - ab + b^{2}=a^{2}-2ab + b^{2}\).

Step2: Combine like - terms in the equation \(c^{2}=2ab+a^{2}-2ab + b^{2}\)

The \(2ab\) and \(-2ab\) terms cancel out. So \(c^{2}=a^{2}+b^{2}\).

For part B:
The first part \(4(\frac{1}{2}ab)\) represents the combined area of four right - angled triangles. The second part \((a - b)(a - b)\) represents the area of the square in the middle. The side length of the larger square formed by the four triangles and the smaller central square is \(c\). So the sum of the area of the four triangles and the smaller central square gives the area of the larger square with side length \(c\), which is \(c^{2}\).

Answer:

For part C: The trinomial \(a^{2}-2ab + b^{2}\) is obtained from \((a - b)(a - b)\) by using the FOIL method to multiply the two binomials.
For part B: The expression represents the sum of the areas of four triangles and the central square which is equal to the area of the larger square with side \(c\).