Question

the figure shows right - triangle abc, which has been copied four times to make the shape of a square with vertices a, b, f, and e. notice the placement of the four triangles also creates a centrally located square. follow the steps below to examine why the pythagorean theorem applies to any right triangle. write an equation to describe the area of square abfe. a = lw a = c*c = c² a. why does this describe the area of square abfe? we can break the area of the square into the area of the individual shapes that make up the square. area of the square = 4(area of triangle abc)+area of central square c² = 4(1/2 ab)+(lw)_{square} c² = 4(1/2 ab)+(a - b)(a - b) b. why does this describe the area of the smaller central square? c² = 2ab+a² - 2ab + b² c. explain how you obtain this trinomial from the product (a - b)(a - b). simplify the equation to c² = a² + b².

Explanation:

Step1: Area of outer - square

The side - length of square \(ABFE\) is \(c\). Using the formula for the area of a square \(A = s^2\) (where \(s\) is the side - length), the area of square \(ABFE\) is \(A=c\times c = c^{2}\). This describes the area of square \(ABFE\) because all sides of a square are equal, and if the side - length is \(c\), the area is the product of the side - length with itself.

Step2: Area of central square

The side - length of the central square is \((a - b)\). Using the formula for the area of a square \(A = s^2\), the area of the central square is \((a - b)(a - b)\). We get this because if we look at the dimensions of the central square, its horizontal and vertical side - lengths are found by subtracting the length of the shorter leg (\(b\)) of the right - triangle from the longer leg (\(a\)) of the right - triangle.

Step3: Expand \((a - b)(a - b)\)

Using the FOIL method (\((x + y)(x + y)=x^{2}+2xy + y^{2}\), here \(x=a\) and \(y=-b\)), we have \((a - b)(a - b)=a^{2}-ab - ab + b^{2}=a^{2}-2ab + b^{2}\).

Step4: Simplify the area equation

We know that \(c^{2}=4\times(\frac{1}{2}ab)+(a - b)(a - b)\). Substituting the expansion of \((a - b)(a - b)\) into the equation, we get \(c^{2}=2ab+a^{2}-2ab + b^{2}\). The \(2ab\) and \(-2ab\) terms cancel out, leaving us with \(c^{2}=a^{2}+b^{2}\).

Answer:

The Pythagorean Theorem \(c^{2}=a^{2}+b^{2}\) is derived for right - triangle \(ABC\) by equating the area of the outer square \(ABFE\) to the sum of the areas of the four right - triangles and the central square.