Betty is writing a proof of the Pythagorean Theorem using the diagram shown. In the diagram, there is a large square. A smaller square is drawn inside the large square, dividing the large square into a smaller square and 4 small triangles.%0D%0A%0D%0AStep 1: First, I know that the area of a triangle is equal to bh2%0D%0A%0D%0Aℎ%0D%0A2%0D%0A. The area of each small triangle would be ab2%0D%0A%0D%0A%0D%0A2%0D%0A. Since there are 4 triangles in the entire square, the area of all 4 triangles together would be equal to 2ab%0D%0A2%0D%0A%0D%0A%0D%0A. The area of the small square in the middle can be represented by c2%0D%0A%0D%0A2%0D%0A since the sides each have a length of c%0D%0A%0D%0A. Therefore, the area of the entire large square can be written as 2ab+c2%0D%0A2%0D%0A%0D%0A%0D%0A+%0D%0A%0D%0A2%0D%0A.%0D%0A%0D%0AStep 2: Another way to think about the area of the entire large square is by looking at its side lengths. The length of each side can be written as a+b%0D%0A%0D%0A+%0D%0A%0D%0A, so the area would be (a+b)×(a+b)%0D%0A(%0D%0A%0D%0A+%0D%0A%0D%0A)%0D%0A×%0D%0A(%0D%0A%0D%0A+%0D%0A%0D%0A)%0D%0A.%0D%0A%0D%0AStep 3: Since we've expressed the area of the large square in two different ways, now set those two areas equal to each other and then simplify using algebra.%0D%0A%0D%0A(a+b)×(a+b)=2ab+c2%0D%0A(%0D%0A%0D%0A+%0D%0A%0D%0A)%0D%0A×%0D%0A(%0D%0A%0D%0A+%0D%0A%0D%0A)%0D%0A=%0D%0A2%0D%0A%0D%0A%0D%0A+%0D%0A%0D%0A2%0D%0Aa2+2ab+b2=2ab+c2%0D%0A%0D%0A2%0D%0A+%0D%0A2%0D%0A%0D%0A%0D%0A+%0D%0A%0D%0A2%0D%0A=%0D%0A2%0D%0A%0D%0A%0D%0A+%0D%0A%0D%0A2%0D%0Aa2+b2=c2%0D%0A%0D%0A2%0D%0A+%0D%0A%0D%0A2%0D%0A=%0D%0A%0D%0A2%0D%0AIn which step of Betty's proof did she make a mistake? Enter 1 for Step 1, enter 2 for Step 2, enter 3 for Step 3, or enter 4 if Betty did not make a mistake in her proof.
