Answer:
B. Simplify both sides of the equation to get [tex]a^2 + 2ab + b^2 - c^2 = 2ab[/tex]; then subtract 2ab and add [tex]c^2[/tex] to both sides of the equation
Step-by-step explanation:
Given
[tex](a + b)^2 - c^2 = 4(\frac{1}{2}ab)[/tex]
Required
Describe the next steps to
First, we open all brackets
[tex](a + b)^2 - c^2 = 4(\frac{1}{2}ab)[/tex]
[tex](a + b)(a + b) - c^2 = 4(\frac{1}{2}ab)[/tex]
[tex]a^2 + 2ab + b^2 - c^2 = 4 * (\frac{1}{2}ab)[/tex]
[tex]a^2 + 2ab + b^2 - c^2 = 2ab[/tex]
At this point, options A and C are incorrect because they didn't present the right result of the expression.
So, we have options B and D to consider
The next step is to subtract 2ab from both sides
[tex]a^2 + 2ab + b^2 - c^2 - 2ab = 2ab - 2ab[/tex]
Collect like terms
[tex]a^2 + b^2 - c^2 +2ab - 2ab = 2ab - 2ab[/tex]
[tex]a^2 + b^2 - c^2 = 0[/tex]
Lastly, to prove the Pythagoras theorem, the equation has to be in form of[tex]a^2 + b^2 = c^2[/tex]; meaning the [tex]c^2[/tex] is added to both sides; See below
[tex]a^2 + b^2 - c^2 = 0[/tex]
Add [tex]c^2[/tex] to both sides
[tex]a^2 + b^2 - c^2 + c^2= 0 + c^2[/tex]
[tex]a^2 + b^2 = c^2[/tex]
At this point, only option B completes Kira's proof
