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Question
what is the result after subtracting the 2ab from both sides? pythagorean theorem a = a1 + a2 + a3 + a4 + a5 (a+b)² = a1 + a2 + a3 + a4 + a5 (a+b)² = a1 + a2 + a3 + a4 + c² (a+b)² = ½ab + ½ab ½ab + ½ab + c² (a+b)(a+b) = 2ab + c² a² + 2ab + b² = 2ab + c² a² + c² = b²; a² + b² = c²; a² + b² = c² + 2ab; (a + b)² = c²
Step1: Start with the equation
We have the equation \( a^{2}+2ab + b^{2}=2ab + c^{2} \).
Step2: Subtract \( 2ab \) from both sides
Subtract \( 2ab \) from the left - hand side (\( a^{2}+2ab + b^{2}-2ab\)) and the right - hand side (\(2ab + c^{2}-2ab\)).
Simplifying the left - hand side: \( a^{2}+2ab + b^{2}-2ab=a^{2}+b^{2} \).
Simplifying the right - hand side: \(2ab + c^{2}-2ab = c^{2} \).
So the resulting equation is \( a^{2}+b^{2}=c^{2} \).
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\( a^{2}+b^{2}=c^{2} \) (corresponding to the option "a² + b² = c²" among the given options)