(a+b)^2 and (a-b)^2 proof | Why (a+b)2=a2+2ab+b2 ? Why (a-b)2=a2+b2-2ab ? — Proof by two different methods

Proof of (a+b)² & (a–b)² by Two Different Methods.








                        We know that
(a+b)² = a²+2ab+b²
(a–b)² = a²–2ab+b²
But when someone asks how did you get these formulas ? Then, we have to show their proof.
                 Let us prove these formulas. Here we will prove these formulas in two different ways one is algebraic method and another is geometric method.

Proof of (a+b)² by method–1 :-


             (a+b)² = (a+b)(a+b)
                         = a(a+b)+b(a+b) [Distributive law] 
                         = a.a+a.b+b.a+b.b [Distributive law]
                         = a²+ab+ba+b²
                         = a²+ab+ab+b² [Commutative law]
                         = a²+2ab+b²
                 ∴ (a+b)² = a²+2ab+b²

Proof of (a+b)² by method–2 :-

 
Square

                 Suppose, ABCD is a square. Take a point E on AD. 
                            Where AE=a & ED=b
                Then, draw a line from the point E, Which will be parallel to CD and it will join F on BC.
                  Again take a point G on AB. 
                             Where AG=b & GB=a
                Then, draw a line from the point G, which will be parallel to AD and meet at H on CD.
                         From the figure :-
                               The area of AEIG = ab
                               The area of EIHD = b²
                               The area of GBFI = a²
                               The area of IFCH = ab
                            & the area of ABCD = (a+b)(a+b)
                                                               = (a+b)²
                 Hence from above, the summation of all small areas = Area of ABCD.
                          Hence (a+b)² = ab+b²+a²+ab
                                                  = a²+2ab+b²
                               ∴ (a+b)² = a²+2ab+b²

Proof of (a–b)² by method–1 :-


            (a–b)² = (a–b)(a–b)
                        = a(a–b)–b(a–b) [Distributive law]
                        = a.a–a.b–b.a+b.b [Distributive law]
                        = a²–ab–ba+b²
                        = a²–ab–ab+b² [Commutative law]
                        = a²–2ab+b²
                     ∴ (a–b)² = a²–2ab+b²

Proof of (a–b)² by method–2 :-


Square


                   Let us consider, ABCD is a square whose measurement on one side is "a". Hence AD=a.
                    Take a point E on the line AD, whose distance is "b" from D. Hence ED=b.
                    Then, obviously
                                 AE=AD–ED
                                      = (a–b)
                     Hence from above
                          AD=a, ED=b & AE=(a–b)
                    Now, draw a line from point E, parallel to CD and join point F on BC.
                   Again take a point G on AB, where AG=b. Now, draw a line from point G, parallel to AD and join at a point H on CD.
                       From the figure :-
                            The area of AGIE = b(a–b)
                                                           = ab–b² = A1
                            The area of IFCH = b(a–b)
                                                           = ab–b² = A2
                            The area of EIHD = b² = A3
                            The area of ABCD = a²
                            The area of GBFI = (a–b)(a–b)
                                                             = (a–b)²
                We can also write that the area of GBFI = Area of ABCD – (A1+A2+A3)
                   => (a–b)² = a² –(ab–b²+ab–b²+b²)
                                    = a² –(2ab–b²)
                                    = a²–2ab+b²
                           ∴  (a–b)² = a²–2ab+b²







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