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let a=b => a^2=a*b subtract by b^2 on both sides => (a^2)-(b^2)=(a*b)-(b^2) (a+b)(a-b)=b(a-b) => a+b=b since a=b 2a=a => 2=1......
Given any number b > 2, let l = 2b/(b-2) then the rectangle with sides of l and b will have an area = l*b = 2b/(b-2) * b = 2b2/(b-2) and perimeter = 2*(l+b) = 2*[2b/(b-2) + b] =2*[2b/(b-2) + b(b-2)/(b-2)] = 2/(b-2)* [2b+b2-2b] = 2b2/(b-2) So that the area and perimeter have the same numeric value.
a^2 + b^2 + 2ab = (a + b)^2
a2 + b2 = (a + b)2 - 2ab = (a - b)2 + 2ab.a2 - b2 = (a + b)*(a - b)
(b + 2)(b + 2) or (b + 2)2