Remove square from this whose side is ‘b’.(b<a)
Consider a square with side ‘a’.
We get the above figure by removing the square with side ‘b’.It consists of 2 parts
I and II.
So a2-b2 = Area of figure I + Area of figure II = a(a-b) + b(a-b) = (a-b)(a + b)
Thus a2-b2 = (a-b)(a + b).
Here a = 3 units and b = 2 units
So,L.H.S = a2-b2 = 32-22 = 5 units
R.H.S = (a-b)(a + b) = (3-2)(3 + 2) = 1(5) = 5 units
Therefore L.H.S = R.H.S
Hence,the identity is verified.
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