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Question

Demonstrate that (3a+2b)2=9a2+4b2+12ab with a figure.

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Solution

Step 1: Draw a line with a point which divides 3a,2b
Step 2: Total distance of this line =3a+2b
Step 3: Now we have to find out the square of 3a+2b i.e., Area of big square, ABCD= (3a+2b)2
Step 4: From the diagram, inside square red, AEFG and yellow square, CHFI be written as (3a)2,(2b)2
Step 5: The remaining corner side will be calculated as rectangular side = length × breadth = 3a×2b
Therefore, Area of the big square, ABCD= Sum of the inside square (AEFG+CHFI)+2 times the corner rectangular side.
(3a+2b)2=9a2+4b2+12ab
Hence, geometrically we proved the identity (3a+2b)2=9a2+4b2+12ab.
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