Demonstrate that $(3 a + 2 b)^{2} = 9 a^{2} + 4 b^{2} + 12 a b$ with a figure.

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Solution

Step 1: Draw a line with a point which divides $3 a, 2 b$
Step 2: Total distance of this line $= 3 a + 2 b$
Step 3: Now we have to find out the square of $3 a + 2 b$ i.e., Area of big square, $A B C D =$ $(3 a + 2 b)^{2}$
Step 4: From the diagram, inside square red, $A E F G$ and yellow square, $C H F I$ be written as $(3 a)^{2}, (2 b)^{2}$
Step 5: The remaining corner side will be calculated as rectangular side $=$ length $\times$ breadth $=$ $3 a \times 2 b$
Therefore, Area of the big square, $A B C D =$ Sum of the inside square $(A E F G + C H F I) + 2$ times the corner rectangular side.
$(3 a + 2 b)^{2} = 9 a^{2} + 4 b^{2} + 12 a b$
Hence, geometrically we proved the identity $(3 a + 2 b)^{2} = 9 a^{2} + 4 b^{2} + 12 a b$ .

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