Verify $(a - b)^{2} = a^{2} - 2 a b + b^{2}$ geometrically.

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Solution

Step 1: Draw a square $A C D F$ with $A C = a$ units.
Step 2: Cut $A B = b$ units so that $B C = (a - b)$ unts.
Step 3: Complete the squares and rectangle as shown in the diagram.
Step 4: Area of yellow square $I D E O =$ Area of square $A C D F -$ Area of rectangle $G O F E -$ Area of rectangle $B C I O -$ Area of red square $A B O G$
Therefore, $(a - b)^{2} = a^{2} - b (a - b) - b (a - b) - b^{2}$
$=$ $a^{2} - a b + b^{2} - a b + b^{2} - b^{2}$
$=$ $a^{2} - 2 a b + b^{2}$
Hence, geometrically we proved the identity $(a - b)^{2} = a^{2} - 2 a b + b^{2}$ .

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