Question

Prove $(x-2y{)}^2=x^2-4xy+2y^2$ geometrically.

Answer 1

Step 1: Draw a square $ACDF$ with $AC=x$.
Step 2: Cut $AB=2y$ so that $BC=(x-2y)$.
Step 3: Complete the squares and rectangle as shown in the diagram.
Step 4: Area of yellow square $IDEO=$ Area of square $ACDF-$ Area of rectangle $GOFE-$ Area of rectangle $BCIO-$ Area of red square $ABOG$
Therefore, $(x-2y{)}^2=x^2-2y(x-2y)-2y(x-2y)-(2y{)}^2$
$=$ $x^2-2xy+4y^2-2xy+4y^2-4y^2$
$=$ $x^2-4xy+2y^2$
Hence, geometrically we proved the identity $(x-2y{)}^2=x^2-4xy+2y^2$.