Demonstrate that $(2 x - y)^{2} = 4 x^{2} + y^{2} - 4 x y$ with a figure.

Open in App

Solution

Step 1: Draw a square $A C D F$ with $A C = 2 x$ .
Step 2: Cut $A B = y$ , so that $B C = (2 x - y)$ .
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, $(2 x - y)^{2} = (2 x)^{2} - y (2 x - y) - y (2 x - y) - (y)^{2}$
$=$ $4 x^{2} - 2 x y + y^{2} - 2 x y + y^{2} - y^{2}$
$=$ $4 x^{2} + y^{2} - 4 x y$
Hence, geometrically we proved the identity $(2 x - y)^{2} = 4 x^{2} + y^{2} - 4 x y$

Suggest Corrections

Similar questions

(2 x + y - 5)^{2}

^{2}

(- 1, - 2)

x - 2 y + 3 = 0

Factorise: $4 x^{2} + 4 x y + y^{2}$

(2 x - y + z)^{2}

= 4 x^{2} + y^{2} + z^{2} - 4 x y - 2 y z + 4 z x

Join BYJU'S Learning Program