Verify $(x y - z)^{2} = (x y)^{2} + z^{2} - 2 x y z$ geometrically.

Open in App

Solution

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

Suggest Corrections

Similar questions

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

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

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

(x + y + z)^{2}

{| z_{1} - z_{2} |}^{2} = {| z_{1} |}^{2} + {| z_{2} |}^{2} - 2 Re (z_{1} ¯ ¯¯¯ ¯ z_{2})

Join BYJU'S Learning Program