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

Open in App

Solution

Step 1: Draw a square $A C D F$ with $A C = 3 a$ .
Step 2: Cut $A B = 2 b$ , so that $B C = (3 a - 2 b)$ .
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, $(3 a - 2 b)^{2} = (3 a)^{2} - 2 b (3 a - 2 b) - 2 b (3 a - 2 b) - (2 b)^{2}$
$=$ $9 a^{2} - 6 a b + 4 b^{2} - 6 a b + 4 b^{2} - 4 b^{2}$
$=$ $36 x^{2} + 5^{2} - 60 x$
Hence, geometrically we proved the identity $(3 a - 2 b)^{2} = 9 a^{2} + 4 b^{2} - 12 a b$ .

Suggest Corrections

Similar questions

(3 a + 2 b)^{2} = 9 a^{2} + 4 b^{2} + 12 a b

9 a^{2} - 12 a b + 4 b^{2} - 9 m^{2} + 6 m n - n^{2}

9 a^{2} + 4 b^{2} + 16 c^{2} + 12 a b - 16 b c - 24 c a

(3 a - 2 b - 5 c)

9 a^{2} + 4 b^{2} + 25 c^{2} - 12 a b + 20 b c - 30 c a

Join BYJU'S Learning Program