Question

Verify the identity (a+b)^2= a^2+2ab+b^2 geometrically by taking a=3 and b=2

Answer 1

$\huge\mathbb{SOLUTION}}$

$\bold{(a+b)^2=a^2+2ab+b^2}}$

Draw a square with the side a+b,I,e 3+2

L.H.S area of whole square

$\large\bold{=(3+2)^2=5^2=25}$

R.H.S. = area of square with 3units + area of square with side 2 units + area of rectangle with sides 3,2units+area of rectangle with sides 2,3 units

$\large\bold{=3^2+2^2+3*2+3*2}$

$\large\bold{=9+4+6+6=25}$

$\large\bold{l.h.s=r.h.s}$

Therefore hence the identity is verified

Answer 2

Answer:

here is your answer

Explanation:

tep 1: Draw a square ACDF with AC=a units.

Step 2: Cut AB=b units so that BC=(a−b) unts.

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, (a−b)  

2

=a  

2

−b(a−b)−b(a−b)−b  

2

 

= a  

2

−ab+b  

2

−ab+b  

2

−b  

2

 

= a  

2

−2ab+b  

2

 

Hence, geometrically we proved the identity (a−b)  

2

=a  

2

−2ab+b  

2

.