Question

Verify the identity: ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ geometrically.

Answer 1

Hint: To prove the given identity, we have to consider a square ABCD of side a units. The area of a square is equal to the square of its side, which will give us the area of the square ABCD equal to ${{a}^{2}}$ square units. Then we need to divide the square ABCD into two rectangles and a square of side b units. Then the area of each of the individual three parts can be found out. Finally, on equating the sum of the areas of the three parts to that of the area of the square ABCD, the given identity will be proved.

Complete step-by-step answer:
Let us consider a square of side a units as shown below.

The area of the above square ABCD can be given as
\[\Rightarrow ar\left( ABCD \right)={{a}^{2}}.......\left( i \right)\]
Now, let us consider divide the above square into three regions as shown below.

From the above figure, the area of the rectangle can be written as
$\begin{align}
  & \Rightarrow ar\left( ADEF \right)=AD\times AF \\
 & \Rightarrow ar\left( ADEF \right)=a\left( a-b \right).......\left( ii \right) \\
\end{align}$
Similarly, the area of the rectangle ECGH can be given as
$\Rightarrow ar\left( ECGH \right)=b\left( a-b \right).......\left( iii \right)$
And the area of the square FBGH can be given as
$\Rightarrow ar\left( FBGH \right)={{b}^{2}}.......\left( iv \right)$
Now, from the above figure we can also observe that
$\Rightarrow ar\left( ABCD \right)=ar\left( ADEF \right)+ar\left( ECGH \right)+ar\left( FBGH \right)$
Putting (i), (ii), (iii), and (iv) into the above equation, we get
$\Rightarrow {{a}^{2}}=a\left( a-b \right)+b\left( a-b \right)+{{b}^{2}}$
Subtracting ${{b}^{2}}$ from both the sides, we get
$\begin{align}
  & \Rightarrow {{a}^{2}}-{{b}^{2}}=a\left( a-b \right)+b\left( a-b \right)+{{b}^{2}}-{{b}^{2}} \\
 & \Rightarrow {{a}^{2}}-{{b}^{2}}=a\left( a-b \right)+b\left( a-b \right) \\
\end{align}$
Finally, taking $\left( a-b \right)$ common in the RHS, we get
$\Rightarrow {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
Hence, we have proved the given identity geometrically.

Note: We can also prove the given identity algebraically. For this we need to consider the RHS of the given identity and expand it using the distributive property given by $a\left( b+c \right)=ab+ac$. On simplifying the obtained expression, we will get the RHS equal to the LHS and hence the given identity will get proved.