Question

Prove geometrically that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$.

Answer 1

Hint: Draw a square of side a and cut a square of length b from one of its sides. Think how will equate its area with the area of a rectangle of length a+b and breadth a-b. Hence prove that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$.

Complete step-by-step solution -

Geometrical proof of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
Before dwelling into the proof, we need to know how can we get the terms ${{a}^{2}}$,${{b}^{2}}$ and $\left( a-b \right)\left( a+b \right)$. It is not difficult to see the area of a square of side a is ${{a}^{2}}$ and the area of a square of side b is ${{b}^{2}}$. Also, the area of a rectangle with length a+b and breadth a-b is$\left( a+b \right)\left( a-b \right)$.
Draw a square ABCD of side length “a” as shown.

Now cut a square of side b from corner B as shown

Produce FG to meet AD at H

Observe that AE = AB – EB = a-b and CF = CB-FB = a-b.
Hence we can attach the rectangle HGAE to DHFC with sides HG and CF kept side by side.
We form the new figure with the same area as above as shown below.

Now DK = DC + CK = a+b
Hence area of rectangle AKJH $=\left( a+b \right)\left( a-b \right)$
Also the same area is equal to area of square ABCD – area of the square GFBE $={{a}^{2}}-{{b}^{2}}$
Hence we have
$\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$.
Hence the result is proved.

Note: [1] Such geometrical proofs in mathematics are known as proof by rearrangement.
[2] Note that the area of DHJK is not equal to area of ABCD as DHJK was formed by rearrangement from ABCD after the removal of the area of the square GFBE.
[3] The above proof proves the identity for $a>0,b>0$ and $a>b$ and not the in general. For a general proof we have to reside to the distributive property of multiplication over addition and subtraction.