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Grade: 12th pass


Using the geometrical process prove that a²-b²=(a+b)(a-b).

Using the geometrical process prove that a²-b²=(a+b)(a-b).

one month ago

Answers : (1)

Anand Kumar Pandey
askIITians Faculty
4522 Points
							Dear Student

Thea2−b2identity represents thedifference of two square quantitiesand it can be written in factoring form as the product of binomialsa+banda−b. The factoring form ofa2−b2formula can be derived in mathematics geometrically on the basis of areas of geometric shapes.

Subtracting areas of squares

Take a square, whose length of each side isa units.
Therefore, the area of the square isa2.
Draw a small square with the side ofbunits at any corner of the square.
So, the area of small square isb2.
Now, subtract the square, whose area isb2 from the square, whose area isa2.
It forms a new geometric shape and its area is equal toa2−b2.

Divide the new subtracted geometric shape as two different rectangles but the length of one of the two rectangles should be equal tobunits.
Look at the upper rectangle.
Geometrically, the length of this rectangle isaunits and its width isa−bunits.
Similarly, look at the lower rectangle.
The length of this rectangle isa−bunits and its width is equal tobunits.

The width of the upper rectangle isa−band the length of the lower rectangle is alsoa−bgeometrically. If the lower rectangle is rotated by90∘, then the widths of both rectangles become same and it is useful to join them together as a rectangle.[factoring a squared minus b squared]Separate both rectangles.
Rotate the lower rectangle by90∘and then join both rectangles. It formed another rectangle.
The length and width of the new rectangle area+banda−brespectively. Therefore, the area of this rectangle is(a+b)(a−b).

In first step, it is derived that the area of subtracted shape isa2−b2and the same shape is now transformed as a rectangle, whose area is(a+b)(a−b).

Therefore, the areas of both shapes should be equal geometrically.


Geometrically, it is proved that thea2subtractedb2is equal to the product of the binomialsa+banda−b.

one month ago
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