By using this concept of multiplication of monomials, let us try to prove the identities geometrically, which are very much useful in solving algebraic problems.

**Geometrical proof of Identities**

By using this concept of multiplication
of monomials, let us try to prove the identities geometrically, which are very much
useful in solving algebraic problems.

** **

__1. Identity-1: ( x + a)(x
+ b) = x^{2} + x(a + b) + ab__

Consider four regions. One region is
square shaped with dimension 3×3 (Grey). Also, the other three regions
are rectangle in shape with dimensions 4×3 (yellow), 3×2
(red) and 4×2 (Blue).

Arrange these four regions to form a
rectangular shape as shown in the Fig. 3.8.

By observing *Fig. 3.8*, we can
note that,

**Area of the bigger rectangle
= Area of a square (Grey ) + Area of Three rectangles**

(3 + 4)(3 +2)=(3×3)+(4×3)+(3×2)+(4×2)

(3 + 4)(3 +2)=(3×3)+3×(4 +2)+(4×2) ...(1)

Where, LHS is (3+ 4)(3+2)=7×5=35

RHS is (3 ×3) + 3 ( 4 + 2) + ( 4 ×2) = 9 +(3×6)+8

= 9 +18+8=35

Therefore, LHS = RHS.

In the similar way as explained above,
let us check for another set of four regions as shown in the *Fig. 3.9.*

**Note**

We know that, Area of rectangle

= *length *× *breadth*

*=l *×*b*

Also, area of square

*=side*×*side*

*=a *×* a *=*a*^{2}

By observing *Fig. 3.9*, we can
note that,

**Area of the bigger rectangle
= Area of a square (Grey ) + Area of Three rectangles**

(5+4) (5+1) = (5×5) + (5×1) + (5×4) + (1×4)

(5+4) (5+1) = (5×5) + 5(1+4) + (1×4) ...(2)

Where, LHS is (5 +1)(5 + 4)=6×9 =54

RHS is 5^{2} +5(1+ 4)+(1×4)=25
+(5×5)+ 4

=25+25+4=54

Therefore, LHS = RHS.

Thus, equation (1) and (2) is true for
given set of any three values.

By generalising those three values as
‘*x’,* ‘*a*’ and ‘*b’*

we get, ( *x* + *a*)( *x* +*b*)= ( *x* × *x*) + *x* (*a* +*b*)+ (*a* ×*b*)

That is**,** ( *x* + *a*)( *x* +*b*)= *x*^{2} + *x* (*a* +*b*)+ *ab*

Hence, (
*x* + *a*)( *x* +*b*)=
*x*^{2} + *x* (*a* +*b*)+ *ab* is an identity.

Now, let us prove this identity, geometrically.

Let one side of a rectangle be (*x* +*a*) and
the other
side be (*x* +
*b*) *units*.

Then, the total area of the rectangle
ABCD = length **× **breadth = (*x*+*a*)(*x*+*b*) ...(3)

From the Fig. 3.10, we can see that the

**Area of the rectangle ABCD
= area of the square AFIE + area of the rectangle FBGI** + **area of the rectangle EIHD
+ area of the rectangle IGCH**

*= x *^{2}* *+* ax *+*
bx *+*ab*

= *x* ^{2} + *x*(*a* + *b*) +*ab* ...(4)

From (3)
and (4) we get, (*x*
+
*a* )(*x* +*b*) = *x* ^{2} +
*x*(*a* + *b*) +*ab*

Hence, (
*x* + *a*)( *x* +*b*) =
*x* ^{2} + *x* (*a* +*b*)+ *ab* is an identity.

**Note**

(i) In case if *b*
= −*b* then the identity is

(*x* + *a* )(*x* +(−*b*)) = *x* ^{2}
+ *x* (*a* +(−*b*)) +*a*(−*b*)

(*x* + *a* )(*x* −*b*) = *x* ^{2}
+ *x*(*a* − *b*) –*ab*

(ii) If *a* =−*a* then the identity is (*x*
+ (−*a*))(*x* + *b* ) = *x* ^{2} + *x* ((−*a* ) + *b*) +(−*a* )*b*

(*x* − *a* )(*x* +*b*) = *x* ^{2}
+ *x*(*b* − *a* ) –*ab*

(iii) If *a* = −*a and b* = −*b* then the identity is

(*x* + (−*a* ))(*x* + (−*b*)) = *x*^{2} + *x*((−*a* ) +(−*b*)) +(−*a* )(−*b*)

(*x* − *a* )(*x* −*b*) = *x* ^{2}
−*x*(*a* + *b*) +*ab*

__Example 3.1 __

Simplify the following using the identity
(*x + a *)(*x +b*) =* x*^{2} + *x*(*a
+ b*) +*ab* :

(i) (*x* + 3) (*x* + 5) (ii) (*y* + 8)(*y+* 6)
(iii) 43×36

**Solution:**

(i) (*x*
+ 3)(*x* + 5)

Let us represent
the expression geometrically, as shown in the Fig. 3.11.

In the rectangle with length (
*x* + 5) and breadth (
*x* + 3), we get,

**Area of bigger rectangle
= Area of a square + Area of three rectangles**

Therefore,

( *x* + 3)( *x* + 5)= *x*^{2} + 5*x* + 3*x* +15

* = x*^{2}* *+(5* *+*
*3)*x *+15

*=x *^{2}* *+8*x *+15*
*.

**(ii) ( y + 6)(y + 8)**

Let us represent the expression geometrically,
as shown in *Fig. 3.12*.

In the rectangle with length (
*y* + 6) and breadth (
*y* + 8) *units*,
we get,

**Area of bigger rectangle
= Area of square + Area of three rectangles**

Therefore,

( *y* + 6)( *y* + 8)= *y* ^{2} + 6 *y* + 8 *y* + 48

( *y* + 6) ( *y* + 8)= *y* ^{2} +(6 + 8)*y* + 48

= *y* ^{2} +14 *y* + 48

(iii) 43×
36 =(
40 +3)×(
40 −4)

We know the identity

( *x* + *a*)( *x* +*b*)= *x* ^{2} + *x* (*a* +*b*)+ *ab*

Taking, *x =* 40, *a*=3 and
b= –4, we get

(40 +3)(40−4)= 40^{2} + 40(3−4)+3(−4)

=1600 +
40(−1)−12

=1600−40−12

=1600−52

Therefore, 43×
36 =1548
.

__2. Identity-2: ____(a + b)____ __^{2}__ ____= a ^{2}__

Consider four regions. Two square shaped
regions with the dimensions of 3 × 3 (Grey) and 2 × 2 (Blue). Also, there are two
rectangle shaped regions and both are in the dimension of 3 × 2 (yellow).

Arrange them in a square shape as shown
in the *Fig. 3.13*.

By observing the *Fig. 3.13*, we
can note that

**Area of the bigger square
= Area of two small squares + Area of two rectangles.**

(3 + 2)^{2} = 3^{2}
+
(2×3)
+
(3×2)
+
2^{2} = 3^{2} + (3× 2) + (3× 2) + 2^{2} [since, 2**×**3 = 3**×**2]

Therefore, (3 +
2)^{2} = 3^{2} +
2 × (3×
2) +2^{2}

where, L.H.S is (3 +
2)^{2} =5^{2} =25 ……(1)

R.H.S is 3^{2} +
2 ×(3×
2) +2^{2}
=
9 +12
+
4 = 25 ……(2)

From (1) and (2), L.H.S = R.H.S.

Now, we can prove this for the variables
*a* and *b*.

Let us take a square ABCD of side (*a*+*b)*,
hence its* *area is (*a* +*b*)(*a* +*b*)= (*a* +*b*)^{2}

By observing the *Fig. 3.14*,

**Area of the bigger square
ABCD = Area of two small squares (Grey and blue) + Area of two rectangles (yellow).**

So, (*a* +
*b*)^{2} =*a* ^{2} + *ba* +
*ab* +*b*^{2}

=*a* ^{2} + *ab* + *ab* +*b*^{2 }(since, *ba* = *ab* )

(*a* +*b*)^{2} = *a* ^{2} + 2*ab* +*b*^{2}

Hence proved.

**Note**

If, we substitute *b=
*a in* *the identity (*x* + *a* )(*x* +*b*) = *x* ^{2}
+ *x*(*a* + *b*) + *ab* , we get,

(*x* + *a* ) (*x* + *a* ) = *x*^{2} + *x*(*a + a* ) + *a × a*

(*x* + *a* )^{2} = *x* ^{2} + *x* (2*a* ) + *a*^{2}

( *x* + *a* )^{2} = *x*^{2} + 2*ax* +*a*^{2} , which is similar to the identity (*a* +*b*)^{2} = *a* ^{2} + 2*ab* +*b*^{2} .

__Example 3.2____ __

Simplify the following using the identity
(*a* + *b*)^{2} =*a* ^{2} + 2*ab* +*b*^{2} .

(i) (2*x* +5)^{2}
(ii) 21^{2}

**Solution:**

(i)** **(2**
***x*** **+** **5)^{2}

Let the side of the square be 2*x*
+5
*units*.

Then its area is *side*×*side* , that is (2
*x* + 5)^{2} .

The geometrical representation of the
given expression is as shown in *Fig. 3.15*.

Area of the bigger square
= Area of two squares + Area of two rectangles.

(2 *x* + 5)^{2} = 4*x*^{2} + 25 + 10*x* +10*x*

= 4*x*^{2} +(10
+10)*x* + 25 (Adding
like terms)

= 4 *x*^{2} + 20*x* +25.

(ii)** **21^{2}

Let the side of the square be 21. Hence,
its area is 21^{2} .

Consider 21^{2} as (20
+1)^{2} which is one of the way to represent
it geometrically as shown in *Fig. 3.16*.

Now,

Area of the bigger square
= Area of two squares + Area of two rectangles.

21^{2 }= 400+1+20+20=441.

**Aliter method:**

We know the identity, (*a* +*b* )^{2} = (*a* +*b*)(*a* +*b*)

= *a* ^{2} + 2*ab* +*b*^{2}

To find the value of 21^{2},

We take, 21^{2} = (20
+1)^{2}

= (20
+1)(20
+1)

Here, *a* = 20 and *b =* 1.

Therefore,

*a*^{2}* *+ 2*ab *+* b*^{2}* *=*
*20^{2}* *+2×*
*20×1+1^{2}

=400+40+1 =441.

__3. Identity-3: (a __

When we replace ‘*b*’ by ‘–*b*’
in ‘identity-2’, we get a new identity.

We know that, (*a* +*b*)^{2} = *a*^{2} +
2*ab* +*b*^{2}

Taking ‘*b*’ as ‘–*b*’ in ‘identity-2’,
we get, [*a* + (−*b*)]^{2} = *a*^{2} +
2*a*(−*b*)+ (−*b*)^{2}

(*a* −*b*)^{2} = *a*^{2} −2*ab* + (−*b*)(−*b*)

Therefore, (*a* −*b*)^{2} = *a*^{2} −2*ab*
+*b*^{2}

Note that, when we change the sign of
*b,* the sign of 2*ab* (second term) alone is changed.

__Example 3.3__

Using the identity (*a* −
*b*)^{2} = *a*^{2} − 2*ab* +*b*^{2} , simplify the following:

(i) (3*x* -5*y*)^{2}

(ii) 47^{2}

**Solution:**

(i)** **(3*x*** **-5*y*)^{2}

Put *a* = 3*x* and *b* = *5y *in the identity* *(*a
*−*
b*)^{2}* *=*a*^{2}* *−* *2*ab *+*b*^{2}* *, we get,

(3*x* -5*y*)^{2} =(3*x*)^{2}
−
2 ×(3*x*)×
(5*y*) +(5*y*)^{2}

=3^{2} ×*x*^{2} − (2×3× 5)*xy* +(5^{2}
×*y*^{2} )

= 9 *x*^{2} −30*xy* +25*y*^{2}.

(ii)** **47^{2}** **=**
**(50** **−3)^{2}**
**, substituting** ***a =*50 and *b = *3 in the identity (*a*
−
*b*)^{2} =*a*^{2} − 2*ab* +*b*^{2}, we get,

(50-3)^{2}=50^{2}−2×50×3+3^{2}

= 2500 − 300 + 9

=2509 − 300 = 2209.

__4. Identity-4: ( a + b)(a − b)= a^{2}–b^{2}__

In the given figure, AB = AD = a.

So, area of square ABCD = a^{2 }.

Also, SB= DP= *b*

Area of the rectangle SBCT = *ab*.

Similarly, area of the rectangle DPRC=
*ab*.

Also, area of the square TQRC= *b*^{2} .

Area of the rectangle DPQT *ab − b*^{2} .

Now, AS = PQ = (*a − b*) and AP = SQ = (a + b).

Hence, area of the rectangle APQS (the
shaded rectangle) = area of square ABCD –
area of rectangle STCB + area of rectangle DPQT

*= a ^{2}–ab + *(

*= a ^{2}–ab + ab – b^{2}*

*= a ^{2} – b^{2}*

Hence, (*a* + *b*) (*a*
– *b*) = *a ^{2}*

__Example 3.4 __

Simplify by using the identity** ****(***a*** ****+**** ***b***)(***a*** ****−**** ***b***)** = *a ^{2}*

(i) (3*x* + 4)(3*x* − 4)

(ii) 53 × 47.

**Solution:**

(i) (3*x* + 4)(3*x* − 4)

Substitute, *a=*3*x* and *b=*4
in the identity (*a* +*b*)×(*a* −*b*)= *a* ^{2}
−*b*^{2}, we get,

(3*x* + 4)(3*x* − 4) =
(3*x*)^{2} −4^{2}

(3^{3} ×
*x*^{3}) −16 = 9 *x*^{2} −16
.

(ii) 53×47 = (50+3)×(50−3).

Take, a =50 and b = 3,

Substituting the values of ‘a’ and ‘b’
in the identity (*a* + *b*)(*a* – *b*) = (*a*^{2}
– *b*^{2}), we get,

= 50^{2} −3^{2}

= 2500 − 9

=2491.

** **

**Try this**

**Consider a square shaped paddy field with side of 48 m. A
pathway with uniform breadth is surrounded the square field and the length of the
outer side is 52 m. Can you find the area of the pathway by using identities?**

**Solution:**

Let *a* = 52

*b* = 4

*(a–b)*^{2} = *a*^{2} – 2*a*b + *b*^{2}

= 52^{2 }–
2 (52) (4) + 4^{2}

= 2704 – 416 + 16
= 2304

Tags : Algebra | Term 3 Chapter 3 | 7th Maths , 7th Maths : Term 3 Unit 3 : Algebra

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