Modeling in Identities

Modeling in Identities

Calculation of area by means of multiplication is an effective visual representation. 

Multiplication and Area 

\( \displaystyle 3.4 \) ; If we multiply 3 with 4, we will have found the area (the number of squares) within a rectangle, which has a length of 3 units.

Now, by using this feature, let’s form an identity using the area calculation. 

🔻🔻Identities are expressions that are equal to each other in different forms.  🔻🔻

Identical of \( \displaystyle (a+b)² \)

The square of anything is the multiplication of any number by itself. Let’s write it down within brackets next to one another. 

\( \displaystyle (a+b)²=(a+b).(a+b) \)

This tells us; 

If you are going to create a rectangular shape, one side will be \( \displaystyle a+b\)
 and the other will also be \( \displaystyle a+b\) . When we multiply them by one another \( \displaystyle (a+b).(a+b) \) the area that is calculated is equivalent to our identity.  

We have placed \( \displaystyle a+b \) on either side, now let’s turn this into a rectangular area.

This area is going to give us our identity.

Let’s divide the area into pieces; 

\( \displaystyle (a+b).(a+b)=(a+b)² \)

\( \displaystyle (a+b)²=a²+2ab+b² \)

Identical of \( \displaystyle (a-b)² \) 

\( \displaystyle (a-b)²=(a-b).(a-b) \)

We are going to create a rectangular area where the length of either side will be \( \displaystyle a-b \).

We need to subtract \( \displaystyle b\) from  \( \displaystyle a \). We can do this by cutting ‘\( \displaystyle b \) ’ length from ‘\( \displaystyle a \)  length. 

 Let us take a square that has the side of  \( \displaystyle a \) , in this way We will have created a side of \( \displaystyle a \)

The lengths of the sides are a and the area is \( \displaystyle a² \). 

Let me now cut ‘\( \displaystyle b \) unit length’ from ‘\( \displaystyle a \) unit length’ to create an (\( \displaystyle a-b \)) unit length. 

Let’s cut this piece and throw it away.

The area that has been cut away:     \( \displaystyle a.b=ab \)

Let’s get rid of the area that has been cut from the complete square. We can reach an identical by;

\( \displaystyle a²-ab \)

We still have been able to obtain the area We want, let’s continue,

Before subtraction let's add an area as big as \( \displaystyle b² \)
 next to the part  we will subtract, this will give us an area that can be  expressed easily \( \displaystyle (ab) \)

The identity now is as follows;

\( \displaystyle a²-ab+b² \)

We have subtracted \( \displaystyle ab \) from  \( \displaystyle a² \)and added an area of \( \displaystyle b² \). 

Now, let’s cut the part we need to subtract from the area including the area that have been added

The identity; 

\( \displaystyle ab \)  area is also to be subtracted from the last identity;

\( \displaystyle a²-ab+b²-ab \)

If we  re-organize 

\( \displaystyle a²-2ab+b² \)

The final area that is achieved;

\( \displaystyle a²-2ab+b² \)

Identity of \( \displaystyle a²+b² \) 

A square that has the length of \( \displaystyle b \) (\( \displaystyle b² \)) is to be added to a square that has the length of \( \displaystyle a \)
(area \( \displaystyle a²\))

We must find another way to express this area. 

Let’s change the other length of this shape, change it to \( \displaystyle a+b \) and turn it into a square.

We must subtract all white areas from the square \( \displaystyle (a+b)\) .

\( \displaystyle (a+b).(a+b)-(ab+b²+ab-b²) \)
The total area of the square - The total area of the white areas

\( \displaystyle +b²\) and \( \displaystyle -b²\) simplifies each other.

\( \displaystyle (a+b).(a+b)-(ab+ab) \)

\( \displaystyle (a+b)²-2ab\)

Thus the  equivalent  identity is ;

\( \displaystyle a²+b²=(a+b)²-2ab \)

Identity of \( \displaystyle a²-b² \)

When we subtract a \( \displaystyle b² \) area from \( \displaystyle a² \) , we must be able to express what is left behind in another way. 

we must find another way to express this area.

We cut the shape from the corner. We have obtained two equal trapezoids. 

By turning a one on its head, we can stuck it next to other one such that. We have obtained a rectangle.

The area of the rectangle :  \( \displaystyle (a-b).(a+b)\)


\( \displaystyle a²- b²=(a-b).(a+b)\)

Let’s find this in another way ;

The area is equal to the total area of these two rectangles. 

\( \displaystyle a.(a-b) +b. (a-b)\)

Let's place the common multiplier in brackets, As \( \displaystyle (a-b) \) is multiplied by both \( \displaystyle a \) and \( \displaystyle b \), it is the ‘common multiplier’.

\( \displaystyle a.(a-b) +b. (a-b) = ( a+ b) . ( a - b)\)

In earlier years, you had learnt the topic of distributive property of multiplication.

\( \displaystyle 3.(8+5) =3.8+3.5\)

We have been given the expression of 3.8 + 3.5 ; it’s as if we have placed the common  multiplier 3 within  parenthesis.

\( \displaystyle a²- b²= ( a+b) . (a -b )\)