Modeling in Identities

Modeling in Identities

By making use of the feature of area calculation through multiplication, this is a visual way of showing that the result of multiplication is equal to the value of the area. 

Multiplication and Area 

\( \displaystyle 3.4 \) ; If we times 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 one another. 🔻🔻

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 \( \displaystyle (a-b)² \) 

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

I’m going to create a rectangular area and the length of either side will be \( \displaystyle a-b \).

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

Let me take a square that has the side of a unit, in this way I 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. I can reach an identical through ;

\( \displaystyle a²-ab \)

I still have been able to obtain the area I want, I continue,

Before I subtract, let me add an area as big as \( \displaystyle b² \)
 next to the part I will be subtracting, this will give me an area that I can express easier \( \displaystyle (ab) \)

My 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 that I need to subtract from the area including the area that I have added

My identity; 

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

\( \displaystyle a²-ab+b²-ab \)

If I organize 

\( \displaystyle a²-2ab+b² \)

The final area that I have 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 \( \displaystyle a²\))

I 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.

I 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 identical is ;

\( \displaystyle a²+b²=(a+b)²-2ab \)

Identity of \( \displaystyle a²-b² \)

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

I must find another way to express this area.

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

By turning one on its head, I have stuck it next to other one. I 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 another way ;

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

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

Let me 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.

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