Geometrical proof of the identity - I — lesson. Mathematics State Board, Class 7.

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When it comes to solving the algebraic equation, identities play a vital role to solve them. In this lesson, we are going to explore the geometrical proof of identities.

Let us first see the varies identities which we will discuss here:

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

$(a + b^2) = a^2 + 2ab + b^2$

$(a - b^2) = a^2 - 2ab + b^2$

$(a + b)(a - b) = a^{2} - b^{2}$

Let us take each identity one by one and discuss the proof of that identity.

Identity

$1$ :

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

Let us construct a rectangular diagram of four regions. One region is square-shaped with a dimension of

$4 × 4$ (Orange). Also, the other three regions are rectangle in shape with dimensions

$4×2$ (Blue),

$3 × 4$ (Yellow) and

$3 × 2$ (Green).

By observing the above rectangle, we can notice that:

$\text{Area of the bigger rectangle}$

$=$

$\text{Area of a square (Orange)}$

$+$

$\text{Area of three rectangles}$

$(4 + 3)$

$(4 + 2)$

$=$

$(4 × 4) + (4 × 2) + (3 × 4) + (3 × 2)$

Now, we simplify the LHS and RHS of the above expression.

LHS

$=$

$(4 + 3)$

$(4 + 2)$

$=$

$7×6 = 42$

RHS

$=$

$(4 × 4) + (4 × 2) + (3 × 4) + (3 × 2)$

RHS

$=$

$16+8+12+6 = 42$

Therefore, LHS

$=$ RHS

Similarly, if we use the variables in this case instead of number we get:

Let one side of a rectangle be

$(x +a)$ , and the other side be

$(x + b)$ units.

Then,

$\text{the total area of the rectangle }AEGI$

$=$

$\text{length } \times \text{breadth}$

$=$

$(x+a)(x+b)$ ………….

$(1)$

$\text{The area of the rectangle }AEGI$

$=$

$\text{The area of the square }ABCD$

$+$

$\text{The area of the rectangle }BEFD$

$+$

$\text{The area of the rectangle }DFGH$

$+$

$\text{The area of the rectangle }CDHI$ .

$\text{The area of the rectangle AEGI}$

$=$

$x^{2} + ax + ab + bx$

$= x^{2} + x (a + b) + ab$ ……………..

$(2)$

From the equation,

$(3)$ and

$(4)$ we get

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

Therefore,

$(x + a) (x + b) = x^{2} + x (a + b) + ab$ is a identity.

Example:

Simplify the expression

$(a + 3) (a + 11)$ using the identity

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

The expression is

$(a + 3) (a + 11)$ .

Now write the given expression

$(a + 3) (a + 11)$ with respect to the given identity

$(x+a)(x+b)$

$=$

$x^2+(a+b)x+ab$ .

$\begin{array}{l} (x + a) (x + b) = (a + 3) (a + 11) \\ = (a^{2} + a (3 + 11) + 3 \times 11) \end{array}$

Now simplify the expression.

$(a^{2} + 14 a + 33)$ .

Therefore,

$(a + 3) (a + 11)$

$=$

$(a^{2} + 14 a + 33)$ .

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3. Geometrical proof of the identity - I

Theory: