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

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Identity \(2\):

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

Let us construct a figure of four regions. The two square shaped regions with the dimensions of \(3 × 3\) (Blue) and \(2 × 2\) (Yellow). Observe the remaining two rectangle shaped regions. Both are in \(3 × 2\) (Green) dimension.

By observing the above rectangle, we can notice that:

\(\text{Area of the bigger square = Area of the two small square + Area of the two rectangles}\)

\(3 + 2\)\(2 + 3\) \(=\) \((3 × 3) + (2 × 3) + (3 × 3) + (2 × 2)\)

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

LHS \(=\) \(3 + 2\)\(2 + 3\) \(=\) \(5×5 = 25\)

RHS \(=\) \((3 × 3) + (2 × 3) + (3 × 3) + (2 × 2)\)

\(=\)

3^{2} + (2 \times 3) + (3 \times 2) + 2^{2}

\(=\) \(9+6+6+4 = 25\)

Therefore, LHS \(=\) RHS

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

Assume the square of ABCD of side \(a + b\). From the above figure, we can get that:

\(\text{The total area of the bigger square = The area of the two small squares × The are of the two rectangles}\)

That is, \((a + b)^2\)

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

Since, \(ba=ab\); \((a + b)^2\)

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

Therefore,

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

is a identity.

Example:

Simplify the expression

(a + 6) (a + 6)

using the identity

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

Now write the given expression

(a + 6) (a + 6)

with respect to the given identity

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

\begin{array}{l} (a + b) (a + b) = (a + 6) (a + 6) \\ {(a + 6)}^{2} = (a^{2} + 2 (6 \times 6) a + 6^{2}) \end{array}

Now simplify the expression.

= (a^{2} + 72 a + 36)

Therefore,

(a + 6) (a + 6)

= (a^{2} + 72 a + 36)

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4. Geometrical proof of identity - II

Theory: