Question

Draw the geometrical figure for the identity ${(a+b)}^2=a^2+2ab+b^2$.

Answer 1

Consider a line segment.

Consider any arbitrary point on the line segment and name the first part as $a$ and the second part as $b$. Please refer to fig $a$.

So the length of the line segment in fig $a$ is now $(a+b)$.

Now, let’s draw a square having length $(a+b)$. Please refer to fig $b$.

Let’s extend the arbitrary point to other sides of the square and draw lines joining the points on the opposite side. Please refer to fib $b$.

As we see, the square has been divided into four parts $(1,2,3,4)$ as seen in fig $b$.

The next step is to calculate the area of the square having length $(a+b)$.

As per fig $b$ , to calculate the area of the square : we need to calculate the area's of parts $1,2,3,4$ and sum up.

Calculation : Please refer to fig $c$.

Area of part $1:$

Part $1$ is a square of length $a$.

$\therefore$ area of part $1=a^2$ ------- $\left(i\right)$

Area of part $2:$

Part $2$ is a rectangle of length $:b$ and width $:a$

$\therefore$ area of part $2=length\times breadth=ba$ ----------$\left(ii\right)$

Area of part $3:$

Part $3$ is a rectangle of length$:b$ and width $:a$

$\therefore$ area of part $3=length\times breadth=ba$ -------------$\left(iii\right)$

Area of part $4:$

Part $4$ is a square of length $:b$

$\therefore$ area of part $4=b^2$ --------$\left(iv\right)$

So, Area of square of length $(a+b)={(a+b)}^2=\left(i\right)+\left(ii\right)+\left(iii\right)+\left(iv\right)$

$\therefore$

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

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

Hence Proved