Question

How do you find the area of a square with side lengths of $x-3$ units?

Answer 1

Hint: We know that the length of each of the sides of a square is the same. Suppose that the length of one side of a square is $a$ units. Then the lengths of all the sides of the square is equal to $a$ units. The area of a square with side lengths of $a$ units is obtained to be ${{a}^{2}}$ square units.

Complete step by step solution:
Let us consider the given problem.
We are given a square. The side lengths of the square are $x-3$ units. We are asked to find the area of the square. To find the area of a square, we need to use a formula.
Let us suppose that we are given a square whose vertices are $A,B,C$ and $D.$ Suppose that the length of a side $AB$ of the square $ABCD$ is given. Let the length be $a$ units. That is, $AB=a.$
We know that the lengths of all the sides of a square are equal. Therefore, we will get $AB=BC=CD=AD=a.$
We know that the area of a square is found by multiplying the side length with itself.
So, the area of the square $ABCD$ is equal to $a\times a={{a}^{2}}$ square units.
Let us use this formula to find the area of the given square with side lengths of $x-3$ units.
We calculate the area of the given square by multiplying $x-3$ with $x-3.$
So, we will get $\left( x-3 \right)\left( x-3 \right)={{\left( x-3 \right)}^{2}}.$
We will get ${{\left( x-3 \right)}^{2}}={{x}^{2}}-6x+9.$
Hence the area of the given square is ${{x}^{2}}-6x+9$ square units.

Note: We know the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}.$ Also, we know that the area of a rectangle is obtained by multiplying its length with its breadth. We know that the length of a square equals its breadth. So, the area is square of the length.