Question

How to find an equation for the area of a square?
If the length of a square is decreased by 1 unit, the area is now 8 square units. What’s the equation for the area of this square?

Answer 1

Hint: This question can be answered by giving the formula for the calculation of the area of a square. We then need to calculate the length for the given side length and area. We assume the original length of the side of the square is x. Then, we substitute the side as x-1 and area as 8 in the formula and simplify to get the solution.

Complete step by step solution:
The formula to calculate the area of a square for a side length given as $a$ units is given by,
$\Rightarrow Area={{a}^{2}}\text{ square units}\ldots \ldots \left( 1 \right)$
Since all the sides of the square are equal, we take the value of any one of the sides and substitute in the above equation to get the area of the square.
We are required to find the equation for the area of a square whose side length is decreased by 1 unit and whose area is 8 square units.
Let us assume that the original side length be x units. Now it is given that the length of the square is decreased by 1 unit. This means that the new length of the side of the square after decreasing 1 unit is x-1 units. The area is given as 8 square units. Substituting these values in equation (1),
$\Rightarrow 8={{\left( x-1 \right)}^{2}}$
This is the required equation for the area of the square with side length of x-1 units and an area of 8 square units.
To simplify this, we take the square root on both sides of the equation.
$\Rightarrow \sqrt{8}=\sqrt{{{\left( x-1 \right)}^{2}}}$
Since the square and root cancels on the right-hand side, and taking only the positive value for the square root of 8 since area is always positive,
$\Rightarrow 2.828=\left( x-1 \right)$
Adding 1 on both sides,
$\Rightarrow 3.828=x$

Hence, the original length of the side was 3.828 units and the equation for this square is given as $8={{\left( x-1 \right)}^{2}}.$

Note: Basic definitions and formulae for the area are to be known well by the students. These form the basis for many questions. This can also be solved graphically by drawing the square of the required dimensions and substituting the obtained values for area and side in the formula to obtain the equation.