Answer:
The area of the original rectangle is \(48in^{2}\)
Step-by-step explanation:
Let's define the following variables :
\(W\) : ''Width''
\(L\) : ''Length''
We know that the area of the original rectangle is :
\(Area=(W).(L)\)
We need to find the values of this variables in order to calculate the original area of the rectangle.
We know that if the width is decreased by 3 inches, then the area of the resulting rectangle is \(24in^{2}\). We can write the following equation :
\((W-3).(L)=24\) (I)
We know that the width and the length are consecutive even integers.
Therefore we have two cases :
\(W+2=L\) (A)
\(W-2=L\) (B)
Let's suppose the case (A). So if we replace the equation of the case (A) in (I) we will obtain :
\((W-3).(L)=24\)
\((W-3).(W+2)=24\)
\(W^{2}+2W-3W-6=24\)
\(W^{2}-W-30=0\)
If we use the quadratic formula we will obtain two possibles values for \(W\) :
\(W_{1}=6\) and \(W_{2}=-5\)
\(W_{2}\) is absurd because a length can't be negative. The value \(W_{1}\) is possible.
If \(W=6\) , then using the equation of the case (A), we obtain that
\(W+2=L\\6+2=L\\L=8\)
The pair :
\(W=6\\L=8\)
is a possible solution for the problem. If we use the equation of the case (B) in (I) we will obtain the following expression :
\(W^{2}-5W-18=0\)
If we use the quadratic formula in this equation we will obtain that
\(W_{3}=\frac{5+\sqrt{97}}{2}\\W_{4}=\frac{5-\sqrt{97}}{2}\)
This expressions are absurd because \(W\) must be an even integer number.
Finally, the solution \(W=6\) , \(L=8\) is the only correct solution.
Calculating the area of the original rectangle :
\(Area=(W).(L)=(6).(8)=48\)
The area of the original rectangle is \(48in^{2}\)
