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