Respuesta :
Answer:
Length of original rectangle: 11 units.
[tex]\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}[/tex]
[tex]\text{Perimeter of new rectangle}=20[/tex]
Step-by-step explanation:
Let x represent width of the original rectangle.
We have been given that a rectangle has a length 6 more than it's width. S the length of the original rectangle would be [tex]x+6[/tex].
We have been given that when the width is decreased by 2 and the length decreased by 4 the resulting has an area of 21 square units.
The width of new rectangle would be [tex]x-2[/tex].
The length of new rectangle would be [tex]x+6-4=x+2[/tex].
The area of new rectangle would be [tex](x+2)(x-2)[/tex].
Now we will equate area of new rectangle with 21 and solve for x as:
[tex](x+2)(x-2)=21[/tex]
Applying difference of squares, we will get:
[tex]x^2-2^2=21[/tex]
[tex]x^2-4=21[/tex]
[tex]x^2-4+4=21+4[/tex]
[tex]x^2=25[/tex]
Since width cannot be negative, so we will take positive square root of both sides.
[tex]\sqrt{x^2}=\sqrt{25}[/tex]
[tex]x=5[/tex]
Therefore, the width of original rectangle is 5 units.
Length of the original rectangle would be [tex]x+6\Rightarrow x+5=11[/tex].
Therefore, the length of original rectangle is 11 units.
[tex]\text{Area of original rectangle}=5\times 11[/tex]
[tex]\text{Area of original rectangle}=55[/tex]
Therefore, area of the original rectangle is 55 square units.
Now we will find ratio of the original rectangle area to the new rectangle area as:
[tex]\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}[/tex]
We know that perimeter of rectangle is two times the sum of length and width.
[tex]\text{Perimeter of new rectangle}=2((x+2)+(x-2))[/tex]
[tex]\text{Perimeter of new rectangle}=2((5+2)+(5-2))[/tex]
[tex]\text{Perimeter of new rectangle}=2(7+3)[/tex]
[tex]\text{Perimeter of new rectangle}=2(10)[/tex]
[tex]\text{Perimeter of new rectangle}=20[/tex]
Therefore, the perimeter of the new rectangle is 20 units.
