Answer:
The width of the original rectangle on the left is [tex]5\ m[/tex]
Step-by-step explanation:
step 1
Find the scale factor
we know that
If two figures are similar, then the ratio of its perimeters is equal to the scale factor
so
Let
z -----> the scale factor
a ----> perimeter of the reduced rectangle on the right
b ----> perimeter of the original rectangle on the left
[tex]z=\frac{a}{b}[/tex]
we have
[tex]a=24\ m[/tex]
[tex]b=30\ m[/tex]
substitute
[tex]z=\frac{24}{30}=0.8[/tex]
step 2
Find the width of the reduced rectangle on the right
we know that
The perimeter of rectangle is equal to
[tex]P=2(L+W)[/tex]
we have
[tex]L=8\ m[/tex]
[tex]P=24\ m[/tex]
substitute and solve for W
[tex]24=2(8+W)[/tex]
[tex]12=(8+W)[/tex]
[tex]W=12-8=4\ m[/tex]
step 3
Find the width of the original rectangle on the left
we know that
If two figures are similar, then the ratio of its corresponding sides is proportional and this ratio is called the scale factor
so
Let
z -----> the scale factor
y ----> the width of the reduced rectangle on the right
x ----> the width of the original rectangle on the left
[tex]z=\frac{y}{x}[/tex]
we have
[tex]y=4\ m[/tex]
[tex]z=0.8[/tex]
substitute and solve for x
[tex]0.8=\frac{4}{x}[/tex]
[tex]x=\frac{4}{0.8}[/tex]
[tex]x=5\ m[/tex]
