[answer asap]the perimeter of the original rectangle on the left is 30 meters. the perimeter of the reduced rectangle on the right is 24 meters.[see image]what is x, the width of the original rectangle on the left? round to the nearest hundredth if necessary.

Answer:The width of the original rectangle on the left is [tex]5\ m[/tex]Step-by-step explanation:step 1Find the scale factorwe know thatIf two figures are similar, then the ratio of its perimeters is equal to the scale factorsoLetz -----> the scale factora ----> 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 2Find the width of the reduced rectangle on the right we know thatThe 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 3Find the width of the original rectangle on the leftwe know thatIf two figures are similar, then the ratio of its corresponding sides is proportional and this ratio is called the scale factorsoLetz -----> the scale factory ----> 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]