Answer:
[tex] \boxed{ \sf{length \: of \: a \: rectangle = 21 \: in.}}[/tex]
[tex] \boxed{ \sf{width \: of \: a \: rectangle = 10.5 \: in.}}[/tex]
Step-by-step explanation:
Let the length of a rectangle be ' l '
Width of a rectangle ( w ) = [tex] \sf{ \frac{1}{2} l}[/tex]
Perimeter of a rectangle ( P ) = 63 in.
First, finding the length of a rectangle ( l )
[tex] \boxed{ \sf{perimeter \: of \: a \: rectangle = 2(l + b)}} \:[/tex]
[tex] \dashrightarrow{ \sf{63 = 2(l + \frac{1}{2} l}})[/tex]
Distribute 2 through the parentheses
[tex] \dashrightarrow{ \sf{63 = 2l + 2 \times \frac{1}{2} l}}[/tex]
[tex] \dashrightarrow{ \sf{63 = 2l + l}}[/tex]
Add the like terms : 2l and 3l
[tex] \dashrightarrow{ \sf{63 = 3l}}[/tex]
Swap the sides of the equation
[tex] \dashrightarrow {\sf{3l = 63}}[/tex]
Divide both sides by 3
[tex] \dashrightarrow{ \sf{ \frac{3l}{3} = \frac{63}{3}}} [/tex]
Calculate
[tex] \dashrightarrow{ \sf{l = 21 \: in.}}[/tex]
Length of a rectangle = 21 in.
Now, replacing / substituting the value of l in 1/2 l in order to find the width of a rectangle
[tex] \dashrightarrow{ \sf{width = \frac{1}{2} l}}[/tex]
[tex] \dashrightarrow{ \sf{width = \frac{1}{2} \times 21}}[/tex]
[tex] \dashrightarrow{ \sf{width = 10.5 \: in.}}[/tex]
Width of a rectangle = 10.5 in.
Hope I helped!
Best regards! :D
