Answer:
The area of rectangle is [tex]2\frac{80}{81}\ cm^2[/tex]
Step-by-step explanation:
we know that
The perimeter of a rectangle is equal to
[tex]P=2(L+W)[/tex]
we have
[tex]P=7\frac{1}{3}\ cm=7+\frac{1}{3}=\frac{22}{3}\ cm[/tex]
so
[tex]\frac{22}{3}=2(L+W)[/tex]
[tex]\frac{11}{3}=(L+W)[/tex] ----> equation A
Remember that
The length of a rectangle is twice its width.
so
[tex]L=2W[/tex] ----> equation B
substitute equation B in equation A
[tex]\frac{11}{3}=(2W+W)[/tex]
solve for W
[tex]W=\frac{11}{9}\ cm[/tex]
Find the value of L
[tex]L=2(\frac{11}{9})[/tex]
[tex]L=\frac{22}{9}\ cm[/tex]
Find the area of rectangle
[tex]A=LW[/tex]
substitute the values
[tex]A=(\frac{22}{9})(\frac{11}{9})[/tex]
[tex]A=\frac{242}{81}\ cm^2[/tex]
Convert to mixed number
[tex]\frac{242}{81}\ cm^2=\frac{162}{81}+\frac{80}{81}=2\frac{80}{81}\ cm^2[/tex]
