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Answer :
Solution
The length of a rectangle is five times its width.
Let the length be represented by L
Let the width be represented by W
The length of the rectangle is five times the width i.e
[tex]L=5W[/tex]To find the perimeter, P, of a rectangle, the formula is
[tex]P=2(L+W)[/tex]Given that the perimeter, P, of the rectangle is 120ft,
Subsitute for length and width into the formula above
[tex]\begin{gathered} P=2(L+W) \\ 120=2(5W+W) \\ 120=2(6W) \\ 120=12W \\ \text{Divide both sides by 12} \\ \frac{12W}{12}=\frac{120}{12} \\ W=10ft \end{gathered}[/tex]Recall that, the length of the rectangle is
[tex]\begin{gathered} L=5W \\ L=5(10) \\ L=50ft \\ W=10ft \end{gathered}[/tex]To find the area, A, of a rectangle, the formula is
[tex]A=LW[/tex]Substitute the values of the length amd width into the formula above
[tex]\begin{gathered} A=(50)(10)=500ft^2 \\ A=500ft^2 \end{gathered}[/tex]Hence, the area of the rectangle is 500ft²
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