# Chapter 3 - Equations and Problem Solving - 3.5 - Problem Solving - Problem Set 3.5: 6

270 square feet

#### Work Step by Step

Since the width is stated in terms of the length, we can let l represent the length. The width is 1 more than one-third of the length, which is l, so the width is 1 + $\frac{1}{3}$l. The formula for the perimeter of the rectangle is width + width + length + length The perimeter of the rectangle is 74, so we set up the following equation: l + l + 1 + $\frac{1}{3}$l + 1 + $\frac{1}{3}$l = 74 2l + 2 + $\frac{2}{3}$l = 74 $\frac{6l}{3}$ + 2 + $\frac{2}{3}$l = 74 Subtract 2 from both sides. $\frac{6}{3}$l + $\frac{2}{3}$l = 72 $\frac{8}{3}$l = 72 Divide both sides by $\frac{8}{3}$ l = 72 $\div$ $\frac{8}{3}$ l = 72 $\times$ $\frac{3}{8}$ l = 27 Since the length is 27, the width is 1 + $\frac{1}{3}$ $\times$ 27 = 10. Area= $l \times w$, so we obtain: $A=27 \times 10 = 270$ square feet.

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