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The length and width of a rectangle are consecutive odd integers. The perimeter of the rectangle is 80 feet. Find the length and width of the rectangle.

" Width "=◻" feet "" Length "=◻" feet "

The length and width of a rectangle are consecutive odd integers. The perimeter of the rectangle is $80$ feet. Find the length and width of the rectangle. $\newline$ $\text { Width }=\square \text { feet } \text { Length }=\square \text { feet }$

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Volume of cubes and rectangular prisms: word problems

Full solution

Q. The length and width of a rectangle are consecutive odd integers. The perimeter of the rectangle is

80

feet. Find the length and width of the rectangle.

\newline

\text { Width }=\square \text { feet } \text { Length }=\square \text { feet }

Denote width and length: Let's denote the width of the rectangle as $W$ (an odd integer) and the length as $L$ (the next consecutive odd integer). Since they are consecutive odd integers, we can express the length as $L = W + 2$ . The perimeter ( $P$ ) of a rectangle is given by the formula $P = 2 \times (W + L)$ . We are given that the perimeter is $80$ feet. So, we can set up the equation: $80 = 2 \times (W + W + 2)$ .
Perimeter formula: Now, let's simplify the equation: $80 = 2 \times (2W + 2)$ . Divide both sides by $2$ to simplify further: $40 = 2W + 2$ .
Set up equation: Next, we subtract $2$ from both sides to solve for $W$ : $40 - 2 = 2W$ . This simplifies to $38 = 2W$ .
Simplify equation: Now, divide both sides by $2$ to find the value of $W$ : $\frac{38}{2} = W$ . $\newline$ This gives us $W = 19$ feet.
Solve for W: Since the length $L$ is the next consecutive odd integer after the width $W$ , we add $2$ to the width to find the length: $L = W + 2 = 19 + 2$ . This gives us $L = 21$ feet.
Find value of $W$ : Finally, let's check our work by calculating the perimeter with the found width and length: $P = 2 \times (W + L) = 2 \times (19 + 21)$ . This simplifies to $P = 2 \times 40 = 80$ feet, which matches the given perimeter.

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