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The length and width of a rectangle are consecutive odd integers. The perimeter of the rectangle is 48 centimeters. Find the length and width of the rectangle.

" Width "=◻" centimeters "quad" Length "=◻" centimeters "

The length and width of a rectangle are consecutive odd integers. The perimeter of the rectangle is $48$ centimeters. Find the length and width of the rectangle. $\newline$ $\text { Width }=\square \text { centimeters } \quad \text { Length }=\square \text { centimeters }$

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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

48

centimeters. Find the length and width of the rectangle.

\newline

\text { Width }=\square \text { centimeters } \quad \text { Length }=\square \text { centimeters }

Denote Width as W: Let's denote the width of the rectangle as $W$ (an odd integer). Since the length and width are consecutive odd integers, the length will be $W + 2$ (the next odd integer).
Perimeter Formula: The perimeter $P$ of a rectangle is given by the formula $P = 2 \times (\text{length} + \text{width})$ . We know the perimeter is $48$ cm, so we can set up the equation: $48 = 2 \times (W + W + 2)$ .
Simplify Equation: Simplify the equation from Step $2$ . This gives us $48 = 2 \times (2W + 2)$ , which simplifies further to $48 = 4W + 4$ .
Isolate Term with W: Subtract $4$ from both sides of the equation to isolate the term with $W$ . This gives us $44 = 4W$ .
Solve for W: Divide both sides of the equation by $4$ to solve for $W$ . This gives us $W = \frac{44}{4}$ , which simplifies to $W = 11$ .
Calculate Length: Since $W$ is the width and is $11$ cm, the length ( $L$ ) will be $W + 2$ . So $L = 11 + 2$ , which gives us $L = 13$ cm.

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