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The length and width of a rectangle are consecutive integers. The perimeter of the rectangle is 34 centimeters. Find the length and width of the rectangle.

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

The length and width of a rectangle are consecutive integers. The perimeter of the rectangle is $34$ 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 integers. The perimeter of the rectangle is

34

centimeters. Find the length and width of the rectangle.

\newline

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

Denote Width and Length: Let's denote the width of the rectangle as $w$ and the length as $w + 1$ since they are consecutive integers. The perimeter of a rectangle is given by the formula $P = 2l + 2w$ , where $l$ is the length and $w$ is the width. We are given that the perimeter $P$ is $34$ centimeters.
Set Up Perimeter Equation: We can set up the equation for the perimeter using the expressions for the width and length: $\newline$ $34 = 2(w + 1) + 2w$
Simplify and Solve: Now, let's simplify and solve for $w$ : $\newline$ $34 = 2w + 2 + 2w$ $\newline$ $34 = 4w + 2$
Isolate Term with w: Subtract $2$ from both sides to isolate the term with $w$ : $\newline$ $34 - 2 = 4w$ $\newline$ $32 = 4w$
Find Width: Divide both sides by $4$ to solve for $w$ : $\newline$ $32 / 4 = w$ $\newline$ $8 = w$ $\newline$ So, the width of the rectangle is $8$ centimeters.
Find Length: Since the length is one more than the width, we can find the length by adding $1$ to the width: $\newline$ Length $= w + 1$ $\newline$ Length $= 8 + 1$ $\newline$ Length $= 9$ $\newline$ So, the length of the rectangle is $9$ centimeters.

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