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2006 CEMC Pascal Problems/Problem 9

Problem

In the diagram, the rectangle has a width of $w$, a length of $8$, and a perimeter of $24$. What is the ratio of its width to its length?

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$\text{ (A) }\ 1:4 \qquad\text{ (B) }\ 1:3 \qquad\text{ (C) }\ 1:2 \qquad\text{ (D) }\ 3:8 \qquad\text{ (E) }\ 2:3$

Solution 1

We can use the fact that for any rectangle, $p = l + l + w + w = 2l + 2w$, where $p$ is the perimeter of the rectangle, $l$ is the length of the rectangle, and $w$ is the width of the rectangle.

Substituting values we know into the equation, we get:

$24 = 2 \times 8 + 2w$

Rearranging this equation and evaluating $2 \times 8$, we get:

$2w + 16 = 24$

Solving this, we get:

$2w = 24 - 16 = 8$

$w = \frac{8}{2} = 4$

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Now, we can find the ratio of the rectangle's width to its length. This will be $w:l$, or $4:8$.

$8 = 4 \times 2$, so the simplified ratio is $\boxed {\textbf {(C) } 1:2}$.

~anabel.disher

Solution 2 (answer choices)

We can use each of the answer choices to solve the problem, and see which ones are correct.

For answer choice C, the ratio being $1:2$ means that $w = l \div 2 = 8 \div 2 = 4$, where $w$ is the width of the rectangle, and $l$ is the length of the rectangle.

Since $p = 2w + 2l$ for any rectangle, where $p$ is the perimeter of the rectangle and $w$ and $l$ are the same as defined above, we can plug-in $w$ and $l$ and see if it matches the perimeter.

$2 \times 4 + 2 \times 8 = p$

$p = 8 + 16$

$p = 24$

This matches our perimeter, so the simplified ratio must be $\boxed {\textbf {(C) } 1:2}$.

~anabel.disher

Solution 2.5 (answer choices)

Instead of seeing if the perimeter is equal to $24$, we can use the fact that $w$ would have to be equal to $l \div 2$ for $w:l$ to be $1:2$.

We can then use $p = 2w + 2l$, and plug-in $w = l \div 2$ and $p = 24$ to see if we get $w = 8$. This gives:

$24 = 2 \times l \div 2 + 2 \times l$

$2l + l = 24$

$3l = 24$

$l = 8$

Since we got $8$ for our solution (which is the same as the actual length given in the problem), the simplified ratio must be $\boxed {\textbf {(C) } 1:2}$.

~anabel.disher

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