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User:Anabel.disher/Sandbox/Prob 7

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Problem

The number $0.2012$ is between

$\text{ (A) }\ 0 \text{ and } \frac{1}{10} \qquad\text{ (B) }\ \frac{1}{10} \text{ and } \frac{1}{5} \qquad\text{ (C) }\ \frac{1}{5} \text{ and } \frac{1}{4} \qquad\text{ (D) }\ \frac{1}{4} \text{ and } \frac{1}{3} \qquad\text{ (E) }\ \frac{1}{3} \text{ and } \frac{1}{2}$

Solution 1

The area of a rectangle is its width multiplied by its height, so the area would be $25 \times 9 \text{cm}^{2}$.

The area of a square is its side length squared, so we can let $s$ be the side length of the square. Since the area of the rectangle is the same as the area of the square, we have:

$s^{2} = 25 \times 9 = 225$

Taking the square root of both sides to remove the squared value, we get:

$s = 15$

Thus, the dimensions of the square are $\boxed {\textbf{(A) } 15 \text{cm by} 15 \text{cm}}$

~anabel.disher

Solution 2

Instead of solving for $25 \times 9$, we can notice that $\sqrt{25 \times 9} = \sqrt{25} \times \sqrt{9}$, so:

$s^2 = 25 \times 9$

Taking the square root of both sides gives:

$s = \sqrt{25 \times 9} = \sqrt{25} \times{9} = 5 \times 3 = 15$

This gives us $\boxed {\textbf{(A) } 15 \text{cm by} 15 \text{cm}}$ as the answer, like in solution 1.

~anabel.disher

Solution 3 (answer choices)

We can square the side lengths of the squares found in the answer choices to see if they are too high or too low, and if they are equal to the area of the rectangle or not.

$25 \times 9$ = $225$, which is the area of the rectangle

$16 \times 16 = 256$, which is too high, so the answer cannot be C or D.

$8 \times 8 = 64$, which is too low, so the answer cannot be B.

Thus, the dimensions of the square are $\boxed {\textbf{(A) } 15 \text{cm by} 15 \text{cm}}$

~anabel.disher

Solution 4 (answer choices again)

We can notice that $25 \times 9 = 225$, which is a multiple of $5$. Since the area of the rectangle and the square are equal and the area of the square is the side length of the square squared, this means that the side length of the square must also be a multiple of $5$.

The only answer choice with a side length divisible by $5$ is $\boxed {\textbf{(A) } 15 \text{cm by} 15 \text{cm}}$.

~anabel.disher

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