User:Anabel.disher/Sandbox/Prob 7

< User:Anabel.disher‎ | Sandbox
Revision as of 13:44, 18 June 2025 by Anabel.disher (talk | contribs) (Created page with "==Problem== The number <math>0.2012</math> is between <math> \text{ (A) }\ 0 \text{ and } \frac{1}{10} \qquad\text{ (B) }\ \frac{1}{10} \text{ and } \frac{1}{5} \qquad\te...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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