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A circle with a radius \(r\) is inscribed in a square with side \(s\). If the ratio of the area of the square to the area of the circle is \(m\), and the ratio of the perimeter of the square to that of the circle is \(n\), which of the following must be true?
A. \(\frac{m}{n} > 1\)
B. \(\frac{m}{n} = 1\)
C. \(1\gt \frac{m}{n} > \frac{1}{2}\)
D. \(\frac{m}{n} = \frac{1}{2}\)
E. \(\frac{m}{n} < \frac{1}{2}\)
A. \(\frac{m}{n} > 1\)
B. \(\frac{m}{n} = 1\)
C. \(1\gt \frac{m}{n} > \frac{1}{2}\)
D. \(\frac{m}{n} = \frac{1}{2}\)
E. \(\frac{m}{n} < \frac{1}{2}\)
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09 Jun 2023, 00:39
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Using a number-picking approach:
Assume \(s = 6\). For the inscribed circle, \(r = \frac{s}{2} = \frac{6}{2} = 3\). Consequently, the area of the square is \(6 * 6 = 36\) and the area of the circle is \(\pi*r^2 = 9\pi\). The ratio \(m\) is \(\frac{4}{\pi}\).
The perimeter of the square is \(4s=24\) and the perimeter of the circle is \(2\pi r = 6\pi\). The ratio \(n\) is \(\frac{4}{\pi}\).
Therefore, \(m = n\), so \(\frac{m}{n} = 1\).
Algebraic approach:
Using an algebraic approach:
"A circle with a radius \(r\) is inscribed in a square with side \(s\)" implies \(s = 2r\);
"The ratio of the area of the square to the area of the circle is \(m\)" can be written as \(m = \frac{s^2}{\pi{r^2} } = \frac{(2r)^2}{\pi{r^2} }=\frac{4}{\pi}\).
"The ratio of the perimeter of the square to that of the circle is \(n\)" can be stated as \(n = \frac{4s}{2\pi{r} }= \frac{4(2r)} { 2\pi{r} }=\frac{4}{\pi}\).
Therefore, \(m = n\), so \(\frac{m}{n} = 1\).
Answer: B
