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Difference between revisions of "2009 CEMC Gauss (Grade 8) Problems/Problem 20"

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{{CEMC box|year=2009|competition=Gauss (Grade 8)|num-b=19|num-a=21}}

Latest revision as of 21:30, 19 October 2025

Problem

A piece of string fits exactly once around the perimeter of a square whose area is $144$. Rounded to the nearest whole number, the area of the largest circle that can be formed from the piece of string is

$\text{ (A) }\ 144 \qquad\text{ (B) }\ 733 \qquad\text{ (C) }\ 113 \qquad\text{ (D) }\ 452 \qquad\text{ (E) }\ 183$

Solution

The area of a square is its side length squared. If $s$ is the side length of the square, we can then find it using an equation:

$s^{2} = 144$

$s = \sqrt{144} = 12$

We now want to find out what the radius is of a circle with the same perimeter as the square, since the same string will be used to make the circle.

The perimeter of a shape is the sum of the shape's side lengths. Since this is a square, all four of its side lengths are the same, and the perimeter is four times the side length of the square:

$12 \times 4 = 48$

We can now set up an equation involving the radius of the circle using its circumference:

$2 \times \pi \times r = 48$

$r = \frac{48}{2\pi} = \frac{24}{\pi}$

Using this radius, we can now find the area of the circle:

$\pi \times r^{2} = \pi \times (\frac{24}{\pi})^{2} = \pi \times \frac{24^2}{\pi^2}$

$=\frac{576}{\pi}$

Rounding this to the nearest whole number, we get $\boxed {\textbf {(E) } 183}$.

~anabel.disher

2009 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
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Problem 21
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CEMC Gauss (Grade 8)
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