Math Reasonings Solved Exercises
Exercise 7
If the radius of a circle measures 6, what is the perimeter of the square inscribed in such circle?
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\(12 \sqrt{2}\)
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12
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24
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36
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\(24 \sqrt{2}\)
Try to solve it before checking the answer.
\(24 \sqrt{2}\)
The diagonal of the square is a diameter of the circumference, which measures 12.
If \(L\) is the side of the square, then, by the Pythagorean theorem, we have:
\[ \begin{aligned} L^2 + L^2 = 12^2 &\Rightarrow 2 L^2 = 144 \\[.5em] &\Rightarrow L^2 = 72 \\[.5em] &\Rightarrow L = \sqrt{72} \\[.5em] &\hspace{2em} = \sqrt{3^2 \times 2^2 \times 2} \\[.5em] &\hspace{2em} = 3 \times 2 \times \sqrt{2} \\[.5em] &\hspace{2em} = 6 \sqrt{2} \end{aligned} \]
If \(L\) is the side of the square, then, by the Pythagorean theorem, we have:
Then:
\[ \begin{aligned} L^2 + L^2 = 12^2 &\Rightarrow 2 L^2 = 144 \\[.5em] &\Rightarrow L^2 = 72 \\[.5em] &\Rightarrow L = \sqrt{72} \\[.5em] &\hspace{2em} = \sqrt{3^2 \times 2^2 \times 2} \\[.5em] &\hspace{2em} = 3 \times 2 \times \sqrt{2} \\[.5em] &\hspace{2em} = 6 \sqrt{2} \end{aligned} \]
\[ \text{Perimeter} = 4L = 4 \left( 6 \sqrt{2} \right) = \boldsymbol{24 \sqrt{2}} \, \]