Exercise 7

If the radius of a circle measures 6 centimeters (cm), then the perimeter of the square inscribed in the circle measures:

  1. \(12 \sqrt{2}\) cm

  2. 12 cm

  3. 24 cm

  4. 36 cm

  5. \(24 \sqrt{2}\) cm

Try to solve it before checking the answer.

  1. \(24 \sqrt{2}\) cm

The diagonal of the square is a diameter of the circumference, which measures 12 cm.

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:

\[ \begin{aligned} L^2 + L^2 = 12^2 &\Rightarrow 2 L^2 = 144 \\[.5em] &\Rightarrow L^2 = 72 \\[.5em] &\Rightarrow L = \sqrt{72} \\[.5em] &\hspace{2em} = \sqrt{3^2 \times 2^2 \times 2} \\[.5em] &\hspace{2em} = 3 \times 2 \times \sqrt{2} \\[.5em] &\hspace{2em} = 6 \sqrt{2} \end{aligned} \]

Then:

\[ \text{Perimeter} = 4L = 4 \left( 6 \sqrt{2} \right) = \boldsymbol{24 \sqrt{2}} \, cm \]

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