Inscribed Squares

Square \(ABCD\) is inscribed in a circle with center at \(O,\) as shown in the figure. If the area of the shaded region is \(25\pi -50\), find the area of the square.

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SOLUTION

Let \(r\) be the radius of the circle, and \(x\) the side length of the square, then the area of the square is \(x^2\). By the Pythagorean theorem, we have \((2r)^2=x^2+x^2.\)

Simplifying further, we get \(x^2=2r^2.\qquad (1)\)

The area of the shaded region is equal to the area of the circle minus the area of the square, so we have

\[\begin{align} 25\pi -50 &=\pi r^2 - 2r^2\\ &=r^2(\pi-2)\\ r^2&=\dfrac{25\pi -50}{\pi -2}\\ &=25.\qquad (2) \end{align}\]

Now substituting (2) into (1) gives \(x^2=2\times 25=50.\ _\square \)

A square with side length \(a\) is inscribed in a circle. Express the radius of the circle in terms of \(a\).

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SOLUTION

The diagonal of the square is the diameter of the circle. Let \(d\) and \(r\) be the diameter and radius of the circle, respectively. Then by the Pythagorean theorem, we have

\[\begin{align} d^2&=a^2+a^2\\ &=2a^2\\ d&=\sqrt{2a^2}\\ &=a\sqrt{2}. \end{align}\]

We know that the diameter is twice the radius, so

\[r=\dfrac{d}{2}=\dfrac{a\sqrt{2}}{2}.\ _\square \]

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A square of perimeter \(16\) is inscribed in a semicircle, as shown.

Find the perimeter of the semicircle rounded to the nearest integer.

Use \(\frac{22}{7}\) for the approximation of \(\pi\).

\[\dfrac{\pi}{4}\big(16+8\sqrt{3}\big)-8+4\sqrt{3}\] \[8+4\sqrt{3}+\pi\big(2\sqrt{3}+2\big)^2\] \[4\pi+2\pi\sqrt{3}-8-4\sqrt{3}\] \[\pi\big(2\sqrt{3}+2\big)-4\sqrt{3}\]

The green square in the diagram is symmetrically placed at the center of the circle. Four red equilateral triangles are drawn such that square \(ABCD\) is formed. Let \(y,b,g,\) and \(r\) be the areas of the yellow, blue, green, and red regions, respectively.

If \(r=4\sqrt{3}\), find \(y+g-b\).