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If a square is inscribed in a semicircle of radius r and the square has an area of 8 square units, find the area of a square inscribed in a circle of radius r.
I started by assuming that the side of the square is 2(root2). But I did not know how this relates to what it's dimensions were to be if it was inscribed in a full circle. Could someone help? Thank you.
Given the small inscribed square has an area of $8$, so a side length of $2\sqrt2$ The radius $r$ of the semicircle and circle is equal to the distance between the midpoint of the bottom side of the small inscribed square and one of the top vertices. This forms a right triangle with side lengths $2\sqrt2$, $\sqrt2$, and hypotenuse $r$. Using the Pythagorean theorem $$(2\sqrt2)^2+(\sqrt2)^2=r^2$$ $$8+2=r^2$$ $$r=\sqrt{10}$$ Now that we have the radius of the circle, we know that the side length of the large inscribed square is $\frac{2r}{\sqrt2} = r\sqrt2$, ($2r$ is the diagonal of the large inscribed square, also the diameter of circle). The side length of the large inscribed square is $\sqrt{20}=2\sqrt5$, so its area is $20$ square units.
