Provided with only the radius of the semicircle (10 cm) and the knowledge that the corners of the square touch the semicircle, how can one find the area of this square?
[Math] How to find the area of a square inside a semicircle using only the radius
circlesgeometry
Related Solutions
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.
N.B. $\operatorname{cs}(ACD)$ is the circle segment defined by the rays, $AC$ and $AD$.
In the piture below, we have that $$\operatorname{cs}(ACD)+\operatorname{cs}(BCD)$$ contains the part were looking for. However we have overcounted. Can you see by how much we overcounted (i.e. what we now need to substract)?
Best Answer
Join the vertices lying on the boundary of the semicircle with it's center.
Now the hypotenuse of the the 2 right triangles formed will be radius to the circle and it's length is $\frac{a}{2}\sqrt5$ (Where a is the length of the square). So we can say
$$r=\frac{a}{2}\sqrt5 \to a=2 \frac{r}{\sqrt5}$$
Hence the area is:
$$A= a^2= \frac{4r^2}{5} = {400 \over 5}=80\,\,\,cm^{2}$$
Hope this helps!!