Square $ABCD$ has side equal to $a$. Points $A$ and $D$ are centers of two Quarter-Circles (see image below), which intersect at point K. Find the area defined by side $CD$ and arcs $KC$ and $KD$.
Here's what I did: The darkened area can be found by Substracting area of figure defined by points $AKD$ from quarter-circle $CAD$. Area of quarter-circle $= \dfrac{a^2\pi}{4}$. Now onto the harder part:
The way I calculated the Area of $AKD$ is by noticing that it's half of an elipse (at least I'm pretty sure it is). With $R1 = \dfrac{a}{2}$ (by symmetry) and $R2=\dfrac{a\sqrt3}{2}$ (by Pythagoras). The area of $AKD$ will be half of an ellipse: $\dfrac{R1R2\pi}{2} = \dfrac{\ a^2\sqrt3}{8}\pi$
The area of darkened figure will be the difference between two areas: $\dfrac{a^2\pi}{4} – \dfrac{\ a^2\sqrt3}{8}\pi $.
But my answer, for some reason, is way off. What am I doing wrong? Does $AKD$ not represent a semi-ellipse?
Best Answer
Find area $S$ first:
$$S=\frac16 a^2\pi-P_{\triangle ADK}$$
Area of ADK is:
$$P_{ADK}=2S+P_{\triangle ADK}=2(\frac16 a^2\pi-P_{\triangle ADK})+P_{\triangle ADK}=\frac13 a^2\pi-P_{\triangle ADK}$$
$$P_{ADK}=\frac13 a^2\pi-\frac14a^2\sqrt3$$
Shaded area is simply:
$$P_{shaded}=P_{ADC}-P_{ADK}=\frac14 a^2\pi-(\frac13 a^2\pi-\frac14a^2\sqrt3)$$
$$P_{shaded}=\frac14a^2\sqrt3-\frac1{12}a^2\pi=\frac1{12}a^2(3\sqrt3-\pi)$$