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let us consider following picture

enter image description here

where $ABCD$ is square, and using $A$ and $C$ as center,there is drawn arc,we should find area of dark part.we know that length of square is $a$,as i see the parts of intersection or $ABK$ and $CBK$ are congruent or equal to each other,so in my opinion area of dark part is equal


because $ACD$ represent as quarter of circle with radius $a$ and it's area is $\pi*a^2/4$,by substraction i will find are of $ABC$ and by dividng by $2$,i will find $BKC$,then idea is substract area of $ABD+BKC$ from $a^2$,is it correct?please help me,answer is $(\sqrt{3}/4-\pi/12)*a^2$ where $\sqrt{3}$ comes from?thanks in advance

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up vote 2 down vote accepted

The triangle $\Delta AKD$ is equilateral. Thus, the area of the shape $AKD$ (with arcs $AK$ and $KD$ as edges) is equal to twice the area of $60^\circ$ sectors of a circle with radius $a$ minus the area of the triangle $AKD$. So area of the shape $AKD$ can be written as $2\pi a^2/6-a^2\sqrt{3}/4$. Subtracting this area from the area of the quarter circle $ACD$ gives us the answer $$\pi a^2/4-(\pi a^2/3-a^2\sqrt{3}/4)=a^2(\sqrt{3}/4-\pi/12), $$ as desired.

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sorry could you explain a little more detailed?why it is $AKD$ equilateral? – dato datuashvili Jul 15 '13 at 14:34
also what about $AK$ and $KD$ arc? – dato datuashvili Jul 15 '13 at 14:36
i can't sector of $60$,could you draw it seperately? – dato datuashvili Jul 15 '13 at 14:40
$AK$ is radius of the quarter circle $ABD$, so it is equal to $AD=a$. Similarly $KD=a$. So $AK=KD=AD =a$ and $AKD$ is equilateral. Now the area enclosed by the arcs $AK$, $DK$, and the side $AD$ is the area of the two $60^\circ$ sectors $DKA$ (centered at $D$) and $AKD$ (centered at $A$) minus the area of $equilateral triangle $AKD$. Basically, we count the area of the triangle twice when we add the area of the two sectors so we have to subtract it. – S.B. Jul 15 '13 at 14:44
but segments $Ak$ and $KD$ are equal right? – dato datuashvili Jul 15 '13 at 14:47

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