Find the area of the region $ABCD$.

Question

In the Figure $\square PQRS$ is a square with side $2\sqrt6$.

By joining the midpoints another square $\square WXYZ$ is formed .

Circles are drawn with $4$ vertices as the center and

radius equal to the side of the square $\square WXYZ$.

Find the area common to all the $4$ circles .

$a.)6\pi\left(\dfrac{\sqrt3-1}{2}\right)\\ b.)4\pi-3\sqrt3\\ c.)\dfrac12\left(\pi -3\sqrt3\right)\\ d.)4\pi-12\left(\sqrt3 -1\right)\quad \LARGE \color{green}{\checkmark}\\$

So i have to find the area of region $ABCD$.

I have found that $WX=WY=YZ=XZ=2\sqrt3.$

Let Region $WBA$=Region $XAD$=Region $ZCD$=Region $YBC=a$

and

Region $WAX$=Region $XDZ$=Region $ZCY$=Region $YBW=b$

and

Region $ABCD=m$

$Area(\square WXYZ)=12\\ m+4a+4b=12$.

Area of sectors.

$Area$(sector $WZY$)=$Area$(sector $XYZ$)=$Area$(sector $WXY$)=$Area$(sector $WZX$)=$m+3a+2b=3\pi$.

Now i m stucked.

I m looking for simple short way.

I have studied maths upto $12th$ grade.

Answer 1

Let $[F]$ be the area of a figure $F$.

Since $\angle{WYA}=30^\circ,\angle{AYZ}=60^\circ$, one has $$\begin{align}a+b&=[YWA]\\&=[\text{sector}\ YWA]-\left([\text{sector}\ ZYA]-[\triangle{ZYA}]\right)\\&=(2\sqrt 3)^2\pi\times\frac{30}{360}-\left((2\sqrt 3)^2\pi\times\frac{60}{360}-\frac{\sqrt 3}{4}\times(2\sqrt 3)^2\right)\\&=3\sqrt 3-\pi\end{align}$$ Hence, one has$$m=12-4(a+b)=12-4(3\sqrt 3-\pi)=4\pi+12-12\sqrt 3.$$

Answer 2

Each side of smaller square WXZY is calculated as $$WX=\sqrt{(WQ)^2+(QX)^2}=\sqrt{(\sqrt{6})^2+(\sqrt{6})^2}=2\sqrt{3}$$

The area common to four circles each having a radius $a$ & center at the vertice of a cube with side $a$ is given by the general expression (obtained by using integration) $$\bbox[4pt, border:1px solid blue;]{A_{common}=a^2\left(\frac{\pi-3(\sqrt{3}-1)}{3}\right)}$$ Hence, by setting $a=2\sqrt{3}$ in the above formula, we get the requited common area ABCD $$=(2\sqrt{3})^2\left(\frac{\pi-3(\sqrt{3}-1)}{3}\right)=4\left(\pi-3(\sqrt{3}-1)\right)=4\pi-12(\sqrt{3}-1)$$

Answer 3

Base on the following simple principle:-

$n(A \cap B) = n(A) + n(B) – n(A \cup B)$,

the said region can be found by a series a additions and subtrations of regions whose areas can easily be found. Example:-

For further example, see my answer in Find the area of the region which is the union of three circles