Area Of Overlapping Leaves

Solution:

I will divide this problem in two parts. First one, find the gray area, second one, find the yellow area. Gray area + yellow area is the solution.

Gray area:

Let's concentrate on point $Q$ (or: $M$ , $N$ or $P$ , conclusion will be the same). I believe, that it can be spotted that $BQ$ and $AQ$ 's length is $a$ . Therefore, triangle $ABQ$ is equilateral. Height of that triangle is $\frac { a\sqrt { 3 } }{ 2 }$ . Since the distance between point $Q$ and segment $CD$ is as same as the distance between point $N$ and segment $AB$ , we can calculate that $QN = a( \sqrt{3} - 1)$ . $QN = MP$ and using Pythagoras's theorem we get that the lenght of $\square MNPQ$ 's side is $a \sqrt{2 - \sqrt{3}}$ . Therefore, the smaller square's area is $a^{2} (2 - \sqrt{3})$ .

Yellow area:

$\angle NCM = 30°$ , which is $\frac{360°}{12}$ , so the area of circular sector is $\frac{a^{2} \pi}{12}$ . If we find the area of triangle $NCM$ then we only need to subtract its area from the area of sector $NCM$ and that will give us 1st part of yellow area. We will use Pythagoras's theoreme once more to find the height of triangle $NCM$ . Height $^{2}$ = $MC^{2} - (\frac{MN}{2})^{2}$ and $Height = \frac{a}{2} \sqrt{2 + \sqrt{3}}$ . So, triangle's area is $\frac{NM}{2} \times Height = \frac{a^{2}}{4}$ . Then, the $\frac{1}{4}$ of the yellow area is $\frac{a^{2}}{12}(\pi - 3)$ . It is obvious, that the yellow area is $\frac{a^{2}}{3}(\pi - 3)$ .

Conclusion: Gray area + yellow area => $a^{2} (2 - \sqrt{3}) + \frac{a^{2}}{3}(\pi - 3) = \frac{a^{2}}{3}(\pi + 3 - 3\sqrt{3})$

$\text{ABCD is a square with sides a. O is the center. It is divided ito 4 equal squares.}\\ \text{A quarter of the flower is shown as area OMN.}\\ Let~ MR \bot CD. ~~~In~ rt. ~\Delta ~MCR, ~~2RM=Hyp.~MC. ~~\implies ~\angle MCR=30^o\\ \because ~of~ symmetry, ~\angle BCN=30^o. ~Clearly~\angle NCM=30^o.\\ \therefore ~\Delta ~ MNC ~is ~a-30^o-a.~~~~~~CR=\dfrac{\sqrt3} 2 a, ~~~\therefore ~\color{#3D99F6}{MO=\dfrac{\sqrt3} 2 a-\dfrac 1 2 a.}\\$ $Required ~area=4*\{area ~ CMN\}\\ =4*\{area ~ of ~sector ~ NCM- (areas~ of ~\Delta s~ MOC ~ and ~NOC )\}\\ =4*\{\pi*a^2*\dfrac{30^o}{360^o}- 2*\frac 1 2 *MO*\dfrac a 2\\ \Huge =\dfrac{ a^2} 3 (\pi +3-3\sqrt3. )$

First, let the area of the square is $\large{A=a^2}$

$\large{\text{Picture 1}}$ .

The blue area is a quilateral with side lenght $a$ . Let the area of the blue area is $x$ , then

$\large{x=\frac{1}{2}\times{a^2}\times{\sin60^{\circ}}=\frac{a^2\sqrt{3}}{4}}$

The yellow area is a sector of a circle with radius $a$ and an angel $30^{\circ}$ . Let the area of the yellow area is $y$ , then

$\large{y=\frac{30^{\circ}}{360^{\circ}}\times{\pi}a^2=\frac{{\pi}a^2}{12}}$

So, we have the area of the white area-let the area of the white area is $z$ - namely:

$\large{z=A-x-2y=\frac{a^2}{12}\left(12-3\sqrt{3}-2\pi\right)}$

$\large{\text{Picture 2}}$ .

Let the area of the gray shaded area is $u$ , then

$\large{u=A-\frac{90^{\circ}}{360^{\circ}}\times{\pi}a^2-2z=\frac{a^2}{12}\left(-12+6\sqrt{3}+\pi\right)}$

$\large{\text{Picture 3}}$ .

Thus, the area of the green area is

$\large{A-4z-4u=\frac{a^2}{3}\left(\pi+3-3\sqrt{3}\right)}_{\square}$