Cute Little Shuriken

$\Large \text{Assume that }n = \text{Circle segment area and}\space x+y+n = \text{Circle sector area}$

$\Large x + y = \text{ Sector Area }- \text{Segment Area}$

$\Large = ( x + y + n ) - n \\ \Large = \frac {30}{360} \times \pi \times 7^2 - ( \frac { 60}{360} \times \pi \times 7^2 - \frac {\sqrt{3}}{4} \times 7^2) \\ \Large = \frac{ 49}{12} \times \frac{22}{7} - ( \frac{49}{6} \times \frac{22}{7} - \frac{49}{4} \times \sqrt{3}) \\ \Large = \frac{77}{6} - \frac{77}{3} + \frac{49}{4} \sqrt{3}$

$\Large \text{Shaded Area} = \text{ Square Area} - 4 ( x+y) \\ \Large = 7^2 - 4(\frac{77}{6} - \frac{77}{3} + \frac{49}{4} \sqrt{3}) \\ \Large = 15.462... \\ \Large \approx 15.5$

$\text{ABCD is a square with sides a. O is the center. It is divided ito 4 equal squares.}\\ \text{A quarter of the flower is shown as area OMN.}\\ Let~ MR \bot CD. ~~~In~ rt. ~\Delta ~MCR, ~~2RM=Hyp.~MC. ~~\implies ~\angle MCR=30^o\\ \because ~of~ symmetry, ~\angle BCN=30^o. ~Clearly~\angle NCM=30^o.\\ \therefore ~\Delta ~ MNC ~is ~a-30^o-a.~~~~~~CR=\dfrac{\sqrt3} 2 a, ~~~\therefore ~\color{#3D99F6}{MO=\dfrac{\sqrt3} 2 a-\dfrac 1 2 a.}\\$ $Required ~area=4*\{area ~ CMN\}\\ =4*\{area ~ of ~sector ~ NCM- (areas~ of ~\Delta s~ MOC ~ and ~NOC )\}\\ =4*\{\pi*a^2*\dfrac{30^o}{360^o}- 2*\frac 1 2 *MO*\dfrac a 2\\ =\dfrac{ a^2} 3 (\pi +3-3\sqrt3. )\\ a=7\ and\ \pi=22/7, \ substituting \ we\ get\ area=15.4628. \\ \text{This is the copy of my solution of the problem 4 months ago.}$

The area of the shaded region is: $\frac{a^{2}}{3}$ $*(22/7+3-3\sqrt{3}$ ), where $a=7$ . We can draw a square inside the orange area whose side is equal to $a\sqrt{2-\sqrt{3}}$ by using Pythagoras's theorem. Hence, the area of the square is $a^{2}(2-\sqrt{3})$ . Also, the area between our square and the orange area is: $4(the\ area\ of\ the four\ circular\ sectors$ = $4(\frac{a^{2}}{12}$ ) $(\frac{22}{7}$ -3)).

Thus, total area of the orange part: $7^{2}(2-\sqrt{3})$ + $4(\frac{7^{2}}{12}$ ) $(\frac{22}{7}$ - 3) = $\frac{7^{2}}{3}$ $(22/7+3-3\sqrt{3}$ )= 15.5