Beauty inside a Square : 2

pic ture

In the 1st picture

The area of the equilateral triangle is $\frac{4^2\sqrt{3}}{4} = 4\sqrt{3}$

The right pink shaded part is a part of a circle. As its internal angle is $30^\circ$ , its area is $\frac{\pi r^2}{12} = \frac{4^2\pi}{12} = \frac{4\pi}{3}$

So, total area of pink shaded parts $= 2 \times \frac{4\pi}{3} = \frac{8\pi}{3}$

Hence the area of the white part $=(16-4\sqrt{3}-\frac{8\pi}{3})$

In the 2nd picture

Total white part $=4(16-4\sqrt{3}-\frac{8\pi}{3})$

Hence, Area of shaded part $=16 - 4(16-4\sqrt{3}-\frac{8\pi}{3})$

$=16\sqrt{3}+\frac{32\pi}{3}-48$

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Consider the Region HEFG, the Area of HEFG can be expressed as $$ $= A_{ABCD} - 4\times A_{DGFCD}$ $= A_{ABCD} - 4\times \left (A_{AEFCDA} - (A_{AEFDA} + A_{DFGD} )\right)$ $= A_{ABCD} - 4\times \left (A_{AEFCDA} - (A_{AEFDA} +\left(A_{AFGDA}-A_{AFD}\right) )\right)$ $=r^2-4\left(\frac{\pi r^2}{4}-\left(\frac{\pi r^2}{6}+\left(\frac{\pi r^2}{6}-\frac{\sqrt{3}}{4}r^2\right)\right)\right)$ $r^2 - 4\left(\frac{\pi r^2}{4}-\frac{\pi r^2}{3}+\frac{\sqrt{3}}{4}r^2\right)=r^2 - 4\left(-\frac{\pi r^2}{12}+\frac{\sqrt{3}}{4}r^2\right)=r^2 + \frac{\pi r^2}{3}-\sqrt{3}r^2$

So

A_{HEFG} = r^2 + \frac{\pi r^2}{3}-\sqrt{3}r^2\tag1

Now,

$A_{AEBFCGDA} = 4\times A_{AEFCA} - A_{HEFG}= 4\times \left(A_{AEFCDA}-A_{ACDA}\right)- A_{HEFG}$ $= 4\times \left(\frac{\pi r^2}{4} - \frac{r^2}{2}\right)- A_{HEFG}= 4\times \left(\frac{\pi r^2}{4} - \frac{r^2}{2}\right)- A_{HEFG}= 4\times \left(\frac{\pi r^2}{4} - \frac{r^2}{2}\right)- A_{HEFG}$ $=\pi r^2 -2r^2-A_{HEFG}$

$=\pi r^2 -2r^2- r^2 - \frac{\pi r^2}{3}+\sqrt{3}r^2$ $=\left(\sqrt{3}+\frac{2\pi}{3} -3\right)r^2$

My solution was exactly the same as Fahim's. This was one of my favourite geometry problems from high school.

This can be solved with analytic geometry and calculus as follows. If we consider the box to be [0, 4]x[0,4] in R^2, the shaded area is the area of the total square (which is 4x4 = 16) minus 8 times the area bounded by y = 4 and y = (16-x^2)^.5 between x = 0 and x = 2 (by symmetry). The latter area can be computed as the integral from x = 0 to x = 2 of f(x) = 4-(16-x^2)^.5. This integral can be computed via the trigonometric substitution x = 4*sin(theta). Simplifying yields the desired answer.

Find Area Shaded into the square with side $L = r = 4$

If we denote with:

$A,\, B,\, C,\, D\,$ the vertices of the square clockwise from bottom left to top left;

$E,\, F,\, G,\, H,\,$ points of intersections between the circles clockwise from top center to right center;

$E_1,\, F_1,\, G_1,\, H_1,\,$ the projections of $E,\, F,\, G,\, H,\,$ on the closest side of

the square, we have:

$A_{sh} = A_s - 8A(E1DE);$

we denote:

$A_{sh}\,$ :=Area Shaded;

$A_s\,$ := Area Square = $4^2 = 16;$

$E_1DE\,:\,$ the surface between perpendicular segments $\overline {E_1D},\, \overline {EE_1}\,$ and $\stackrel \frown {DE}$

we have:

$A(E_1DE) = A(AG_1E_1D) - A(AG_1ED) - A(AG_1E);$

$\angle {DAE} = \frac \pi 6\,$ , because $|\overline{AE}| = |\overline{BE}| = |\overline{AB}| = 4,\,$ and $\triangle {(ABE)}\,$ is equilater, so $\angle {BAE} = \frac \pi 3,$

and $\angle {DAE} = {\theta}_0 = \frac \pi 2 - \frac \pi 3 = \frac \pi 6;\,$ thus:

$A(AG_1ED) = {\frac 1 2}{\theta}_0r^2 = {\frac 1 2}\frac \pi 6 \times {16} = \frac 4 3 \pi$

$A(AG1E1D) = 2 \times 4 = 8;$

$A(AG_1E) = {\frac 1 2}|\overline{AG_1}||\overline{G_1E}| = {\frac 1 2}{\times 2{\times 4 {\times {\sin {(\frac \pi 3)}}}}} = 4{\frac {\sqrt 3} 2} = 2\sqrt 3;$

$A(E_1DE) = 8 - 2\sqrt 3 - {\frac 4 3}\pi;$

$A_{sh} = 16 - 8(8 - 2\sqrt 3 - \frac 4 3 \pi);$

$A_{sh} = 16\sqrt 3 + 32\frac \pi 3 - 48 = 13.22313456...$

13 square unit or 13.20 square unit

area =integral {4-sqrt(16-x^2)} 0 to 2 = 16-4 area 2