Как найти площадь тени квадрата

Let $a$ be a side of the square. Consider the following diagram

The area we need to calculate is as follows.
$$begin{eqnarray} color{Black}{text{Black}}=(color{blue}{text{Blue}}+color{black}{text{Black}})-color{blue}{text{Blue}}. end{eqnarray}$$

Note that the blue area can be calculated as
$$begin{eqnarray}color{blue}{text{Blue}}=frac14a^2pi-2cdotleft(color{orange}{text{Yellow}}+color{red}{text{Red}}right).end{eqnarray}$$

We already know most of the lengths. What’s stopping us from calculating the black area is lack of known angles. Because of symmetry, almost any angle would do the trick.

It’s fairly easy to calculate angles of triangle $begin{eqnarray}color{orange}{triangle POA}end{eqnarray}$, if we use cosine rule.

$$begin{eqnarray}
|PA|^2&=&|AO|^2+|PO|^2-2cdot|AO|cdot|PO|cosangle POA\
a^2&=&frac{a^2}{4}+frac{2a^2}{4}-2cdotfrac a2cdotfrac{asqrt2}{2}cdotcosangle POA\
4a^2&=&3a^2-2a^2sqrt2cosangle POA\
1&=&-2sqrt2cosangle POA\
cosangle POA&=&-frac{1}{2sqrt2}=-frac{sqrt2}{4}.
end{eqnarray}$$

Now, because of symmetry, we have $angle POA=angle POB$, so $angle AOB=360^circ-2angle POA$. So the cosine of angle $angle AOB$ can be calculated as follows:
$$begin{eqnarray}
cosangle AOB&=&cos(360^circ-2angle POA)=cos(2pi-2angle POA)\
cosangle AOB&=&cos(-2angle POA)=cos(2angle POA)\
cosangle AOB&=&cos^2(angle POA)-sin^2(angle POA)\
cosangle AOB&=&cos^2(angle POA)-(1-cos^2(angle POA))\
cosangle AOB&=&2cos^2(angle POA)-1\
cosangle AOB&=&2cdotleft(-frac{sqrt2}{4}right)^2-1=-frac34\
end{eqnarray}$$

From this, we can easily calculate the sine of angle $angle AOB$, using Pythagorean identity.
$$ sinangle AOB=sqrt{1-frac9{16}}=sqrtfrac{16-9}{16}=frac{sqrt7}4 $$

Going this way, I believe it’s not hard to calculate other angles and use known trigonometry-like formulas for area. Then you can easily pack it together using the first equation with colors.

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