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Don't let the simplicity of this diagram fool you. I have been wondering about this for quite some time, but I can't think of an easy/smart way of finding it.
Any ideas?
For reference, the Area is:
$$\bbox[10pt, border:2pt solid grey]{90−18.75\pi−25\cdot \arctan\left(\frac 12\right)}$$
We observe that $\triangle PRT$ can be partitioned into five congruent sub-triangles. Therefore, the entire shaded region has area given by ... $$\begin{align} 3 u + |\text{region}\; PAT| &= 3u + |\square OAPT| - |\text{sector}\;OAT| \\[6pt] &= 3u + \frac{3}{5}\,|\triangle PRT| - |\text{sector}\;OAT| \\[6pt] &= 3\cdot\frac{1}{4} r^2 \left( 4 - \pi \right) \;+\; \frac{3}{5}\cdot r^2 \;-\; \frac{1}{2}r^2\cdot 2\theta \end{align}$$ Since $\theta = \operatorname{atan}\frac{1}{2}$, this becomes
$$r^2\left(\; \frac{18}{5} - \frac{3}{4}\pi - \operatorname{atan}\frac{1}{2} \;\right) \qquad\stackrel{r=5}{\to}\qquad 90 - \frac{75}{4}\pi - 25\;\operatorname{atan}\frac{1}{2}$$
[Note: My second answer is much better.]
I'll focus on the unshaded region at the bottom-left.
By an aspect of the Inscribed Angle Theorem, we know that $\angle AOB = 2\;\angle ABP$ (justifying marking these $\theta$ and $2\theta$). By a related result, we have that $$\phi = \frac{1}{2}\left(\angle BOC - \angle AOB\right) = 45^\circ - \theta$$ Moreover, we know that $$\phi = \operatorname{atan}\frac{1}{2} \approx 26.56^\circ \qquad\to\qquad \theta = 45^\circ - \operatorname{atan}\frac{1}{2} \approx 18.43^\circ$$
From here, knowing the circle's radius, one may calculate the lower-left area as ... $$\begin{align} &|\triangle PAB| + |\triangle OAB| - |\text{sector } OAB| \\ \end{align}$$ ... from which we readily derive the area in the original question. For now, I'll leave these details to the reader.
Put $\arctan{1\over2}=:\alpha$. Then $$\sin(2\alpha)={2\tan\alpha\over1+\tan^2\alpha}={4\over5}\ .$$ The area $A$ in question consists of three "arrow heads" plus the area shaded in the following figure. The latter is a right triangle minus a sector and a smaller triangle. We therefore obtain $$A={3\over4}(10^2- 25\pi)+{25\over2}\bigl(2-2\alpha-\sin(2\alpha)\bigr)=90-{75\over4}\pi-25\alpha\ .$$
The big black roundy-corner on the bottom right has area $(10^2 - \pi\cdot 5^2)/4$ and there are 3 complete copies of it and one copy trimmed by a small roundy-triangle. We will focus on this roundy-triangle which is the same as the one in the bottom left.
So the key is to compute the area of the small white roundy-triangle at the bottom left.
To this end we must find the intersection of the diagonal and the circle that create the top of this triangle.
The equation of the circle and of the diagonal are $$ y = 1/2 x \\ (x-5)^2 + (y-5)^2 = 25 $$ Solving that with WA gives : $x=2,\ y=1$.
So now we can decompose this roundy-triangle into two parts by drawing a vertical line that goes through this intersection. This gives a true triangle (the left part) which has area $1$ and another roundy-triangle (the right part).
To compute the area of the roundy-right-triangle, we can use integration : $$ \int_2^5 -\sqrt{25 - (x-5)^2} + 5\ dx $$ See WA for the plot of this function.
Considering the right triangle with legs $1-s\sin(\theta)$ and $1-s\cos(\theta)$, we have $$ 1=(1-s\cos(\theta))^2+(1-s\sin(\theta))^2\tag{1} $$ Solving for $s$, we get $$ s=\sin(\theta)+\cos(\theta)-\sqrt{2\sin(\theta)\cos(\theta)}\tag{2} $$ Setting $t=\tan(\theta)$ yields $$ \begin{align} s\sin(\theta)&=\frac{t}{1+t^2}\left(t-\sqrt{2t}+1\right)\\ s\cos(\theta)&=\frac{1}{1+t^2}\left(t-\sqrt{2t}+1\right)\\ \tan(\phi/2)&=\frac{1-s\cos(\theta)}{2-s\sin(\theta)} \end{align}\tag{3} $$ Setting $\tan(\theta)=\frac12$ gives $$ \begin{align} s\sin(\theta)&=\frac15\\ s\cos(\theta)&=\frac25\\ \tan(\phi/2)&=\frac13 \end{align}\tag{4} $$ The area of the green piece is $$ \frac\phi2-\frac12(1-s\cos(\theta))=\tan^{-1}\left(\frac13\right)-\frac3{10}\tag{5} $$ The sum of the areas of the purple and green pieces is $$ \frac12s\sin(\theta)=\frac1{10}\tag{6} $$ Therefore, the area of the purple piece is $$ \frac25-\tan^{-1}\left(\frac13\right)\tag{7} $$
$(7)$ used a circle with radius $1$. For a circle with radius $5$, we get an area of $$ 10-25\tan^{-1}\left(\frac13\right)\tag{8} $$ The area of half the $10\times20$ rectangle minus the two circles of radius $5$ is $$ 100-25\pi\tag{9} $$ Therefore, the area in the given image is the difference of $(9)$ and $(8)$: $$ \bbox[5px,border:2px solid #C0A000]{90+25\tan^{-1}\left(\frac13\right)-25\pi}\tag{10} $$
$(10)$ is a different form of the same area: $$ \begin{align} 90+25\tan^{-1}\left(\frac13\right)-25\pi &=90+25\tan^{-1}\left(\frac13\right)-25\tan^{-1}\left(1\right)-\frac{75}4\pi\\ &=90+25\tan^{-1}\left(\frac{\frac13-1}{1+\frac13\cdot1}\right)-\frac{75}4\pi\\ &=\bbox[5px,border:2px solid #C0A000]{90-25\tan^{-1}\left(\frac12\right)-\frac{75}4\pi}\tag{11} \end{align} $$
Well, I'm assuming you have problems only with the upper right corner, since the rest is just $\frac 34 (10² - \pi5²)$. I also assume you can calculate the area of the small part of the circle, so we have
$$ A = \frac{5 \cdot 10}2 - A_1$$ $$ A_1 = \pi \cdot 5^2 - \pi \cdot (\frac52)^2 - A_{small}$$
where $A$ is the shaded area we didn't know yet, $A_1$ is the area of the triangle intersected with the circle and $A_{small}$ the area of the small portion of the circle.
Hope this helps.
By the look of the answer, it would appear that there is no avoiding calculating the area of the roundy white triangle, or at least indirectly.
Call this area $x$ and the area of the lowest right hand corner region $a$, where $4a=100-25\pi$. Let the area of the smaller segment of the circle be $s$. The chord of this segment subtends an angle $\pi-2\arctan\frac 12$ at the centre of the circle. In which case
$$s=\frac 12\cdot25(\pi-2\arctan\frac 12-\sin(2\arctan\frac 12))$$ $$\Rightarrow s=\frac{25}{2}\pi-25\arctan\frac 12-10$$
Meanwhile, $x=25-a-s$ and the required region is $R=4a-x=5a-25+s$
Substituting in the value of $s$ gives the required answer $$R=90-18.75\pi-25\arctan \frac 12$$
$$\small \begin{align} A_\textsf{Shaded} &= A_\textsf{Square} - \color{darkorange}{A_\textsf{Circle}} - (\color{green}{A_\textsf{4-gon}}+\color{red}{A_\textsf{Triangle}})+\color{blue}{A_\textsf{CircleSegment}}\\&= 10^2 - \color{darkorange}{\pi \times 5^2} - \left(\color{green}{9}+\color{red}{1}\right)+\color{blue}{\frac{\arctan{(3/4)}}{2\pi}\pi \times 5^2} \\ &= 100 - \color{darkorange}{25\pi}-10+\color{blue}{12.5\arctan{\frac {3}{4}}}\\ &\approx 19.504…\end{align}$$
$\blacksquare $
My method is just an ordinary one but it shows how the suggest answer is obtained.
It should be clear that how the lengths and angles are defined as shown.
$\theta = \tan^{-1}(\dfrac {1}{2})$ in radians.
By power$(A, C_2)$, $z = \dfrac {100}{\sqrt {125}}$. Then, by power $(P, C_3)$, $y = \dfrac {25}{\sqrt {125}}$.
By similar triangles, $t = … = 2$.
After transferring the leftmost black shaded part to its more suitable position (above X), the required area
$= [(10 \times 10) square] - [C_2] - [red]$
$= 100 - 25 \pi - ([purple] - [\dfrac {C_1}{4}] - [yellow])$
$= 100 - 25\pi – 25 + \dfrac {25\pi}{4} + [yellow]$
$= 75 - 18.75\pi + [yellow]$
$= 75 - 18.75\pi + ([⊿XCQ] - [grey])$
$=75 - 18.75\pi + 25 - [grey]$
$= 100 - 18.75\pi - ([⊿XYQ] + [section YPQ] $
$= 100 - 18.75\pi - ([⊿XYQ] + [sector OPQ] $
$= 100 - 18.75\pi - (10 + \dfrac {(25)2\theta}{2})$
$= 90 - 18.75\pi - 25\tan^{-1}(\dfrac {1}{2})$
If $A=B+X$ where $B=\frac 12+\int_2^5(5-\sqrt{10x-x^2})dx$ (small white area below the diagonal) and $X$ the asked area then one has by symmetry $$2A+2(\pi\cdot5^2)=20\cdot 10\iff A+25\cdot\pi=100$$ We have$\int_2^5(5-\sqrt{10x-x^2})dx\approx 0.95624$ so $$X=100-25\cdot\pi-0.5-0.95624\approx 20.003943$$
I posted a solution directly on 9gag:
"Easy"? What's easy? My math/algebra skills are very rusty, but I'd set the problem up this way:
Drop a perpendicular from the intersection of the circle and the straight line.
Take the area of the triangle @30 deg.
(Moving to the right) Do an integral from the perpendicular to the osculation point on the rectangle given the equation of the circle.
Add the two areas: the triangle and the integral
Question is: is doing an integral "easy". Conceptually yes, mathematically ... I don't know. Depends.
