The figure is a square of side 2$a$. It is requested to find the striped area, but no information is given of the position of the lines that intersect the sides and that makes the area vary greatly. There is a special case for the two lines bisecting sides, but I I imagine that the problem points to a general solution with those mobile sides which complicates me just enough to do it.

For conventional geometry if there is a way to do it I do not see it, for integration I imagine defining possible points that I will call on the left side and y on the right x where $0 <x <= 2a$ e $0 <y <=2a$

Any idea how to raise it?

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  • 2
    $\begingroup$ Look at what shapes you can make out in the square - I see triangles, semi-circles and a quarter circle, all of which you can find the area of without integration (I hope). $\endgroup$ – MrMazgari Dec 18 '19 at 4:29
  • $\begingroup$ Hint: $\frac{a^2}{10} \left(6-5 \sin ^{-1}\left(\frac{4}{5}\right)\right)$ $\endgroup$ – David G. Stork Dec 18 '19 at 4:47

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Assume that A and D are midpoints. Let $(x,y)$ be the location of the point Y, which is the intersection of the circle $x^2+y^2 = 4a^2$ and the line $y=2(x-a)$. Solve to get $x=\frac85a$ and $y=\frac65a$.

Let [.] denote areas. Then the striped area is

$$A = 2[OAY]+[OYZ]- [OAXD] - 2[DXC]$$ $$=2\left(\frac12 \frac65a^2\right)+\frac12 (2a)^2\alpha -a^2 -2\left( \frac12 a^2\beta\right) = \left( \frac15+2\alpha-\beta\right)a^2$$

$\beta$ satisfies $\tan\beta = \frac12$ and $\alpha$ is derived from $\tan\theta = \frac yx = \frac34$ as follows,

$$\tan\alpha = \tan(90-2\theta) = \cot(2\theta) = \frac{1-\tan^2\theta}{2\tan\theta} = \frac7{24}$$

Thus, the area is

$$A= \left( \frac15+2\tan^{-1}\frac7{24}-\tan^{-1}\frac12\right)a^2$$

or about $0.3a^2$, which is verified from numerical integration.

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  • $\begingroup$ Thanks for answering, could you review this ibb.co/K7B99F6 thanks $\endgroup$ – wally Dec 18 '19 at 17:26
  • $\begingroup$ ahhh how do you get to $\tan\alpha = \tan(90-2\theta)$ $\endgroup$ – wally Dec 19 '19 at 3:48
  • $\begingroup$ @wally - from the diagram, $ \alpha+2\theta=\angle O=90$ $\endgroup$ – Quanto Dec 19 '19 at 4:00
  • $\begingroup$ Ups, a mistake from me,thanks $\endgroup$ – wally Dec 19 '19 at 23:14

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The area of the region $OMNDP$ is found by subtracting the sum of the areas of semicircle $AOPD$ and half of the area of the square $AEOF$ from the sum of $45^\circ$ sector of the circle ($AMND$) and the area of the sector $EOA$: \begin{align} [OMNDP]&= \tfrac12\,\pi a^2 + \tfrac14\,\pi a^2 -\tfrac12\,\pi a^2 -\tfrac12\, a^2 =\tfrac14\,a^2\,(\pi-2) \tag{1}\label{1} . \end{align}

The area of the half of the region in question can be found by subtracting the area $[DPN]$ from \eqref{1}. Using the coordinate system with the origin $F$, $x$-axis emanating from it along the direction $QD$ and $y$-axis along $FC$,

\begin{align} [DPN]&=\int_0^{|DQ|} \int_{f_1(x)}^{f_2(x)}\,dy\,dx \tag{2}\label{2} ,\\ f_1(x)&=\sqrt{a^2-x^2} ,\\ f_2(x)&=A_y+\sqrt{4a^2-(x-A_x)^2} . \end{align}

We can easily find that

\begin{align} |DQ|&=\tfrac{2\sqrt5}5\,a ,\\ A_x&=-D_x =-\tfrac{2\sqrt5}5\,a ,\\ A_y&=-D_y =-\tfrac{\sqrt5}5\,a \end{align} and \eqref{2} becomes

\begin{align} [DPN]&= \tfrac1{10}\,a^2(35\arctan(2)-6-10\pi) \end{align}

\begin{align} [OMNP]&=[OMNDP]-[DPN] = \tfrac1{20}\,a^2(25\pi+2-70\arctan(2)) , \end{align}


\begin{align} [OKLMNP]= 2[OMNP]&= \tfrac1{10}\,a^2(25\pi+2-70\arctan(2)) \approx .30394\,a^2 . \end{align}

Note, that despite it looks slightly different, the result is the same as in the @Quanto's answer.

Another way could be a standard integration using coordinate system at the origin $O$ and $OC$ and $OD$ as $x$ and $y$ axes respectively.

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  • $\begingroup$ thanks ,even analyzing it, but it is understood , There is a general solution for mobile points E and F $\endgroup$ – wally Dec 19 '19 at 23:17

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