# Find area of striped region enclosed by a quarter and two semi circles along with two lines inside a square

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 $$0 e $$0

Any idea how to raise it?

• 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). – MrMazgari Dec 18 '19 at 4:29
• Hint: $\frac{a^2}{10} \left(6-5 \sin ^{-1}\left(\frac{4}{5}\right)\right)$ – David G. Stork Dec 18 '19 at 4:47

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.

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

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}
and

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

and

\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.

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