1
$\begingroup$

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?

enter image description here

| cite | improve this question | | | | |
$\endgroup$
  • 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
2
$\begingroup$

enter image description here

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.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

enter image description here

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.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.