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let us consider following picture

enter image description here

we have following informations.we have circular sector,central angle is $90$,and in this sector there is inscribed small circle ,which touches arcs of sectors and radius,radius of this small circle is equal to $\sqrt{2}$,we should find area of dark part.

my approaches is following,let us connect radius $\sqrt{2}$ to intersection points of small circle with radius of big sector, we will get square with length $\sqrt{2}$,clearly area of dark part is area of sector -area of square-area of small sector(inside small circle)and minus also area of small part below,which i think represent also sector with central angle $90$,but my question is what is radius of big sector?is it $2*\sqrt{2}$?or does radius of small circle divide given radius of big sector into two part?also please give me hint if my approaches is wrong or good.thanks in advance

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  • $\begingroup$ The radius of the big sector is equal to "distance of small circle's center from origin" + "radius of small circle" = $\sqrt{2}\sqrt{2}+\sqrt{2}=2+\sqrt{2}$. $\endgroup$
    – zuggg
    Jul 16, 2013 at 5:39
  • $\begingroup$ The more interesting question is the case where the corner is not a right angle, but $\theta \le \pi$ $\endgroup$
    – Ali
    Jul 16, 2013 at 8:11

3 Answers 3

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Hints:

  • Let $r = \sqrt{2}$ be the small radius.
  • Big radius is equal to diagonal of small $r^2$ square plus $r$, that is $\sqrt{2}\cdot \sqrt{2} + \sqrt{2} = 2+ \sqrt{2}$.
  • The small area near the origin has area of a square minus area of a circle divided by four.
  • The dark area is whatever is left divided by two (the figure is symmetric).

I hope this helps ;-)

Edit: For check: $\frac{1}{2}\Bigg(\frac{\pi R^2}{4}-\pi r^2-\Big(r^2-\frac{\pi r^2}{4}\Big)\Bigg)$, where $R = (\sqrt{2}+1)r$.

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  • $\begingroup$ sorry square-circle/4 right?(square-circle)/4 how? $\endgroup$ Jul 16, 2013 at 5:46
  • $\begingroup$ small square - circle/4 or (big square-circle)/4. $r^2-\pi r^2/4$ or $\frac{(2r)^2-\pi r^2}{4}$. $\endgroup$
    – dtldarek
    Jul 16, 2013 at 5:48
  • $\begingroup$ aaa you mean create big square?and it means that dark side and white which left below is equal to each other? $\endgroup$ Jul 16, 2013 at 5:49
  • $\begingroup$ @dato You don't have to, with small square it is simpler. $\endgroup$
    – dtldarek
    Jul 16, 2013 at 5:50
  • $\begingroup$ yes,and my final question is that dark side and white side are equal to each other? $\endgroup$ Jul 16, 2013 at 5:52
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enter image description here

Skeleton:

Denote small radius as $r$, large radius as $R$. Then (looking at the image, diagonal) $2R = r+\sqrt{2}\cdot 2r + r = 2r+2\sqrt{2}r$, $\:$ so $$R=(1+\sqrt{2})r.$$

Square of black figure: $$ S = \dfrac{1}{8} (S_{\mathrm{large}} - 4\cdot\dfrac{3}{4}S_{\mathrm{small}} - S_{\Box}) = \dfrac{1}{8}(\pi R^2 - 3\pi r^2 - 4r^2) = \dfrac{\pi\sqrt{2}-2}{4}r^2. $$

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Let $Y$ be the upper left corner of the picture, and let $O$ be the lower left corner, the centre of the quarter-circle.

Let $C$ be the centre of the small circle. Let the line $OC$ meet the big circle at $M$.

Draw a perpendicular from $C$ to $OY$, meeting $OY$ at $P$,

Note that by the Pythagorean Theorem, we have $OC=2$. Thus $OM=2+\sqrt{2}$. This is the radius of the big quarter-circle.

Now we know almost everything. Our target region is one-half of our quarter circle, minus triangle $OCP$, minus the little circle's circular sector $CMP$.

The area of half our quarter-circle is easily found. So is the area of $\triangle OCP$.

Finally, $\angle PCM$ is $135^\circ$, so the circular sector $CMP$ has area three-eighths of the area of the small circle.

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  • $\begingroup$ thanks for help @André Nicolas $\endgroup$ Jul 16, 2013 at 5:53
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    $\begingroup$ You are welcome. Your picture helped a lot. I hope that even without a picture in my solution, it is clear where the extra lines are supposed to be. $\endgroup$ Jul 16, 2013 at 6:00
  • $\begingroup$ ok good lucks in your life .thanks ones again $\endgroup$ Jul 16, 2013 at 7:01

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