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enter image description here

Find the area of the shaded region. (Each arcs of circles in the figure are assumed to be $\frac{1}{4}$ of a full circle)

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  • $\begingroup$ This problem is classic...... $\endgroup$ – Brethlosze Aug 21 '17 at 16:57
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You need to find the area of 4 remaining parts and subtract them from the area of the square. enter image description here

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    $\begingroup$ Well done, that is an olympic degree question. $\endgroup$ – Kitiara Aug 21 '17 at 19:20
  • $\begingroup$ @Kitiara, Thank you very much. $\endgroup$ – Seyed Aug 21 '17 at 20:13
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$FC=2x\sin 15°$

$\sin 15°=\sqrt{\dfrac{1-\cos 30°}{2}}=\sqrt{\dfrac{1-\frac{\sqrt 3}{2}}{2}}=\dfrac{1}{2}\,\dfrac{\sqrt{3}-1}{\sqrt{2}}$

$FC=2x\dfrac{\sqrt{3}-1}{2 \sqrt{2}}=x\dfrac{\sqrt{3}-1}{ \sqrt{2}}\\ Area_{FHGC}=FC^2=\left(x\dfrac{\sqrt{3}-1}{ \sqrt{2}}\right)^2=x^2(2-\sqrt 3)$

$area_{red}=\dfrac{1}{2} x^2 (t-\sin t)\\ area_{red}=\dfrac{1}{2}x^2\left(\dfrac{\pi}{6}-\dfrac{1}{2}\right) $

$Area=x^2\left[2-\sqrt 3+2\left(\dfrac{\pi}{6}-\dfrac{1}{2}\right)\right]\\ Area=x^2\left(1+\dfrac{\pi}{3}-\sqrt{3}\right)$

enter image description here

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  • $\begingroup$ There is a mistake, i don't get the correct answer when i replace X with another number. $\endgroup$ – Kitiara Aug 21 '17 at 19:36
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    $\begingroup$ Thank you. Now it works $\endgroup$ – Raffaele Aug 21 '17 at 19:58
  • $\begingroup$ Now it is correct, the second way of solving the question. Well done. $\endgroup$ – Kitiara Aug 21 '17 at 20:07
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HINT: $$ A=4\int_{1/2}^{\sqrt 3/2}\sqrt{1-x^2}-1/2dx = 1-\sqrt 3-\frac \pi 3 = 0.31515 $$

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    $\begingroup$ Your answer is the best and should be the one that is accepted. The result is the same as that one and, oh, so much better! $\endgroup$ – Cye Waldman Aug 21 '17 at 20:08
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    $\begingroup$ I'm not getting correct results with this one.[link](wolframalpha.com/input/?i=4*int+1%2F2+to+sqrt(3)%2F2+(sqrt(1-(10)%5E2)-1%2F2+dx) $\endgroup$ – Kitiara Aug 21 '17 at 20:13
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    $\begingroup$ @CyeWaldman, And oh! here is categorized for geometry :) $\endgroup$ – Seyed Aug 21 '17 at 21:55
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    $\begingroup$ You are assuming x = 1 in this form. $\endgroup$ – Kitiara Aug 21 '17 at 22:51
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    $\begingroup$ @Kitiara I doesn't matter. The area must necessarily scale as $x^2$. $\endgroup$ – Cye Waldman Aug 21 '17 at 23:18

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