# Find the area of the shaded region of this figure 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)

• This problem is classic...... – Brethlosze Aug 21 '17 at 16:57

You need to find the area of 4 remaining parts and subtract them from the area of the square. • Well done, that is an olympic degree question. – Kitiara Aug 21 '17 at 19:20
• @Kitiara, Thank you very much. – Seyed Aug 21 '17 at 20:13

$FC=2x\sin 15°$

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

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

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

$Area=x^2\left[2-\sqrt 3+2\left(\dfrac{\pi}{6}-\dfrac{1}{2}\right)\right]\\ Area=x^2\left(1+\dfrac{\pi}{3}-\sqrt{3}\right)$ • There is a mistake, i don't get the correct answer when i replace X with another number. – Kitiara Aug 21 '17 at 19:36
• Thank you. Now it works – Raffaele Aug 21 '17 at 19:58
• Now it is correct, the second way of solving the question. Well done. – Kitiara Aug 21 '17 at 20:07

HINT: $$A=4\int_{1/2}^{\sqrt 3/2}\sqrt{1-x^2}-1/2dx = 1-\sqrt 3-\frac \pi 3 = 0.31515$$

• 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! – Cye Waldman Aug 21 '17 at 20:08
• 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) – Kitiara Aug 21 '17 at 20:13
• @CyeWaldman, And oh! here is categorized for geometry :) – Seyed Aug 21 '17 at 21:55
• You are assuming x = 1 in this form. – Kitiara Aug 21 '17 at 22:51
• @Kitiara I doesn't matter. The area must necessarily scale as $x^2$. – Cye Waldman Aug 21 '17 at 23:18