What is the area of the hatched region, knowing that the arc AC is 1/4 of the circumference with a center in D?
I've tried using algebra to solve this but it seemed insufficient, I thought of using integrals to find the area, by finding the analytic geometry equations for the circles but it went nowhere.
On
Sector 1 Angle $= 2\cdot \sin^{-1}(\sqrt(\frac{31}{128})\cdot 2)= 2.78617$ rad
Sector 2 Angle $= 2\cdot \sin^{-1}(\sqrt(\frac{31}{128})) = 1.02906$ rad
Shaded Area = Area Chord Segment 1 – Area Chord Segment 2
Area Seg 1 = Area Sector 1 - Area Triangle 1
$\frac{2.78617}{2π}\cdot π(\frac{a}{2})^2 - (\sqrt(\frac{31}{128})a)( \sqrt(\frac{97}{128})a - \sqrt(2)\frac{a}{2})$ $= .26785 a^2$
A Seg 2 = Area Sector 2 - Area Triangle 2
$\frac{1.02906}{2π}\cdot πa^2 – (\sqrt(\frac{31}{128})a(\sqrt(\frac{97}{128})a)$ $= .08612 a^2$
Shaded Area $= (.26785 - .08612) a^2 = .18173 a^2$
Consider this image:
Think that the area of square $ABCD$ can be expressed as the sum of
In other words, this can be written as:
$$A_{ABCD}=\text{shaded}+\frac14a^2\pi+\frac14\left(a^2-\left(\frac{a}{2}\right)^2\pi\right)+2CHF\tag{1}$$
The challenge now is to solve for the area cutout in $CHF$, but this can be done if we visualize the problem like this:
We have rectangle $ACFJ$ with an area $\frac{a^2}{2}$ broken down as: $$A_{ACFJ}=A_{\text{sector} \,CAH}+A_{\text{sector} \,HEF}+A_{\triangle AEJ}+A_{\triangle HAE}+CHF=\frac{a^2}{2}\tag{2}$$
To find the area of sector $CAH$, we need to solve for $\angle CAH$. We know that $AC=AH=a$. We also know that $H$ is the intersection of two circles defined by: $$x^2+y^2=a^2\\ \left(x-\frac a2\right)^2+\left(y-\frac a 2\right)^2=\left(\frac{a}{2}\right)^2\\ H:\left(\frac{1}{8} \left(5 a-\sqrt{7} a\right),\sqrt{a^2-\frac{1}{64} \left(5 a-\sqrt{7} a\right)^2}\right)$$ Using the distance formula, we have: $$CH:\frac{1}{2} \sqrt{-a \left(\sqrt{10 \sqrt{7}+32} \sqrt{a^2}-8 a\right)}$$ Using the cosine law, we have: $$\angle\text{CAH}=\cos ^{-1}\left(\frac{\text{AC}^2+\text{AH}^2-\text{CH}^2}{2 \text{AC} \text{AH}}\right)=\cos ^{-1}\left(\frac{1}{4} \sqrt{\frac{5 \sqrt{7}}{2}+8}\right)$$ Which gives us: $$A_{\text{sector}\,CAH}=\frac{a^2 \angle\text{CAH}}{2}=\frac{1}{2} a^2 \cos ^{-1}\left(\frac{1}{4} \sqrt{\frac{5 \sqrt{7}}{2}+8}\right)$$
Repeating the same process, we get $A_{\text{sector}\,HEF}$ as: $$HE=FE=\frac a2\\ HF=\frac{\sqrt{a \left(\left(\sqrt{7}+13\right) a-2 \sqrt{10 \sqrt{7}+32} \sqrt{a^2}\right)}}{2 \sqrt{2}}\\ \angle\text{HEF}=\cos ^{-1}\left(\frac{\text{FE}^2+\text{HE}^2-\text{HF}^2}{2 \text{FE} \text{HE}}\right)=\cos ^{-1}\left(\frac{1}{4} \left(-\sqrt{7}+2 \sqrt{10 \sqrt{7}+32}-9\right)\right)\\ A_{\text{sector}\,HEF}=\frac{1}{2} \left(\frac{a}{2}\right)^2 \text{HEF}=\frac{1}{8} a^2 \cos ^{-1}\left(\frac{2 \sqrt{10 \sqrt{7}+32} \sqrt{a^2}-\left(\sqrt{7}+9\right) a}{4 a}\right)$$
Solving for $A_{\triangle AEJ}$, we know that $AJ=EJ=\frac a2$, therefore $$A_{\triangle AEJ}=\frac{1}{2} \left(\frac{a}{2}\right)^2$$ We know that $AJ=EJ=\frac a2$, therefore $AE=\frac{a\sqrt2}2$, and since we already know $\angle CAH$ and $\angle EAJ=\frac\pi4$ then: $$\angle\text{HAE}=-\angle\text{CAH}-\frac{\pi }{4}+\frac{\pi }{2}=\frac{\pi }{4}-\cos ^{-1}\left(\frac{1}{4} \sqrt{\frac{5 \sqrt{7}}{2}+8}\right)$$
And thus: $$A_{\triangle HAE}=\frac{\left(\sqrt{2} a\right) \text{AH} \sin (\text{HAE})}{2\ 2}$$
From $(2)$, we can now solve the area of $CHF$: $$CHF=\frac{1}{8} \left(3 a^2+a^2 \left(-\cos ^{-1}\left(\frac{2 \left(\frac{a^2}{2}-\frac{1}{8} a \left(\left(\sqrt{7}+13\right) a-2 \sqrt{10 \sqrt{7}+32} \sqrt{a^2}\right)\right)}{a^2}\right)\right)-4 a^2 \cos ^{-1}\left(\frac{1}{4} \sqrt{\frac{5 \sqrt{7}}{2}+8}\right)-2 \sqrt{2} a^2 \sin \left(\frac{\pi }{4}-\cos ^{-1}\left(\frac{1}{4} \sqrt{\frac{5 \sqrt{7}}{2}+8}\right)\right)\right)$$
From $(1)$ we can now solve for the shaded region, and we get: $$\text{shaded}\to-\frac{1}{16} a^2 \left(-4 \cos ^{-1}\left(\frac{2 \sqrt{10 \sqrt{7}+32} \sqrt{a^2}-\left(\sqrt{7}+9\right) a}{4 a}\right)+3 \pi -8 \left(2 \cos ^{-1}\left(\frac{1}{4} \sqrt{\frac{5 \sqrt{7}}{2}+8}\right)+\sqrt{2} \cos \left(\frac{\pi }{4}+\cos ^{-1}\left(\frac{1}{4} \sqrt{\frac{5 \sqrt{7}}{2}+8}\right)\right)\right)\right)$$