square & semicircle

$\angle BCE=2\times arctan(\frac12)\approx53.13^\circ$

Distance from point E to the top $=sin(53.13^\circ)=0.8$

Distance from point E to the bottom $=1-0.8=0.2=\frac15$

Area of $\triangle AED=\frac12\times1\times\frac15=\boxed{\frac1{10}}$

This question looks like the sort of thing I'd see on my maths HSC, so I think I'll do a full proof for practice, along with some other comments.

Construct the lines CM, EM, EN and EO where M is the midpoint of AB, N lies on AB so that EN is perpendicular to AB and O lies on AD so that EO is perpendicular to AD: Note that the sides of the square are all 1, so AM and BM are both 1/2. ME is also 1/2, as the semicircle has a constant radius of 1/2.

$let \angle BMC=\theta$

$\tan \theta =\frac{1}{1/2} =2$

$\theta =\arctan2\,(where\,\theta\,is\,acute)$

Now we prove congruency of the triangles BCM and ECM to find angle CME, and in turn angle NME:

$\angle CBM=\angle CEM=90^{\circ}$

$MC\,is\,common$

$BM=EM\,(radius\,of\,semicircle)$

$∴\triangle BCM \equiv \triangle ECM\,(RHS)$

This is one of the few times in my entire study of mathematics that I have found the "Right angle, Hypotenuse, Side" rule useful.

$∴\angle CME = \theta\,(corresponding\,angles\,in\,congruent\,triangles\,are\,equal)$

$∴\angle NME = 180^{\circ}-2\theta\,(angles\,on\,a\,straight\,line\,sum\,to\,180^{\circ})$

Now we can use angle NME to find the length of NM, which will help us find the perpendicular height, EO, of the triangle we want.

$in \triangle EMN:\,\cos (180^{\circ}-2\theta )=\frac{NM}{EM}$

$\cos (180^{\circ})\cos (2\theta )+\sin (180^{\circ})\sin (2\theta )=\frac{NM}{1/2}$

$-1\cos (2\theta )+0\sin (2\theta)=2NM$

$1-2\cos ^2(2\theta )=2NM$

$NM=\frac{1}{2}-\cos^{2}\theta$

There are several techniques to simplify inverse trig functions inside normal ones. Here, I do it algebraically. There should be no issues with negatives, as theta is acute.

$NM=\frac{1}{2}-\frac{1}{\sec^{2}\theta}$

$=\frac{1}{2}-\frac{1}{1+\tan^{2}\theta}$

$=\frac{1}{2}-\frac{1}{1+(\tan(\arctan2))^{2}}$

$=\frac{1}{2}-\frac{1}{1+2^{2}}$

$=\frac{3}{10}$

$AN=AM-MN=\frac{1}{2}-\frac{3}{10}=\frac{1}{5}$

$EO=AN=\frac{1}{5}$

Now that we know the length of the base and the perpendicular height of triangle AED, we can find n:

$Area=\frac{1}{2}bh$

$\frac{1}{n}=\frac{1}{2}(1)(\frac{1}{5})$

$\frac{1}{n}=\frac{1}{10}$

$n=10$