Finding the angles and area of unusual shapes My friend sent me the following geometry problems. I think I have the first one, but I think the 2nd and 3rd are unsolvable, although I could be missing something.

My attempt:


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*I'm pretty sure this one is 1080. I'm having a hard time writing out my explanation here, but I can justify it on paper. The idea is that the outer shape (if we ignore the triangles) is an octagon, and the sum of its angles is 1080, and I can show that the sum of the marked angles is also 1080.

*I don't think this one gives us enough information. If we moved the point D left or right, the angle of $x$ would change and the constraints would still be satisfied.

*I can't figure out the answer to this one. I think it's unsolvable as well, but I can't prove it like #2. We can't find the area of anything in this picture. The shaded region and the whole shape are both close to being a trapezoid, but they aren't.

Here are my thoughts on your friend's questions which are a lot of fun, thank you!

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*Ignore the 4 reversed triangles then the sum of the interior angles of the octagon that is left is 1080 degrees but each concave angle is equal to the sum of the 2 reversed triangle angles plus 180. Hence the total sum is 1080-4x180=360 degrees. (Thanks to Smash for pointing out the error in my previous attempt)


*The answer is 40 degrees since A, B and C lie in a circle centered at D. The chord AB subtends the angle 25 at C on the circle, hence it must subtend 50 degrees at the circle centre, D. Hence angle DAC = 90-50=40 degrees.


*Let x and y be the horizontal and vertical dimensions of the rectangle formed by the right angle and see the points marked on the image below.Then triangle ABO is similar to triangle BCD. Hence $y/4=4/x$ so $xy=16$. Also $EF=y-2$ and $AO=6-x$.
Now shaded area ABDF = area BDEO + area ABO - area AFE $=xy+y(6-x)/2-6(y-2)/2.$ So shaded area $= xy+3y-xy/2-3y+6=xy/2+6=8+6=14.$
Note that it is possible to find $x$ and $y$ since triangle ABO is right angled so $4^2=(6-x)^2+y^2$. Together with $xy=16$ we have two equations we can solve in 2 variables.
Wolfram Alpha gives two possible pairs, x=9.63935, y=1.65986 and x=4.37664, y=3.65578. Clearly only the second pair is consistent with the diagram.

