Finding the angles and area of unusual shapes

Question

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:

1. 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.

2. 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.

3. 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.

Answer 1

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

1. 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)

2. 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.

3. 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.

Answer 2

$(1)$ It looks unsolvable. Yes, breaking it into the "outer octagon" and the "inner triangles" is a good idea. But think, you can shift the dots to matching the intersection angles right? but you are missing the large angle in each triangle towards the "whole" angle (the sum of 3 components of the intersection) for the octagon.

$(2)$ moving $D$ messes with the equality $AD=BD$ as "the edges will stretch disproportionately." Use the fact that $AD=BD$!

$(3)$ at the risk of being rude, the question is a mess while at heart a good start, you're correct. Maybe adding an edge length or two, or an angle is all that is needed

Answer 3

2) Note $ \angle ABD = 90 -25 =65$, which yields $x=65 -25 =40$ due to isosceles triangle.

3) The shaded area is equal to the area of the quadrilateral minus the areas of the two small right triangles. Thus,

$$A= \frac 12\cdot 4(x+4) + \frac12 \cdot 6(2+y) -\frac12 \cdot 4x -\frac12\cdot 6y=14$$

where $x$ and $y$ are the unknown segment lengths that cancel out in the difference.