Is 158 close?

Area of square minus area of circel is ~2150
Divided by 4 is 537.5

This is divided by the diagonal of the large rectangle at an angle of 60 degrees.

So.... (dunno)

Deviltube?

This is correct

hmm i feel like i am an idiot.. 26.57/90 is 0,2952

so 536.5 x 0,2952 = 158,36

i know its wrong but i dont kown how to calculate it

If you approximate the blue shape to be a triangle, you can decompose it in two smaller and roughly equal 25 * 12.5 triangles. So the answer should be a bit less than 300.

But incomplete

So like .4x537.5

And here's the rest of it.

around 200

close

dead on

...

100*100=10000 (square area)
pi(50)^2=7853.98... (circle area)

10000 - 7853.98 = 2146.02 (corner areas)
2146.02/4=536.255 (single corner area)

easy.

Consider one square of 100x100, minus the circle of 50^2 x pi=~7,850, so 2,150 for 4 corners, or 538 for each corner; blue area is less than half of 538. But how do you determine the angle/percentage?

>overthinking.

simple area subtraction

SOH CAH TOA
knowing the opposite and the adjacent we can use tan

so tan^-1(100/200) = 26.565 degrees

find area of the square, find area of the quarter circle, do something with the 30degree Angle?

Can't we integrate the area under the curve? Since we know that the segment of circumference in the triangle will have a changing angle, we could find a formula that discribes the curve. :v)

that or 45/2 is the angle, aka: 22.5
only way for line to hit center and corners.

too lazy to count

>26.565 degrees
Not 30?

Bingo.

...

...

/thread

Or you could just integrate with boundaries
Y=0
Y=0.5x
Y=sqrt(sqrt (50,000)-x^2)

wolframalpha.com/input/?i=area bounded by y=-sqrt(2500-x^2)+50 and y=0.5(x+50) and y=0

this will probably be the integral from 0 to the point of intersection of the line and circle + the integral from that point to the end of the area (50 length units), the first function is
y=x/2
the second is something like y=sqrt(50-(x+a)^2) +b

radius is 50
so just doing pir^2 gets
7854
thats for one circle
times that by 2 to get 15708
subtract that from the total area of rectangle
200*100 = 20000

20000 - 15708 = 4292
thats the area left over, now divide that by 8 as equal areas for the grey bits
to get 536.5 per grey area
now as we can work out the angle from SOH CAH TOA
knowing the opposite and the adjacent we can use tan

so tan^-1(100/200) = 26.565 degrees

now cutting that up into a more manageable piece
there is a small triangle.
Now what I did was use the angle, subtract that from 90 to get 63.435 degrees.
Use SOH CAH TOA again to find the missing radius part.
TOA

tan (26.565*50) = 25 approx
63.435 degrees is around 0.176 of a circle
so pir^2 of r 25 * 0.176 = 345
then 536.5 - 345 = 190.52

Probably fucked it but here you go guys

Winrar

just realized I fucked it haha, oh well

trick question, the answer is: not very. large is a descriptor and not a mathematic term.

The missing part here in all the previous comments is the circle segment. Sure integration will work...
Red triangle = 1/8 of area of rectangle (or 1/4 of the square if you want to simplify)
Subtract the green section (has been calculated many times above)
and the srea of the circle segment: A = R²/2 * {([(180-(2*26.5)]*pi/180) - sin(180)}
(en.wikipedia.org/wiki/Circular_segment)

= 195.62

This is the way I was trying to figure it out.

I couldn't work out the angle of the segment.

How did you come to the conclusion it was 180 - (2*26.56)?

I'd stick Cartesian coordinates on it and use integration.
Use mid point of image as (0,0) and gradient of line is 1/2
I can't be arsed though

Can I just say what a rare pleasure it was to work this through without the conversation devolving into insults.

536.5

26.56 as 1 of the angles of the triangle is know, right?
Look at the dotted black lines I drew, the horizontal one is indeed exactly horizontal.
From that follows that the angle between the horizontal dotted line and the hypothenuse of the triangle ist also 26.56, ok?
Both the dotted lines are equal in length (both are the radius of the circle) and form an isosceles triangle with the part of the hypothenuse which intersects the circle, therefore the 2 angles must be the same.
Sum of all angles in a triangle equals 180°.
All good?

I get it!

Thanks man, I was losing my mind trying to work that out