inkyvoyd
  • inkyvoyd
Find the area of the shaded region (inkyvoyd's variation of "FoolForMath's problem(s) of the day) There's multiple ways to solve this. Anyone who has learned geometry should be capable of solving it, but it is not a particularly easy question. Have fun!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
inkyvoyd
  • inkyvoyd
1 Attachment
inkyvoyd
  • inkyvoyd
Other notes: Arcs are circular with radii of 2, and the largest polygon is a square with a side length of two.
inkyvoyd
  • inkyvoyd
Hint: Equilateral triangle.

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More answers

inkyvoyd
  • inkyvoyd
By the way, I have many more where these came from. I quite literally have 4 or 5 books of math problems in which 10% of them are challenging (this one's especially hard though).
KingGeorge
  • KingGeorge
Can I have the titles of some of those books that you have?
inkyvoyd
  • inkyvoyd
They are in Chinese. I can assure you that I know how to do this problem, if you want proof, just ask.
KingGeorge
  • KingGeorge
I just wanted the books myself. But I can't understand Chinese, so....
inkyvoyd
  • inkyvoyd
Well, in that case, I'll try my best to scan a few onto open study every day...
perl
  • perl
do you have detailed solutions?
perl
  • perl
in english or hindi
inkyvoyd
  • inkyvoyd
I can show you 2 different solutions.
perl
  • perl
show me one
inkyvoyd
  • inkyvoyd
I'll put the general solution in a message.
anonymous
  • anonymous
"evilasianmath" ? lol
inkyvoyd
  • inkyvoyd
lol.
dumbcow
  • dumbcow
I found the area to be 0.6942 i used calculus though :|
dumbcow
  • dumbcow
|dw:1333534070710:dw| given : a+b = pi x+2a+b = 4 find area of region a by integrating over difference between 2 circles from 0 to 1 \[a = \int\limits_{0}^{1}\sqrt{4-x^{2}}-\sqrt{4-(x-2)^{2}} dx = \sqrt{3}-\pi/3 = 0.685\] then b = pi - a = 2.456 x = .1735 Area of shaded region is 4*x --> 0.6942
anonymous
  • anonymous
There seems to be multiple curves here. I don't think we can solve this without using calculus. PS: My actual problem involves calculus too.
inkyvoyd
  • inkyvoyd
Fool, it's just cutting. Cutting portions.
dumbcow
  • dumbcow
inkyvoyd, can you solve it strictly using geometry ?
inkyvoyd
  • inkyvoyd
This is a wonderful example of why we invented calculus, though.
anonymous
  • anonymous
Hm I can understand but then you will be attempting to made it planer which might involve some variation.
inkyvoyd
  • inkyvoyd
The two solutions that I first saw (one I thought of myself, the other a classmate figured out) were geometry. In 8th grade.
inkyvoyd
  • inkyvoyd
let me do the cutting, one sec.
anonymous
  • anonymous
Okay cut them off! :D
inkyvoyd
  • inkyvoyd
Actually, I think I'ma eat first.
anonymous
  • anonymous
Bon appetite.
dumbcow
  • dumbcow
haha
inkyvoyd
  • inkyvoyd
Alright, done with dinner.
anonymous
  • anonymous
That was fast lol
inkyvoyd
  • inkyvoyd
inkyvoyd
  • inkyvoyd
Now, assume the area of the center arched shape with 4 sides has area x
inkyvoyd
  • inkyvoyd
and, the area of one of the "flower petals" has area y.
inkyvoyd
  • inkyvoyd
Now, 2y+x=2*circle-isoceles right triangle
anonymous
  • anonymous
I can smell Trigonometry.
inkyvoyd
  • inkyvoyd
No trig.
inkyvoyd
  • inkyvoyd
*90 degree sector
anonymous
  • anonymous
Okay so you are going to use the area of the sector of a circle formula isn't?
inkyvoyd
  • inkyvoyd
x+y=two 60 degree sectors minus equilateral triangle, plus30 degree sector.
inkyvoyd
  • inkyvoyd
You can prove that these triangles have these degrees by first showing that the big triangle is equilateral, and doing some angle bisection.
inkyvoyd
  • inkyvoyd
That was my brilliant solution. What i'm about to do next will make you all feel stupid.
inkyvoyd
  • inkyvoyd
inkyvoyd
  • inkyvoyd
@FoolForMath ,@dumbcow ,@perl ,@dpaInc
inkyvoyd
  • inkyvoyd
@dumbcow
inkyvoyd
  • inkyvoyd
@perl
inkyvoyd
  • inkyvoyd
@dpaInc
dumbcow
  • dumbcow
Ahh...very nice
inkyvoyd
  • inkyvoyd
When there's a circle, always use geometry.
inkyvoyd
  • inkyvoyd
btw, what I tried to explain was my first solution
inkyvoyd
  • inkyvoyd
Hope you guys understood that solution too, :D
anonymous
  • anonymous
|dw:1333536217802:dw| area of the rectangle - area of the sector will give you half that shaded region. wahtching dumbcow do it with calculus and his picture actually gave me the idea.
dumbcow
  • dumbcow
i don't see things like that, prob why geometry isn't my fav math topic
inkyvoyd
  • inkyvoyd
I used to hate geometry, but now I love it, cause I'm better than most white people at it.

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