Area

[Anonymous No.6714406]

I need the green area of this shit, supossing that every side of the square is 20 cm

[Anonymous No.6714425]

>>6714406
>>6646405
>Anonymous ## Mod 07/15/14(Tue)15:16:15 No.6646405▶
...
>Reminder: /sci/ is for discussing topics pertaining to science and mathematics, not for helping you with your homework. See the rules page for details.

[Anonymous No.6714435]

>>6714406
I hope this is bait. that's like algebra I shit.

[Anonymous No.6714438]

R*(1/2*R) + [R*(1/2*R) - 1/4*(pi*R^2)]

[Anonymous No.6714441]

how about this

[Anonymous No.6714463]

Is it bad that I want to use calculus to find the answer?
What is the geometric solution, anyway?

[Anonymous No.6714470]

>>6714441
This is good. Do the same thing with the red lines to the other side, then you can use formulas for areas of triangles and areas of sectors.

[Anonymous No.6714481]

>>6714470
except the best you'll be able to come up with is an approximation since you'll never be able to precisely describe the area of circular segments with triangles

[Anonymous No.6714484]

>>6714481
Take a circle then subtract the triangles.

[Anonymous No.6714492]

>>6714441
SP=2L
4(L)^2-4*30(L)^2?

[Anonymous No.6714499]

Let <span class="math">R_1 = \{ (x,y) \in \mathbb{R}^2 : 0 \leq x \leq 20,\, 10 \leq y \leq 20\}[/spoiler]

Let <span class="math">C_1: (x-10)^2 + y^2 \leq 10^2[/spoiler] and <span class="math">C_2: x^2 + y^2 \leq 10^2[/spoiler]

Then <span class="math">y = \pm \sqrt{20 x-x^2}[/spoiler] and <span class="math">y = \pm \sqrt{100-x^2}[/spoiler]

So the intercept happen at <span class="math">\sqrt{100-x^2}=\sqrt{20 x-x^2} \rightarrow (x',y') = \left( 5,5\sqrt{3} \right)[/spoiler]

So:
<span class="math">R_2 = \{ (x,y) \in \mathbb{R}^2 : 5\sqrt{3} \leq y \leq 10,\, \sqrt{100-y^2} \leq x \leq 10-\sqrt{100-y^2} \}[/spoiler]

<div class="math">A = \int\int_R dydx = \int\int_{R_1} dydx + 2\int\int_{R_2} dxdy</div>

[Anonymous No.6714503]

>>6714499
ohh this is perfect. i can't stop... lel

[Anonymous No.6714506]

>>6714481
that is why I said you can the area for sectors. If you fill red lines in on the rest of the image, you will see that the white part consists solely of 3 sectors and 2 equilateral triangles. Since you the side length of the big square and the angles, you get an exact solution.

[Anonymous No.6714535]

Only because I am fucking bored.

[Anonymous No.6714549]

integrate the equation for the top half of a unit circle from 0 to 5, multiply that by 4, subtract that from the area of 1 unit circle, multiply that by 10cm, subtract that from the area of the green square.

[Anonymous No.6714552]

>>6714549
poster here.

disregard that, I suck dicks

you're supposed to integrate from 0.5 to 1, not 0 to 5.

[Anonymous No.6714559]

>>6714549
>>6714552
Also, don't use a unit circle, just use a circle of radius 10cm and dont multiply by 10cm afterwards

[Anonymous No.6714566]

>>6714499

[Anonymous No.6714588]

>>6714470
yo man i made the red lines but i'm not op

i was just trying to help op along
sometimes you need a little push

[Anonymous No.6714643]

>>6714406
You need more information. There's no knowing if those arcs are part of a circle or not, they could be just willy nilly.

[Anonymous No.6714660]

>>6714406
If the length of the squarre is 1, then the area of the green shit is <span class="math"> 1 -\frac{\pi}{12} -\frac{\sqrt{3}}{8} [/spoiler].

Thus here, if the length is 20 cm, the area will be <span class="math"> \approx 208.67 cm2 <span class="math">[/spoiler][/spoiler]

[Anonymous No.6714666]

get in here guys we can do this together

http://whiteboardfox.com/14822-4111-8525