/sci/ - Science & Math » Thread #11801699

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Anonymous Mon 15 Jun 18:53:15 2020 No.11801699 View Reply Original Report

Quoted By: >>11801700 >>11801720 >>11801722 >>11801745 >>11801754 >>11801765 >>11801787 >>11801838 >>11802082 >>11802341 >>11802387 >>11802746

What the fuck is the answer to this and why am I such a fucking brainlet

Anonymous

Anonymous Mon 15 Jun 2020 18:53:34 No.11801700 Report

Quoted By:

>>11801699
Find the area btw

Anonymous

Anonymous Mon 15 Jun 2020 18:54:47 No.11801705 Report

Quoted By: >>11801708 >>11801720

Integrate

Anonymous

Anonymous Mon 15 Jun 2020 18:55:22 No.11801708 Report

Quoted By: >>11801712

>>11801705
How?

Anonymous

Anonymous Mon 15 Jun 2020 18:57:30 No.11801712 Report

Quoted By:

>>11801708
Between the curves

Anonymous

Anonymous Mon 15 Jun 2020 18:59:52 No.11801720 Report

Quoted By: >>11801730 >>11801739 >>11801743 >>11802105

>>11801699
Fine faggot, next time >>>/wsr/

Looking at the question to the left, the right one is clearly just the overlap of two circles. It would be

Area_square - 2 * Area_excluded from the first shape.

Area_square is 64cm^2
Area_excluded is 64 - 50.27 = 13.7345 cm^2

Total shaded = 64 - 2(13.7345) = 36.5310cm^2

>>11801705
>Integrate
>How?
>I dunno, between the curves maybe?
dumbass Calc II dropout. Do you not know the formula for a circle?

Anonymous

Anonymous Mon 15 Jun 2020 19:00:03 No.11801722 Report

Quoted By: >>11801756

>>11801699
add two quarter circles of radius 6. the result is a full square where you double count the region that is highlighted.
subtract the area of the square and you are left with just the yellow region.

Anonymous

Anonymous Mon 15 Jun 2020 19:01:03 No.11801724 Report

Quoted By:

Maybe you can find the intersection área between two circles.

Anonymous

Anonymous Mon 15 Jun 2020 19:02:25 No.11801730 Report

Quoted By:

>>11801720
Dumbass. I wasn't trying to help OP you faggot

Anonymous

Anonymous Mon 15 Jun 2020 19:03:54 No.11801735 Report

Quoted By: >>11802692

It really bothers me that there is no specification of which part of the circle we are seeing, we are to asume its 1/4? If it werent is there any way to solve it?

Anonymous

Anonymous Mon 15 Jun 2020 19:04:58 No.11801739 Report

Quoted By: >>11801756

>>11801720
>36.53
That’s not right according to the thing though. Keeps saying try again.

Anonymous

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Anonymous Mon 15 Jun 2020 19:06:31 No.11801743 Report

Quoted By:

>>11801720
This is how I tried to work it out. I am not good at maths.

Anonymous

Anonymous Mon 15 Jun 2020 19:06:55 No.11801745 Report

Quoted By: >>11801750

>>11801699
just do an or operation on two quads of the area of a the circle

Anonymous

Anonymous Mon 15 Jun 2020 19:08:27 No.11801750 Report

Quoted By:

>>11801745
LITERALLY JUST FUCKING TELL ME HOW I DONT KNOW WHAT IM FUCKING DOING I HAVENT DONE MATHS FOR 20 FUCKING YEARS AND NOW HAVE BEEN ASKED TO HELP A TEENAGER WITH THEIR FUCKING HOMEWORK

Anonymous

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Anonymous Mon 15 Jun 2020 19:09:43 No.11801754 Report

Quoted By:

>>11801699

Anonymous

Anonymous Mon 15 Jun 2020 19:09:47 No.11801756 Report

Quoted By: >>11801767

>>11801739
yeah it's not right because he did it for a square of side 8 instead of 6.
see my answer >>11801722

Anonymous

Anonymous Mon 15 Jun 2020 19:13:05 No.11801765 Report

Quoted By: >>11801771

>>11801699
checked
area of a circle: pi*r^2
since there's on a fourth of a circle the area is: 1/4*pi*r^2
=28.27cm^2
to find white area: 36cm^2-28.87cm^2
=7.725
multiply by 2
=15.45
subtract total area by white area: 36cm^2-15.45cm^2
=20.55 cm^2

Anonymous

Anonymous Mon 15 Jun 2020 19:13:46 No.11801767 Report

Quoted By: >>11801772

>>11801756
How do I work it out for a square with a size of 6?

Anonymous

Anonymous Mon 15 Jun 2020 19:14:49 No.11801771 Report

Quoted By: >>11801773

>>11801765
This works thanks

Anonymous

Anonymous Mon 15 Jun 2020 19:14:58 No.11801772 Report

Quoted By:

>>11801767
you've got to be trolling.
>here's someone holding my hand to get the answer for a square of side length 8
>but how do I use this to get the answer for a square of side length 6?
if you're not OP congrats you got me

Anonymous

Anonymous Mon 15 Jun 2020 19:15:06 No.11801773 Report

Quoted By: >>11801784

>>11801771
got u, what class is this for?

Anonymous

Anonymous Mon 15 Jun 2020 19:17:08 No.11801784 Report

Quoted By:

>>11801773
Daughters GCSE maths

Anonymous

Anonymous Mon 15 Jun 2020 19:18:32 No.11801787 Report

Quoted By: >>11801820

>>11801699
36(pi-1)/2
Consider a circle of radius 6. Inscribe a square of width 6 inside and subtract its area. You will be left with the area of four slivers. The shaded region is equivalent in area to two such slivers, so divided your remaining area by two

Anonymous

Anonymous Mon 15 Jun 2020 19:28:28 No.11801820 Report

Quoted By:

>>11801787
36(pi - sqrt(2))/2 is the actual answer. I was mistaken about the side length of the inscribing square.

Anonymous

Anonymous Mon 15 Jun 2020 19:34:14 No.11801838 Report

Quoted By:

>>11801699
Anyone who wants to use calc is a brainlet. Find the area of a quarter circle and subtract the area of the (6x6) right triangle. Then multiply by two.

Anonymous

Anonymous Mon 15 Jun 2020 20:13:17 No.11801991 Report

Quoted By:

Area of the quarter circle = pi x 9cm^2
Area of triangle that subtends the quarter circle = 18 cm^2
Quarter circle - triangle = pi x 9cm^2 - 18cm^2
then double
~20.54

Anonymous

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Anonymous Mon 15 Jun 2020 20:33:05 No.11802082 Report

Quoted By:

>>11801699
lmao
let side edge be $a$ , then area of square is $a^2$
area of shaded segment in pic related is then $a^2 - \frac{1}{4} \pi a^2$
then area of orange segment in OP's pic is $<br />a^2 - 2(a^2 - \frac{1}{4} \pi a^2)$
simplifying this is $<br />a^2 (\frac{\pi}{2} - 1)$
for $a= 6 \mathrm{cm}$ this is 20.55 cm2.

Anonymous

Anonymous Mon 15 Jun 2020 20:39:17 No.11802105 Report

Quoted By:

>>11801720
>being this much of an obnoxious cocksure faggot while also being wrong
cringe

Anonymous

Anonymous Mon 15 Jun 2020 21:33:45 No.11802341 Report

Quoted By: >>11802349

>>11801699
We know the radius of the quarter circle and simultainously the length of the side of the square. That means we can know the (area of square) - (area of quarter circle) = area between sides of the square and perimeter of the circle, call it nigger. By the symmetry of the problem, we can get the result by substracting 2*nigger from the area of the quarter circle.

Anonymous

Anonymous Mon 15 Jun 2020 21:35:38 No.11802349 Report

Quoted By:

>>11802341
* 2 times nigger from the area of the square of 1 times nigger from the area of qrt circle.

Anonymous

Anonymous Mon 15 Jun 2020 21:43:57 No.11802387 Report

Quoted By:

>>11801699
20.55 cm2
its not that hard dude

Anonymous

Anonymous Mon 15 Jun 2020 22:53:06 No.11802692 Report

Quoted By:

>>11801735
I think we can safely assume those are 90° squares

Anonymous

Anonymous Mon 15 Jun 2020 23:06:21 No.11802746 Report

Quoted By:

>>11801699
Either propose a 2x2 equation stystem that one derives area from square and the other one from circlee, or use integration technique if you already have calculus knowledge.

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