You should be able to solve this

You should be able to solve this.

You dont know me or what I 'should be able' to do.

EZ PZ if you use integrals
/thread

thought it was quite easy but then I saw the bottom left segment is white
no idea lad, I haven't done any maths since really easy, basically A-level shit in undergrad

there is not enough information provided to solve the problem
so I can solve this one ;)

Maybe if i had my math notes and 10 min of spare time user.

bump

Dont blame the pupils. Blame the governmenet for the schoolsystem in many countrys especially in north America. I heard its really fucked up in some regions.

thers' anoter (intelligent) whay?

Do it then newfriend

Is it 21.46?
>tfw sucked at math in school and still do

i could, but it would be a pain in the ass

not because of picattached

No.

Then stfu.

If the lower left cornor was filled out then yes. Since it is not, you have to do actual maths. Nothing too hard, but hard enough that i don't want to waste time on it.

10 x 20 = 200
π x 52 ≈ 78.54
78.52 : 2 ≈ 39.27
dunno about that white corner to the left

>just realized I didn't even think of taking out the corner
Brainlet feels

no u

Red Area (Ra) = Triangle Area (Ta) - Circle Area (Ca)

Ta = 20 • 10 • ½ = 100 a.u
Ca = 52 • π = 25π

Ra = Ta - Ca
Ra = 100 - 25 • 3,14
Ra = 100 - 78,5 = 21,5 u.a

That's how I solved it

Find half of rectangle area.
Subtragt circle area.

Then you just have to find a way of figuring out the missing spot. No idea about how you do that.

lower left corner little brainlet

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2+2=4-1=3 quick mefs

Red area = Triangle Area - 1 Circle Area - (10 Square Area - Circle Area ) / 4

good try user

You're missing that small area tho

Approximately 19. Kinda dumb how the area in the bottom left was left unshaded.

Nice try. But look again

its 20

I missed a /2 on everithing!

Best try so far, but this is also wrong.

Note that the answer must be a bit under 4(200-50π)/8

Still too big.

About the little white corner:

We can take half of the figure and make a square. So we have:

4 Little Corners (Lc) = Square Area (Sa) - Circle Area (Ca)

4Lc = 100 - 78,5
4Lc = 21,5
Lc = 21,5 • ¼
Lc = 5,375

Now that we've found tha missing corner, we just take it away from the total.

21,5 - 5,375 = 18,125

ANSWER: The area of the red region is 18,125 a.u

i don't know but i assume the answer is somewhere around 10

Too low and not rigorous. But it looks like you have the basic idea of a Riemann sum floating around in your head somewhere.

I am dumbfounded by the fact that you guys can make the same mistake 5 times in a row.

LOOK AGAIN

>4(200-50π)/8
>4(200-50π)/8

My solution is 19,50

100-95×pi nigger ?

Damn!

>4(200-50π)/8
>Still too big.

My solution () is 19,50

>> 752608786 #
Wut?

AB=((10^2+20^2)^(1/2))/2=5*5^(1/2)
AC=((AB)^2-(10/2)^2)^(1/2)=10
R=AC/2=5
Sc=π*R^2=25π
S=(10*20-2*Sc)/2≈21.4602

I...

I can't even...

What ?

>19,50

Closest so far. Now give me a closed form.

Hey guys, can help me real quick? How do I answer someone? Like, directly to another user?

>yfw people don't fail so much at the math but fail at looking at a fucking picture

just use πr^2 with 5 as your radius
times this by two, as we have two circles, giving around 157.08
200 (the area of the rectangle) minus 157.08 gives 42.92
divide by two for 21.46, the shaded region

nigger did you just use the pythagorean theorem to find out the circles radius is 5? how stupid are you that you can't divine that instantly from the picture?

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THANK YOU SO MUCH! HAHAH

Yust put a 2 under The first integral is 1, but I can only approximate the second one.

>Hey guys, can help me real quick? How do I answer someone? Like, directly to another user?

Negative area?

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you fools, the rectangle has a third dimension

>but I can only approximate the second one.

Looks like you're boned then.

Now select part of the text in a post before clicking the postnumber. Prove you tried by quoting this post.

15

Intersecting line is y = x/2. Circle is diameter of 10, you can use that information to plot the circumference of a quarter of the circle as a line. The area under the line (integral) created by their intersection (y = min(both functions)) is the area of the not red bottom left zone. You could do some flips and calculate the top right red zone by itself if you wanted to instead. Probably easier that way since you can just add that to 0.75 * (100 - pi * 25) for the answer.

I dun wanna do the integrals because I don't want to turn my brain on, so you can just have the logic for the correct answer instead.

I got 175-π225/4

It's around 21.5

Its really not that hard

20 * 10 / 2
-
1 whole circle
-
that little bit I can't calculate bottom left

gg

AB and CD immediately prevent one another so yeah ok

I'm an engineer user, I don't need closed forms. ;)
I will monitor this thread for the exact solution.

that's a negative number my guy

>Now select part of the text in a post before clicking the postnumber. Prove you tried by quoting this post.

Did it, i guess

This. Without calculus, it's actually pretty cumbersome.

All of the area that's not a white region.

21,46?????

>Now select part of the text in a post before clicking the postnumber. Prove you tried by quoting this post.

Did it, I guess hah

>I don't need closed forms.
Then get out of my class. TBH, you did good though.

there i even made it harder

100 (half the area of the rectangle) - pi r ^2 where r is 5 (half of the height of the rectangle)

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Lines do not need to be straight

is that right?

68.6?

What is wrong, man?

too ez, gimme another one

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...

>This

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pretty sure u can do this with basic geometry and trig, dont have time to try it atm tho.

2 lines is enough

Easy.
Find the area of the rectangle which is 10×20 so 200, then we can assume that the diameter of the circle is 5 each since they take half of the rectangle each. Then we can do py × r (2.5) squared which is 19.6 . Then we can do 200-19.6-19.6 since their are two circles. That would leave you with 160.8. . We can divide that by 6 since there is 6 segments that arent cricles. That would be 26.8. Then since there is three red segments we'd do 26.8 × 3 = 80.4.
Solved.
The answer is 80.4 (whatever measure) squared

Not OP, but this is right. Area of triangle minus area of circle minus one of the small corners. 100-78.5-2.6875=18.81

To be honest I'm only here for the Kurisu pics

18,7775????????

Am a maths noob, don't know if this makes sense but:
Area of triangle-circle area=100-25(pi)=21.46. So area of bottom right corner red is 21.46/4=5.365.
use tan(x)=o/a to get top right angle 63.435. As a fraction of 90 degrees the red area (bullshit potentially imminent here) is (63.435/90)x5.365=3.781.
If that's true, then total red area is (3x5.365)+3.781=19.876.

As stated, I don't know what I'm fucking doing.

>19.67
Because 20 is double 10.

You didn't use three lines, therefore it's wrong user. IT"S ABOUT THE METHOD NOT THE ANSWER. I don't care if you can do every derivative in your head, you need to write down the stupid long way formula or you don't get points!

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Way off. The way that the hypotenuse bisects the circle allows you to take the area of one circle and subtract it from the area of the triangle. Also, you calculated the area of a circle like a dipshit. 2.5 is not the radius

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as you say

Fine, here it is with 3 lines.

better?

Dank.

how to calc area of one of the small circle segments, thats all i need to solve