Are you?

Are you?

>>/sci/

Maths are created by man, therefore not 100% accurate.

24

>by intuition

it's impossible

36

Assuming a perfect square, area is n*n.
n = sqrt((20+32+16)+x), where x is at least the average of the other three. Hence n ~
sqrt((20+32+16)+(20+32+16)/3) ~ 9.52 ~ 10
(20+32+16)+x = 10*10
=> x = 32cm2

What do the = signs all around the square mean?

parallel to opposite side, implying its a perfect square
(i think, i'm just guessing but its seems lojikol)

28

now fuck off

X is clearly smaller than 32.

All lines with = have equal length to each other.

entryways and fire exits

why isnt the water leaking out then

>comic sans

25.6

>knowing what comic sans looks like

nothing wrong with gomic sans

My nigga

total area = x.
20^2+32^2+16^2 = 3/4 x.
20^2+32^2+16^2 = 1680.
1680/3 = 560.
sqrt 560 = 23.66.
23.66 X 4 = 94.65.
94.65-(32+20+16) = 26.65

A = 26.65^2

>the side is sqrt(96) long

this is wrong, you cant just assume that it is equal to 3/4ths.

>3/4 x

Sorry but you don't know that

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jesus christ, you're retarded

Here are some of the possible answers. All of them are correct.

lines in the center are all comeing from the midpoints of the outside lines.

go to bed user

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pre sure all steins gate questions are supposed to be a paradox

Holy shit you're all retarded, only one person got it so far.
Since the area of the triangle is base*height/2, all the triangles marked with the same letter have the same area.
We know that a+b=16, b+c=20 and c+d=32, and we want to find a+d, so we calculate (a+b)+(c+d)-(b+c)=a+d=16+32-20=28.

Is this a bait?

Thanks smart user I understand it now
>mfw

how do you know that c=c,a=a,b=b and d=d?

They don't, same base length and one side is the same so since they are different shapes they can't have the same area, it was bait probably

yeah, i figured. since he just said that only one guy got the answer right and didn't even tell who it was

>mfw I was dicking around in Matlab for 30 minutes and came up with 9 quadratic equations and a solution of 11

because they have the same base and height
the base is the same because the problem tells us so (the || sign represents congruence), the height is the same because the bases lay over the same line (you measure the base as the lenght of the segment orthogonal to the base that touches the origin of the red segments)

He's a witch and solved it with his evil witch magic.Bring forth the stakes, you shall burn for your beliefs, witch!

nice bait. are you referring to the line going from the midpoint as the height? height should be perpendicular

What this user said. I was slow with my gimping, here's an illustration anyway.

>Since the area of the triangle is base*height/2

i'm out. fuck you retards.

Kek, retard

>but I was only pretending

height is measured like this, you prolong the line of what you took as base and draw an orthogonal line which goes through the opposite angle

was for

Cool

>tfw math god
i was the first one to solve it and did it within the first 3 minutes of the thread

Let the length of the =ed intervals be a. Draw lines between the halfway points on each side, so the sectors are split into two triangles, of size a2/2 and unknown. Suppose the coordinates of the midpoint are x and y.

The three known sectors then give rise to equations:
2*16 = a2 + a (x+y-a)
= a(a + (x+y-a))

2*20 = a(a + (x+(2a-y)-a)
2*32 = a(a + ((2a-x)+(2a-y)-a))

Per wolfram alpha, we get a=2sqrt(6), x=sqrt(6), y=5sqrt(2/3). So the total area is 4*4*6=96, and hence the missing piece is 28.

That's clever. And I'm slightly mad I didn't figure it out myself.

Stop with the samefagging retard

(slight trick involved in seeing that the unknown piece of the bottom left sector is of size a(x+y-a)/2 and similarly for the others - the more direct way of measuring its area would give sqrt(2)a*h for some height, but you can parallel-transfer the center point to the bottom edge of the big square (it will wind up at x+y) to get a different triangle of same area, and use the (x+y-a)-length piece as the base and a (the bottom half of the left edge of the square) as the height)

Ah, that's much nicer than my solution (54442535). Feeling kind of dense for not seeing it now, but it's been ages since I last touched plane geometry.

Hello, desktopfags.

Pretty sure all weebs need a good ol lynching

if you're thinking about the probability questions, almost none of them was a paradox

what's the answer you non-retarded asshole?

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As both A's (or whatever) have the same base and share a side, the area must be difference as the length of the third side is different, I hope you leave

>Holy shit you're all retarded
>all the triangles marked with the same letter have the same area.

except the center point of that image is -x, -y (not center) you theory is wrong, not one of those letters share a value.

This is why you can't get into a decent college, Sup Forums.

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British?

The triangle in the red box on the right is not half of the area of the box you mongol

Australian.

Ah, you draw your xes like )( down there too?

If you are referencing Heron's formula, then you should consider that for a triangle with two sides of given length, and given area, you get a quartic eqaution for the third side, meaning there are multiple solutions for the third side's length.

It's not meant to be the center, since then all triangles would be of the same area anyway. See and .

As long as the third side is different so is the area, which is the case

You better be trolling.

Am I the only one who gets infuriated by this maymay?

yes, but the rectangle on the right should be bigger than the one on the left (it isn't because I didn't draw the bases properly, let me redo it)

No all the triangles marked with the same letter do no have the same area

Math is like the one accurate thing we do.

Our interpretation can be shit but we tend to get it eventually.

See image for a counterexample to your argument. Try thinking about it long and hard before you post again.

Krazer is that you ?

Here's a counterexample. Both the ABC_1 and ABC_2 triangles have c*h_c/2 area, and they both have sides of length a and b, they only differ in the third side, yet they have the same area.

A does not equal the other A

and so on

this is incorrect, you can't assume a=b
imagine if the bit where the lines meet in the centre was close to the top left of the diamond you drew

I literally just thought

16 is 75% of 20

24 is 75% of 32

Why go that far?

Which || correspond to which ||? Question should be more clear.

Now you're just making stuff up, the ones in the picture are not right angled

Bothered by the fact that you didn't label it a,b,c,d and then have a+c = b+d (because, on the diagram, the two triangles labelled 'a' aren't equal and same with the 'b's).
Otherwise... yeah, I'm ashamed I couldn't solve it.

No, he's actually right, I think he means that the sum of the areas marked with "a" is the same as the sum of the areas marked with "b", which is correct,and implies that 20+x=32+16.

I assume he means the total area of everything marked a and b respectively. That checks out - the heights of each pair of triangles sum to the same thing.

5/10. You made me waste some time drawing that picture, but this followup was a bit too obvious.

They're all equal.

Threads like this are the best way to see Dunning-Kruger in action. The sad thing is that the actual Dunning-Krugerites will read the preceding remark and it will not even occur to them that they are the incompetent ones.

properties of a triangle: Median
>A median of a triangle is the line segment that joins any vertex of the triangle with the mid-point of its opposite side.
and
>A median divides the area of the triangle in half.

Right angles or not, if the thing you said about two triangles where two out of three sides are the same between them only having the same area if the third side also is of the same length were true, wouldn't that necessarily imply that the two triangles at the bottom right can't both have area 2? Their baselines are both of length 2, and they share an edge, exactly as in OP's problem.

Eh I don't know about that one. That seems a little weird. I don't see how you can make the assumption that you did

>100 hours in paint

fuck you Sup Forums

So how do you find out the answer?

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But people didn't want to believe that indeed sections a equals a in terms of area and b = b and so forth.