I'm Indian

I'm Indian

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22

>22
Explain

36

34cm^2

Should be fairly easy to solve with these variables; three unknows (x,y and z) and three equations.
The first one would be x.y+(z-x)*y/2+(z-y)*x/2=16
Can't be bothered to work it out desu it's 2:33 in the morning.

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brainlet answer coming through, but I'm saying 35

40

y and x are irrelvant to the other polygons. you can't come up with other equations for them

Are you retarded? X,Y and Z completely define the whole problem. The one with area 20cm^2 for example you can devide into:
A rectangle with are y.z
A triangle with area y*(z-x)/2
A triangle with area (2.z-x)*(z-y)/2

Not to the upper right one I think
z*y+ (z-y)*z/2 + (z-x)*y/2= 20 right?
Or am I just a brainlet?

Or am i getting trolled because it's not necessarily a square?

Kek. I was too slow

also fucked up the lower triangle as I was typing it out

Might as well work it out now while you're at it.

Just to be sure i'm not a retard the lower left one is: (z*z) + (2z-y)*(z-x)/2 + (z-y)*x/2 = 28 right?

that's not how you solve it

Attached: the solution.png (1105x750, 73K)

Noice!
You solved it yourself?

My approach would also get you the solution but it would be less elegant i admit.

z*x+(2.z-y)*(z-x)/2+x*(z-y)/2
But this kraut bastard has already posted the solution so why bother.

I am not a kr*ut

pretty sure it wouldn't.

you'd just end up with variables you coudln't solve because you couldn't equate it with anything like he did with heights and the base line of the splitting triangles.

>r*skie
Even worse. Did you solve this just now or have you seen it before?

? You have three equations with three unknowns. It would be messy to solve and probably would require a numerical solver to make your life a lot easier but in the end you'd get the solution.

this is the first time I saw the problem

32

Good job Ivan. It's a bit cleaner than my way.

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Can you draw the partition?

>You have three equations with three unknowns.
That doesn't mean there is a solution.

You could connect it the same way I did. Imagine an extension of X in the 20cm polygon, called X1
and an extension of Y in the 28cm polygon, called Y1.

In that case, we get X+X1=Y+Y1=2Z
plus three equations for each polygon that capture the areas, like his original one, and there you go

Jealous of my BBSB (Big Beautiful Slav Brain)?

...

>being jealous of a slav
>ever
H-haha. Besides i graduated years ago i can be retarded now.

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It is a proportion problem:
16 20
--- = ---
28 x

solve for x
16x = 20 * 28
x = 20 * 28 / 16
x = 35

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