How smart are you, /b/?

the result is 32

More than 20cm². Less than 32cm². Impossible to say more without some extra information.

>>717106494
All the necessary information is there.

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>>717105172
28.

>>717105172
13

File: easier-to-solve.png (14KB, 375x375px)

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you can make 45, 45, 90 right triangles to make it easier to solve.

The father answered the phone, people usually have two parents.

>>717106573
Hardly. Not without whipping out a ruler anyway.

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>>717107212
You're just incompetent at math.

>>717107158
rule number one in math, don't trust the picture

>>717107212
Right you need like at least one angle or a side you could measure it with a protractor I guess, then solve it, but if i'ts not to scale then you're fucked.

>>717107422
then enlighten us

>>717107461
In this case you can. The = signs along the edge of the square mean that all 8 segments are equal lengths.

>>717107614
no shit but we need angles

If I had angles for 32cm^2 that would tell you if it's a kite (well I guess it applies to all) and that'd give you a side length

>>717106104
this

>>717107759
No need to be rude. All I was doing was saying that the picture could be trusted in that regard.

you guys are sad

>>717106585
I think you are right. I got 28 as well.

>>717105172
It's 42, the answer is always 42

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>>717108033
So these angles can be measured with a protractor?

its 40cm^2

The top and bottom of each side are congruent to a ratio of 5/4

ok mathfags correct me if I'm wrong I'm going on a stretch for this one

all areas equal x so

4x=4x

but we know 3

78+x=4x

-x

78=3x / 3

x= 26

since all sides are equal this should work...

It's 24cm^2 also OP is a fag. And because of that I won't show you my work.

>>717105172
221 newfags

34 /end thread

>>717108428
Fucked up my own math

25.6cm^2 because I switched which ones I was using

I'm very smart. (You never said "pic related")

not doing your homework for you fam

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26

>>717108476
If you break it down into 4 squares, each is 28cm ^2

>>717108476
No.

The left-right ratio is the same on the top and at the bottom.
Grey Area :
G = 32 / 20 * 16 = 25.6 cm2

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22

>>717107158
Give us the length of a side, or one =

>>717108900
you can't use ratios you fucking cuck maybe if we had TWO shapes with a different scale

well, it has 68cm2 already and is quadratic,

we know, that the triangle is smaller than 16cm2

so: (h/2*h/2/)/2 <16
h^2/8<16
h^2<128
so its something smaller than 11,2 or so. as a length

>>717109455
You don't solve shit by finding out what it can't be. You solve it by finding the actual value(s).

File: protractor-a-shit.png (101KB, 375x375px)

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Protractor incoming.

the total area must be the result of a number squared, use subtraction. 20+32+16=68 so missing area is 32 and pic is deceiving. 32 +68 =100=10^2

>>717109772
doesnt have to be an integer squared you fucking degenerate

>>717109772
Do tell how you pulled 32 out of your ass for the missing area.

>>717109772
You are baselessly assuming the sides are some whole number of centimetres in length.
As far as we know, they could be any real number.

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So this is measured in pixels, using this and the protractor image you might be able to solve it, if you are going entirely off the image, and the area will be in pixels, if you ignore the cm unit entirely.

>>717108166

I'll share my solution.

Since division of the square always start in the middle of one of the sides areas of the opposite parts always sum to half of the total area.

Therefore 32+16 = x + 20, e.g. x = 28.

You can see for yourself that if we started with a square cut directly through the middle from both sides and then moved the point that all the parts share, any part would gain all the area its opposite part loses and vice versa.

i smell bullshit

>>717109772
is it even possible for area x and y to be of the same value if a and b are different while there is one middle?

>>717105172
... at least I know that I'm not your personal army.

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>>717110202
you can't fucking assume anything from the picture with out proper math and angles we don't know shit besides its a square how the hell are you this incompetent?

Is everyone just gonna ignore >>717110239 ?

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>>717110202
hey retard this is what I mean by you assume anything did you even pass high school math? they stress like hell do not trust the picture

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>>717105172
hmm 16^2 thats 256... there are RAM 256MB.
hmm 32^2 thats 1024 thats 1GB... there are RAM 1GB.
hmm 20^2 thats 400... theres thee file name

>>717110239
can't do that both halves will end in a different solution 20&32 = 16+x

x= 36

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>>717110902

>>717111573
is it even worth explaining how retarded you are? cause I feel like you wouldn't understand how retarded you are if I were to explain it...

refer to
>>717111301

so we know, the area is smaller then 32 cm2 obviously.

68cm2 is less than 3/4 so its at least 23cm2

the quadratic sites are between 10 and 11,31 cm long.

>>717111463
Anon meant you add opposite corners.

>>717111688
Well to be fair, if OP is gonna ask a retarded question he deserves a retarded answer.

>>717111688
Meh its a semantic argument at this point.

>>717110239
>Since division of the square always start in the middle of one of the sides areas of the opposite parts always sum to half of the total area.

Prove this and you have a valid answer.

>>717111688
I remember in math books there were a few occasions when they said it was alright to measure angles with a protractor, and that the images were to scale, in order to practice using a protractor, in that context doing something like this is acceptable.

>>717105172
I got approximately 18 but I could have typed something wrong in the calculator.

>>717108385
You would have to verify the area of each section independently. Otherwise you could write the same numbers and move the center point around anywhere else.

The answer is 28
The sum of the areas in two opposite shapes is always half the area on the square.
Can be seen by drawing the smaller inner square, and drawing the heights of all inner triangles. Every two opposing inner triangles add up to half the inner square.
Therefore 32+16 = 20 + x
X = 28

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A hunch tells me the fact that the bottom left is half the area of the top right could be useful.

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>>717112348
You will need more information, this is the poper way to solve it, there are no side measurements, or angles it is impossible. How do you find the length of one side of a right triangle if you have one of it's angles and one of it's sides?

68 = (3/4) whole

(68/3)*4 = 51

which looks wrong idk

>>717113000
it is wrong, in every way.

you have to divide by far and multiple by 3, and its still wrong

>>717113000
it is wrong.

you have to divide my 4 and multiple by 3

>>717112666
could you draw this one out? I have a hard time understanding what you mean with the small square part.

and checked

File: another-way.png (19KB, 375x375px)

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Another way to ask this question is what is the area of the yellow section? or any of the green or blue sections, the answer is always unknown.

>>717112953
Faggot, this picture solves it.
It's 28.

24

>>717113973
>>717113502

See the image in 717112953
Look at the inner red square. Now think of the area of two opposing light-green triangles.
The sum of their heights equals the side of the red square, and their bases are each the side of the same square. Therefore their area is half that of the red square.
Keep in mind the the red square is exactly half the area of the big square, and that each of the teal triangles have the same area, equaling 1/8 of the original square.
Therefore the sum of areas of each pair of opposing shapes is 1/2x1/2+1/8+1/8 = 1/2 of the original square.

well, i try again.

we now every green area is

20-x
32-x
16-x
y-x
(20-x+32-x+16-x+y-x)=(sqrt(2*(l/2)^2))^2
(68+y-4x)=(2*(l/2)^2)
(34-2x +y/2)=(l/2)^2

as you clearly see wee have two unknown thinks we can't solve any further. its just not possible.

thanks for wasting 1,5 hours of my life OP

>>717114130
Are you asking srsly?
Your asumption is false, it's completly different thing. Also, the correct answer is already up there.

>>717105172
solved step by step :
https://www.youtube.com/watch?v=SiMHTK15Pik

>>717114321
>>717113502
All of the areas of all the sections except the blue are different.

>>717114425
1.5 hours and still couldn't find the answer, kek. It's perfectly possible to solve it, faggot.

>>717114569
What if you get rid of the numbers from the original picture and replace them with whatever you want will it still be able to be solved?

It's primary school level exercise. You faggots must be really retarded, if you can't solve it.

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>>717113000
>>717112666
checked

>>717114701
Yes it will.

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>>717110239
I proved the result you used to come up to that answer. The result is valid, therefore your answer is correct.

>>717114876
>>717114769
How long would it take a primary school level student to solve this problem?

>>717105172
No enough information, and this

>>717107158
is still not enough as no angles and/or lengths are known

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>>717115338
>Correct solution has already been posted
>Still claims the puzzle is unsolvable

>>717115079
Right there is no way to find a unit measurement, but several proofs of equality can be made.

>>717115114
Not more than a 45 minute long lesson, that's for sure. +with that hint >>717107158 it's easy AF

>>717112788
Dubs never lie

>>717115079
Nice.

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No idea if I got this correct, but maybe? Done in autocad.

>>717115079
>>717115510

now please calculate h, so
h=sqrt((l/2)^2*2))

where l fulfills l^2=A+B+C+D

and A,B,C are 32,20,16

>>717116025
Yep this is the most correct answer in the context that the picture is to scale.

28 is the correct answer

>>717116298
>>717116025

yep, every fucking primary pupil could calculate 27,6009.

diphit

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The triangles with the same letter in this diagram have equivalent areas as they both have the same base and height, so now we can express the areas algebraically.

A+B=20
B+C=32
C+D=?
D+A=16

Rearranging the second and fourth equations gives

C=32-B
D=16-A

Substituting into the third equation gives

32-B+16-A=?
48=A+B+?

But A+B=20, so 48=20+? and therefore ?=28.

>>717116234
Sure.
B+C=32+16=48=A+D.
l²=A+B+C+D=48+48=96
l=4sqrt6
h=sqrt2×l=sqrt2×4sqrt6
=8sqrt3

>>717116993

Actually seems like the right answer here. Mine was only to scale of what the drawing was, so the numbers are slightly off (in the ten thousandths or something). GG.

Now find the lengths of all the line segments?

>>717116025

>>717116993
The logic checks out and you arrived to the correct answer too. Well done.

>>717118785
>>717116993
>>717113502
Followup question what is the area of the yellow section?

>>717115079
What is x in this, you never specified it?

>>717119893
The height of B'.

File: B0QjURiCAAA6ELY.jpg (11KB, 312x247px) Image search: [Google]

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THE PICTURE IS NOT TO SCALE.

My answer for the unknown area is approximately 46.575

this is a really cheeky one but it's not that difficult if you find yourself a solid assumption to follow.

>>717119329
Since the whole area is 96, each blue section must be 12. That means the green triangles are 8, 20, 4, and 16.
The yellow triangles are a bit trickier. They require you to expand the green triangles into overlapping rectangles all with the same width, and from the overlapping you figure out the area of each yellow triangle. Not going to be pretty, though. You'll need to manipulate an 8×9 matrix to solve them all.

>>717105172
28 obviously
have you never had high school education

>>717121481
Well done, you read the thread.

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>>717120471

Give me some guys, i'm composing my full proof right now. The answer is definitely and certainly 46.575

>>717119329
(16-12)×(20-12)/(20-12+28-12)=4×8/24=4×1/3=4/3.

That one in particular is 4/3.

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>>717115079
/Thread

thanks!!!!!!!!!!!!! i really apprieciate your proof - it is very, very easy to follow - although i could not come up with it on my own (i didn't attend to the math class for a long time). at first i though one has to calculate the angles which is not the case.

>>717107158
nice 3d effect

>>717121751
kek
Are you seriously implying that this question can't be done by the majority of people without getting help