Autism test: If you cannot solve this simple math problem, then you have autism

Autism test: If you cannot solve this simple math problem, then you have autism.

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cymath.com/answer.php?q=x^2+(x/3)^2=30
cymath.com/answer.php?q=x^2+(x/3)^2=30^2
youtube.com/watch?v=rs66DAJVQQY
twitter.com/SFWRedditGifs

x-(30-x)

There isn't enough information to get a numerical value for x

15sqrt(2) = x

28,46

Does AE=CF?

yes

route of 27

There is plenty of information, you autist.

It asked for an exact value.

x = (30sqr) - (FCsqr)

Sqrt 810

sorry was wrong

x = 3*sqrt(3)

26

9*squrt(10)

that's already the right answer

Please explain

You need to know the measure of an angle

x^2 = sqrt (30^2 * 9 / 10)
x = sqrt 810

Angle DCF = 90 degrees

Top kek

i meant
squrt (x^2) = sqrt (30^2 * 9 / 10)

uff typing is pain today

120

That's backwards. Autists should have an easier time doing math than normal people.

sqrt x^2???

shhh....
let OP feel smart and special

each diagonal line is the same length as the outer ones

sqrt (x^2) is x

Since the square's sides are cut symmetrically by E and F in 1/3rd parts, FC is equal 1/3x.

So:

30 = 1/3x * x
90 = x^2
x = sqrt(90)
x = 9.49

It's a square, you nerd virgin

The exact value is sqrt(810) = 9*sqrt(10)

all right
AE = 3AB
let AB = alpha
so alpha² +(1/3aplha)² = 30²
10/9*alpha² = 900
alpha² = 810
alpha = sqrt(810) = x

4√6

Because the values of the known lines are the same and in a square, Line FC is 1/3 of x.

Using the Pythagorean theorem yields
(x)^2 + (x/3)^2 = 30^2
x^2 + x^2/9 = 900
(10/9)(x^2) = 900
x^2 = 810
x = sqrt(810)

>square
>length

x = AB

ups, it's ofc:

30^2 = (1/3x)^2 * x^2

How can we be sure AE = 3AB

since it's a square from the beginning
you take the lower triangle
one side 30
second side x
third side 1/3 * x (because it's divided evenly)

after that it's all phytagoras
30^2 = (1/3 * x)^2 + x^2
900 = 10/9 * x^2
810 = x^2
28,46 = x

(are we doing your math homework and you are to afraid say you aren't able to do it on your own?)

Man... I hate math

26

pythagoras
(have i mentioned typing is pain today?)

But there's no information informing AE is 1/3x

because math, the square is made of six congruent triangles

Man... I still hate math

it has to be because all lines in the box are the same length

evenly in a way 1/3 and 2/3

9*SQRT(10)

45

that's more or less the same like 28,46

triangles!

Yeah I see it now... Honestly I'm destined to fail math

i still think that we are doing his homework...

You can use the Pythagorean theorem to solve right angles.
Because the angles are all the same length (30), and they split the square 3 times, you can assume that lines AE and FC are 1/3 the length of x, so they are x/3.
Now for the math...
x^2+(x/3)^2 = 30^2
Simplifying this gives you this:
(10/9)x^2 = 900
Now divide both sides by 10/9
x^2=810
Square root it all and your done
x= √810

it's 3√3. simplify your expressions kiddos. here's a proof:

cymath.com/answer.php?q=x^2+(x/3)^2=30

didn't i just write the same thing?

The confused guy here (me) isn't even OP... I'm just relearning math

28.284271247461900976033774484194

oh thanks, that sounds really graceful like a true mathematician

u can use the angles if want get it easily

9 root 10, you forgot to square the 30

Pythagorean theorem bitch

You used x^2+(x/3)^2=30

You should have used x^2+(x/3)^2=30^2

shit you're right

cymath.com/answer.php?q=x^2+(x/3)^2=30^2

cos(30)*30=x
x=26

hihi. i knew something was wrong. it just sounded to neat and the value was not enough

AB=x, AE = AB/3, therefore AE=x/3

Using Pythagorean theorem
(x/3)^2+x^2=30^2

(x^2/9)+x^2=30^2

(x^2)[(1/9)+1]=30^2

x^2=(30^2)/(10/9)

x=[(30^2)/(10/9)]^1/2

x=28.46

Women's studies major.

Yes, you did

lol angle isn't 30 degrees mang

This was probably OP's math hw and you faggots just did it for him.

when i was in school we were taught to 'use our resources'. i looked shit up online all the time and argued that was what i was doing. worked

lol

Just take the triangle DEF, since DF and EF have the same length, all the angles have to be the same (60°). So now we know the angle CDF (30°) and solve it by simply using pythagoras.

x=25,98

ok op just bookmark cymath.com and never post a thread like this again lol

so another question to you op:
what's the angle between BE and EF?

this is what I got.

The whole term x/3 is squared, not just the x part.
Apply yourself.

3AE=AB you mong

post another problem faggot

lol nope, try again
workin on it, one sec

i already did it for him. angle between BE and EF. do you know it?

if b is the angle between B and E coming from F,
b+b+c=180
with c being the angle between 30 and 30
in the corners we have
b+a=90
cut a line in between angle c and you get
c/2=a
solve system to get b=60 and c=60
you can figure out the rest.

wait that doesn't prove shit

nope, it's something between 31 and 39 degrees. take sin or something similar

you cant apply pythagoras because we only have one side, therefore theres an insufficient amount of information to work out x.

x*2 + 1/3 * x^2 = 30^2
4/3 * x^2 = 900
x^2 = 675
x = sqrt(675) = 25.98

already close?

3*sqrt(3)
AD and BC are parallel, and, if you take the angle of DF makes with DC, take the secant of it, and multiply by x, you get 30, but secant is one to one on [0,pi/2], and secant is an even function, so we can infer that triangle DCF is congruent to triangle ABE, and triangle BEF is congruent to triangle EFD. Not only that, but the right triangles produced by cutting the isosceles triangles in half produces 4 right triangles that are congruent to ABE and DCF. This shows that FC must be 1/3 of x, so x^2/9 + 9x^2/9 = 30, so x^2 = 27, x = sqrt(27) = 3sqrt(3).

you can find the other one if you try

it is sovleable use some theums niggguh youtube.com/watch?v=rs66DAJVQQY

should i tell you?

go for it, don't have a calc on me and cant remember the sin cos tan tables lol. it's been far too long since school

fun problem.
split the sides up showing that one side of each right triangle is x/3

i.e. CF = 2 times the length of BF

then pythogrram that shit and it comes out to
x^2 = 810
which is like x = 9root10 or something ugly.

Way to start a thread OP.
ur still a fag tho

30^2 - x^2 = 900 - 675 = 225 = 15^2
FC and AE must be 15, because squares have right angles. But BF and ED must be sqrt(675) - 15 = 10.98. 15>10.98, but x*sec(a) is not equal to x*sec(b), where a and b are two different, non-negative angles on [0,pi/2], so 30 must not be equal to 30. If a and b are two different angles with different signs, like the picture suggests, then BF is larger than FC, so 10.98 is larger than 15.

you already have some information from your question/answer before. take the divided triangle. one side is 30 the other one sqrt(810), the short one one third of sqrt(810).

after that
2 * sin^(-1)[1/3 * sqrt(810) / 30] = 36.87°

Yup, it's sqrt(810)
smallest angle is about 18 and the angle between the 30s is about 36

i'm not op btw lol, i'm an old man who likes math problems

do your own math homework faggot

This is one cheeky ass question, and y'all got baited.
Triangle DEF is an angular bisector, all corners are 60°. By using the law of Sine, sin(alpha) /a = sin(beta)/b =sin(gamma)/c. We can deduct that DE is 30.
The question stated that it gave a square, meaning AD=DC=x.
AD is made up of DE + EA. So x>30.
This will deem impossible when we refer to the Pythagoras. For DF has to be longer than AD.
The question is therefore incorrect.

30