Find X

Do your own homework brainlet

Alternatively an elementary schooler can imagine that there are two 3x3 squares in the middle, calculate that 12-3 = 9 and 9-3 = 6, and sum that it comes out to 15.

20 hours in MS paint.

It's there on the left side and on the bottom.

what makes me even more curious...why would one want to know? Is there a practical example of making use of it?

the lines aren't parallel dummy

It's a square you retards
You cannot draw two parallel diagonals
The diagram is wrong

OPs pic is wrong too. those angles can't be 90*

>2 right angles
>not perpendicular

>Basic neurology
kek

who said about diagonals?

Fixed

How would you go about drawing two parralel lines using a straightedge ruler?

>using approximations

Use the corner of the ruler to draw a right angle to the line. Then draw another line at a right angle to that line; the resulting line is parallel to the first line.

Holy shit, dumbest post on Veeky Forums

I truncated for the sake of drawing, but not on the Google calculator. It gave me 15 exactly.

Calculators use approximations too

Good job. Now you know the lines are parallel.

what if a right angle is 91 degrees in this space?

really trigons my metrics.

12 *[v1,v2] + 9*[v1,v2] + 3*[-v2,v1] = [x,x]
21*[v1,v2] + 3*[-v2,v1] = [x,x]
(21)v1 - (3)v2 = (21)v2 + (3)v1
(0.75)v1 = v2

sqrt(v1*v1 + v2*v2 ) = 1
=> v1 = 0.8
=> v2 = 0.6
=> x = (21)*0.8 - (3)*0.6 = 15

x = 15

How exactly do you know that the sum is x?

>brainlet here

All of these solutions just seem to be nuking the issue to me. Can't you just do,
(sqrt(12^2+1.5^2)+sqrt(9^2+1.5^2))cos45
which comes out to 15.003

That's just an approximation. The diagonal doesn't bisect the 3cm line. The actual ratio was calculated by

what is Y?

>Find X
Here it is.

ITT: Brainlets

The angles are both 90 degrees, the diagonal must perfectly bisect the 3cm line.

no body promised that it is a square.

Where did the 21 come from?

Look at the problem again, but closer, nimrod

>doesnt know that 12 + 9 = 21
>calls me nimrod
what are you stupid

12+9=21

Was just about to post this

Hilarious how Veeky Forums tries to solve this by fucking trigonometry, it is just simple geometry.

maybe he didn't instantly realize it

btw if it wasn't a square the sides would have been called x and 'another letter'

Line '9' and line '12' are not parallel. No line '3' exists that is perpendicular to both. The diagram is impossible as presented.

Thank you

Nah, I drew it IRL with a ruler, it checks out

Y is the length of the short side of the top triangle created by connecting the square's corners.
If both diagonals were the same length, yes... but they're not. There's a lot of confusion in this thread because as drawn, OP's diagram is misleading. It's actually drawn as a rectangle and taller than it is wide, which is why both diagonals can be depicted as parallel.

It's basically the same as except dumbed down for children who don't understand how to solve systems of equations/what they're doing.

Or wait, maybe it's even clearer like this.

Here's the video where the image came from
youtube.com/watch?v=v80jDho0emQ

Okay, so in the question the numbers are all clearly multiples of 3, so you can divide through to get the numbers on the zigzag diagonal as (4,1,3) giving x=5.

My question to you is, what is the next smallest integer x such that you can have diagonal (a,1,b), with a and b both integers?

If I'm any good at pattern recognition, you can get x=4 with (3,1,2).

Wait no, that's x=3, I'm retarded

Is this a troll? The shape cannot be constructed
Draw a diagonal and break it like in the OP image, there is no instance of those angles are equal to 90

You can get x=4 with (4,1,2) then.

No.
[math](3,1,2) \implies x=\sqrt{13}[/math]
[math](4,1,2) \implies x=\frac{1}{2}\sqrt{74}[/math]

I tried to solve it like pic related and got x=15. I'm not sure if it is calculated in the correct way.

That's great. So is AndIntuitive physically-correct explanations.

There is, do it yourself, draw the center lines first and then the square around it

I tried to draw it after calculating it like . Sorry, it is quite messy, but I hope you can see that it's 90 degrees.

Assuming this is a flat square and not some bullshit:

The red diagonal can be found using Pythagoras and is equal to root(3^2 + 21 ^2)

this is also the hypotenuse of the right triangle with sides x and x therefore

(3^2 + 21 ^2) = 2x^2
225 = x^2
15 = x

amirite?

not a square?

Yes, that is correct.

it can be a square or it can not be a square, to find x you have to assume it is a square because there isn't enough angle information otherwise

it CAN be a square,and 3 lines with measurements on them can be simplified to a diagonal line across that square, you can make a square with any measurements on those lines in the middle

if you assume this is a square, you can find that x= EXACTLY 15

Yeah, I realized after posting that point E could be positioned differently.

Imagine the point A as a pivot for the three segments AC, CB, BD. As you rotate them down, AE(=i) will decrease and ED(=h) will increase.

At some point i will equal h

ye, that works.

>mfw this asshole has tons of videos and views

...

x/y + y/x = (x^2 + y^2)/xy = m
(n/m * x/y)^2 , (n/m * y/x)^2
^ s , t
s + r^2 , t + p^2
^ a , b
sqrt(a) + sqrt(b) / sqrt(2)
^ all possible desc. x

...

those lines CANNOT be parallel if they both start for opposite vertices, therefore the angles CANNOT be 90 degrees

Yes they can.

Proof: Think.

Nigga. 18. It is 18.

>18

Fucking finally. OP here, study guide agrees. Still asking how.

It is 12

Anyone who still thinks it's not 15 look at

obviously he doesn't accept the parallel axiom you brainlet

Sorry, I forgot it was drawn on a non-Euclidean piece of paper.

14.859

15*sqrt(2)=x*sqrt(2) x=15

wtf (3,1,2) means?

that's what i thought at first but yeah, they can. if you break the diagonal at some point in the middle and pivot each line around their respective corner, they can be made to have right angles by increasing their lengths. the problem is figuring out how much extra length.

t. brainlet with poor spatial reasoning.

Your study guide is wrong.

sum the squares together to get a rectangle with length=21 and width=3 and then find the diagonal of the rectangle (which is also the diagonal of the giant square) using Pythagoras. And then again use Pythagoras to find the sides of the giant square.

pic related took me ages on paint, you brainlets better appreciate

The lengths of the three segments in the diagonal.
In the original question they would be (12,3,9) with x=15, but obviously you can divide everything by 3 to get (4,1,3) with x=5.

The question is, assuming the middle segment still has length 1, what is the next possible integer value of x that can have (a,1,b) with a and b integers?

63

Sqrt(909)

15 is correct

>The shape cannot be constructed
let [math]x = 15[/math]
then it's easy to see
[eqn]
x + i x = 12e^{i\phi} + i 3 e^{i \phi} + 9 e^{i\phi}
[/eqn]
where
[eqn]
\phi = \arctan\left(\frac{3}{4}\right)
[/eqn]
which completes the construction in the complex plane

(a,1,b) would have
[eqn] x = \sqrt{\frac{(a+b)^2 + 1}{2}} [/eqn]

True, but that's not the question.

Now if you define y=a+b you get
y^2 - 2 x^2 = -1
which is the negative Pell equation with n=2 which only has the solutions
(x,y) = (1,1)
(x,y) = (5,7)
(x,y) = (29,41)

Visually you get this, pure mathematically you get exactly 15.

It is 15 niggas, obviously

Correct. x=29 is the next number.

Although there are actually an infinite number of solutions. The next one after that is x=169.

Lmao i did this after i demonstrated in my head that the middle point of such segment made by two paralel lines and a sq angle, the diagonal of a square always coincides with the middle point of such segment, while walking my pibull on the park. I'm superior to all of you eurangutan whitey boys.

This is the only solution I didnt understand.

Easy. 12 + 9 divided by root 2 of course.