Why is the UK so cucked?
>need tv license
>can't defend self with weapon legally
>tv station is called bbc
Pic not related so don't argue over the pic this time faggots, just discuss the topic at hand
>>70247792
get a better pic moron. Common core isn't even UK.
It's area is 14x5 anyway, I don't know how common core would change that. 2^2 + 5^2 != 8^8, but so what. the 8s might as well be Xs
>>70247792
>posts common core
>Calls UK cucked
Atleast use a proxy to change to a non-cucked country before posting.
Common core is literally cuck maths.
>>70247792
There is no trapezoid with those measures.
>>70247792
At what age do you typically solve such things in Hamerica? Just asking out of curiosity
~160u^2, btw
>>70248662
I said pic unrelated and no it's not fucking 14x5 faggot, you are a retard
>>70248840
fuck you faggot
>>70248968
No one said it's a trapezium, 12 and 16 aren't parallel and 5 does not make a right angle with 16
There is insufficient information to solve and the scale is fucked
>>70249052
Never because it has insufficient information to solve
If it was cyclic meaning use Brahmagupta's formula then probably college math
Also that pic was taken from a college worksheet but whoever made it is an idiot because there's not enough information to solve
>>70249646
no
>>70247792
Just did that in my head real quick. It's 80, isn't it?
>>70249958
>not enough information to solve
I hope this is bait, if not your a retard
>>70250191
Ah crap, meant to write 70.
>>70248840
I side with the Anglo on this for once.
America=cuck:the nation
>>70250242
Go on then solve it you fucking faggot maple nigger
>>70250191
nope
>>70250303
nope
>>70249958
>NOT ENOUGH INFORMATION TO SOLVE
My sides tbqh
>>70247792
>Pic not related so don't argue over the pic this time faggots, just discuss the topic at hand
You opened a can of worms on this one.
>>70250303
5 x 12 = 60 ( main '' squad '')
8 x 2 x 2 = 32 (the 2 sides)
60 + 32 = 92
>>70249958
>not enough information to solve
i just did it with a calculator... it's 91.22
god help america's schools
>>70250596
Isn't it (5*2) / 2 = 5 per side triangle?
>>70247792
It can't be '5' for the height because 8 on both sides = same angle = same length of 2 for the bottom.
2^2+x^2 = 8^2 --->
squareroot of 60 is the height, not 5.
>>70250368
>>70250303
5 x 12 = 60 ( main '' squad '')
8 x 2 x 2 = 32 (the 2 sides)
60 + 32 = 92
Fuck Americans r studied fucks. No wonder you greasy fucks always do bad in international rankings.
>>70250867
>that pic
>>70250368
>>70247792
Too sleepy to do 7th Grade Geometry
Just did a quicky
Not sure if it is correct
>>70247792
>2016
>watching TV
If you are that stupid that you need TV, you deserve to lose money to the license merchant
>>70250596
You're way off. It's just 8*2. Not 8*2*2. They were halves
>>70250991
2^2 + 5^2 != 8^2
This interpretation is impossible unless those aren't right triangles.
>>70247792
Main rectangle = 5 * 12 = 60
The side triangles = 5 * 2 * 2 = 10
Total = 70
>>70251409
Shit I meant:
Main rectangle = 5 * 12 = 60
The side triangles = (5 * 2)/2 * 2 = 10
Total = 70
>>70251471
This Is correct
The answer Is 70
>>70250382
Solve it then faggot
>>70250596
>5 x 12 = 60 ( main '' squad '')
Prove that 5 is perpendicular to 16 and that 12 is parallel to 16.
>8 x 2 x 2 = 32 (the 2 sides)
Literally what?
>>70250739
nope
>>70250830
Height what? What defines up and down in such a quadrilateral? There's no parallel sides or right angles or anything. Up and down are subjective.
>>70250867
nope
>>70250991
Yes you are correct but that still isn't enough info to solve
>>70251305
That would only be if it's equivilent right triangles with the two shorter sides being 8 and 2 on each you idiot and that's not the case>>70251409
>>70251471
Texas being RETARDED as always
>pic related
>it's you, Texas
>>70251659
no you stupid fucking bong idiot
>>70251726
Dude I think you are actually retarded
my condolences to your family
>>70251356
its like these people here dont even common core
baka
Groogle says 70
http://i.imgur.com/nrcPB8Q.png
>>70247792
Its seventy
14*5=70
If you do not agree you are literally Retarded
>>70247792
Ignoring the fact that the problem makes no sense as the hypotenuse can't be 8, the correct answer would be 70
A triangle with the sides being 4cm and 5cm can not have a hypothenuse that is 8cm, that's the problem for you retards who don't understand.
>>70252153
I guess Pythagoras was literally retarded then.
What's the problem here?
A: stands for area.
If you don't know the formula to calculate the area of a trapezoid then use basic geometry to break it up into two right triangles and a square and add the collective areas together.
It's not rocket science people.
>>70252681
How is this difficult
>>70252450
This.
>>70252493
Sven got it right (except one side being 2, not 4). Theoretically is the area of trapezius 70, but since every two sides of triangle combined have to be more than third side, said trapezius doesn't exist.Now stop arguing over first grade math you cucks.
>it's another tv license thread
You don't need a TV license if you don't watch live broadcasts. You can watch catch up tv without one
>>70251683
you can call it whatever you want you fucking mongrel.
There are parallel sides and right angles
We know there are parallel sides because the length of the top would have to be longer to compensate for a tilt leaving the hypotenuse of the right side as 8.
We know there is a right angle because the dotted line is what we use to indicate right angle over here in common core.
Face still stands, the height(or perpendicular) is not 5. It's squareroot of 60.
>>70252153
The area is only = 70 is the lines marked 12 and 16 are parallel. We have no way of knowing that they are parallel.
>>70252846
That only works if the lines are parallel, you Islamic Dutch monkey.
>>70252153
you are literally retarded.
See >>70250830
and
>>70253157
after you read OP
>>70253242
true, that and the given lengths of the sides (8) and the one given side of the height doesn't even the logic of the missing side out with the Pythagorean theorem.
>>70253155
>actually read the thread
What the fuck is going on.
>>70253340
How are they not parallel
>>70253520
that image does not reflect the reality. The values in the pic do not form a trapezoid
>>70253692
That why i ignored the 8's and did it in a way the 8's are not necessary for the answer
>>70253520
They might be parallel, but we're not given that information.
If they are parallel, then your answer works. If they are not parallel, then it does not work.
It never said it was a trapezoid
I dont understand people's problem
/r/-ing the pic of the mathanon who ran some kind of program to solve this
>>70253520
>>70253941
dotted line in common core indicate perpendicular
>makes them parallel,equal angles
The But you choose if you want to say the 5 or the 8 is wrong.
>>70254216
It can't be solved, there are infinity possible solutions. If it was a cycling quadraliteral then Brahmagupta's formula would work.
>>70254197
There is insufficient information to solve (and some fucktards think it's a trapezium).
>>70253941
So its basically Nitpicking what you are doing.
This is a really basic problem looks like something a kid would do.
So i doubt they would make parallel an issue
>>70253917
You also assumed that 12 parallels 16 and that 5 is perpendicular to both. None of that is given. And the 8s are fucking there so it's confirmed not a trapezium you absolute fucktarded dumbass fuckhead. Go fucking kill yourself you incompetent fuck.
>>70254355
There is sufficient info!
The 5 / 8 is wrong though.
>>70247792
Where's A?
>>70254241
So 8 is wrong I guess
common core is really weird
>>70254399
Except it's clearly NOT a trapezium and 12 is clearly NOT parallel to 16 seeing as the values prove that, fucktarded dumbass. Fuck, it seems the Dutch are even more incompetent than the Claps.
>>70254480
Nevermind, I guess Claps are just as incompetent. There is not sufficient info and you have no grounds for claiming that the information provided is 'wrong'.
>>70254464
Still they are squares and triangles so this method works.
>>70254530
It is an abbreviation of area fucktard
>>70254592
The value does not define if it is parallel or not.
>>70254727
Are you fucking retarded? A square is a quadrilateral with 4 sides of equal length and 4 90 degree angles. If by "this method" you mean splitting it into two triangles and a smaller quadrilateral then sure those will add up to the total area but it just further complicates a problem for which there is already insufficient information to solve.
>>70254216
I remember that pic. Didn't save it though.
Honestly just fire whover drew up the picture. You cant expect kids to handle this as not a trapezoid
>>70254906
Yes it does you absolute fucktard. show me a quadrilateral with a side of length 12, the opposite side being parallel and of length 16, and the other two sides being of length 8 where the distance between the 12 and 16 sides is 5.
Protip: YOU FUCKING CAN'T YOU FUCKING DEGENERATE POTHEAD.
>>70254727
let's try your method.
If we were to assume it to be a trapezoid, we'd find that the area of the triangle that can be calculated would be...
sqrt(8^2 - 5^2) = 6.2...
so if both triangles have this, the bottom side would be 6.2 + 6.2 + 12 = 24.4, but it clearly says 16!!
KANKER
>>70254355
>insufficient information to solve
what a faggot
l2 triangle fags
>>70254934
How does it complicate any of it? it just gives the area.
>>70255057
>You cant expect kids to handle this as not a trapezoid
Eh, if this was 2nd grade I'd be inclined to agree with you, but this is from a college math class (according to the anon who originally posted a worksheet that included this problem). Still fire the person because the person is a fucktard. But I think that expecting it to be handled as a trapezium is treating college students as 8 year olds.
>>70255154
>>70255064
This is my method
>>70252846
>>70255306
how?
>>70255200
How fucking incompetent are you? You can't find the area because you don't know the third side length of the left triangle (you only have 5 and 8), you don't know the second or third side lengths of the right triangle (you only have 8) and you don't know the third or fourth side length of the middle quadrilateral (you only have 5 and 12), plus you have no known angles to work with. Now instead of one unsolvable figure you've got 3. CONGRATU-FUCKING-LATIONS!
>>70254592
see
>>70250830 for proof
and dotted line indicates perpendicular in common core -=-> parallel across
=equal angles
>>70255060
Well you assume that the 8 are the right answer. I say the 5 is right.
Parallel only means that they would never cross each other if the lines where continued
>>70255423
Your 6.2 assumes that the 5 line is perpendicular to the 16 line and I don't even know where you're going with the rest of that, try writing it out more clearly.
>>70255453
btw you choose if 5 or 8 is wrong.
the dotted line indicated perpendicular so mystery length has to = 2.
but a^2 + b^2 =/ c^2 here.
>>70255453
Jesus you are really salty about math.
I just do it in a different way.
A way i do not need to calculate the triangles
70 units^2
This problem is perfectly fine, stop being retarded.
>>70255479
>implying sqrt60 is the height
No you fucktarded fucking dumbass, the line you are referring to as height is 5 because it is defined as 5, that's not subject to fucking debate. What's next? The 16 represents an 18? The 8 represents a 12 just because you don't like the number 8? Fuck off to hell you incompetent dumbass, you give all Claps a bad name. The "stupid Americans" meme is LITERALLY because of you. Personally you.
The object in question is clearly not drawn to scale and there are infinity possible solutions. Perhaps one could find an upper and lower limit to the possible solutions.
>-=-> parallel across
>=equal angles
What the FUCK are -=-> and =equal supposed to mean? Learn to fucking English.
>>70255504
>Well you assume that the 8 are the right answer. I say the 5 is right.
What the fuck are you talking about? Those are measurements, not the area of the object.
>Parallel only means that they would never cross each other if the lines where continued
Thanks for telling me the definition of parallel, teacher! It's almost like I didn't already know!
>>70255858
try a^2 + b^2 = c^2
for the triangle
>>70255659
correct
its assuming the 5 is the height of the left triangle (which if it wasnt then it's literally a dummy variable and you could solve it as a trapezoid)
9.8's derived from the remainder
that along with the 5 (again assuming its perpendicular) gives you 10.98 for the center line
then the top triangle is an 11x12x8
solving for area's a bitch and used wolfram to do it
>>70255947
There is no hope for this one.
But if you are not trolling, I'll give you this hint.. ( again)
>Who is Pythagoras?
>>70255659
No you fucking dumbass, you can't fucking choose that some measurement is wrong. That would make all math problems pointless.
>what's 3+3?
>um... I think the second 3 should be a 4 instead so the answer is 7!
>>70255683
>I just do it in a different way.
That implies that I am doing it a certain way. I'm not because it's not fucking solvable.
>A way i do not need to calculate the triangles
It's a quadrilateral fucktard dumbass, triangles are only a part of it if you choose to split it up into smaller shapes (which I already fucking said is pointless since it's unsolvable either way).
>>70255858
Kill yourself. Literally kill yourself.
>>70255947
Well you said that the value defines if it is parallel or not and I simply state that it has nothing to do what it. So yeah i get the impression you do not know what it means
>>70256105
meant to quote
>>70255632
>>70255957
Shiiiiiit. You're right. This problem may be setup wrong but it doesn't appear to have any bullshit doctrine or pants-on-head retarded tricks like that dumb cunt telling kids pemdas is old.
>>70249958
>Lol, the UK is cucked
>Posts American common core
>Then says "there is insufficient information to solve"
This is why I love Americans.
>>70256107
No hope for you, fucktard dumbass. I guarantee I'm better at math than you.
>>Who is Pythagoras?
A Mathematician (Greek I believe? I'm not big on history) who discovered the Pythagorean Theorem, which states that, given a right triangle, the squares of the two shorter side lengths add to the square of the longest side length.
>>70247792
Let me guess, not 70 somehow?
>>70256161
Tell me what is wrong with this answer and how it is not 70.
>>70252846
Try to do it without cursing this time
>>70256298
I changed my mind, forgot to correctly look at the problem. The area is unknown/impossible
>>70252021
>linking to an image
You're on an image board.
Maybe the shape is not on a flat surface.
I think that could that explain why the sides of the triangle break Pythagoras' Theorem?
OP, I just realized, why the FUCK are you doing middle school tier shit in COLLEGE?
>>70256183
Are you literally this retarded or are you trolling?
A quadrilateral with side lengths 8, 12, 8, and 16, with the 12 and 16 sides parallel, is a trapezium (or trapezoid for americlaps). In such an object, the distance between the 12 and 16 lines is greater than 5. Therefore, it is impossible to draw a line of length 5 from one of the corners adjacent to the 12 line to any point along the 16 line.
Therefore, it is clear that in the object pictured in the OP pic related is not a trapezium, the 12 line and 16 line are not parallel.
Then again, if you were too incompetent to figure this out on your own, you're probably too incompetent to understand the explanation too. I guess all that marijuana goes to your head, you filthy degenerate Dutchman.
>>70247792
>>70247792
5(16+12)
-------------
2
>>70256105
>its assuming the 5 is the height of the left triangle
You used the Pythagorean Theorem. Protip: That only works on a right triangle.
>(which if it wasnt then it's literally a dummy variable and you could solve it as a trapezoid)
Except the object isn't a trapezium, dumbfuck.
>>70247792
Anyone who says the area isn't 70 is a fucking nigger
>>70256272
OP here and I said Bongistan is cucked I didn't say it's stupid. Bongistan is cucked, Clapistan is stupid. There's a difference (however slim) between cucked and stupid.
>>70256684
Then again you assume the 8's to be correct instead of the 5. You clearly state that they contradict each other. I assume the 5 to be correct and I come out with an Answer 70. Seems like my way is better
>>70256333
You assumed that the 12 line is parallel to the 16 line and you assumed that the 5 line is perpendicular to the 16 and 12 lines.
>>70256572
Valid point.
>>70256644
Another anon posted a worksheet that included this problem a while back, supposedly from his college math class. I'm not even in school anymore.
>>70256906
Please post something else, holy shit. I love you America!
You're all wrong.
Those symbols at the sides aren't 8s, they're ∞s.
>>70256970
I didn't say the 5 is incorrect fucktarded dumbass. The values given are BY FUCKING DEFINITION correct since they are fucking GIVEN. Otherwise I can take any math problem and say "I don't like that number"
>Find the integral of sin(2x^3-2)*cos^-1(14x^4) from 12 to 16
Um... I don't like those numbers so I'm going to pretend that the problem actually says "find the integral of x from 12 to 13" oh cool that's easy the answer is x^2+c, oh wait I don't even like finding it from 12 to 13 so I'm going to pretend it's an indefinite integral, okay the answer is x^2+C
Cool I solved it by throwing out stuff I didn't like, I get an A, right teacher?
Back to reality... the 5 and the 8s are correct, as are the 12 and 16. There is insufficient information to solve but I can sure as hell prove it's not a trapezium.
>>70256821
I thought it was 70, too, but that can't be a trapezoid because the dimensions disallow for the top and bottom to be parallel.
You can't form a right triangle with sides 5, 8, and 2.
>>70256791
are you literally retarded
what do you think the 5 means?
is it how many times i fucked your mom?
no, its the height of the 12/8 intersection
If it wasnt dashed i maybe could agree with you
but its clearly a standard for something, and the only reasonable thing is height
and i didnt solve it as a trapezoid
I didnt know /pol/ was full of retards
i thought it was a prank
>>70257388
>I didn't say the 5 is incorrect fucktarded dumbass.
> Therefore, it is impossible to draw a line of length 5 from one of the corners adjacent to the 12 line to any point along the 16 line.
Do you even know what you are doing here
Its 70 m8 just give up
>>70257449
>are you literally retarded
Nope, you are.
>what do you think the 5 means?
It's the length of the line segment.
>is it how many times i fucked your mom?
Why do you like fantasizing about fucking women twice your age?
>no, its the height of the 12/8 intersection
It's the length of a line segment drawn from said intersection to some point along the 16 line (which, by the way, proves that the 12 and 16 lines are not parallel).
>but its clearly a standard for something, and the only reasonable thing is height
What the fuck is "height" supposed to mean? If the 12 and 16 lines were parallel I might be inclined to agree with you, but since they aren't, it's pretty ambiguous what height is. I guess you could define the 16 as the "bottom" and the highest point from there as the "top" but then the top would actually be the right 12/8 intersection, not the left 12/8 intersection, so the line segment of length 5 would still not be the "height".
>and i didnt solve it as a trapezoid
Good, considering it's not one.
>>70256161
>>70256290
5^2 + x^2 = 8^2
64-25=x^2=39
x= 6.2
if x is anything but 2, it's not perpendicular.
But the problem states that the length( of 5 )is perpendicular by utilizing the dotted line.
>I'm better at math
Kek
I have proof that I am better than you
>>70257051
Gotcha.
>>70247792
They took our guns, our knives, our culture and our bike tires.
>>70257756
stay bad
>>70250867
I dunno leaf bro. I thought the base of the triangle was 6.2 when multiplied by 5= 31. you don't have to half 31 because their are 2 right triangles. Add that to 60 and I got 91.
>>70257634
>Do you even know what you are doing here
Yes, I'm trying to educate fucktards like you but obviously you are too fucking incompetent to understand basic logic.
>Its 70 m8 just give up
Prove it. Protip: You can't.
>>70257765
If the problem indeed stated that 5 is perpendicular to 12 and to 16 then the problem would dictate an impossible shape (assuming we are dealing with a flat 2-dimensional plane in Euclidian space). However, I don't recall ever learning that a dotted line automatically denotes the line segments being perpendicular.
>>70257886
Prove me wrong faggot.
>>70257894
>this guy
>>70257995
I've already Proven it
>>70252846
You have just called me names and said I was wrong.Yet you never gave a different solution
>>70257995
Ahh.. Well it's kind of autistic, but we did
Well.. THEREIN lies our problems boys! Common core not being such a standard, now is it?
The answer is 108.4.
I think in a way English people are the ultimate hybrid of european greatness. Germans, Romans, Vikings, French and a little celt.
I think in a way, we're the master race. And just look at the sons of the Anglo-Suprior, the North Americans, the Austrialians, New Zealand. We more or less used our skills to build up India and China's a child of Anglo-Suprior teachings.
there it only took 5 secs
>>70247792
This problem is fucking retarded. No indication of:
- Whether the base and the top are parallel.
- No right angle specified near the dotted line.
- If the base and top are parallel, and the dashed line is a right angle then the trapezoid is not symmetrical as drawn in the diagram.
I can tell you what the upper bound of the area is. That's about it. The area of this shape is less than or equal to 121(units^2), and greater than 44.
Without more info this problem could be anything in that range.
>>70247792
the answer is 70
>>70247792
What is this supposed to represent? It's just a simple math problem. Answer is 80 btw
>>70258255
I also explained why you're wrong. I didn't give a solution because there is not enough information to solve. I already made that clear. Fuck off faggot.
>>70258821
Why did you post a picture of a trapezium? OP pic related is not a trapezium.
>>70247792
Area of a triangle is 1/2 base x height. In this case the area of each triangle is 5.
>>70260294
Sadly, my guess is- this is the only acceptable answer. Despite the dipshit who put this question together being potentially braindead.
>>70260242
Oh look a smart person! Congratulations anon, you are one of very few competent people in this thread. Here, have a cute Japanese girl.
>>70247792
>>70260745
T-thanks anon. Actually given the info, there is probably a more accurate way to bound this area.
But that's the upper limit on a four-sided shape with that circumference. (The area of a square).
>>70261246
Yeah, I've wondered myself how to find the upper and lower limits of such a figure. I think there is a way, but I don't know what formulas to use.
The shape would look roughly like pic related, but by changing the point where the 5 line meets the 16 line (and therefore the angle those lines make) will also affect the angles the 8 lines make with the 16 line (and therefore also move the 12 line), resulting in infinity possible solutions. But there's definitely zonzero finite upper and lower limits to the object's area.
>>70261606
You have a disturbing amount of Harry Styles pictures.
Is this now a Harry Styles thread?
>>70261846
>>70261807
It's certainly not a /pol/ thread.
>>70261807
Barely scratched the surface
ITT /pol/ goes full /sci/
>>70261736
If 3 of the four inner angles of the (corners) were given, or the diagonal intersect. It would probably be more doable.
The thing that makes me sad about this question, is that smarter kids are probably being marked wrong for being right.
>>70253941
its not going to be faget because its 2 units per sdie anyway. so wether its symmetrical or not, its not going to be a real representation
>>70261736
Thinking about this... the upper limit would be having the intersection of the 12 line and the left 8 line being as far from the 16 line as possible. At least I'm pretty sure. So in that case we'd have a right triangle with sides 8, 5, and sqrt39 along with a quadrilateral with sides 5, 12, 8, and 16-sqrt39. So then the question is can we find the greatest possible area of a quadrilateral with sides 5, 12, 8, and sqrt(39).
As for the minimum, we'd make the angle between the left 8 line and the 12 line 180 degrees (or the limit as x approaches 180 from below, since technically exactly 180 makes it a single line segment). The 5 line could be discarded in that. So then we'd basically turn this quadrilateral into a triangle of sides 20, 16, and 8. We can solve this with Heron's formula.
So now the only question is how do we find the maximum possible area of a quadrilateral given the 4 side lengths and nothing else.
>>70262548
Yeah, it's not possible to solve definitively but we can find an upper and lower limit. See >>70262923 for what I've got so far.
>>70261736
do me a favor, go buy some wooden rods, measure them out and cut them at the lengths in this image. try to construct what you've just drawn, realize it's an impossible shape and that you're terrible at math, and then walk out into oncoming traffic.
Could it exist as is if we assume that the actual lines are going on the Z axis so that the 5 is a line drawn from the top onto the XZ plane.
>>70262923
Lower limit:
A=sqrt[s(s-a)(s-b)(s-c)]
s=(a+b+c)/2
s=22
A=sqrt(22*2*6*14)=sqrt(3696)=60.794736614
And yes technically it's the limit as we approach sqrt(3696) from above.
>>70263208
Except that's wrong you fucking retard.
>>70263318
Perhaps. So in that case just ignore the 5 line and treat it as a trapezium. lulz, that will make the children ITT happy
>>70262923
According to http://mathworld.wolfram.com/CyclicQuadrilateral.html
The maximum possible area for a quadrilateral, given side lengths, is the area of a cyclic quadrilateral given those lengths. Of course because of the line segment of length 5 we can't use it for the shape as a whole unless we prove that it can be cyclic, but we CAN use it for the 5 12 8 16-sqrt39 quadrilateral.
So, let's take out Brahmagupta's formula and solve that, then add to the right triangle of side lengths 8, 5, and sqrt39. Will post back after solving that
>>70250368
You could solve the triangles first using pithagoream theorem to get the lenght of the unknown side and then get the área of both triangles and add it to the área of the square in the middle. Its fairly easy.
>>70263849
Sorry i.meant the área of the rectangle MY bad..
>So then the question is can we find the greatest possible area of a quadrilateral with sides 5, 12, 8, and sqrt(39).
Should have said 16-sqrt39 instead of just sqrt39
>>70263849
Sorry i meant the área of the rectangle MY bad
>>70261736
>>70262923
You're over think this. You're not wrong, any retard could tell in one look that the "16" line is like twice the size of the 12 line and that the shape is bullshit. But the answer isn't to solve the area of the shape as specified by the numbers, it's to take the shape by it's face value and hope you come up with the answer the retard who wrote the question came up with.
I have a lot of experience with common core and I guarantee you the answer is either 70 or 91.2 depending on what breed of retard wrote it.
>>70251683
Yeah this picture is pretty horrid. The shape violates Euclidian space if we assume 16 || 12 and 8 || 8 with height = 5
That's not really a common core problem; even doug west's books have errors and he's one of the most precise writers there is.
>>70263849
you can't it is not a right triangle
>>70263700
are you from /sci/
>>/sci/7979720
https://en.wikipedia.org/wiki/Bretschneider%27s_formula
works for a general quadrilateral
although it requires the diagonals and i don't know if it is possible to calculate them
http://www.wolframalpha.com/input/?i=quadrilateral+edge+lengths+8++12++8+16
states that the area would equal
1/2 sqrt(p^2 q^2-18496) where p and q are diagonals
and so the area might be anything
>>70263849
>>70264005
Listen Pedro, if you do this the unknown length is 6.2. Which means the triangles together are 12.4, which means the distance between them is only 3.6, which means the they're not right triangles, which means the Pythagorean theorem doesn't work.
>>70264765
>it's to take the shape by it's face value and hope you come up with the answer the retard who wrote the question came up with.
Which is why public education is just fucking stupid these days. "Hey kids, jump through a hoop and do a tuck and roll for us (the Government), otherwise you'll end up bagging groceries."
>>70247792
>need tv license
nope, get new material.
>can't defend self with weapon legally
we can legally defend ourselves with whatever happens to be close to hand and there is clearly no other option, people have gotten away with shooting intruders and stabbing them to death whereas others have got into trouble for clubbing someone because they pursued them into the garden chasing them down.
>tv station is called bbc
and what a BBC it sucks too
>>70264044
You know.. I'm trying really hard to not...
>Mexican intellectuals
..because /pol/.
But it's what >>70265120 and >>70265203 said.
>>70250991
With this kind of math skill, there is no doubt why are you in debt
>>70265495
Says the guy whose leader goes cap in hand to the Euro for billions in bailouts whenever he breaks a fingernail.
>>70265639
*EU, not Euro.
>>70250303
Yeah it's 70. All the other retards writing bullshit.
>>70265120
>and so the area might be anything
Not anything. 44 < A < 121. The addition of the 5 unit line has to constrain the problem further somehow. But fuck if I know how. I'll leave that to the mathematicians.
>>70265439
No problem senpai i noticed the issue now thanks.
But some americans are really salty
this is the worst thread on all of 4chan including /b/
>>70266156
Right?
>>70263700
A=sqrt[(s-a)(s-b)(s-c)(s-d)]
s=(a+b+c+d)/2
s=(5+12+8+16-sqrt39)/2=(41-sqrt39)/2
A=sqrt[ (31-sqrt39)/2 * (17-sqrt39)/2 * (25-sqrt39)/2 * (9+sqrt39)/2 ]
A=sqrt[(1/16)(31-sqrt39)(17-sqrt39)(25-sqrt39)(9+sqrt39)]
A=68.976155184
Then we still have to get the area of the triangle. The sides are 8, 5, and sqrt39. It is a right triangle. Therefore A=(1/2)bh=2.5sqrt39=15.612494996.
So the total area is 15.612494994+68.978257444=84.590752438
Or in exact terms 2.5sqrt39 + sqrt[(1/16)(31-sqrt39)(17-sqrt39)(25-sqrt39)(9+sqrt39)]
The lower limit approaches A=sqrt(3696)=60.794736614 from above. The upper limit is A=2.5sqrt(39)+sqrt[(1/16)(31-sqrt39)(17-sqrt39)(25-sqrt39)(9+sqrt39)]=84.59075244
>>70265120
No I am not from /sci/ and I don't think that formula works for all quadrilaterals
>>70264765
>I have a lot of experience with common core and I guarantee you the answer is either 70 or 91.2 depending on what breed of retard wrote it.
Correction: The retard that wrote it erroneously believes that the answer to his/her problem is 70 or 91.2
>>70266021
i don't think the 5 affects anything as there are no angles showing where a line with a length of 5 must be drawn from
we can't assume that it intersects at a right angle and we can't assume where it begins as there are no angles
>>70266349
https://en.wikipedia.org/wiki/Bretschneider%27s_formula
>Bretschneider's formula works on any convex quadrilateral, whether it is cyclic or not.
http://mathworld.wolfram.com/BretschneidersFormula.html
>Given a general quadrilateral with sides of lengths a, b, c, and d, the area is given by
>>70266348
are those the one d guys?
>>70265120
To clarify, I posted this image on /pol/ a day or two ago and someone else reposted it on /sci/. I did post in that /sci/ thread but I am not the OP and I rarely browse /sci/
>over 3 hours later and nearing 200 posts, /pol/ is still too retarded to see that the scale in the picture literally doesn't make fucking sense
just fucking draw a trapezial with those measurements and you will see it doesn't fucking work, jesus fuck
>>70265255
>we can legally defend ourselves with whatever happens to be close to hand and there is clearly no other option, people have gotten away with shooting intruders and stabbing them to death whereas others have got into trouble for clubbing someone because they pursued them into the garden chasing them down.
Another bong anon said that you can't legally seek out a weapon for self defense, so the only way you can use a weapon for self defense is if you just HAPPENED to be cleaning your gun or playing ninja with your sword or whatever right when an intruder HAPPENED to break in. If an intruder breaks in, and you then seek out a gun or knife or whatever for self defense, you get v&. Is the anon who said that wrong?
>>70266585
Yeah. Best thing in this thread.
>>70266021
>The addition of the 5 unit line has to constrain the problem further somehow
I've got it all figured out now, see >>70266349 and my earlier posts
Pretty sure I got it all right. Not promising.
>Qualifications: Good at math, learned calculus when I was 15, but have been out of school for years so a bit rusty
>>70250596
Your an idiot. Each of the two triangles on the side have to be taken separately.
Also the formula for the area of a triangle doesn't require you to multiply anything by the hypotenuse. Shit and they make fun of the American school system. You regards can't even find the area of a trapezoid. That was if it was a trapezoid which it isn't even that. Peep the dotted line there's no square at the bottom to confirm its a 90 degree angle
>>70266509
What we can assume about the line segment of length 5 is one end meets the 16 line, and the other line meets the junction of the 12 line and one of the 8 lines. So it is limiting on the upper end (though not on the lower end) of possible areas. For example, in a trapezium with side lenths 8, 12, 8, and 16, the 12 and 16 lines are greater than 7 units apart, so it would be impossible for said line segment of length 5 to exist. Further, Brahmagupta's formula for a cyclic quadrilateral of sides 8, 12, 8, and 16 gives an area of approximately 108 if I remember correctly. Again, this can't exist because the line segment of length 5 is not of sufficient length to reach from the 8/12 angle to any point along the 16 line.
>Bretschneider
Okay, fair enough. Looks like we don't have enough information to solve using that though.
>>70247792
1/2(a+b)h
I think I've worked out the actual correct answer.
I started with scale diagram.
Which was very useful because every time I made a mistake with my calculator it was obvious that a mistake had been made.
x=base of the triangle on the left=sqrt(8^2-5^2)
y=length of the rest of the base=16-sqrt(8^2-5^2)
z=length of the diagonal form the bottom right to top left=sqrt(5^2+y^2)
Now the shape has been divide into triangles for which we know the length of each side.
Using Heron's formula we can calculate the area of each triangle and add them together to get area of the whole shape.
A=82.68
>>70267621
>7979720
Turns out you are correct for the upper limit. Where I went wrong is forgetting that I had defined the angle the 5 line makes with the 16-sqrt39 line as 90 degrees, and thereforetook the quadrilateral with side lengths of 5, 12, 8, and 16-sqrt39 to be cyclic. Your answer is definitely the correct upper limit.
The lower limit is still what I said in >>70263502
>>70268080
Why isn't there only a single answer?
We seem to have all the information we need.
>>70268498
you assumed the line of 5 formed a right triangle with the line of 16 which is the upper limit
>>70248840
>14x5
>not 70
Do you even O Level, faggot?
>>70268628
I was told that's what the dotted line meant.
I can see how there would be a range of answers if this is not the case.
Using a dotted line to indicate perpendicularity doesn't seem very clever.
Do you just guess which other line it is perpendicular to?
When I was at school we used a little square instead of and arc to mark a right angle.
Nice to know I did it right, thought. Thanks.
I had a feeling the people saying 'not enough info' were wrong. (assuming a right angle)
>amerifat """examiners"""
>>70267621
Fine, you hooked me.
> Not knowing Area = (a+b/2)h
That's the last time I reply to bait...
>this entire thread
I guess /pol/ didn't fall for the STEM meme.
>>70269250
It 2016 so everyone gets prizes just for trying.
Have a gold star.
>>70269413
>STEM meme.
>Implying being a welfare nigger is a preferable alternative.
I studied finance, not STEM. Physics, mathematics, chemistry etc. All solid subjects though.
>>70268498
>>70269067
If the 5 is perpendicular to the 16 then yes your answer is the only answer.
>When I was at school we used a little square instead of and arc to mark a right angle.
That's still standard notation.
If it's a right angle then you are right. If it's not a right angle then your answer is the upper limit to all possible solutions.
I will post the final answer (after fixing my mistake in my previous answer) in another post (character limit)
>>70270185
The lower limit occurs when the angle between the left 8 line and the 12 line is 180 degrees (technically when the angle is the limit as x approaches 180 from below, since an angle of exactly 180 degrees turns the two line segments into a single line segment). In this situation, the 5-unit line segment can be ignored. We therefore essentially turn the quadrilateral into a triangle of sides 20, 16, and 8. We can solve this with Heron's formula.
A=sqrt[s(s-a)(s-b)(s-c)]
s=(a+b+c)/2
s=22
A=sqrt(22*2*6*14)=sqrt(3696)=60.794736614
Since we are going with an angle approaching 180 degrees from below, the area is actually the limit approaching sqrt(3696) from above.
Cont.
>>70270234
The upper limit occurs when the intersection of the 12 line and the left 8 line are as far from the 16 line as possible. Therefore we have a right triangle with sides 8, 5, and sqrt39 along with a quadrilateral with sides 5, 12, 8, and 16-sqrt39 where the sides of length 5 and 16-sqrt39 form a right angle. We can split the quadrilateral into two triangles, drawing a line segment from the upper left corner to the lower right corner.
By combining the lower left of these two triangles with the triangle we initially separated, we have a single triangle of base 16 and height 5, giving us an area of 40.
We are now left with one triangle of area 40 and one triangle with a side length of 12 and a side length of 8. In order to solve the area of the latter triangle, we need to determine either the remaining side or an angle. We cannot solve for any of the angles without first solving for the remaining side, so we'll just solve for the remaining side. To do this, we need to take the right triangle that we created in the middle of the figure whose hypotenuse is the same as the unknown side on the triangle we still have to solve. Given that we know the other two sides are 5 and 16-sqrt39, we can use the Pythagorean Theorem to determine the length of the hypotenuse as sqrt(25+(16-sqrt39)^2). We now have all three side lengths for the triangle of unknown area and can use Heron's formula to solve.
a=sqrt(25+(16-sqrt39)^2)
b=12
c=8
A=sqrt(s(s-a)(s-b)(s-c))
s=(a+b+c)/2
s=(20+sqrt(25+(16-sqrt39)^2))/2
A=sqrt[((20+sqrt(25+(16-sqrt39)^2))/2)(((20+sqrt(25+(16-sqrt39)^2))/2)-(sqrt(25+(16-sqrt39)^2)))(((20+sqrt(25+(16-sqrt39)^2))/2)-12)(((20+sqrt(25+(16-sqrt39)^2))/2)-8)]=42.684891941
Now add to the triangle of area 40 to obtain the area of the quadrilateral.
A=40+sqrt[((20+sqrt(25+(16-sqrt39)^2))/2)(((20+sqrt(25+(16-sqrt39)^2))/2)-(sqrt(25+(16-sqrt39)^2)))(((20+sqrt(25+(16-sqrt39)^2))/2)-12)(((20+sqrt(25+(16-sqrt39)^2))/2)-8)]=82.684891941.
cont.
>>70270279
What the fuck am I reading?
>>70270279
The lower limit approaches A=sqrt(3696)=60.794736614 from above. The upper limit is A=40+sqrt[((20+sqrt(25+(16-sqrt39)^2))/2)(((20+sqrt(25+(16-sqrt39)^2))/2)-(sqrt(25+(16-sqrt39)^2)))(((20+sqrt(25+(16-sqrt39)^2))/2)-12)(((20+sqrt(25+(16-sqrt39)^2))/2)-8)]=82.684891941.
>>70270234
>>70270279
Its nice to know that its not only me that values being right above being sane.
You deserve 3 gold stars.
>>70270234
Actually I was wrong about the lower limit
This post: >>70270279 is correct
But this post: >>70270234 is wrong, because you can make the lower left angle approach 0 so you've basically just got a triangle of sides 8, 8, and 12. So then Heron's formula gives an area of roughly 31.75
>>70270650
Thank you, kind anon, have a cute Aryan girl
Here is the fixed version of the lower limit: The lower limit occurs when the angle between the left 8 line and the 16 line approaches 0. In this situation, the 5-unit line segment can be ignored. We therefore essentially turn the quadrilateral into a triangle of sides 12, 8, and 8. We can solve this with Heron's formula.
A=sqrt[s(s-a)(s-b)(s-c)]
s=(a+b+c)/2
s=14
a=sqrt(14*2*6*6)=sqrt(1008)=31.749015733
Since we are making the angle approach 0 rather than be equal to 0, the area is not technically equal to sqrt(1008), rather it approaches sqrt(1008) from above.
So the FINAL SOLUTION (smilingfuhrer.jpg) is:
The lower limit approaches A=sqrt(1008)=31.749015733 from above. The upper limit is A=40+sqrt[((20+sqrt(25+(16-sqrt39)^2))/2)(((20+sqrt(25+(16-sqrt39)^2))/2)-(sqrt(25+(16-sqrt39)^2)))(((20+sqrt(25+(16-sqrt39)^2))/2)-12)(((20+sqrt(25+(16-sqrt39)^2))/2)-8)]=82.684891941.
>>70270916
Thanks Burgerbro.
I think they should keep this question, but set it for a higher age group.
Its misleading diagram is beautifully perverse and removing the right angle and asking for a range of possible areas makes for an interesting challenge.
Shit, I just realised you're OP.
Did you know you thread was going to get so badly derailed?
