What is wrong with this picture?

A isn't there

>>7979723
A stands for area.

>>7979726
oh man I'm really tired

>>7979720
Probably not proportioned right which is literally shit seen in school over the last 3000 years. No one gives a shit.

>>7979720
12x5=60 area of rectangle
2[(5*2)/2]=10 area of both triangles
A = 70

>>7979720
That can't have those dimensions and be a trapezoid.
[math] \sqrt{ 9^2 - 5^2 } \approx 6.25 [/math]
If the length of the bottom of that triangle is 6.25, just subtracting one of those triangles from the base would make the length less than the top side. You have to assume therefore that this is not a trapezoid.

If it's not a trapezoid, there's no easy way to solve this with high school math.

>>7979735
>Geometry is the science of correct reasoning on incorrect figures.
>George Polya (1887 - 1985)

>>7979720
I guess 5 isn't the height. The trapeze is not well proportionated.

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>>7979720
sqrt(5^2+2^2) != 8

>>7979741
Sorry, I wrote 9 in the radical but that's supposed to be an 8. I did the calculation with 8.

>>7979720
The base of the triangle would have to be 2, which cannot be true by Pythagorean's theorem

>>7979720
Is this possible if don't know that the segment of length 5 is perpendicular to the base with length 16? I mean there is no indication that it is perpendicular.

>>7979750
It doesn't indicate it's a right triangle.

>>7979735
>>7979742
>>7979743
>>7979761
>>7979775
We live in an age of computers with software than can draw a god damned trapezoid correctly. This is middle-school geometry. I highly doubt the intention was to trick the student into thinking the shape is accurately represented by those numbers when it actually isn't. Some retard just designed a problem and didn't check the math.

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>>7979720
The shape is possible; the problem doesn't say it's a trapezoid, and it isn't- it says "object". It's a quadrilateral.

The top side (AB) is 12.
The right side (BC) is 8
5 is not the height, it's the second side (AD) of the triangle (ADE).
16 is the length of DC. You need to solve for ED.

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take your CCposting elsewhere OP

>>7979846
I hate you.
I was in the middle of doing this >>7979823
and dividing it into arbitrary triangles to solve and add up for the area.

>>7979741
The figure is obviously not supposed to be to scale.

It's like you don't remember grade school test formatting.

>>7979846
How can you prove it's cyclic?

>>7979867
draw it and try to inscribe it in a circle or use some other obscure geometric property im too lazy to look up for you

the quadrilateral with the dimensions in OP doesnt look anything like the picture

>>7979866
It's not the scale that's the problem. It's that you can't have a right triangle with side lengths of 2, 5, and 8.

>>7979876
prove it

>>7979906
(2)^2 + (5)^2 =/= (8)^2

4 + 25 =/= 64

29 =/= 64

wow amazing. where is my fields medal

>>7979720
The answer is 70.

>>7979740
[math]((a+b)/2)*h[/math]

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>>7979740
>>7979924

A=(1/2)b*h

A=(1/2)12*5


A=30

Approximately A=82.68

>>7979720
2 5 8 are not acceptable dimensions for a right triangle

>>7979720

Make a third triangle on the RHS. Calculate the areas of the two triangles and the rectangle in the middle separately and then add them up. It's pretty straight forward what's the big deal?

ITT retards assuming 5 is indicating the side of a right angle triangle even though nothing in the diagram shows that.

>>7979775
It's implied, especially considering this is like low level highschool stuff

>>7979720
2 5 8 Triangle dont work

>>7980247
ITT: autists that dont understand it is implied by the simple face it is 10 year old shit

>>7980247
If it isn't a right angled triangle, 5 isn't the height so you can't use 0.5bh

>>7980269
5 is the base. 2 is the height. Hypotenuse is 8. You faggots are overthinking this.

>>7980273
For either 2 or 5 to be the height, it has to be a right angles triangle

Otherwise, the line goes slightly to the left or right so you have to calculate the height using the cosine rule or something

>>7979720
8 should be sqrt(29).

However this is just a bad question. Doesn't impune common core. ( which I think is mostly a buzzword to hate so people don't have to say it's actually math they hate and suck at )

>>7980275

I don't think it's a cosine rule problem. They haven't marked a right angle, but that doesn't mean it isn't one. At this level of work the assumption is always that straight lines form 180 angles. And therefore a perpendicular will form 90 degree angles.

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The shape is possible.
Pic related.

>>7979720
It has nothing to do with common core and the shape isn't really that shape with those dimensions.

>>7980310

Nothing is drawn to scale at kiddy level maths. Fucking hell, you faggots are so smart that you're dumb.

>>7980289
And how do you compute the area ?

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>>7979720
It's a fake image.

>>7980314
i think you are missing the point
its a sketch obviously but with those measurements it cant be a trapezoid like the sketch indicates but instead becomes this (>>7980289) which is non specific

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>>7980405
Correct, however your image is also not the right one

>>7980443

No, you guys are missing the point. Nothing is proportioned or drawn to scale at kiddie maths. They just attach arbitrary values and what you can see, you can assume. If you see a trapezoid, it's a trapezoid, even if it can't be a trapezoid for the reasons you've explained.

>>7980405
this is why I don't form opinions on things, it's way too fucking hard to make sure that the evidence you're basing your position on is genuine

>>7980289
>>7980459
Wondrous, we're teaching kids to disassociate from reality and trick them into thinking they're dumb.

If there's any reason why so many people hate math, it's because of stupid shit like this that teaches them math has no real world application.

>>7980255
Just ignore the 5 and find the actual height.

>>7980481
>dude just ignore given dimensions lmao

This thread is nothing but pure, unabridled autism. Does it work for teaching 8th graders how to use a formula? Yes. It doesn't matter. This isn't your Real Analysis class where you have to rigorously prove everything is possible before finding the answer.

>>7979823
>>7979862
Solve it then if you're so smart faggots. The shape is indeed possible (and looks roughly like pic related) but it is impossible to solve because insufficient information is provided.
>>7979846
Prove that it's cyclic dumbfuck.
>>7979870
It's not (necessarily) cyclic though so fuck off, Brahmagupta's formula doesn't work. There are infinite possible solutions. (the solution provided by Brahmagupta's formula may be one of those solutions, too lazy to check)
Hmmm.. it would be interesting to put together a formula to determine the upper and lower limits of all possible solutions.

File: common core trapezoid fixed.png (4KB, 462x228px) Image search: [Google]

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>>7980791
>pic related
Forgot pic because I'm a dumbass. At least not as much of a dumbass as a lot of the retards ITT

>>7979720
The triangle is a Pythagorean triple. The triple is 5,7,8. The 5 side is longer than the 7 side.

>>7980797
So its not cyclic. Lol

>>7980804
Not only that, but there's infinite possible solutions. I want to find the upper and lower limits of the possible solutions. New thread for that >>7980801

>>7979740
>2[(5*2)/2]
Really mother fucker?

>>7980247
5 is the height, retard.

>>7979720
The triangles are impossible to make as the 2 sides are smaller than the hypotenouse

>>7980318
triangulate it.

>>7979720
The figure doesn't look like that (isn't possible, perhaps?). Look at the right triangle on the left side. Using the pythagorean theorem, the length of the bottom part of the triangle is sqrt(39) = 6.3. Same on the other side, so 16 - 2*6.3 = 3.4. This is less than the top (12).

>>7979823
>you need to solve for ED
easy. (16-12)/2 = 2

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>>7979720
yet another thread where /sci/ lacks basic geometry skills

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i probably made a mistake somewhere, but i think this is about right. made the interval x=(sqrt(39), 13)

>>7982754
y(7.669547726) = 63.26675057

>>7979720
The trapezoid isn't correct
From the figure we should get the base of both triangles as 2 but from Pythagoras theorem we won't

The greatest minds of /sci/ have gathered together in order to solve a 7th grade geometry question.
So far each of them have proposed numerous theories as to what the answer might be but no truly uniform agreement has been reached yet

>>7982839
The only reason there is any argument over this is because the teacher who constructed this figure is retarded

>>7979786
Or people just don't care because it doesn't matter if it's well-proportioned, or not, nothing in math says that it has to be well-proportioned.

>>7979720
>What is wrong with this picture?
Common core
What if the kid is an artist? He needs no conception of area, volume, or geometry

>>7980478
Math doesn't always need to describe reality, number theory is an example. Also, many times in real life things will appear to be larger than some things, but then when we actually literally measure them the opposite seems to be true. So these sorts of problems do have applications in the real world.

>>7982851
>nothing in math says that it has to be well-proportioned.
Except this is supposed to prepare for real life applications and should hence reflect real life

>>7979720
It is a quadrilateral with two parallel sides [math]S_1[/math] and [math]S_2[/math]; and a given average distance [math]H[/math]
[math]H=5[/math]
[math]S_1 = 12[/math]
[math]S_2 = 16[/math]
[math]Average S =14[/math]
[math]S_average = 5 *14 = 70[/math]

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The figure is impossible

>>7980791
a dumbfuck learned something today, thanks

>>7982901
>Except this is supposed to prepare for real life applications and should hence reflect real life

>people will actively lie to you and blame you for it when you act wrongly on that false information
I dunno anon, seems useful real world information to me.

>>7979906
>>7979866
how's retard school treating you?

>>7979720
that it was posted for the 10.000th time.

>>7982901
Making correct inferences by isolating essential information is important for any real life application of anything

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This is my answer, j is the area
I could have a precise number if I were arsed to be precise with moving the side but who cares about the exact number anyway :^)

I think the webm is self-explanatory but I can clarify further

>>7983974
sorry for the really small webm but I suck at this shit, hope you've got magnifying glasses

>>7979732
*braindead

There are no units making it impossible to answer correctly.

>>7983974
>>7983977

why is your bin in the down left?

That is triggering my ptsd

>>7984566
I put my bin in the bottom right corner as well, why does this trigger you?

What is the point of the two eights? You obviously only need the bases and height?

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>>7980478
>no real world application

>>7979720
Its just a stupid diagram that someone didnt think about before drawing, but it doesnt matter, all it wants you to do is plug in the numbers to the equation and write down the answer

>>7979720
It has nothing to do with common core.

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>>7979846
>>7979823
>>7980289
>>7980797
>>7982754
>>7982906
>>7983974
This was discussed many times before, you know

>>7985975
There's nothing to indicate that that's a right angle.

>>7979720

Pythagorean theorem reveals the base is too narrow, or the height is too short.

>>7979866

It's not even out of scale, brohan. The thing isn't a trapezoid or any-kind-of-zoid with those dimensions.

This entire damn thread could've been avoided if the problem designer hadn't specified the length of the legs. Seriously, why did they do that?

>>7986088
So the kids would learn to pick out the relevant information

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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 we have to find the greatest possible area of a quadrilateral with sides 5, 12, 8, and 16-sqrt(39). The greatest possible area of a quadrilateral given all side lengths is the area of a cyclic quadrilateral with those sides. So we use Brahmagupta's formula.
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)]

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.
A=sqrt[s(s-a)(s-b)(s-c)]
s=(a+b+c)/2
s=22
A=sqrt(22*2*6*14)=60.794736614

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

Regards,

A /pol/ack who is smarter than /sci/fags. Sieg heil!

>>7986609
I like how you think being able to do some elementary geometry means you're smarter than everyone on this board.

>>7986614
considering how many retards got it wrong already he's not far off

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>>7986609
Actually, I take back what I said for the upper limit. I forgot that I had defined the angle that the 5 line makes with the 16-sqrt39 line as 90 degrees. Which therefore means Brahmagupta's formula cannot be used because it is not a cyclic quadrilateral. But we do not have enough information to split the remaining quadrilateral into two triangles and solve that way.
Pic related is courtesy of another /pol/ack, not myself. The upper limit is A=82.68.

The lower limit is what I said in >>7986609
that is A=60.79

>>7986630
(((4*5)/2)*2)+(12*5) = 80

?

>>7986630
(4^2)+(5^2)=(8^2)
16+25=64
41=/=64

OH SHIT BRUHHH

>>7986624
The minimum area is much lower. What if the quadrilateral is concave? The minimum area approaches that of a triangle with 12, 8, and 8 long sides. That area is 42.33

File: math diagram chinese.gif (24KB, 400x279px) Image search: [Google]

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Foreign geometry classes appear to be an advanced meme.

>>7986649
Yeah, you're right, but that would be 31.75 not 42.33 would it not?

>>7980464
This

>>7980456
Wouldn't the answer then be [math]80-5\sqrt{39}[/math] assuming the top and bottom are parallel and the vertical line is perpendicular to the bottom?