If Sup Forums didn't exist then who would currently be president-elect of the United States?

If Sup Forums didn't exist then who would currently be president-elect of the United States?

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Donald Trump

Prollly Jeb. Keep that little spic boipussy flowing Mexicans.

Don't you just do the square, then add the triangles?
That seems easy

area is 70 i don't get it

Cause you can't just figure the area of two identical triangles and a rectangle can you?

Point of it is the shapes are impossible at those dimensions

I don't get Op's pic

Literally 14 * 5 you noob

None. There would be a third world war. Obongo would stay in office until he dies by a cardiac arrest.

Impossible how?

sqrt(8^2-5^2)=6.2449979984 which gives you that bottom part. Multiply it by two and subtract it from 16 to, supposedly, get 12. Instead you get 3.5100040032.

It doesn't matter that I say this, this thread is going to get 300 replies and get archived anyway.

height*(bottom+top)/2 = 70

durr so hard to use formulerr

Donald Trump

memesters had nothing to do with him being president, if they had any real world influence ron paul would have served two terms

ron paul support was literally twice the meme as trump support

Jews

>tfw you realize it's been so fucking long since you did any form of geometry you're not sure what the formula for the surface of a triangle is

That's for a trapezoid. We don't know what OPs shape is.

With the dimensions noted the actual shape should be much wider than it is tall. OP is an autist, basically.

The shape is impossible. Look at the triangle, the base of the triangle has length 6

5^2 + 2^2 =\= 8^2

The sides of triangle are 1.5, 5, 8.
Imposibru

Seems pretty obvious that the base of the triangle has length 2.

area is 70
whats wrong with this picture idk why they are even complaining, that is a perfectly normal and straightforward question.

do Pythagorean theorem

first one 2, sorry. still can't exist.

That only works for right triangles. We don't know anything about the angles here.

And by the pythagorean theorum, the square of the two sides must equal the square of the hypotenuse, which isn't the case here.

This isn't really Sup Forums material though. Slide thread?

...

It's adjacent to a square

IT'S NOT TO SCALE YOU FUCKING MONGOLOIDS. THIS IS WHY YOU'RE RETARDED.

The 'trapezoid' is sectioned into a rectangle and two triangles with right angles.

What square? I see no angles anywhere. For a quadrilateral to be a square, the angles should all be 90 degrees.

the base of the triangle is 2 dipshit
12 is across from 16 before the bases of the triangles start. This leaves 4 for 2 triangles of equal base, making each have a a base of 2. The area of a triangle is 1/2b*h so its (1/2(2))*5 which equals 5, 2 triangles means 10, and then 5*12=60+10=70

it is explicitly said in every test involving shapes I've seen that i've had that nothing is dra

never mind I just realized that triangle is impossible ignore this entire post

You have no information about the angles though.

Prove it you dumb fuck

It's a right angle, by implication.

I've written these problems all the time and I've always implied that strongly, unless otherwise noted. It's supposed to psychologically train the student that they must be as perceptive as possible and identify concepts/ideas on their own without needing someone to tell them.

t. Math Instructor.

the total area is 70 unit^2

here's a simple check with my shit-tier math skills
> pythagorean theorem: a^2 + b^2 = c^2
>a=?
>b=5
>c=8

a^2 + 25 = 64
64 - 25 = a^2 39 = a^2
square root of 39 is 6.24, so a = 6.24

6.24 + 6.24 + X = 16
X = 3.52 (this is the distance between the two triangles)

on top, they are clearly 12 units apart, which means this trapezoid is absolute bullshit

I don't see that in OPs pic. We don't even know if the top and bottom are parallel.

Wow.

Only smart people ITT. Probably both vegan cause non-vegans are idiots.

wait what the hell?

8^2 = 5^2 + base^2
64=25 + base^2
39=base^2
base=~6.25

so the base would have to be 24.5 for those measurements to be correct

why you do this

This is middle school math
You need to assume the trapezoid has a perfect square at the center
And even if you were correct, you need to have the other dimensions to solve it, which aren't going to be on a basic middle school geometry test you cucks

Fuck off back to your tumblr safe space you fucktarded cunt.

>by implication
This is the problem.

>be perceptive as possible
>don't be perceptive enough to notice that assumptions are being made that can lead to error.

Also as other anons have point out, if those triangles are adjacent to a "square", it would give us a triangle in which 5^2+2^2=8^2

LOL holy shit. Are you upset because you don't know basic math? Talk about REEEE.

Try picking up a book.

Prove it's a trapezoid you stupid fuck
Then tell me why a trapezoid has a "perfect square at the center"

nevermind. this trapezoid is bullshit

>you need to ASSUME the trapezoid has a perfect square at the center
Common core everyone.

>assume a trapezoid
>assume right angles
>assume parallelism
>assume square
>also I'm not going to point this out in the problem statement and I'm going to give you a "implied" trapezoid that is literally impossible if you assume these things.

You're the one claiming two non-parallel lines are parallel you stupid fuck

if it is in centimeter

we got to split it into 2 right trengle and one squah.

the squah is 12 on one side and 16 on the other... the sides are 5 so the squah is actuale a trans-squah.

If you call it her-squa or hi-squa then you violate the laws of pyfagorus. Now you need to use gender neutral shapes like circle

So the sides of the circle are 12, 16, 5, 5 since diversity is strength we're just going to mix them all together now. ok so 12 + 16 + 5 +5 is uh 28 29 30 31 32 33... 38.


Mkkay.

So the area of the circle is 38 cm.

Moving onto the Trans-angles we can see by the acute right hand that they are.


Hrrrm wait...the bottom of the squarhab- I mean circle isn't 16.

Ok class dismissed, we will resume SJW class tomorrow homework is to force a conservative to admit voter ID law is racist because black people are dumb and can't get ID because they're poor.

it's not a trapezoid, it's an imaginary geometric shape that is completely impossible

It doesn't make sense even then.
the bases of each triangle needs to be 6.25 units. which means that the bases together are 12.5 unites and that leaves just 3.5 units between the two at the bottom of the trapezoid, meanwhile the top clearly says there's 12 units between the two.

It would be impossible for it to be a perfect square in this scenario no matter what way it's drawn. You cannot have a perfect square if the top is 12, the sides are 8, and the bottom is 3.5.

about 140

CALM THE FUCK DOWN GUYS

I SOLVED IT

>A trapezoid is a 4-sided figure with one pair of parallel sides. For example, in the diagram to the right, the bases are parallel. To find the area of a trapezoid, take the sum of its bases, multiply the sum by the height of the trapezoid, and then divide the result by 2, The formula for the area of a trapezoid is:
This is a trapezoid problem, the kids had probably just learned about it in geometry
just because you think you're so smart because you know more advanced geometry doesn't mean you guys can use common sense whatsoever
What makes more sense for a kids to learn in middle school:
complex area calculations
or
trapezoid equations
We were always taught trapezoids in middle school and never got into more advanced calculations with imperfect calculations in high school
use common sense boys

Its fucking 70.

i think closer to 110 or 120

I think you guys are missing the point here.

This is middle school math. Our goal, as an educator, is not to trick them. At this stage, they're learning about basic problem development and how to setup problems. Before they can walk, they must learn how to crawl.

I've created several of these problems, but with correct dimensions. A fucking 11 year old kid isn't going to automatically make all of these assumptions or consider Euclidean geometry, the axioms, etc. They need to be trained to look at something, identify the basic ideas and apply basic concepts.

At this stage, we're not trying to trick you. In HS, you have to start thinking that way, i.e., questions on parallelism, angles, etc.

If I gave you "1+1 = ?" and you answered 2, I could say, "hah! I meant mod 2, so the answer is 0!!" You assumed incorrectly!"

Be reasonable here. They're kids. They just need to apply the fundamentals.

its clearly a trapaziod you fuck.

Ahhh...I see...

If you make the shape into a rectangle and 2 right triangles, the rectangle is 5 by 12, then the 2 triangles have 8 by 5 by 2 (16-12=4 and 4/2=2). The 2 base has to be about 6~ to be possible.

du må ikke nodvendighvis ha vinkelene.
Hypotenusen2 =katet2+katet2
Hypotenusen=kvadratroten av hypotenusen

Yeah I'm retarded, disregard my retarded post, I made a dumbfuck mistake

Mostly four

youtube.com/watch?v=SzzrZ_xAzUY&list=PLnBLKEtH52wlKmC2s5lYYSiIiIrkZBvij&index=3

It doesn't claim the thing is a trapezoid you stupid fuck. And regardless, this is from a college math class so fuck off
You're obviously fucking 70 and senile since you can't fucking solve this you stupid fuck
less
It's from college math you stupid fuck
No it's fucking not, go an hero

They taught us this in middle school. The area of a trapezoid is not a hard concept for a child to understand.

i'm starting to regret following you niggers for politics

go read a book.

How do you know its from a college math class?
Are you assuming?
:^)

>use common sense
No actually what you're advocating is teaching kids to memorize and regurgitate equations they don't conceptually understand. They aren't learning anything. The way you're teaching them leads to error in the real world when they get a real life problem that isn't hand held from what they just had you cram down their head.

This isn't even a possible trapezoid as other anons have pointed out. The kids don't even realize how wrong they are to use the equation for a trapezoid here because you haven't taught them to think conceptually, so they don't even question how or why the equation should work...in which case the training of them to use logic would lead them to realize how messed up the problem is. I can count the number of times I've used the equation for a trapezoid in my higher level courses and it's zero. It's a glorified playstation trophy that is meaningless. Meanwhile if you conceptually understand the use of pathagorean theorem and basic geometric concepts, you can arrive at the same answer...give you are actually given a trapezoid unlike the picture in OP. If you can conceptually work the problem out chances are higher you will notice something wrong which is the whole point of teaching kids these skills. You're just training them to not think.

RUBIO

ITS RUBIO

The right-hand side of the centre object is not 5, so it's not a square.

>this problem is college math
>common core is K-12
>this problem is middle school tier.

Are you a community college dropout? Or are you taking remedials still?

It's not. Read:
And learn to accept being wrong from your betters.

Ted Cruz

Please speak English, dork. If for no other reason just so everyone can see how wrong you are.

Hint: There is no hypotenuse if there's no 90 degree angle.

...

100

Jeb!
SL000000W and STEADY, my boy, slooow and steady

FYI for everyone saying this is an impossible shape, it's not.
It's just drawn very misleadingly.
Note that nowhere does it say there are right angles.
Just redraw the shape.
Of course it becomes much more difficult to calculate the area using the formulas a child would know.

Im gettinf about 108 doing this math on my phone. i assumed the 5 length line was a red herring meant to confuse

A trapezoid is "a quadrilateral with only one pair of parallel sides. "
That image is clearly a "a quadrilateral with only one pair of parallel sides."
The formula for finding the area of a trapezoid is a=1/2*h(b1+b2).

Its not open for discussion.

I know because the user that first posted it like a year ago posted a picture of his entire worksheet asking for help and he was in college, he didn't even realize how fucked up this problem is
So no I'm not assuming you stupid fuck
And I went to university so fuck off
No fuck off
Go back to tumblr

>That image is clearly a "a quadrilateral with only one pair of parallel sides."
Actually it isn't. There is no indication that the top and bottom are parallel and for the shape to actually work given the dimensions, they necessarily can't be parallel. That's not open for discussion. That's math and that's objective.

>Of course it becomes much more difficult to calculate the area using the formulas a child would know.
>much more difficult
*IMPOSSIBLE

The clearly are parallel.

sure pal

"Only one" doesn't mean "at most one" it means "exactly one" you fucktarded moron
An fucking hero
Thank you for not being retarded, my fellow American
Your dick is clearly parallel to your faggot tranny boyfriend's dick, faggot

Slightly more accurate angles.

They can't even be parallel. Clearly from the image, a 12 unit line is shorter than an 8 unit line...the entire picture and problem is fucked up.

So, I did a sketch in a program, and the figure gets all distorted as fuck and the area is around 83.

If we assume, that the sides are indeed 12,8,16,8 (going clockwise) and that the line of length 5 is indeed a perpendicular, then the following solution holds -

1) Since we have a right triangle on the left, we calculate its area - sqrt(8^2 - 5^2) = sqrt(39), which is the base. And the area is 5*sqrt(39)/2 ~ 15.612 .

Since it gets fuzzy and I am lazy to MSPAINT, we will denote AB = 12, BC = 12, CQ = 16 - sqrt(39), QD = sqrt(39) DA = 8.

2) we have the figure ABCQ, with a right angle AQC. Without much thought we draw a line from A to C, we get a right triangle ACQ and a fuck knows what triangle ABC.

the area of ACQ is 5*(16-sqrt(39))/2 ~ 24.388

3)the area of ABC is a whole different matter, but however we know all the lengths:
AB= 12, BC = 8 and AC = sqrt(5^2 + (16-sqrt(39))^2) ~ 10.961

We will use Herons formula (read: wolfram alpha), we get sqrt( i am not writing this, just write "area Triangle with lengths 12,8, sqrt(5^2 + (16-sqrt(39))^2)" into wolfram) which is about 42.682

So we add up the areas to see if I got something close to what the computer program told me I should get -

15.612 + 24.388 + 42.682 ~ 82.682.

Yipekayay motherfuckers.

and the only way for them to be out of parallel would be if one of the sides was another angle than the other?

...

im talking about the sides with length 8.

I seem to remember it can be solved, but maybe that's only if you assume the angles are 90 degrees.

because you wont fuckign read the solution without the picture

>If we assume ... that the line of length 5 is indeed a perpendicular

It can be solved for a range of possible solutions but not for a single solution.

Occams razor your retarded nigger faggots. Fucking both liberals and conservatives hate common core, thats becuase sometimes it's retarded but most times its not.