Common core, not even once

Common core, not even once
>amerifat education

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en.wikipedia.org/wiki/Quadrilateral#Non-trigonometric_formulas
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A = 5*12 + 5*4 = 60 + 20 = 80
What's wrong with the image exactly? Are you just retarded and don't know capital A is used for area?

I'm also retarded, but I'll leave the error as it stands.

When the fuck did I imply that A doesn't refer to area, you absolute fucking incompetent moron?

>5151
>66366
Lord Kek agrees that Amerifats are fucking retards.

pythagorean theorem
The shape literally can't exist

the length of the bottom that applies to the right triangles is 4. both of those triangles must have a bottom length of sqrt(39) ~= 6

The answer is 70, it would have to be a rectangle of 5x16 to be 80

The only problem with the image is the "8" side lengths. They are unnecessary, and also impossible (they should be sqrt(29) due to the Pythagorean theorem).

The area, if you ignore the 8's, is (b1+b2)/2 * H

12+16=28
28/2 = 14
14*5 = 70

You don't even need trig.

Yes thats right get rid of common core and all mistakes in test papers will disappear.

Every wrong thing is correct if you ignore what is wrong about it.

In one of these threads some user made OP's image to scale. I probably have it saved somewhere

The image is because there is no right angle.
I spent about 20 minutes arguing over it, but the other person was right. It's not solvable without more information.

the answer is ~91.22 you mongoloids

Never said they were excusable, just that he did the math wrong, and I was also explaining what was wrong with the picture.

This picture makes no sense at all. The scale is really off, it is not just an object but a trapezoid, you learn that in 1st grade, and they don't define A.

t. mathematician

...

My nigga

There are no right angles or parallel lines implied. You can't really guess it from the picture since the scale is so bad.

It's obviously supposed to trip up people who don't know what they're doing.

>Math pictures have to be drawn in the correct shape/scale

but now it isn't a trapezoid, there are no parallel sides.

jesus this board is full of retards. now i know how they come up with conspiracy theories. oh you cant solve it huh huh huh not enough information. the scale is fine and the angles are implied thats all you need to know

they never called it one

There's really no reason to follow the Pythagorean theorem if you're just trying to teach grade schoolers how to find the area of a trapezoid. It doesn't change the process at all.

I don't know much about the American education system, what sort of age would the kids be doing these sort of questions?

basic geometry is ~middle school iirc
more advanced geometry w/ trig is ~mid high-school

No but it isn't exactly hard to use a real trapezoid isn't it?

Most of the simple geometry like OPs pic is stuff for like late elementary school (9-12 years old).

What age would you be mid high school?

OK not as bad as I thought then

Around middle school so about ages 9-12

>I can't fucking even begin to understand geometry.

This.

all they had to do was literally not include the 8s.

In highschool your classes are a bit more personalized.

For me I took the more advanced geometry Freshman year (so 13-14), but I had seniors in my class (17-18).

If dumb weebs can solve this, surely you can too?

15-16, but this seems more basic

middle school is ~10-14 iirc

75%

A capital 'A' is the variable for area you stupid fuck

It's 70.
What the fuck is so difficult about it?

The scaling is off.

If you wanted it to look how it's drawn, you'd have to make the 8s into sqrt(29), so somewhere between 5 and 6.

If you want the number to hold true, but you can change the shape, see

One gold key

Ah, I see.

The 8 side triangle length is impossible with 2 as the base and 5 as the height. It should be 5.385. Sad!

thats not an acceptable answer mohamed.

0.844 with 2 silver right?

2C3*0.75^2*0,25+3C3*0,75^3

3 silver

25% chance of a given key failing * 2 keys needing to fail for a permanant lockout
6.25% chance of 2 keys failing and being permanently locked out

...

silver keys are better

pretend they are 4 sided dice.

you take one gold die and if you roll a 4 you lose.
75% chance win

you take 3 silver dice and if you roll two 4's you lose
~84% chance win

(10/64 outcomes are losses see pic related)

Obviously, when the fuck did I imply the contrary you absolute fucking moron?

I don't remember the formulae so I'm gonna work it out the simplest way

3 keys,

Outcomes are

P(000) = 0.25^3
P(001) = 0.25^2*0.75
P(010) = 0.25^2*0.75
P(011) = 0.25*0.75^2
P(100) = 0.25^2*0.75
P(101) = 0.25*0.75^2
P(110) = 0.25*0.75^2
P(111) = 0.75^3


And only 011, 110, 101 and 111 grant you your waifu.

Total probability:

P(011)+P(110)+P(101)+P(111) = 3*0.25*0.75^2 + 0.75^3 = 0.84375

You have an 84% chance of getting your waifu if you go for the 3 keys.

All you gotta do is cut the triangles off, and create a square. Where its sides are impossible but are 12x5. So this impossible cube's area is 60.

Then combine the two impossible triangles to form a rectangle (16 minus 12, which is leftover with 4 and divide that by 2. Now we know the length of the two triangle bases), The small rectangle will measure 2x5. Who's area is 10.

Add them together, and its 70 for the total area.

I did this in my head, its not hard just using the numbers and rearranging the image.

same logic against it:

75% chance of a key working * 2 keys need to work to get in
56.25% chance of getting in

Both of us made made the same math error.

Looks pretty straightforward. Quick formula for a trapezoid

A = (a+b/2)h
A = (16+12/2)5
A = 14*5
A = 70

Notice that neither 8 is used in this equation.

Also, although it *looks* like this can be analized as a 12*5 square with a couple of right triangles attached, notice that there are no compasses confirming the degrees. Therefore you cannot assume they are actually right triangles.

It is a bit of a trick question for sure, but I don't see anything wrong with it.

that's not how it works tho
besides, you have three keys, not two
my math is fine

The triangles must be right, because in order to use the formula you used, the 5 must be the height, and if the 5 is height, and the 12 and 16 are the bases, then the bases and the height are perpendicular.

Literally has to be a right triangle, that's why this shape can not exist as a trapezoid.

/thread

>having to use math to solve this easy logic puzzle.

The more things added into probability, the less change of it happening. The gold key is the only logical answer.

then tell us what you're bitching about fukboi

>American Education
stop please

You forgot there is only one base of the triangle when you combine them together. so its 2x5.

The answer is 70.

91.25
The center is 5x12=60
the ends together are 6.25x5=31.25

Of course, they're right triangles. This is middle school geometry.

Probability theory does not work that way.

Guy I replied to said you cannot assume they are right triangles, I corrected him.

>2016
>still believing euclidean lies

Oh it doesnt? You have a higher chance of getting struck by lighting 1 time, than you do 3 times. This can apply to literally anything. RNG included. I didn't say it was part of "Theory" I only stated that it just tends to work that way based on logical observation.

They are not. Hypotenuse must be sqrt(29) for them to be right angles.

That's why the picture is retarded. They must be right triangles because what I said hereBut they can't be right triangles because the numbers don't work.

So this shape can't exist.

What else did I expect from an American flag?

>"Are you retarded?"
>Gets easy question wrong

OP may be retarded but you're not much better

Sorry, I misunderstood. Also it can exist, just not as a trapezium.

>middle school geometry

I was good at math and didn't learn geometry until sophomore year of high school. I have friends who graduated without getting that far.

16-12=4
4/2 = 2
2^2 + 5^2 != 8^2

What the flying fuck is this? You're interpretation is completely wrong. OP's image is 100% correct and solvable.

I just now realize I am sharing this board with morons.

You guys tards. 70.

carpenter here

this shape is symmetrical therefore its a simple matter of cutting and pasting triangles and dividing the remainder of the lengths by 2.
dune

14*5

>"move" the left triangle to the right side to make a rectangle
>multiply14 by 5 to get 70
this is how I did it in my head in elementary school

on a side note this shapes measurements dont make any sense, the bottom of the triangle is 2 and the side is 5, that means the hypotenuse should be about 5.385

On review, good sir, I believe you are correct. The formula I used implies right triangles, therefore the alternate method should work and confirm, and the fact it does not proves that some element of what was given here is just plain wrong.

So I'll revise my opinion, it's a trick question, and a shitty one. I see what they were thinking - throw in a couple numbers that don't need to be used to make question tricky. That's the wrong kind of tricky though. It will grade people down for understanding the math in depth rather than applying the rote formula they were given.

Which is exactly what I expected, I just haven't done math in years and missed the trick there when I looked at it the first time. Thanks for pointing it out.

This is why we have to keep common core.

You retards are just too much.

With the given dimensions, this is the shape it has to be. Work out that area, math geniuses.

are you retarded?

Can I use Region+Massprop?

same area b/c symmetry of triangles and lines

Oh fuck this retarded problem. That 8 they threw in there isn't actually necessary to solve this problem if it was a regular trapezoid, but it turns out that the 8 means it isn't a regular trapezoid.

Its impossible to have a triangle with an hypotenuse = 8 having a height of 5 and a base of 4. Due to Pythagoras, the hypotenuse should be 6.403. The problem itself it's easy, but the values given are stupid. The only way of having that hypotenuse and height would be if the base was 6.24.
Tldr: the values given make the figure impossible to exist.

It's a base of 2.

16-12 is 4, but there are 2 triangles, so each triangle has a base of 2.

Read this post and check it for yourself,
the trapezoid as it's pictured it's made to be tricky and it's impossible to exist the way you're thinking it to be. If the eight there was gone, then you could reason it that way, but originally you can't solve it by your method.

the triangles cant exist with those dimensions

Also, if the base was indeed two, then Its even harder to get that hypothenuse, my negro.

>makes fun of ""amerifats""
>lives in the usa

The formula for finding the area of a trapezoid is A=1/2h(b1+b2).
The answer is 70. But I think you already knew that.

Common Core was actually based on a European creation, and I really wish all you dumb cunts would accept this fact.

It's the product of retarded leftists who found it "necessary" to change the way kids accept what they're told by making them re-think simple mathematics.

Let's put it this way - it took this sort of thing to cuck most of Europe by making the eurotrash question all which they'd known for centuries, when you control the learning process, you can influence people any way you want. Now, you're sucking Muslim cock over there, all because (((they))) convinced you dumb goyim that if you can find a crazy way to make 2 and 3 add up to 9, you can convince people of anything.

None of this shit came to be until we had our "liberal intellectuals" claiming that "MUH EUROPEAN KIDS ARE GETTING AHEAD OF MUH AMERICAN KIDS, BECAUSE THEY'RE LEARNING IN MUH NEW WAYS" shit.

What you'll find here are conservatives wanting to kill Common Core and leftists who want it to be pushed harder. Think about that for a moment, and we'll talk later once it settles in.

its solvable if you ignore the 8's

>66.66
this shape is evil I'm not solving it

I understand where you're coming from, but a lot of trigonometry questions are actually nonsensical. It doesn't matter how it should look, it's just an example so that someone can practice applying the formulas.

This has nothing to do with Common Core.

Need to get my ruler

>used to be super good at solving math problems
>can't even resolve the simple ones now

(0.75^3) + (0.75^2)(0.25)3 = .84375
silver keys are better

couldn't be arsed to remember the formula for combinations

Oh, cool. I love autism detection threads.

Using the general formula for a quadrilateral: sqrt((s-a)(s-b)(s-c)(s-d)-1/4(ac+bd+pq)(ac+bd-pq)), where a, b, c, and d are the side lengths, s is the semiperimeter, and p and q are diagonal lengths. the semiperimeter is 22. Using the law of cosines, the diagonals are ~17.114 and ~17.115. Note how they are nearly the same, making the shape a quasi-trapezoid. Plugging all that in, we get a final area of ~129.72.

Check my work, Sup Forums

General area formula found here: en.wikipedia.org/wiki/Quadrilateral#Non-trigonometric_formulas

it may be impossible but it exists