How long until multi-stall gender-neutral bathrooms are commonplace at United States universities?
>>71964871
They will be once Trump gets elected president
>>71964871
8 + 8 + 12 + 16 = 44.
I really can't believe minorities struggle with this shit.
>>71964955
Why do you say that?
>>71965021
44 days?
>>71965021
Nice.
>>71965021
thats the length of the perimeter dumbass, the area is 80
>>71966621
>80
What?
80 Apples?
>>71966848
80 squared units of whatever those lengths are in. If its inches, its 80 square inches, if its miles then its 80 square miles
>>71964871
What's wrong with those? They are fairly practical
>>71968769
Nothing is wrong with them. I fully agree with you. They are especially convenient for parents with young kids of the other gender, and also for shady areas and late at night when someone might want to go in the bathroom with their girlfriend/daughter/whatever in case of some weird fag.
Rapists won't obey "women only" signs.
While I recognize it's not a huge issue, I am all for making ALL public bathrooms gender-neutral. And it's not for the trannies, idgaf about trannies. It just makes sense.
>>71965021
>hate shitskins cause they can do 1st grade level math,and he cannot
>>71966621
it's actually 70
>>71965021
>>71966621
>>71970380
Are you niggers serious? Those lengths are impossible. A trapezoid cannot exist within those parameters.
(B+b)/2+H
fucking dumbasses
Can someone please post the ACTUAL solution? I can't work out more than a couple steps.
>>71970791
oopsie, it's (B+b)/2 * H.
>(12+16)/2 * 5
>(28/2) * 5
>14 * 5
>70 units
>>71970663
wat
>>71970663
Interesting, can you help me understand your postulation?
>>71971196
I've never heard of a 2-5-8 right triangle, because it's impossible.
>>71971104
Retard is [(B1+B2)*H]/2
>>71971356
So? The point of the question is whether you can correctly apply the formula to calculate the area, the actual trapezoid's shape is an abstraction.
>>71970888
You have to use reverse pythag and you will end up with a surd
>>71971400
PEMDAS, it doesn't matter if it's /2*H or *H/2
>>71970663
>trapezoid
>>71971512
>trapezoid
IT DOESN'T SAY IT'S A TRAPEZOID IN THE QUESTION YOU FUCKS
Dear groids,
The triangle on the left has to follow the pythagorean theorem.
so 64=25+39
Implies
12+2*sqrt(39) = 16
12+12.49=16
24.49=16
But this is impossible.
And it's impossible because a trapezoid cannot be of those dimensions.
>>71971650
Does the formula change?
>>71971738
So you just write 'you suck teacher' as the answer? Sorry anon, you failed.
>>71964871
>Gender neutral bathrooms should require you to solve math problem
Please no shills.
>>71971650
It doesn't explicitly state that it is a trapezoid, but it's implied by the picture and the fact that they expect you to use the formula for calculating the area of a trapezoid to solve.
>>71971861
DESIGNATED
>>71971858
That's why this meme attacks common core. Because it's asking school kids to solve problems that are formulated incorrectly. Of course people botching questions and then putting them on tests and quizzes is not exclusive to common core.
>>71972162
SHILL
H
I
L
L
>>71972279
SHITTING
>>71971842
>>71972057
iirc, /sci/ solved it about a month ago using some complex shit.
>>71972408
CURRY
>>71956939
>>71972173
Wait, so the "correct" answer to this question, as would be graded on a test, would be to point out that it is impossible?
>>71972484
Do you know where I can find the solution?
Shill thread
Exact pic and op posted yesterday >>71861644
SH&R
Retards.
16 *5 = 80
12* 5 = 60
80 - 60 = 20 Difference between a 5*12 and a 5*16 rectangle
20/2 = 10 Only half of the difference area exists since triangles
Area = 60 + 10 = 70
Why are right-wingers so stupid? Can't even solve a basic geometry problem. This is, of course, ignoring the geometric impossibility of a triangle with the width and height of 2 and 5 having a hypotenuse of length 8.
>>71972560
My understanding of common core mathematics is that the process of achieving the answer is emphasized more than the solution itself-- there are many similar common core questions that are impossible to solve, but they expect you to demonstrate your ability to utilize their odd brand of mathematical induction to come up with an answer, even if it is technically incorrect.
>>71972581
Search the image or the word "trapezoid" on whatever the /sci/ archive is
>>71964871
Hey retards, that is not the height, it does not say it is perpendicular so do not assume it. Also, that is not a trapezoid, it does not say the "bases" are parallel so do not assume it. You it is a composite figure of a quadrilateral so you need to solve it as such.
>>71972764
>Patting yourself on the back over solving some really easy shit
>Going so far as to display your thought process to people that really don't care
Looks like I've found the true retard ITT
>>71964871
(12*5)+(4*5)=80
>>71973185
eat a dick, fascist
>>71964871
That triangle on the right isn't necessarily a right angle triangle. It needs the box to truly be considered right angled.
>>71973219
It's ok German bro, we cant all be engineers.
>>71971650
Also, you just need to use the sine and cosine rules to solve this. Nothing too complicated.
>>71970663
Okay? Based on the given values, the area is still 70. Nobody gives a fuck if some theoretical trapezoid on some 11-year-old's math homework could actually exist in the physical world. Fucking autism.
>>71970663
Who cares, you autistic faggot? The teacher wrote that test. Teachers are human, and are therefore not infallible. Besides, at that level kids don't need to know whether or not a trapezoid with those dimensions can exist or not; it's not the point of the exercise.
>>71970888
The top and bottom lines are not defined as parallel and the 5-length line is not defined as perpendicular to anything. The most we can do is find the upper and lower limits of all possible areas.
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.
>>71970888
>>71973924
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.
>>71970888
>>71973924
>>71973962
Therefore:
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.
>>71964871
12 x 5 = 60
16 - 12 = 4
Area of a triangle = 1/2 base x height
4 x 5 = 20
60 + 20 = 80
What's with all the people over complicating this? All that matters is what it would be on paper.
>>71972484
They thought they solved it with Brahmagupta's formula but they falsely assumed that the shape can be assumed to be a cyclic quadrilateral.
>>71972581
>>71971842
See >>71973924 >>71973962 >>71974020
>>71972764
Are you fucking retarded?
>>71972560
As it would be graded on a test? No. Some fucktard intended the shape to be a trapezoid, but made several mistakes:
>did not define the 5 line as perpendicular to the 12 line or the 16 line
>did not define the 12 line as parallel to the 16 line or call the shape a trapezoid
>gave the shape impossible dimensions (impossible on a 2-dimensional Euclidian plane, anyway)
However, the ACTUAL correct answer (not what the person that made the problem intended) is that the shape is not a trapezoid and that there is insufficient information to solve, but we can obtain the limits of all possible areas.
>>71971861
Okay I already solved it, see my previous posts. Give me gender neutral bathrooms now, you designated faggot.
>>71973046
Then solve it with a definitive area faggot,protip: you can't
>>71971650
