What's the answer, Veeky Forums?

What's the answer, Veeky Forums?

>mfw most of you can't do this

Nice facebook-tier thread and thumbnail you fucking faggot.

t. brainlet

what does this have to do with common core

Solve it.

I would try to solve it it if this image consisted of more than twelve pixels. Is the left side length "B"?

well I can barely fucking read it, but assuming the height is 5 and top and bottom are 12 and 16, it'd be 70.

nope

full res, my bad.

alright then what's your trick here? .5(12+16)(5)=70

It's impossible for the hypotenuse of a right triangle to be 8 while a leg is 5. Still seems more like this teacher needs to retake their own class than anything to do with Common Core

>right triangle
What right triangle?

The right triangle that must be formed by the "8" and the "5". If it's not a right triangle then this isn't a trapezoid and the area formula doesn't work

>It's impossible for the hypotenuse of a right triangle to be 8 while a leg is 5
What the fuck are you talking about? Those numbers are perfectly valid you brainlet.

>muh formula doesn't work
The absolute state of American education.

That triangle isn't possible

No they aren't. The sides of the triangle are 4 and 5. The hypotenuse should be 41^0.5, which is 6.something, not 8.

Excuse me, the sides are 2 and 5, so the hypotenuse should be 19^0.5, which is 4.something

I can't believe I have to defend this shit on Veeky Forums

Same poster. Try and change the blue length to be anything. Triangle doesn't work

>the sides are 2
lol, idiot.

Same poster again. Just making sure this poster sees this image and realizes why the fucking formula doesn't work

Yes it is.

>without muh formula I can't think
Veeky Forums confirmed for brainlets.

Would any of you like to take a crack at physically constructing a right triangle with leg 5 and hypotenuse 8? I'd love to see it

It's a magical 2 5 8 triangle. Doesn't work

a^2+5^2 = 8^2
a^2 = 64-25
a^2 = 39
a~=6.245

75

Except A is known. It's half the distance of the bottom minus the distance of the top
What the hell is the matter with you

lol

? where was this notated?

Because it's a fucking trapezoid. By definition a rectangle with two triangles on opposite sides.

70*

>Veeky Forums
>can't into math

I'd disagree with the other guy that this is DEFINITIONAL of a trapezoid, but he's still correct because both slants are of length 8, so both triangles must be equal, meaning the remaining 4 units from the line of length 16 must be split evenly between the two.

Where did it say, ANYWHERE on that page, that it was a trapezoid?
Or that it was symmetric about any of the axes?

In addition, that right triangle only works if the trapezoid's height is 6.whatever. Again, maybe think outside the formulas for a bit

What if it's the 2d profile of a 3d object?
What if this is the front view of a blueprint with the dotted line denoting a hidden feature?

Fuck me. I guess you got me there

I didn't think it would be this bad. Damn.

impossiburu

...

Now unless I'm a complete brain dead retard the formula is 1/2*h*(a+b)
So it's 52

(16-12)/2=2
2^2+5^2!=8^2
Fuck off with your kikebook memes op

quick scribble why that shape doesnt work

diameter^2 x 1.67

[math]4(10+\sqrt{28\sqrt{39}-61}) \approx 82.682070044[/math]
Easy.

Impossible shape, [math]2^2 + 5^2 \not= 8^2 [/math]

>ignore the 8s
>cut into two triangles (top left vertex to >bottom right vertex)
>use some trig
Donezooooo

Just move the left triangle on the right to make a rectangle.

This.

16 12 8 8a
(16,12,8,8)
96-16-0-0

I get the common core meme but this is bad tier. It's not convoluted; it's basic geometry. Fuck sake, people.

I agree it's a tricky question that forces you to make assumptions, but did anyone else get 88.734? Assuming the object is a tracksuit body with two triangles.

trapezoid*

Are you from an area that has a pixel shortage?

It doesn't have any right angles where the "5" is labelled, so you can't assume it's the vertical height. The vertical height is actually sqrt(64-(0.5*(16-12))^2) = sqrt(60). So the total area is 0.5*(16+12)*sqrt(60) = 14sqrt(60) ~ 108.4 square units.

You can't even have the 5 at ANY angle, since 5 < sqrt(60)

It's an impossible trapezoid.

someone find the average of all the replies, it just might be the answer :^)

add my 100!

Not Assuming the Height of the trapezoid is 5, and that 12 and 16 are parallel lines, the shortest distance is a perpendicular, forming a right angle at the intersection of 5 and 16. However, if the lines are not parallel, or 5 is not the perpendicular "Height", then 5 is as useful as a liberal arts degree.
Moving along, assuming 12 and 16 are parallel, and that 5 is perpendicular, then the triangle would have a right angle, but be mathematically inept. The 8 would have to be relabeled as sqrt(29) in order to keep 5 and 16 the same length.
Assuming again that 12,16 are parallel and 5 is perpendicular, it can be answered by completely ignoring the 8, using 's method, which would come out to be 16*5=80. However, if either assumption is false (not a trapezoid, more information would be needed to understand the irregular shape.