If person 1 saves person 2's life, is person 2 morally bound to be person 1's slave for the rest of their lives if that is the wish of person 1?
>>70405860
Obligatory
>never start a land war in Asia
who cares
dubs thread
>EMTs, Doctors, Nurses.
>Firefighters.
>Police.
Precedent says no.
>>70405860
No, person 2 can choose to return the favor by guarding person 1 until such a time that he is able to repay the lifedebt. Person 2 will sacrifice their life to save person 1 if necessary. It's the honorable thing to do.
>>70405860
I'm an American.
All I have to do is die.
>>70405860
No.
>>70405860
Just let person 1 fuck you
>>70405931
That image is wrong fucktard, otherwise prove that the fucking 5 line is perpendicular to the fucking 16 line you fucking faggot. That is just the upper limit 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.
>>70406182
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.
>>70406208
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.
>>70406182
>>70406208
>>70406223
its a minimum bound
i don't have to prove perpendiculars, I just assume it for minimum area calculation
Its the minimum area, not the only solution
>>70405931
Btw even the upper limit is wrong, it's ~82.6849 not 82.6771
>>70406264
Maximum*
>>70406274
thats an acceptable answer from truncation
what bug is up your ass?
>>70405986
They get paid to do it though. And the person they save is a taxpayer and tax money is used to pay them.
>>70406042
Is person 2 be morally bound to guard person 1 if person 1 wishes it?
>>70406148
Why?
>>70406173
If I'm person 1 then you'd better hope you're a cute Japanese girl. Otherwise I expect something other than sex in return.
>>70406345
Are you fucking retarded? First of all, truncation is not an acceptable answer. Second of all, even if it was, the answer (with 6 significant figures) would be 82.6848 if truncated (as opposed to 82.6849 if rounded), not the fucking shit 82.6771 that you gave. Learn what the fuck significant figures are. You used approximations in your calculations that resulted in you being unable to accurately claim your answer to have 6 figures of accuracy, therefore your answer should not and cannot contain 6 significant figures.
Now go fuck yourself you incompetent fuck.
>>70406358
>Why?
You need to say why they should, first
>>70406506
Because it is only because of person 1's action(s) that person 2 is alive. Therefore, person 2 owes a life debt to person 1. Prove me wrong.
>>70406480
the test it was on uses a fucking single function calculator, they can't use wolfram you dipshit
>>70405860
no
why would they be?
dumb question
>>70405860
Absolutely not. In a just society there are no obligations without express prior agreement.
If we permit your proposition, we will soon find ourselves being ruled by racketeers.
>>70406358
>Is person 2 be morally bound to guard person 1 if person 1 wishes it?
I suppose it would depend on the culture. In honor bound societies it would probably be expected of person 2 to be in a lifedebt. Asian master saves mortal enemy and enemy is so moved by the gesture they vow to a lifedebt. Vikings too maybe, probably any hardcore warrior sect like Sparta, Samaurai. I think it's more a bromantic story than an actual unspoken pact or law. It's always a choice. Unless the culture demands a lifedebt. I don't know of any cultures that do.
>>70406182
>>70406208
>>70406264
Poor, poor, lads. They think beta mathfags are welcome here. This is /pol/ we don't like numbers, almost as much as we don't like facts.
>>70406358
>cute Japenese girl
>this fucking weeaboo
get off my fucking board.
hey can we be real for a second isn't the area just
5*12 (the inner square)
+
5*2
= 70?
or am i wrongly assuming that the dotted line (5) is perpendicular to the long base-line (16)
>>70407569
its a physically impossible to form a trapezoid
did you even look at the first fucking response?
>>70406480
>>70406345
>>70406274
>>70406223
>>70406208
>>70406182
>>70405931
>>70405860
Is this entire thread just a replay of one from earlier?
>>70405860
Isn't that a 12x5 square plus 2 8x5x4 triangles?
>>70407569
If you assume trapezoid then things are perpendicular and the little triangles will be 2 wide.
2 wide 5 tall right angle triangle does not have hypotenuse of 8 therefore this shape cannot be a trapezoid.
>>70405860
No, but person 2 should rightfully repay the debt in some kind in anyway. If person 2 wants repay the debt by being a slave, sure, but there's no obligation. It should be a trade, so if person 1 ever needs help, person 2 is morally obligated to help him, but there shouldn't be any kind of legal binding.
I think honestly person 1 shouldn't expect a reward, he should save person 2's life regardless.
Also, here in Australia we were taught shit like that in high school. It's a+b/2 x h or something like that. Answer is 70, Google says it is too.
>>70407687
oh i see, if you actually size out the known segments it's not actually a trapezoid
>>70405860
how did common core fuck this up? I can't tell from the image
A= H(W1+W2)/2
did they make some new retarded formula for it?
>>70408360
pythagoras is rolling in his grave
>>70408438
Now I feel retarded. I completely ignored the 8s since they weren't needed
