1 million views!

@uq1149

4 жыл бұрын

👍

@youtunenaam144

4 жыл бұрын

Congratulations

@mfwc

4 жыл бұрын

nice

@banderfargoyl

4 жыл бұрын

Stir controversy, get views.

@user-yq7he1cm3s

4 жыл бұрын

Gooood👍

I find it disturbing for a teacher to give a question and expect a person to answer, "SYNTAX ERROR!"

@nairobi203

3 жыл бұрын

The correct answer would be: It Is not possible and explain why...

@freddyrosenberg9288

3 жыл бұрын

@@nairobi203 I have been in situations where I found an error in the test. The professor was not amused.

@nairobi203

3 жыл бұрын

@@freddyrosenberg9288 oh .. so not a deliberate "error" to see if student apply logical thinking, versus just executing a formula... Do things make sense at all should also be taught at school... Critical thinking.... Should It Not?

@acount4473

3 жыл бұрын

More like an array index out of bounds exception.

@outflame3040

3 жыл бұрын

Stack Overflow

if there is an error in the question everyone gets full points for that question.

@itsyourenotyour9101

2 жыл бұрын

This is the first problem I got right. And I just guessed. Time to retire.

@luck3949

2 жыл бұрын

In my school only those who were able to prove that there's an error in question received the full points. Nevertheless it made every physics problem funny, because every question followed things like "ignore the losses 9n heat; ignore the friction force; ignore the .. ". And a ton more of ignore every other miniscular force. I always tried to find some other force that wasn't mentioned in the long list of ignored forces, and sometimes successfully managed to find one. After that teacher gave me full points for a task that basically was skipped, and added one more "ignore" statement.

@AvoidsPikes-

2 жыл бұрын

Those were some rare, but appreciated, points.

@whitefox3189

2 жыл бұрын

That's only in cases the exam makers made a mistake. For those in the upper grades writing "There is no correct answer" is the only right answer.

@pacenal_18

2 жыл бұрын

Its not an error i think. It seems more like the teacher tricking the students to see their thinking, like how heu branch out.

The question should have been, "Prove that this triangle is impossible".

@jazzling

3 жыл бұрын

Not everything in real life has a solution. This is a problem that requires critical thinking instead of a formula.

@christopherwellman2364

Жыл бұрын

Nah

@keldonmcfarland2969

Жыл бұрын

@snarl banarl Not exactly. The students would have to know the difference between the base, hight, and the altitude. Most students probably drew the hight as 6 instead of the altitude. If they got the altitude at 6, They would have to know to use the unit circle and not bother too much with Pythagorean theorem. Getting the proof would still be a good problem.

@Sir_Isaac_Newton_

Жыл бұрын

@@jazzling What you say is contradictory in my opinion. If the problem tells me that the triangle has a height of 6, then I take that as true and look for a solution to it. I don't go "surely, the problem must've lied to me" because why would it. If it were real life, it'd be different, because there, we can actually make measurement mistakes and whatnot.

@KW-ox6dz

Жыл бұрын

@@Sir_Isaac_Newton_ Agree, the statement of a problem to solve should never have a lying information. Unless it asks for what is wrong in the statement.

School- 1+1=2 Home work- 9x9=81 Exam- John as two apple's, his train is 7mins late. Calculate the mass of sun.

@accidentallyaj5138

3 жыл бұрын

I remember this meme from 5 years ago

@valvaraad

3 жыл бұрын

Meme from Russian community: two ostriches were flying, one was red, the other to the right, what is the mass of kilogram of asphalt if a hedgehog is twenty-four years old?

@seventeen777

3 жыл бұрын

@@valvaraad the mass is 1kg

@valvaraad

3 жыл бұрын

@@seventeen777 correct

@LeTtRrZ

3 жыл бұрын

Mass of earth = 5.97e+24 kg Distance to sun = 8 light minutes + 32 light seconds = 299792458 * 512 = 1.5349e+11 m G = 6.67e-11 v = C/t = (2 * 3.14 *1.5349e+11)/(365.25*24*3600) = 3.0698e+11/31,557,600 = 30,530 m/s Fg = (GMm)/(r^2) = (mv^2)/r (GM)/r = v^2 M = (rv^2)/G M = (1.5349e+11 * 30530^2)/6.67e-11 M = 2.145e+30 kg Note that this is a rough estimate assuming a circular orbit, when in fact the orbit is elliptical.

there can't be a correct answer to an incorrect question

@lexgotham

7 жыл бұрын

The correct answer to an incorrect question is that the question is incorrect. You know that the teacher was expecting the students to notice that such a triangle doesn't exist. That's the only way she could expect her students to notice it without giving them a hint. If she had asked "Does this triangle exist?", it would have been way to obvious. And the only students who complain about the question in a philosophical manner (as you do) are the ones who didn't notice that this was an impossible triangle in the first place. It's not about the teacher, it's about them trying to get away with their ignorance.

@rarerol

7 жыл бұрын

Nice to meet you Mr. Selfproclaimed Knowitall

@lexgotham

7 жыл бұрын

Rarerol I don't know it all. I really don't. But when I don't know something, I don't pretend I do and that I just have been played with. I just take it as an opportunity to learn something. The question is legitimate and it has been asked for a reason. Bitching about it is just making noise and missing the point. Your answer is so expected as you do the same thing over and over: putting your pride first and blaming the others, whatever it takes, instead of questioning yourself. It's easier that way. And I'm sure, this answer will automatically trigger a response of self-defense, once again. You will just not think about it, will you?

@kalithp5397

7 жыл бұрын

Technically if we are nitpicking. In a formal logic kind of way all answers to a question with contradictory premises are valid. Since the triangle in question does not exist, it is perfectly valid to say that all of them have an area of 30 or 40 or are always pink.

@wmobberley4416

7 жыл бұрын

I did too. I guess that must be because the question was crap, therefore it must have been a female teacher suffering from PMT who set it. See? Logical deduction.

The values are clearly given in hex, allowing a max altitude of 8. Thus the answer is indeed 30(hex) or 48(dec).

@MetaKnight68

8 жыл бұрын

+Lord Balder This deserves top comment. XD

@snuffeldjuret

8 жыл бұрын

The world is basist though, if nothing is specified the base is assumed to be 10.

@Qermaq

7 жыл бұрын

snuffeldjuret I'm a bassist too! Nice.

@verigone2677

7 жыл бұрын

The world is not bassist, Flea and Les Claypool are bassists...the world as a whole is not smart enough to assume any numbering system more simple to understand than base 10. You would think base 2 would be easy, but no, they don't like the whole multiplying thing unless it must means keeping track of the number of zeros.

@gk1O1

7 жыл бұрын

genius!

I remember this exact problem from my 9th grade geometry class in 1972. I'm 60 years old now. It's funny how this is still in my memory.

@deadskinrippers

3 жыл бұрын

Ok I did it thru a whole other method (evil indeed). I compared the areas of 30units sq to 1/2 X bh. Get a relation bh=60 Now taking the opposite side to hyp of big triangle as 'a' and corresponding adjacent side as 'b' we get (a +b) = 8√5, sub for b from above equation b=60/a We have a quadratic equation a^2 -8√5a + 60 = 0 Solving for a the two roots are 2√5 & 6√5 If a =2√5 then b =6√5 and if a=6√5 then b=2√5 Squaring of these two should add upto the square of hypotenuse, bit it it doesn't meaning ERROR! This right angle triangle is not possible to have this altitude of 6 units with hypotenuse as 10.

@niveknospmoht8743

3 жыл бұрын

Same here but in 1975. I X'd out the problem on my test and didn't answer it. I knew within about 10 seconds the question was bogus having grown up with my father being a contractor where we used quite a bit of math, but this question stood out using the 3,4,5 method we used laying out walls to make sure they were square or doubled up for the longer walls. So I knew right away the six had to be a bogus or made up. One of the legs had to be the same length which would be impossible. Teacher was impressed that I caught this in such a short time

@deadskinrippers

3 жыл бұрын

@ansel blessonbrilliant. However I think the point is' i' is not taught at 10th grade level. Quadratics is ok. Concept of i is simply ignored or told be left aside and not expect questions on it at at that level we deal with determinants either equal or greater than 0. If less at 10th grade level it's ignored.

@sandipnayee5296

3 жыл бұрын

@ansel blesson isn’t that the way to find the area of a SCALENE triangle. I am pretty sure that it is the herons formula

@soniaskolnick3868

3 жыл бұрын

You remember because you were traumatized by it.

I would have answered: "Listen, teacher. You told us how to calculate the area of a triangle, and then asked us to calculate the area of this triangle. You never specified the system, and the operating assumption on us answering *any of your questions* is that the realm in which you are working is correct and we are to extrapolate upon it. So really our answer is correct because you gave us a *non-euclidean* right triangle to begin with, and you are wrong because you tried to force previously unspecified euclidean space upon our non-euclidean answers."

@canadiangopnik7007

Жыл бұрын

based

I tried to do it by cutting the hypotenuse into x and 10-x, and applying the Pythagoras theorem to form a quadratic in x, and kept getting complex numbers.

@erikthegodeatingpenguin2335

8 жыл бұрын

+jfb-1337 lol

@Justin-dk9rl

8 жыл бұрын

+jfb-1337 Funny thing is that even via complex numbers you get to an area of 30

@nendwr

8 жыл бұрын

+jfb-1337 I went along similar lines, but less elegant - I just gave up when I got a sine that was >1: Equation A: tan θ = x/6 x = 6 tan θ Equation B: tan ((π/2)-θ) = (10-x)/6 cot θ = (10-x)/6 6 cot θ = 10-x x = 10 - 6 cot θ Substituting the simultaneous equations into one another: 10 - 6 cot θ = 6 tan θ 10 = 6 tan θ + 6 cot θ 10/6 = tan θ + cot θ 10/6 = 1/(sin θ cos θ) 10 sin θ cos θ /6 = 1 10 sin θ cos θ = 6 2 sin θ cos θ = 6/5 sin 2θ = 6/5 [uh oh...] If I'd thought of complex numbers, I could then have unhelpfully solved for the angle as: θ = (((1/2)+c) π - i ln((6/5)±sqrt(1/5)))/2 Where c is a real integer. Unfortunately this has infinitely many solutions (all of which are complex). So giving up was the sensible solution.

@drizzy8450

8 жыл бұрын

+jfb-1337 I ended up with the equation 6tan(theta)+6cos(theta)=10 which there are no REAL solutions to...

@simleek6766

8 жыл бұрын

+jfb-1337 Yeah, the triangle's complex, but it still has a base of 10 and a height of 6. Solving for your x for your mini right triangles, we have: - __ C1/ | \__C2 - / |6 \__ -/___|__________ \ - x 10-x That gives us 3 equations: C1^2 = x^2 + 6^2 C2^2 = (10-x)^2 + 6^2 and C1^2 + C2^2 = 10 eventually, that turns into: 2x^2 - 20x + 162 = 0 and the quadratic equation solves for x as: x = 5(+-)2√14 i So, the base size isn't real... however, that doesn't mean it doesn't exist. If you choose your plus or minus and solve for the total size of both mini-triangles, you get: [(10-(5+2√14 i))*6 + (5+2√14 i)*6 ]/2 = [60 - 30 - 12√14 i + 30 + 12√14 i]/2 = 60/2 = 30 So, the containing triangle and the mini triangles exist, but have complex dimensions. Also, the triangle's area is still 30. The teacher needs to go back to school. :P

This video is the answer for those that say "there are no stupid questions"

You can be a smarty-pants even further, and say that in a non-flat space you can have such right-angle triangle with a surface of 30.

@Crowsinger

4 жыл бұрын

It's not a triangle anymore, it's a pyramid. 😃

@angelmendez-rivera351

3 жыл бұрын

Crowsinger No, it isn't. Pyramids are 3D shapes embedded in 3D space, not 2D shapes embedded 3D space.

@Crowsinger

3 жыл бұрын

@@angelmendez-rivera351 He's talking about 3D space. "Non-flat".

@angelmendez-rivera351

3 жыл бұрын

Crowsinger No, he isn't. Non-flat does not mean 3D space. It implies non-Euclidean space. In other words, it implies spherical geometry, or hyperbolic geometry. For example, in Euclidean geometry, the angles of a triangle always sum to π, but in spherical geometry, they always sum to a number that lies in the interval (π, 3π/2]. And that is not the same as a pyramid, because in a pyramid, all the angles sum to 4π or some higher multiple of π, depending on the base, assuming Euclidean space, of course.

@angelmendez-rivera351

3 жыл бұрын

Crowsinger In other words, non-flat is simply a synonym for curved space.

I encountered this question in one of my MTAP test.. *Now I know why I'm correct although I left the question blank.*

@sabyasachipattanayek3748

4 жыл бұрын

Ha ha ha

@sheriff332

3 жыл бұрын

Now I leave the question paper blank and get full marks

well now i know what an altitude is.

@PtolemyJones

7 жыл бұрын

It is totally possible my memory has faded, but I suspect there was a different term when I was in high school...

@meghanto

7 жыл бұрын

Perpendicular?

@UTU49

7 жыл бұрын

When I was in school, I always thought of it as "purple-dickular". Having a dirty mind makes many things easier to remember. ;)

@geoffgwyther7269

7 жыл бұрын

yes,, I had to look that up, in my day it would have been called a perpendicular.. Learning all the time.

@m2kopter840

5 жыл бұрын

height

I think the answer 30 is correct... The statement in the problem ("there is a triangle ...") is clearly false, and anything follows from a false statement :)

@p0gr

8 жыл бұрын

+F Viktor exactly, i was going to write that.

@wurttmapper2200

6 жыл бұрын

Then all answers are correct, incluiding the answer "there is no answer"

@andreasnaidero7472

5 жыл бұрын

@@wurttmapper2200 Yes but it is not correct to say that 30 is an incorrect answer.

@wurttmapper2200

5 жыл бұрын

Andrea Snaidero Indeed

@andreasnaidero7472

5 жыл бұрын

@@wurttmapper2200 I'm sorry I take you back one year but I'm right😋. If you say that you contradict yourself. You said that all answers are correct so it's incorrect saying that there'are some wrong answers. You see, saying that an answer is incorrect isn't an answer, so it is not true that all sentences about answers are valid. All answers are valid but not all phrases about them.

Teacher : do this children ! This is evil. Student : i got this mam ! Teacher : its wrong ! Student : answer ? Teacher : no . Question is

I hold my position that every triangle in the set of right triangles with hypotenuse 10 and altitude to hypotenuse 6 do indeed have area 30.

@noex100

8 жыл бұрын

So you think that triangles that don't exist have an area? WTF?

@andrewsauer2729

8 жыл бұрын

noex100 It was half joke, half set theory. If you look at exactly how my statement was phrased, you will see that I was claiming that every triangle like he described had an area of 30. Since there are no triangles that are like he described, I am then quantifying over the empty set, and therefore technically anything I say about "every element in the empty set" is true. Think about it this way: you can't find a counterexample to my claim.

@noex100

8 жыл бұрын

andrew sauer Well, technically what you said is incorrect if you acknowledge the existence of the complex plane. The area can be calculated as an imaginary number. But strictly in the Euclidean plane, that's true.

@johnchessant3012

7 жыл бұрын

Exactly what I thought! :D It's like saying: if this triangle exists, then its area has to be 30. Kind of like saying that if 1 + 2 + 3 + ... exists, then its value has to be -1/12.

@urinelakeurinalcake

7 жыл бұрын

+Maurice Scholz So did you

am i the only one who cant solve those exercises because i dont know the english mathematical terms?

@feraudyh

7 жыл бұрын

Yes, I was not sure about the meaning of "altitude" either.

@kuhlde1337

7 жыл бұрын

As someone who speaks English fluently, I also assumed altitude equals height.

@nicolaspallauf2901

7 жыл бұрын

***** i think hypothenuse is universal though xD

@thewolfdoctor761

6 жыл бұрын

I got my BS in math in 1974. I have never heard of an altitude to the hypotenuse before now.

@ha.ha.ha.ha.9125

5 жыл бұрын

@@feraudyh altitude is height brudda

The lesson here is: do not trust the textbook, and do not trust the teacher.

@nairobi203

3 жыл бұрын

The lesson Is... Think...

@Utkarsh.22013

Ай бұрын

😂😂 3:22

I got complex roots after solving the simple quadratic equation when trying to figure out sides of sub-triangles.

@xouthos

3 жыл бұрын

Same here, I ended up with this unsolvable quadratic equation 2z^2*-20z+72=0 to solve z which is part of the hypotenuse, that's when I started suspecting that there was no solution to this.

@user-cz7bu5qk8w

3 жыл бұрын

Pretty easy. The hypo is 10 so the sides must be 6 and 8. But each of the sides is a hypo to the 6 length altitude and a hypo must be longer than a leg. Not possible.

@BrutalBeast666

3 жыл бұрын

@@user-cz7bu5qk8w Actually the sides can be anything in the range ]0, 10[. For example when both sides are length √50 you get C = √((√50)²+(√50)²) = √(50+50) = √(100) = 10.

@monnamonsta

3 жыл бұрын

@@user-cz7bu5qk8w your guess is only correct when the sides must be natural numbers.

@deadskinrippers

3 жыл бұрын

Ok I did it thru a whole other method (evil indeed). I compared the areas of 30units sq to 1/2 X bh. Get a relation bh=60 Now taking the opposite side to hyp of big triangle as 'a' and corresponding adjacent side as 'b' we get (a +b) = 8√5, sub for b from above equation b=60/a We have a quadratic equation a^2 -8√5a + 60 = 0 Solving for a the two roots are 2√5 & 6√5 If a =2√5 then b =6√5 and if a=6√5 then b=2√5 Squaring of these two should add upto the square of hypotenuse, bit it it doesn't meaning ERROR! This right angle triangle is not possible to have this altitude of 6 units with hypotenuse as 10.

the question is "there is a triangle..." so the question is wrong?...

Since I'm geometrically challenged, I solved it algebraically... Call one leg of the triangle A; the other leg of the triangle B. So A squared + B squared = 10 squared = 100. The height of the triangle splits the triangle into two sub-triangles. One sub-triangle has a leg of length 6 and a hypotenuse of length A. Call its other leg G. G squared + 6 squared = A squared. The other sub-triangle has a leg of length 6 and a hypotenuse of length B. Call its other leg H. H squared + 6 squared = B squared. Since A squared + B squared = 100, then (G squared + 6 squared) + (H squared + 6 squared) = 100, or G squared + H squared = 28. Since G and H together form the triangle's hypotenuse, G + H = 10. Or G = 10 - H. So, (10 - H) squared + H squared = 28. Expanding, (100 - 20H + H squared) + H squared = 28. Combining, 2 (H Squared) - 20H + 72 = 0. Dividing each side by 2, H squared - 10H + 36 = 0. But this quadratic has no roots! No solutions! The quadratic formula's (b squared - 4ac) is negative, giving no real solutions! So, this triangle cannot exist! And if the triangle cannot exist, it can have no area. Take that, textbook!

@MejiUlises

8 жыл бұрын

+jkarafin Followed the same path :P

@jimmysyar889

8 жыл бұрын

You could do it easier by doing X and 10-X where the altitude splits the hypotenuse and doing x(10-x)=6^2 -x^2+10x=36 X^2-10x+36=0 Solve and find that there are no real roots

@amaansajid3651

8 жыл бұрын

The quadratic formula is actually X=-b +/- (the root of b^2-4ac)/2a

@amaansajid3651

8 жыл бұрын

But yeah your method was actually really good

@zxwy37

8 жыл бұрын

+jkarafin It's that, or it's an imaginary triangle! :-D

I'm happy and angry at the same time. If a teacher does this to our class all of us will probably never Go back to that class again

@nairobi203

3 жыл бұрын

He teaches to use your brain. I think this is a good thing. Have a nice day.

@agentEE233

3 жыл бұрын

Oh the smart ones are here I'm scared

@agentEE233

3 жыл бұрын

@@nairobi203 and why are you dissing me? I think I didn't say anything that may offended you

@nairobi203

3 жыл бұрын

@@agentEE233 hi, sorry if I exagerated... I modified my comment

@agentEE233

3 жыл бұрын

@@nairobi203 oh ok

Me: Finally! I think I finally solved one of these problems on my own! Presh: Guess what. Trick question. Me: FUUUUU-

if all else fails, the answer would always be 42.

@738polarbear

7 жыл бұрын

Why are you talking such rubbish .It's not even funny ,and neither are you.

@YaeBocchi

7 жыл бұрын

738polarbear What I'm saying here is that absurdity is something that you could laugh about. Dude, you need more than intelligence quotient in life or you'll turn yourself into a humorless loner.

@stevepqrclaret

7 жыл бұрын

Your advice, I'm afraid, seems to be too late...

@elmermartinez7645

7 жыл бұрын

en.wikipedia.org/wiki/The_Hitchhiker%27s_Guide_to_the_Galaxy + ctrl-f 42, the answer to life, the universe and everything.

@Xenosanta

7 жыл бұрын

Read the 5 books in the trilogy. Brain the size of a planet and they have me parking cars. Do you call that job satisfaction? Cause I don't. The first 1 million years were the worst, then the next million years were the worst. Then I went into a decline.

I knew it right away because this was one of the 1st things we learned regarding right triangles, that the height of a right triangle can be a maximum of the hypotenuese/2. And we had in in the test also, where we had to calculate it.

I very vaguely remember a problem similar to this in a math team meet I was in back in high school. I couldn't figure the damn thing out, and when the teacher went through the problem, I noticed that the triangle the problem ended up with was literally impossible. (The resulting triangle had the sum of its legs equal to the length of the third side.) One challenge later, and everyone got those two points.

When I was in elementary school in the 200Xs, we were taught to calculate the areas and perimeters of right triangles given all three side lengths. Not only did the worksheets regularly give incorrect hypotenuse values (we were only on single digit multiplication, and there’s only one Pythagorean triple that fits that limitation,) but sometimes they weren’t even triangles. I distinctly remember being asked to find the perimeter and area of a right triangle with legs 2 and 10, and with hypotenuse _12._ Why they tried to integrate the two concepts together when the students lacked the necessary skill level to do so properly is beyond me; They could easily have used right triangles to teach the area formula and stuck to generic triangles for perimeter. When I quite earnestly told my teacher that I couldn’t solve the problem because of this, she sent me to the principal’s office for “being disrespectful. Elementary school f***ing sucked. ...why, yes, this _was_ in America. How did you guess?

@sunilkumarpandey8336

2 жыл бұрын

I feel sad for u

@keldonmcfarland2969

Жыл бұрын

Wow! ‼️ What state was this in? The bid for that text book must have been extremely low and the school probably wanted to save the money.

Nah, there's an answer. Going back to your circle construction, you can identify the third point of the triangle with angle θ, treating the circle like a unit circle. The area of a triangle is then: ½bh = ½(10)(5sinθ) = 25sinθ. We want the θ where sinθ=6/5. By Euler's relationships, this is θ = π/2 - i ln[(6+√11)/5]. This means that the triangle's area is indeed 30. It's just that to draw the triangle, so long as the base runs from (-5,0) to (5,0), the third point needs to be at coordinates: x = ±i√11 y = 6 Good luck sketching that ;)

@nadni1

8 жыл бұрын

+tibschris now THIS is a correct answer

@pookie9461

8 жыл бұрын

I would love to see the look on that math teacher's face if someone in the class pulled this out of their ass...

@Scorch428

8 жыл бұрын

I see you take your KZread comments seriously. P.S. Asian, right?

@tibschris

8 жыл бұрын

Scorch428 No, Euler was Swiss.

@kaevondong

8 жыл бұрын

Best. Reply. Ever.

i had an immediate feeling: "the altitude is too much, it cant be a right triangle" before realizing exactly why

I know most people in the comments don’t like that this question was given, and fair enough, it would be harsh to penalise students in a test for a misleading question like this, but I also think it’s a good demonstration to show the importance of critical thinking in problem solving and not to just plug in formulas without considering what you’re doing

Watching the miniature I thought the 10 was only the part of the hypotenuse at one side of the height (the leg of one of the two small triangles formed). If I had taken the test based on that miniature (rather than the phrasing of the problem), this reasoning could have actually helped me make the teacher accept that my interpretation was correct, because it was not absurd as the intended take on the problem. And I would most likely know it was absurd if he gloated about it in class.

Such a triangle may not exist, but this was not the question. If you answer "THIS triangle has an area of 30", you are correct, because if such a thing exist, it has to have an area of 30 and if not your answer is irrelevant, because you can conclude anything from a wrong premise. So you can also conclude an area of 30.

@Cannongabang

8 жыл бұрын

well the question regards a "right triangle" but such triangle is not right so... well, it depends on the tracher i guess

@edvink8766

8 жыл бұрын

+Supremebubble The question is based on false premises. Therefor anything concluded from question will be incorrect.

@justinhall3073

8 жыл бұрын

+Supremebubble I agree. Math is a tool we use to explain our physical world and its laws extend to things we do not understand. It is important that students learn the language of math and its laws. Then they can learn our physical boundaries.

@MrTheboffin

8 жыл бұрын

+Edvin Karahoda no the only question there is, is what the area of the triangle. we don't need to know the triangle is right to work that out. its just a trick question to wich the resulte is debatable a best

@michelsfeir1127

8 жыл бұрын

+Supremebubble *Applause* Reminds me of an xkcd xkcd.com/169/

I'm at 0:30 in the video and I got that the answer to the problem is 30 square units. Divide the hypotenuse into two smaller segments based on where it intersects the altitude line, and give these two segments length a and b respectively. For each smaller triangle created by the altitude, A = bh / 2. You can say the base for both of these triangles is 6, and the height is a and b respectively. The area of the bigger triangle is the area of the two smaller triangles added together so A = 6a/2 + 6b / 2 A = (6a + 6b) / 2 A = 6(a + b) / 2 We know that a + b = 10 because they are the two segments that make up the hypotenuse, whose length is 10 so A = 6(10) / 2 A = 60 / 2 A = 30 well shit

@darcraven01

8 жыл бұрын

+Neil Gupta *will assume the "well shit" was an edit after you watched the rest of the video... laughs a lot* lol

@__nog642

8 жыл бұрын

darcraven01 yup.

@BorussiaMG4ever

8 жыл бұрын

but u can work with that way, too. if u have the segments a and b = 10 - a. call the two other sides of the triangle c and d. now u can find three equations: (1) c^2 + d^2 = 10^2 (2) 6^2+ a^2 = c^2 (3) 6^2 + (10 - a)^2 = d^2 (2) in (1): 36 + a^2 + d^2 = 100 with (3): 36 + a^2 + 36 + (10 - a)^2 = 100 a^2 - 10a + 36 = 0 => a = 5 +- sqrt(-11) and there is no solution for that in R

@Sandokiri

7 жыл бұрын

That's normally true. However, no right triangle can exist, such that the altitude to the hypotenuse is greater than half of the hypotenuse.

This solution opened up my mind in several ways, I had never thought of a right triangle and altitude etc. in such a way

your videos are so interesting, keep up the good content

Im pretty sure that fucking triangle in the triangle isnt 90 degrees

@SomeRandomFellow

8 жыл бұрын

altitudes are always perpendicular. That's part of their definition

@spammie1234

7 жыл бұрын

The top was.

@zwz.zdenek

7 жыл бұрын

The drawings were all done by hand, it would seem. The Thales circle diameter wasn't going through the center.

@PtolemyJones

7 жыл бұрын

Never judge the question by the diagrams, but by the numbers given. I noticed the diagram was off, and it annoyed me, but the diagram is less a representation of the true image then a placeholder for the various numbers.

@Cr8Tron

7 жыл бұрын

+Some Random Fellow Not when they're in poorly displayed diagrams.

This is a perfect example of cheap deception. We weren't wrong for thinking it is 30. Well, the answer was wrong, But a typo cannot be an answer in terms of tne problem at hand. You see, The questions says it IS a right triangle and it DOES have the altitude of 6. Regardless of if the question is screwed up, this is DEFINITE fact in the hypothetical situation. If the hypothetical situation is impossible, than no answer is possible, therefore this isnt a question at all. its a lie. Because the question lies, it isnt possible answer, therefore we cant be expected to be correct. The class, if anyone drew it out, could note the problem, but in this scenario, the student has to assume the teacher is doing a type of pre-test. If not, than it is work a teacher wont help on, so even if this is wrong, of course nobody answered correctly! Nobody would sit there and draw a diagram with the assumed time limit the teacher had to test out the legitimacy, as that is stupid! Also, as a side comment, I know you said Right triangle, but clearly, if you wrote it out in any way other than the intended right angle, you wouldn't have an altitude of 5! or at least last time I checked, a right triangle had to have a 90 Degree angle, not the possibility of two 90 Degree angles if you cut it down the altitude, so the diagram you originally presented that is also in the thumbnail is wrong. Not confirming this, need to look back at the image before I can confirm it isn't.

@Zackattackplayspokemon

8 жыл бұрын

Ok, side comment is incorrect, the original point still stands

@davidvogel2786

8 жыл бұрын

Not all problems need to have a real number solution for it to be a valid test question. Test questions are supposed to test your understanding of the concepts of the material not just rote memorization. And before you think that the students had to understand anything about circles to figure this problem out, the entire problem can be solved with basic algebra and the Pythagorean theorem (which I'm assuming they would know if their test was on right triangles) All they would have to do is draw the triangle to prove they know where an altitude would go, know that the altitude creates 2 smaller right triangles within the big right triangle, set up Pythagorean equations for all three triangles, and then combine them to solve for the two lengths of the hypotenuse that the altitude separates(I'll call these lengths y and 10-y). If they get this far, then they can substitute these values back into the smaller right triangles to get the base and height of the big right triangle and get the area from there. The beauty of this question is that by proving that y and 10-y are not real numbers, the problem ends here and the solution is simply no real solution and your work proves why there is no real solution. This is actually shorter than if the altitude was less than 5 as the students would then have to calculate the base and height from y and 10-y to then find the area which is more steps and therefore longer.

@Zackattackplayspokemon

8 жыл бұрын

David Vogel I dont understand why a teacher would give a Quiz, which typically has a grade, for a false answer. Typically, teachers giving tests check the questions for these false things. Also, popping a test out of nowhere and flat out hiding the concepts you are supposed to learn is just pointless. We know there is no answer, but to the kids and my point of view, no kid would think "Okay, before I do the problem, let me make sure this question is possible" because when given a normal test setting, they would run out of time. Even the tests that try to trick people try to make the questions have at least some chance to be solved. Think about it. If you were told to do this with a similar question, having not seen this before, you would try to solve it, not prove the teacher wrong. Giving questions with no real answer due to the fault of the question itself doesnt work, because if the question isn't specifying a situation irl where this happens, you can always assume the question is telling the truth. If this question was a usable question, it might need to be worded as "Sally drew a right triangle and labeled its altitude 6 and its base 10. Sally said that her triangle had an area of 30 units and could be recreated in real life with real units. Is Sally correct?" However this is not what is presented. The question guarantees that 6 and 10 are the genuine lengths, but by them being impossible it contradicts itself and ends up invalid to use for a test, something the creators should find and fix, not the people answering the question.

@Schmidtelpunkt

8 жыл бұрын

It is from a textbook, not a quiz. So it might make sense to revisit the derivation of the formula and also fooling the pupils, in order to start the discussion. A meaningful pedagogic excercise however isn't necessarily a meaningful mathematical as well. In a quiz it would make sense in the same light, although I have to agree that for giving grades, the wording would have to be different in the way that it allows giving the correct answer (e.g. Which area of the triangle meets the given conditions? Answer: None, because...). Else pupils would have either to break the laws of the test or the laws of maths, so a lack of care about the logical implications of the wording would get rewarded, essentially punishing precision.

@LordSantiagor

7 жыл бұрын

Detecting deception is one of the most important things you can learn in life :)

Well, I figured something was up when I did some algebraic substitutions to solve for the sides and was getting quadratics that had only imaginary solutions...

@-kat

4 жыл бұрын

John David Omg ikr, I did some crazy algebra and plugged in a bunch of variables and got some messed up -i(sqrt)some weird decimal

I just spent a couple of hours trying to remember how triconometry works in order to solve this (now that I remember, math clases back then never covered the "height

I thought the textbook had intended it to be a hypotenuse of 10, and one of the other sides is 6, in which case you have a 6-8-10 triangle with area 24...

@omp199

8 жыл бұрын

manudude02 If this problem really _was_ in a textbook (which I doubt), then it is obvious that that was not the intention of the writer. Nobody would _accidentally_ write "an altitude to the hypotenuse" if they intended to refer to one of the sides. That is not even remotely plausible. Nobody would accidentally introduce the notion of an altitude if it was not in their mind in the first place.

@johnconway6267

8 жыл бұрын

+manudude02 In this instance though, the 6 is an altitude to the base of 8 . The question however states that the 6 is an altitude to the hypotenuse of 10 . A totally different scenario .

@MaGaO

8 жыл бұрын

+manudude02 I find the thmbnail image much more telling. It is the image of a right triangle with a height to the hypotenuse of 6, and a distance from the projection of that height to one of the vertices of 10.

@user-zu1ix3yq2w

8 жыл бұрын

+omp199 Holy crap. It's very possible. Maybe not for you. Have you ever heard of dyslexia?

@omp199

8 жыл бұрын

Sarek Mather Ha ha! I have heard the term. I have often heard people claim dyslexia as the reason for their not being able to read very well, or spell very well. I have _never_ heard anyone claim dyslexia as the reason for introducing an entirely extraneous and abstruse concept into a discussion. Are you really being serious? Can you imagine what the world would be like if all those people had a condition that made them do things like that? Alice: Do you spell your name with two bs? Bob: Let me see. I think you write it in Arabic script. Alice: Like this? Bob: Oh, no. I meant with two bs. I'm sorry that I said to write it in Arabic script. I have dyslexia.

The question is wrong, not the answers.

It's specified that an altitude is 6. It cannot be the one dropped to the hypoteneuse (which has to be

I figured it out the moment you flipped the triangle. Awesome!

Maybe if the triangle is laying in a non-flat surface...

@user-tf6pg7jj6c

8 жыл бұрын

+x nick In mathematics the triangle always lay on a plain, and the plain is always flat. But in physics it is posible, so i like your thinking. Can you give an accurate answer if the triangle is non flat ?

@xnick_uy

8 жыл бұрын

Петър Петров I think that "mathematics" would actually mean "euclidean/flat geometry". You have also differential geometry, projective geometry, and so on, all in the realm of mathematics. We have also to remember that a triangle is an abstract object that can only approximately be physically achieved. I'm not going to work out the details --I'm not expert-- but I think that if you chose the triangle to lie near a saddle point in a hyperbolic paraboloid you can mess around with its lengths pretty much as you wish (I guess).

@JordanMetroidManiac

8 жыл бұрын

You might want to use Calculus for that. Maybe not though...

@DrCruel

8 жыл бұрын

+Петър Петров: You could give a bounded answer with a hard minimum.

@NickCaux

8 жыл бұрын

+x nick Elliptic Geometry is the best geometry

If we have inconsistent data, then formal logic allows us to deduce whatever we want. And if the solution of a problem follows logically from the information given to us, then it´s a valid solution. So in this particular problem, any answer one could come up with would be a correct answer, since any answer one could come up with follows logically from the inconsistent premises.

@tjl102

8 жыл бұрын

sounds like someone knows common core

@DanTheLogicMan

8 жыл бұрын

+Myren Eario Not quite. In order to preserve validity, the conclusion would need to follow from the false premises. If the conclusion does not follow from the premises, then the argument fails both validity and soundness rather than just soundness.

@myreneario7216

8 жыл бұрын

TheismIsUntenable Yes, if you just have false but consistent premises then you can make an invalid answer. But if you really have inconsistent premises, then you can´t make an invalid answer (assuming you´re using classical and not using some kind of paraconsistent logic). So the question is: Are the premises from this video merely wrong, or are they also inconsistent? Now since the video was using some basic geometric axioms for its solution, I would argue that these basic geometric axioms are part of the premises, and then together with the axiom that there exists a triangle with these weird properties, we can derive a contradiction like the video did. However you could argue that the only premise we had, is that a certain right-angled triangle with height 6 and so on exists, and then the problem is just unsolvable, because we have no idea what 6 means.

@lazbn90

3 жыл бұрын

​@@DanTheLogicMan Not really, this is not Philosophy, this is a tautology in mathematics. False implies anything within the claims that can be constructed from the logical operators. So, "The area equals 30", "The area equals i the imaginary unit", "The area equals a smooth manifold" are all True under the hypothesis of the problem, which is the existence of something that does not exists in math, hence, False premise.

@DanTheLogicMan

3 жыл бұрын

@@lazbn90 You might be the dumbest person on the planet replying to a comment from 5 years ago.

the virtue of this question is that it helps me to identify an incorrect question.I think it takes courage to admit the question has errors as we prefer to assume the question does have a definite answer.

Thanks for making me fall in love with maths

I assumed that the "altitude" of the hypotenuse was the maximum height it obtained with one end on the base and the other directly above the right angle - what I would call the "rise". In this case, using Pythagoras theorem I calculated the two sides intersecting the right angle were 6 (given) and 8 i.e. SQRT(SQR(10)-SQR(6)). The area of this triangle is obviously 24. This fuzzy thinking problem is what comes from mixing English with math. "Altitude" is not a mathematical term, and as such is open to interpretation. Since equating perpendicular height (the correct mathematical term for what was attempted in this video) with "altitude" produces nonsense, the only other interpretation remaining to "altitude" must be rise (as in the rise and run of a gradient in an Euclidean geometric plane).

@thewolfdoctor761

6 жыл бұрын

Yes, the usage of the term "altitude" made the problem confusing. I had never heard of that term and came up with 24 as the answer within 5 seconds. Maybe young people are taught about altitude to the hypotenuse now, but that was not a term used in the 60s.

But *any* triangle with a hypothenuse equal to 10 and an altitude to the hypotenuse equal to 6 has an area of 30. The fact that the problem stated that it was a right triangle is not your problem. It's the one who made the task who was the idiot.

@user-uq4cp3id9e

8 жыл бұрын

+MrBaronmoll ummm.. does the word "hypothenuse" ring any bells 2 u ? not any triangle has it, but only a right one...

@VulpeculaJoy

8 жыл бұрын

хунь выебин Ok, ok, just replace "hypothenuse" with the side that is the longest and it still works.

@VulpeculaJoy

8 жыл бұрын

illusion466 What? Why 10 and 1? Provided you get the length of *the longest side* and the altitude above it you can always determine the area no matter the shape. If you don't get provided values that refer to the longest side it's only possible if it's a blunt triangle.

@VulpeculaJoy

8 жыл бұрын

illusion466 If you get provided the longest side the altitude simply *has to be smaller* else you don't get provided the longest side. So I don't know what you're talking about.

@gerojohn

8 жыл бұрын

+MrBaronmoll Well, the last time I got involved with math (and it was some 20 years ago), hypotenuse was called the right triangle's side that is adjacent to the non-right angles. If the triangle is not a right one then it's just called... large side, not hypotenuse!

The first problem on this channel I could solve completly ! Nice !

I solved the problem before progressing forward with the video, and when I graphed two functions and saw they never touched, I figured that the "evilness" of the problem was that there was *no solution* to the problem.

so the correct answer is "does it though?"

@rawaaabd9126

4 жыл бұрын

ممكن حدا يشرح بالعربي

Textbooks are full of errors such as this one:) They often have the wrong results in the back too.

Me, who writes my homework on a sphere: look what they need to mimic a fraction of my power!

In addition to my own proof below as to why this is impossible, really love the proof given in the video!

I am tired of stupid smart*** riddles. This is less mathematical thinking and more "art of deception". We use maths to help us in real life problems. A real life problem would never deceive you into "take the 10, take the 6, what's the area". It would either be an already existent triangle which you'd have to find its area, or a triangle you'd have to research if it were possible.

@George-iz2ce

8 жыл бұрын

Shall I write a book as an answer to this? First, I have graduated physics, so I don't think there is too much space for someone to tell me what mathematical thinking is, and I am mostly talking about this specific riddle. Second, the reason one would not think about is deception: you do not expect riddles in exams. When experimenting, physicists do not meet smartazz riddles. They measure the (honorable?) data mother nature provides, and then we can get together and make a result or so. This would never have been posed in nature; but it could, from an attention herder.

@snuffeldjuret

8 жыл бұрын

I would rather say critical thinking.

@friendlyguy308

8 жыл бұрын

+George Menoutis Well, sometimes you could get data that doesn't make sense due to some new theory no one has discovered yet. But you would think it makes sense and go on your merry way to use it, which would be bad

@sonalidasgupta3562

5 жыл бұрын

...i think its much more fair from physics point of view...where you would say replication of such triangle failed under lab conditions and ask OP to re-measure...

@somewhitepunk

3 жыл бұрын

he presents the story of the teacher within the video. this hints to you that there is a deception to uncover. i agree that this video is deceptive if it didn't include the teacher story. I actually just took this video as "solve for possible values of altitude given a hypotenuse", which was actually lead to some interesting algebraa.

idk if anyone's said this already but I'm in an advanced college mathematics class.... and the problem with this question is the wording. The term "altitude to the hypotenuse" is very confusing. Altitude does not necessarily have to bisect the right angle, correct? I believe that altitude refers to the height from the base to any point on the hypotenuse.

@whallynarwhal6019

7 жыл бұрын

You're on the money. The question is clearly worded incorrectly - an altitude does not have to bisect the angle and therefore the "correct" answer is wrong. The straight forward answer is the correct answer. Half of the videos this guy makes are very misleading in premise and the clickbait titles are starting to bother me.

@nikolura

7 жыл бұрын

an altitude falls square to the opposite side (that base of the triangle). The axis that bisects the angle is not an altitude but a bisector. Hence that triangle which has a "hypotenuse" of 10 and altitude of 6 is simply not a "right" triangle but a generic one, with base 10 and height (altitude) 6.

@akronymus

7 жыл бұрын

Altitude is in fact a correct term synonymous to the height of a triangle [and, of course, area = 1/2 * base * height] . By definition, the hypotenuse is the longest side of the triangle, which is always opposite to the biggest angle. As the sum of all 3 angles == 180°, the 90° angle of a right triangle must always be the biggest one -> means, the hypotenuse must be opposite to the 90° angle. Then comes the sentence of Thales saying that the height of a right triangle with the hypotenuse as base-line can as maximum be half of the hypotenuse's length (because it must be positioned on a circle with hypotenuse as diameter). Which means: a triangle with longest line as base = 10 units and height of 6 units is NOT a 90° one. I think, most grown-up mathematicians can be tricked with such a question -- just because such classical geometric things lie so deep in the personal past (first year of geometry at school), and the terms 'hypotenuse' / 'catheti' are rarely used in practice -- a triangle has 3 sides and 3 angles with size units, that's it. I admit that I'd probably have answered »30 of course« ...

@akronymus

7 жыл бұрын

addendum: 'altitude' is a line starting 90° from base and ending at the top of the triangle. No, it does NOT cut the top angle into two equal halfs (except for a symmetric 'isosceles' one with 45° 90° 45°) -- this would be called 'angle bisector' or 'bisectrix' (no chance to google for it, there will only be slimy stuff). Main problem is that many of the classical terms are out-dated concerning practical use. So this original question smells a little moldy.

@akronymus

7 жыл бұрын

(imagining google) BISECTRIX community ... oh my Spock ...

It's really thinkable question. Thanks sharing such basic and very useful questions. Th ks alot.

By using the second euclidian theorem you get AH^2 = BH × CH. Substituting CH with "10 - BH" you get a second degree equation whose delta is negative

I figured it out too, but with a totally different way. we name the two segments of the 10-unit-long side seg1 and seg2. then we have tan(alpha)=6/seg1 in the one triangle (alpha being the angle opposing the 6unit long side), and and tan(alpha)=seg2/6 in the other side and seg2 = 10-seg1. from not on seg1 will be calles S to keep things more readable. so we have 6/S = (10-S)/6. we multiply by 6 and by S and get 36 = 10S - A². then we rearrange some numbers and get A²-10A+36=0. and our trustworthy pq-formula then says A = 5 ± Sqrt(25-36). and as we need the root of -11 in this calculation, it's pretty ovvious something has to be terribly wrong.

@FF-jf8yg

8 жыл бұрын

+Kronus Exodues I dare say that was clever!

@101perspective

8 жыл бұрын

+Kronus Exodues I figured it out also. I simply waited until no one was looking and then glanced at Kronus's notes.

@TheGeefriend

8 жыл бұрын

+Kronus Exodues Nothing is "terribly wrong". We just need to use Complex numbers. Numbers, which have negative values when squared. The most common example of those, is i. i is most commonly described as the "root of -1", which is wrong. i is defined as a number, which is -1 when it's squared. Now, a complex number can be described as z=x+iy, where x and y are both real numbers, with the i defined above. the root of -11 does exist, and it's exactly Sqrt(11)*i, since you never defined in which domain you were solving the problem. It just doesn't exist within real numbers. And complex numbers are very important, in many fields. Physics, engineering, economics, chemistry, and so on. But I'm sure you already knew all that, if not, then I'd recommend reading into the complex numbers. They're interesting, and very useful.

@JM-lh8rl

8 жыл бұрын

I'm sorry, I got lost when you did "tan(alpha)=seg2/6". When you said alpha was the angle opposing the side of length 6 -and assuming that both angles of the smaller triangles that are opposite to the height are equal- the tangent of alpha, referring to segment 2, would be 6/seg2, not seg2/6. This would throw off your whole reasoning, but maybe there's something I'm not seeing here. Either way, great way to look at the problem.

@verigone2677

7 жыл бұрын

+Kronus Exodues ... Well, you did actually figure this out using the circle method...you just used the tan function instead of the simpler draw it version of the circle method. If the expected tan yields an answer requiring Euler 's constant (i or sqrt(-1)), then the point you are looking for exists outside the circle. Basically, you just did the trig proof that the point of altitude 6 above hypotenuse 10 does not exist in the circle of radius 5...albeit, it was the hard way, but a damn good proof none the less. Kind of like bringing a bazooka to a dart match.

Doing top mathematics in year 12 and never heard of altitude to a line. I have only heard altitude relative to sea level.

@sciencemanguy

8 жыл бұрын

+Joshua Williams ... Which highschool senior has never heard of a mathematical altitude? I'm guessing solely from this information that you are going to go to community college unless if you are pretending.

@joshuawilliams994

8 жыл бұрын

sciencemanguy I'm from South Australia. I never really hear altitude when it comes to shapes, rather just height. I do hear it in physics though. Another thing to note though is that we haven't done geometry yet this year.

@sciencemanguy

8 жыл бұрын

wow. South Africa really is behind in education. Wow.

@joshuawilliams994

8 жыл бұрын

sciencemanguy Where did you get South Africa from?

@sciencemanguy

8 жыл бұрын

I meant south Australia, whoops

Kindly help with the solutions for these maths questions 1. A theatre has 2100 seats. All of the rows of the seats in the theater have either 45 seats or 40 seats. If there are three times as many rows with 45 seats as there are with 40 seats, how many rows are there? 2. A circle has radius 7cm. Calculate for the area of the maximum interior equilateral triangle that can exist in the circle. 3. AC and BC are tangents to the circle with the center and radius 7cm. ABC is an equilateral triangle and AB is a chord. Find, correct to the nearest whole number, the area of the portion of triangle ABC outside the circle.

you can also use the 3-4-5 method. it's pretty clear to see that the larger triangle's side can't be 8 while the other triangle's would have to be 2.

I didn't even try the Area = b*h / 2, because I figured it was probably a trap. My plan was to find all the other side lengths to figure out first if it was even a legal right triangle. I started with a theorem about altitude with the hypotenuse of a right triangle. The theorem states that the altitude = sqrt (x * y) where x and y are the segment lengths of the hypotenuse that are formed by the altitude. Simplifying this gives altitude^2 = xy. I know the altitude = 6 and also that x + y = 10 (because x and y together form the hypotenuse). I now have 2 equations. xy = 36 and x+y = 10. I substututed x = 10 - y into the first equation and got y(10 - y) = 36. Simplifying gives 10y - y^2 = 36, which has no real solutions. Therefore the dimension cannot describe a right triangle and the problem is invalid. I did it in a more roundabout way, but if it works...

@nychold

8 жыл бұрын

+Vincent Killion I did it similarly, pun intended. The altitude line separates the right triangle into 2 similar triangles, and you can use fractions to solve the lengths of similar triangles. In my case, I used 6/(10-x) = x/6, which lead me to 36 = 10x - x^2. Suffering a bit of bad math (4*36 = 84), I wandered down a difficult road of sqrt hell, but eventually came back to double check myself and spotted the error. :)

@Zebo12345678

8 жыл бұрын

+Vincent Killion I used the Pythagorean Theorem to find the other side, then did the area formula and got 24. I didn't even think of it not even being a real triangle.

that doesn't sound like a riddle, rather a rule.

You can also answer the question like this. Think of the two right sides of the triangle as x and y, supposedly the area of this triangle is 30 which means xy/2=30 and we know since the hypotenuse is 10 that x^2+y^2=10^2=100 or (x-y)^2+2xy=100 (x-y)^2+4×30=100 (x-y)^2+120=100 (x-y)^2=100-120 (x-y)^2=-20 Which mean that there are no vulues for x and y that can solve this equation so therefore there is no triangle which its hypotenuse is 10 and its altitude on the hypotenuse is 6.

I've been teached that, if you see a right triangle, you should use 1/2*ab formula. It helped me

That teacher was incorrect. All of the answers submitted were correct answers. Any of them were true. Regardless of what number they put for the area. The problem states "There is a right triangle with hypotenuse 10 and altitude 6". This is false. Ex falso quodlibet. Anything follows from that. It follows from that that 1 = 2. You cannot possibly submit an incorrect answer to this problem.

@hilarymanuel

8 жыл бұрын

Interesting. I've never encountered this concept before but it's something. My initial reaction was that of gigo, faulty questions beget faulty answers.

@christiandinkel8481

8 жыл бұрын

Hilary Manuel Ex falso quodlibet is just the Latin name for "Faulty premises beget any conclusions"[1], which is very similar to your rule, but also happens to be one of the fundamental rules of western logic since at least ancient Greece.[2] It is also the name of a formal axiom in most systems of formal logic in use today, where it refers to a specific rule that says that if FALSE then FALSE is just as TRUE as if FALSE then TRUE.[3] Last Edit: Some ancient forms of traditional logic in China apparently did not have this rule, but instead had the opposite rule (ex falso nelibet). The western way of doing it has some nice qualities in terms of expressiveness and seems to be the standard these days. [citation needed]

@christiandinkel8481

8 жыл бұрын

Exactly what I was saying, Melvin. P=>Q is always true if P is false, regardless of the truth of Q. So since the question implies something false (that there is such a triangle), any answer based on that premise must be a correct answer, since it always implies "Given that there is a triangle such that..." and so on. That's why that is called the "given" part of the question. It is literally given.

@christiandinkel8481

8 жыл бұрын

When the students are saying "The sides of this triangle have these lengths" when they say "this triangle", as in "the given triangle" they do what in linguistics (semantics) is often called a "presupposition", which logically behaves identically to an implied premise, which is "There is such a triangle." Thus their answer semantically is validly representable by the form p therefore q, which is what all formal calculus, whether axiomatic or rule based, fundamentally compiles to, through its respective interpretation function, which you can most easily see in a representation like a sequent calculus. Thus all these answers have the truth value T. EDIT: Or, more simply put, just because the premise of the answer was given in the question itself doesn't make it any less of a premise. And all logical argument is of the form "If these premises hold, these conclusions hold."

@pascalschackmann3936

8 жыл бұрын

no man, just no.

the video starts with this written: 'A textbook has the following PROBLEM'. Well, everything is correct, it does have a problem :)

I used the height 6^2=axb, where a and b are the segments the height determines on the hypothenuse. a+b =10, the product ab is maximum when they are equal to 5 each. So the product at its maximum is 25, less than 36; hence impossible triangle.

Before viewing this video it never occurred to me that you would be lying to me. Thank you for the education.

this IS a right triangle, just so long as you let some of its dimensions be complex. Also, the algebra works perfectly saying the area is 30 even if the sides have to be complex numbers

@RoboticsNShenanigans

8 жыл бұрын

this triangle cannot be a right triangle. as mentioned in the video, a right triangle's altitude cannot be more than half the length of the hypotenuse.

@Double-Negative

8 жыл бұрын

+RoboticsNShenanigans It can so long as the sides are complex lengths

@JNCressey

8 жыл бұрын

Let the hypotenuse/base of the triangle have endpoints (0,0) and (10,0). [satisfies hypotenuse length 10] Let the third vertex be (c,6), c∈ℂ. [satisfies altitude is 6] Need to solve (c,6)•((10,0)-(c,6))=0 for right angle. c=5+i×√11 or c=5-i×√11. Take c=5+i×√11. We now have a right triangle (0,0), (5+i×√11,6), (10,0) with hypotenuse 10 and altitude to the hypotenuse 6. The area of a triangle is half the area of a parallelogram. We can get the area of a parallelogram from the magnitude of the cross product of two edges. (5+i×√11,6,0) cross (10,0,0) = (0, 0, -60). So the triangle area is *30.* So it's all good. TBH, I wasn't expecting the answer to just be 30, I was expecting a complex number. but anyway, turns out all the students were right all along.

@JNCressey

8 жыл бұрын

Hmm, I kinda overlooked complex inner product should be used instead of dot product. Better rework it out. Maybe, this time, I'll get a complex answer. .... But then I get Im(c)=0, Re(c) = 5+i×√11, which doesn't work. So I guess the video was right all along, there is no such triangle.

@Double-Negative

8 жыл бұрын

+JNCressey I calculated it using similar triangles and breaking up the 10 into something like 5-i√11 and 5+i√11. Using this I calculated the sides that make a right angle using big messy square roots and then multiplied them. I got 60. of course A=B×W/2, so it was 30 after all.

It's a Cthulhu triangle!

@fllthdcrb

7 жыл бұрын

That's a very interesting way of putting it. Check the comment I left earlier. If you accept the premise, and think about what the rest of the triangle could be like, you are forced to conclude that the "lengths" of the other two sides are complex numbers! This is not a triangle as we know it. But maybe someone could come up with an interpretation that makes sense. Or maybe not.

Three possible answers: 25, 30, 36, depending on whether you accept "right triangle and hypotenuse 10", or "altitude 6 and hypotenuse 10", or "right triangle and altitude 6". The last one assumes hypotenuse 12, but could be anything greater than or equal to 12, which means the answer is unknown in that case.

What you can do is set up 3 equations where you draw the triangle as if it did exist, then separate the hypotenuse to x and 10-x. Then come up with 3 sets of equations, where a is the base of the triangle, b is the height, and x is the distance from the altitude/hypotenuse intersection to where b and the hypotenuse intersect (or you could flip them) a^2+b^2=10^2 6^2+x^2=b^2 (10-x)^2+6^2=a^2 You can use these equations to solve for all three variables, however, what you end up with is x = 0 or 10, so basically, the only way for the altitude to be 6 is if it's also one of the legs of the triangles, which makes no sense since then you would have two right angles in your triangle, which is impossible, also proving that such a triangle does not exist.

while this is somewhat clever, it is cheating to give false information in the question. Math is about logic and this question fails as soon as it contradicts itself

@angelmendez-rivera351

3 жыл бұрын

Well, no. The whole point of the question is to realize that the question contradicts itself.

@somewhitepunk

3 жыл бұрын

I think it is cheating (or at least tricky) for the teacher to do this to her students. However, we are give this story as part of the setup of the video, and the context should lead you to investigate some kind of contradiction or rules violation. In other words, the point of the video was to hint to the viewer a problem within a question and let them try to determine it. If he didn't include the teacher story, then I'd agree it would be a deceptive video

@redgrey1453

3 жыл бұрын

Perhaps the problem is that the word "right" was inserted incorrectly. Given a triangle of such dimensions, the area is 30. That such a triangle may not be "right" is immaterial.

@lazbn90

3 жыл бұрын

There is no cheating here, just that the KZreadr and the alleged professor didn't know the basics of Mathematical Logic. ​ This is a tautology in mathematics: False implies anything within the claims that can be constructed from the logical operators. So, "The area equals 30", "The area equals i the imaginary unit", "The area equals a smooth manifold" are all True under the hypothesis of the problem, which is the existence of something that does not exists in math, hence, a False premise.

Okay... but in this hypothetical problem, that doesn't mean the students got the answer wrong. The question was wrong. And if the students weren't taught the rules of the altitude of a right-angled triangle-- I know I wasn't-- then it can't exactly be given to those same students as an awareness test. And if the rule of max altitude is as simple "1/2 the hypotenuse's length", a good amount of students presented with the problem would very easily spot the issue if they've been taught the rule. Hypothetically.

@lokalnyork

8 жыл бұрын

+MegaKaitouKID1412 No. If You would ask students "how long it take Sun to orbit around Earth", You shouldn't expect any other answer than "it's other way around", typo or not. The correct answer to that is "that triangle can't exist, so it's area is immeasurable". Not even "just 0", but undetermined.

@MegaKaitouKID1412

8 жыл бұрын

Local Ork However, if you hadn't taught kids that the sun doesn't orbit the Earth (and it wasn't something that is considered everyman's knowledge common sense that their parents and television would have taught them) then the question doesn't belong on the test-- unless you specify that the student should use the tools available to them to solve the problem. ("Using your ruler and protractor, draw a right-angled triangle with an altitude of six and a hypotenuse of ten. What is the area of this triangle?" would be fair to mark them on not realizing that it's impossible, because it prompts the kids into a process where they should realize it's impossible even if not already taught it's impossible.) Just providing a question with impossible information that students can't be reasonably expected to know is impossible because you didn't teach them that-- which the fact that not a single student got the question right implies (and seriously, I was not taught this in any level of geometry ever)-- is unfair. Because you never taught it and you aren't providing the tools on the test for them to think it through on the test.

@lokalnyork

8 жыл бұрын

MegaKaitouKID1412 I think You are missing whole point of this question. It's actually trick question with purpose of checking if You know Pythagorean theorem (which is math 101, often taught along area equation), and not area equation problem. And I'm speaking from perspective of person that also got it wrong (yay me). This is wonderful question that require thinking outside of the box and checking Your facts first. Unless You want to have generation of engineers that make cars that explode, bridges that fall, doctors that cut out wrong parts of Your body and whole lot od mistakes, simply because people don't check if they can do things in the first place.

@MegaKaitouKID1412

8 жыл бұрын

Local Ork That's great and all, but making a trick problem that tricks people with something they were never taught without giving them the tools to realize this thing that was never taught all the while making the problem look like it is something they were taught is more akin to handing a guy off the street a wrench and asking him to fix a car, tell him to change the tire, and then when the car explodes, blaming him for not realizing that there was actually an issue with the engine. *shrugs* That type of thing would be great in an in-class problem that's taken up, but it's not exactly suitable to a measurement of how much someone has LEARNED when they're seeing that type of problem, that type of trick, on a test. Especially when the problem doesn't actually point out in any way that there's more going on than what they actually did learn. A person doesn't come out of education with knowledge they were never taught. The point of education is to teach people what they need to know, not not teaching them and expecting them to know. You know what I mean?

@KWGTech

8 жыл бұрын

+MegaKaitouKID1412 Our math teacher gave us two problems like that in abstract algebra. Most of the class ended up getting those wrong, but with the "generous bonus points" it actually turned out to be the highest score weve ever gotten. Good teacher he is.

Hey Presh your videos are really awesome I want to buy your books. From where I shall get your books.

A vessel is filled with liquid, 3 parts of it which are water and 5 parts syrup. Jow much of the mixture mustbe drawn off and replaceed with water so that mixture may be half each of water and syrup?

The lengths of the sides, *A* and *B*, would have to be described using complex numbers. If we begin with the premise that A^2 + B^2 = 100, and also that A x B x 1/2 = 30, the result is that the length of side *A* is [sqrt(50 +/- 10i * sqrt(11))] and the length of side *B* is [60 / sqrt(50 +/- 10i * sqrt(11))]. Thus, the *teacher* was actually the incorrect one in this case and any student who got an answer of "30 square units" as the area of the triangle was correct because the necessity that the length of all the sides of the triangle were described by real numbers was never formally established in the question.

@brentdaignault7565

8 жыл бұрын

+Omar Goodman I think it could be assumed that in a Geometry class, a problem from a Geometry textbook would involve the limited rules of geometry just taught. In other words, Algebra would not be allowed as a solution technique. The point is, that within the rules of Geometry there is no solution because the triangles cannot be drawn. See the discussion with omp199. Even with Algebraic tools, we have to evaluate the full solution set and toss the ones that don't have real solutions (only a real component). This depends on the problem of course, many physics problems have valid solutions that have complex components.

@omargoodman2999

8 жыл бұрын

Brent Daignault None of that matters. The question established the parameters. The fact that those parameters exist outside the scope of "normal geometry" is inconsequential to the problem. So, for the teacher to claim that the answers were "incorrect" because they didn't call the original parameters into question is wrong. No where did it state that this triangle is based in non-complex geometry and, the fact that the parameters given prove it cannot be based in non-complex geometry adequately demonstrates that it isn't. Now, had the question been, "does this represent a real triangle", then that would be different. The answer of 30 square units is correct for this problem, complex side lengths notwithstanding. You might as well say that a triangle with hypot. 10 and altitude 5 giving an area of 25 was "incorrect" because the actual (and unreported) total lengths of the sides were less than the hypotenuse and, thus, could not make a complete triangle. So, how the teacher *should* have gone about this is to say that while, technically, the answer of 30 square units is mathematically correct, the parameters given indicate that this triangle could only exist in complex dimensions and then go on to demonstrate how you can differentiate a "real" triangle from one based in complex dimensions; *not* to call out students that got a mathematically correct answer as "wrong".

@randomwindowsstuffz

8 жыл бұрын

+Omar Goodman You are mistaken because there is no geometric/algebraic system in which lengths (metrics) can be complex. For example, in C (the complex plane), the distance between two points a+bi and c+di is sqrt((a-c)^2 + (b-d)^2), which is always a real number. Even C^n (a complex/vector space with n complex dimensions), the distance between two points/vectors is the square root of the sum of squares of the absolute value of the point's coordinate point for each complex dimension (it's hard to word it well in text - you may want to read about complex vector spaces), which is also real-valued. The point is that in EVERY algebraic structure which is metrizable, distance is always a real number larger or equal to 0, because of the definition of "metric" in mathematics, as that is the definition which has shown itself to be useful in many geometric and algebraic systems. Therefore, unfortunately, the triangle you mentioned is impossible in any complex geometry by the mathematical convention of "distance" unless we find new geometries where it is useful to have complex-valued distances (and in that case we may change its definition). Of course there is always an option of trying finding a geometry, without complex numbers, where a triangle can bend in strange ways to make this happen, but I don't know if that's possible: I'm not an expert in that field.

@omargoodman2999

8 жыл бұрын

randomwindowsstuffz do you have exhaustive knowledge of every mathematical system that could possibly exist? Because, if not, you really have no grounds on which to claim that no such system exists. Your inability to conceive of it doesn't adequately disprove it. The triangle in question, essentially, represents a "hypertriangle" anchored to a real axis along the hypotenuse and "flexing" into a complex axis along the other two sides Other than that, all the math works out. The two (complex) sides, squared, add up to the square of the hypotenuse and, multiplied, produce a product of 60, half of which corresponds with the area of the triangle and matching the result of the calculation using the hypotenuse and altitude. So, unless you can provide a *mathematic* counterproof, you've really contributed nothing useful here.

@randomwindowsstuffz

8 жыл бұрын

Omar Goodman I don't need to have an exhaustive knowledge of every system. If you read proofs in abstract algebra proving that a system cannot exist, they all target some fundamental property of algebraic structures or topology that creates a contradiction. But that's unneeded here because distance can only be real based on the DEFINITION of the "metric function" aka "distance function".

I kept confusing the question with the Pythagorean triplets 6,8 and 10 🤣😂

@GroundThing

3 жыл бұрын

Almost certainly why the hypotenuse was given as 10. The question was seemingly made to have a few false answers, to ensure the students would not rely on unreliable heuristics, without working out the math for themselves

@Meop79

3 жыл бұрын

No. The 6 8 10 rule or 3 4 5 rule is an easy proof to mention when saying the problem is impossible.

I just came across this puzzle a month ago, 5 years after it was posted, and thinking, without watching the video to the end, that there was some ingenious and complicated solution about to be presented, I decided to hold off and check for myself whether a triangle of this specification can exist on the surface of a sphere, and if so, what would be its new angular dimensions. I note that a few people commenting here have also mentioned non-flat surfaces. In fact it does. The triangle as given here has two right angles, including that for the altitude perpendicular to the base. When placed on a sphere, in certain cases, only one of these right angles can be conserved by adjusting the sphere's radius. The simplest triangle of this class which works is isosceles, and the easiest to solve. I proceeded along these lines and calculated the parameters that satisfy an altitude of 6 and a base of 10 for an isosceles triangle. The new base corner angle for this 'sphericalised' triangle is about 56 degrees, giving an internal angular sum of 202 degrees. The sphere has a radius of approx. 9.09 units and the area of the triangle is 31.85 square units. Another example: the perpendicular divides the base into 4 units and 6 units. A spherical radius of 8.787 will allow the apex angle to be a right angle, whereas on a flat plane it is 78.7 degrees. This spherical triangle's area is 32.08. Very interesting problem, and worth having a deeper look into.

This is how I realized that this triangle is an impossibility: Because a perpendicular has been dropped from the vertex of the triangle which splits the hypotenuse of length 10 into two parts, this process creates 2 similar triangles on either side of the perpendicular length 6. Because the triangles are similar they must have the same ratios, but because we don't know what sum of the split hypotenuse was necessary to get 10, we have to use a little algebra where (10-y)/6 = 6/(10-x). This gives us (10-y)(10-x) = 36 Now the only way a pair of positive integers can sum to be 10 is in this manner: Either the side lengths are 1+ 9; 2+8; 3+7; 4+6; or 5+5 Now in addition to the pairs having to sum to 10, they've also got to multiply to give us 36. The prime factorization of 36 is 2*3*2*3 and this'll give us these possibilities: 18*2; 12*3; 9*4; or 6*6 note that none of these products sum to 10, and thus this is an impossible triangle because there is no way for x and y to satisfy our demands of a sum of two positive integers that sum to 10 that can be multiplied to get 36. In the video it was said that the maximum altitude can only be 5 given that the hyp was 10, and this proof shows that where 5+5 = 10 and 5*5 = 25! So there's our sum that adds to 10 and our product which gives a perfect square.

Hp= 10 so Pythagoras triplet is 6,8,10 so area of r8 ∆ = (6×8)÷2=24

I have taken calculus 3 and I have NEVER heard the word "altitude" in math....

@XYXZProductions

8 жыл бұрын

seriously?... back in middle school geometry is when I learned.

@asmcriminaL

8 жыл бұрын

I have never taken formal geometry. I guess it's kind of implied that we know it.

@XYXZProductions

8 жыл бұрын

+asmcriminaL lol so you just skipped? geometry and went to calc

@asmcriminaL

8 жыл бұрын

Geometry is not part of the curriculum at our school. I taken trig. You can do a lot with triangles. I just looked up the curriculum for geometry. All that stuff that is done in geometry is broken up through our curriculum. Trig has the unit circle, triangles, radians and all of that, pre-calculus does hyperbolas, ellipses, parabolas are in algebra 2 and pre-calculus.

@SlipperyTeeth

8 жыл бұрын

+asmcriminaL Wait. Wait. Wait. You took trig, but they never told you what an altitude was. One of the coolest parts about triangles. The orthocenter. The "Euler line". Even if it was public education, you should get a refund.

I am pretty sure the answer of the triangle based on the hypotenuse 10(not including the altitude 6) is 24 by Pythagoras Theorem and Heron's formula I guess.

Looking through the comments, I have seen 3 possible solutions. (1) Use Hex numbers (2) Project onto a sphere (3) Use complex numbers to find a solution.

I was so happy when I saw this bc i thought that i could solve it. My way was a bit different than his but at the end i got unreal numbers and all my hopes shattered

@HypnosisBear

3 жыл бұрын

Don't worry, Actually that triangle does exist in the Complex plane.👍👍

The triangle requirements are correct if placed on the surface of a sphere.

@muhammadalumar

4 жыл бұрын

It cannot be called triangle any more a triangle is plane figure

@stixoimatizontas

3 жыл бұрын

@@muhammadalumar A euclidean triangle is a plane figure, not any triangle.

It has an altitude of 6? Which altitude though? The hypotenuse is 10 yes, but there are 3 altitudes on a triangle it gave us one and didnt specify which one, I would have said this problem was impossible, there were 3 possible answers. I haven’t worked out what the answers would be for the other two altitudes but could there be an actual answer there?

Nice problem and good explanation

I paused it at 0:14 and came up with an answer of 24. Now I'll continue the video and see if the answer is correct.

@carlo1831

7 жыл бұрын

I'm not the greatest at geometry and was assuming that the height of the hypotenuse on a right triangle was the distance straight up from the lowest point to the highest point; in his case 6, which would make the upright 6. With side A being equal to 6 and the hypotenuse equal to 10, side b would have to be 8. 6^2 = 36, Hypotenuse is the square root of A squared + B squared. Since A squared is 36, 100, which is the square of 10 has to be the solution of A squared + B squared. Since A squared is 36, 64 has to be B squared. So side A is 6; side B is 8. The formula for the area of a right triangle is (A*B) / 2. (6*8) / 2 is 24.

@daniluzzu

7 жыл бұрын

I came up with the same solution! I think the issue here is that the explanation in the video complicates things horribly. The altitude of a right triangle is simply the vertical leg. In that case you have a hypotenuse of 10 and the legs are 6 and 8. I don't understand why this isn't obvious.

@carlo1831

7 жыл бұрын

Danilo Bonina Comment deleted by poster.

@UTU49

7 жыл бұрын

Unfortunately that's not what "altitude to the hypotenuse" means. Other than that your logic and calculations are correct.

@carlo1831

7 жыл бұрын

Actually you can't solve this problem with the inputs given. Had they said the height or altitude of the triangle, yes. But the altitude of the hypotenuse? Sorry. The question was stated incorrectly.

this reminded me when i was asked in a test in 3rd grade to divide a 52 card deck by some number or other, can't remember. what i know is that i got a remainder from that division. on all the questions on the textbook where there supposedly was a remainder left from division you were supposed to present it together with the final answer. i did this in my test answer and when the teacher checked it she said my answer was wrong. i asked how (was a cheeky brat, but i checked with calculator and parents later and saw that i was right). she said "you gave too much information, nobody need that much without asking for it. i took a point because of that". the hate i still feel for that woman is bottomless. This was not the first time she fucked me over like this (specially in my best and favourite subject) and it was not the last. she even did this type of shit to my older sister on a daily basis. i would have had a perfect score had it not been for the mega bitch.

@animegirl17

4 жыл бұрын

I feel your pain

Another reason why it is often useful to try and draw the figure approximately to scale. It reveals the problem for this question.

I am a beginner please help me out If Maximum height is 5, then the side would be 5√2. so can we say that that is the maximum side this triangle with hypotenuse 10 could have?? In that case what about triangles like .. the one with hypotenuse 10 and other two sides besides being 6 and 8., And many other such possibilities. So are you talking only about cyclic triangle here?? And the one which I am talking about with sides 6 8 10 are not cyclic. Is that it or am I going wrong anywhere???

It is not required that a valid argument have premises that are actually true, but to have premises that, if they were true, would guarantee the truth of the argument's conclusion. Thus, the area is 10*6/2 = 30.

@gingganggoolie

8 жыл бұрын

+Paulo Bouhid the premises have to be consistent with each other though, otherwise the problem can't be solved

@RoboticsNShenanigans

8 жыл бұрын

+Iona Cloran The only premise that matters in the context of the question is understanding the A=1/2(bh) formula. The triangle doesn't have to be real to understand that premise.

@GrzesiuG44

8 жыл бұрын

+RoboticsNShenanigans Why do you assume 1/2(bh) formula works for "not real" triangles?

@paulobouhid6648

8 жыл бұрын

+Grzegorz Graczyk If the premise of a statement is false, the whole statement is true, regardless the conclusion is true or false. Thus, this allow me to assume that the area is whatever I want. I assume 1/2(bh). You can assume whatever you want, and your whole statement will be true, also.

@RoboticsNShenanigans

8 жыл бұрын

+Paulo Bouhid your last comment makes no sense. the only error in this problem is saying the triangle os a right triangle. there's no reason to assume anything.

Obviously this triangle exists on a curved surface...

@J7Handle

5 жыл бұрын

You mean, non-Euclidean space?

Can you unscrew all of those flathead screws? - hands me a phillips head screwdriver

So, now you're supposed to double check all the information in a story problem? I'm glad I'm not in school anymore.