Geometry

>>8011185
Damn, it was a good problem. Too bad it is a troll.

Here is what could be done, and I did. Label all the unnamed sides. The one below I called x, the one to the left I called w and the one that is part of the "10" in the hypotenuse I called w.

Then you can find an expression for both x and y in terms of w.

After that x squared + y squared = 10 and by expanding this you get a second degree polynomial.

After that it was fun but then I found out that the roots of the polynomial were complex numbers.

At least in the geometry I know, a triangle cannot have complex sides.

>>8011185
8 times root 17?

30

>>8011185
25?

>>8011185
19.21

45

impossible to have an right triangle with those dimensions.

Better question:
What if it's a spherical triangle?

eq1.
36 + b^2 = c^2

eq2.
36 + (10 - b)^2 = d^2

eq3.
d^2 + c^2 = 100

From eq2:
36 + 100 20b + b^2 = d^2

b^2 - 20b - d^2 + 136 = 0

b = (20 +- sqrt(400 - 4d^2 + 544))/2

b = 10 +- sqrt(100 - d^2 + 136)

Plug into eq1:
c^2 = 36 + 100 +- 20sqrt(100 - d^2 + 136) + 100 - d^2 + 136

Plug into eq3:

100 = 372 +- 20sqrt(100 - d^2 + 136)

d = sqrt(1276/25)

Plug into eq3:

1276/25 + c^2 = 100

c = sqrt(1224/25)

Area = sqrt(1561824/625)/2

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>>8011279

it's impossible for that height to be longer than five you fucking retards

>>8011185
There is no solution.

Or perhaps more accurately, given the nature of vacuous truth, the solution can be anything.

The altitude of a triangle with hypotenuse 10 is at most 5 (the altitude when it is isosceles).

please tell me all of you are joking
please

>>8011295

If you're so smart why dont you show us the solution

By Thales theorem the altitude can't be larger than 5 if the hypotenuse is 10.

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>>8011307

fine
the guy who said that a triangle with these dimensions can't happen was right
there is either a typo in the question, or it's just evil. if you wanna ignore the impossibility of the triangle and just answer the question it's obviously 30, as 10 is the base and 6 is the altitude to it (10 * 6)/2 = 30.
The triangle is impossible because(refer to pic)
according to the question AD = 6 and created a 90 degree angle with BC. lets draw a circle which blocks the triangle, and name it's center O
BC = 10 (given) so AO =5(radius)
look at the triangle ODA. according to the question it's hypotenuse is 5 but one of the other side is 6, and that can't be.

the altitude to hypotenuse can't be equal to more than half it's length.

>>8011343
Why is this >>8011279
wrong?

>>8011343
>>8011315
>>8011290
>>8011288

topkek

>>8011185
The first reply already said it was impossible but all the faggots gotta come up and pretend they are the geniuses who figured it all out.

10/10 /sci/ everyday material.

>>8011348
Tell that fucker to label what the fuck his variables are and what sides are his equations representing and then I'll tell you why it's wrong.

>>8011355
You should be able to figure it out from the first three equations.

c is the bottom of the big triangle.

d is the left side of the big triangle.

>>8011353
I jus wanted to show an easier way to prove it. The first guy didnt show the whole process

>>8011348
not sure cuz shit looks messy but in

b= (20 +- sqrt(400 - 4d^2 + 544))/2

seems to be where he messed up since he is counting -d^2 and 136 as your constant he didn't reverse the sign
-4(a)(c) c is (-d^2 +136) and a is 1 so it should be -4(-d^2+136) = 4d^2-136*4

>>8011365
How can anything be simpler than
>At least in the geometry I know, a triangle cannot have complex sides.

Heh, I was asked this question on a test recently. I guess being familiar with the maymays isn't too bad

>>8011371
I mean he didn't show the algebric process, instead he just posted the results. So people had their doubts.
The conclusion he came with is perfect, I just wanted to contribute

>>8011353
What was dumb about the first guy is that he had to actually work through the problem to realize it, when it's obvious from what >>8011288 >>8011290 said.

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T=1/2*b*h
30

>>8011741
Anon please dont be dumb

>>8011753
Not that guy but, why can't you do this?
Just flip the triangle over so the hypotenuse is the base and use the simpler formula. Fuck all that extra jazz. I thought everyone in this thread was retarded until I saw this >>8011741 solution.

>>8011185
last side is 8
40

>>8011816
That formula only works for euclidean triangles. A shape with hypotenuse 10 and altitude 6 is not a euclidean triangle.

The image in the OP is a troll. It would be like drawing a square and labeling two of the sides as having lengths 5 and 6.

>>8013319
just realized anon
normally it'd be 30 because .5 b h
actually fell for this meme and thought it was some weird special case

>>8011290
>given the nature of vacuous truth, the solution can be anything.

So I can claim that the solution is 42, and it's "true" just because nobody can construct such a triangle and show that it isn't 42?

>>8011290
>There is no solution.
>Or perhaps more accurately, given the nature of vacuous truth, the solution can be anything.

But "there is no solution" and "the solution can be anything" are not equivalent at all.

It's like saying that any real number is the solution of an equation which has no solutions. The equation [math]3x + 2 = \frac{12x + 7}{4}[/math] has no solutions, but according to you the claim that any number is the solution is a true statement. Yet, substitiuting any number for [math]x[/math] reduces to [math]1 = 0[/math]. Thus, it's not true that any number is the solution, while it's true that no number is a solution.

>>8013449
It's true because "For every triangle with hypotenuse 10 and altitude 6, the area of that triangle is 42" is a true statement.

This is because [math] \text{F} \rightarrow \text{T} [/math], i.e. false implies true.

>>8013468
It depends on how the question is phrased. "[math] x \neq x [/math]" has no solution. However, for every [math]x[/math] such that [math]x \neq x[/math], [math] x = 42 [/math].

>>8013471
>"For every triangle with hypotenuse 10 and altitude 6, the area of that triangle is 42" is a true statement.

Oh really? It seems to me that "For every triangle with hypotenuse 10 and altitude 6, the area of that triangle is 42" is false, because "For every triangle with hypotenuse 10 and altitude 6, the area of that triangle is 30" is true, and "42 does not equal 30" ist true also.

>>8011185

Uh... 30?
Basic maths? Lol

>>8013507
It's not a right triangle though. The problem as stated requires it to be a right triangle, which isn't possible to satisfy.

>>8013471
>This is because F→T, i.e. false implies true.

But false implies anything, and anything implies true. A proven theorem in the form of an implicaton is useful only either if the premise is known to be true (then the conclusion must be true also, this method is known as "direct proof"), or the conclusion is false (then the premise must be false also, this method is known as "proof by contradiction"). A false premise says nothing about the conclusion, just as a true conclusion says nothing about the premise.

>>8013510
> A right triangle has a hypotenuse equal to 10 and an altitude equal to 6

>>8013535
This is possible, given you didn't specify _which_ height must be 6. However a right triangle with hypotenuse equal 10 and height towards hypotenuse equal 6 is _not_ possible.

>>8013471
An equation explicitly stating the solution of an equation (such as [math]x = 42[/math] must be _equivalent_ to the original equation, (i.e. it must imply back at the original equation). For instance, oftentimes a surd equation implies a quadratic equation (after squaring both sides), but usually not all of the latter's solutions are solutions to the original surd equation. Given that [math]x \neq x \iff x = 42[/math], we cannot say that [math]A = {42}[/math] is the set of solutions to [math]x \neq x[/math]. As a matter of fact, [math]x \neq x \iff x \in \emptyset[/math], and [math]A = {42} \neq \emptyset[/math].

Futhermore, the concept of "vacuous truth" seems to be rather flaky and not standing up to scrutiny. According to https://en.wikipedia.org/wiki/Vacuous_truth
>a vacuous truth is a statement that asserts that all members of the empty set have a certain property
Then I may make the "vacuously true" claim that "all members of the empty set are equal to the number 1", which means that the number 1 is en element of the empty set, which is obviously false.

>>8013633
>[math]A = 42[/math]
Any of these above shoudl be [math]A = \{42\}[/math] (forgot to escape the curly braces with backslashes and the LaTeX interpreter dropped them).

If the triangle is noneuclidean, it could indeed have an area of literally anything.
Just define a plane where it is so.

>>8013633
>Then I may make the "vacuously true" claim that "all members of the empty set are equal to the number 1", which means that the number 1 is en element of the empty set, which is obviously false.
It is true that all members of the empty set are equal to 1. This does not imply that the empty set contains 1.

Every instance of vacuous truth is a mathematical theorem.

>>8011290
>Or perhaps more accurately, given the nature of vacuous truth, the solution can be anything.

It doesn't quite work that way. Vacuous truth essentially means that "nonsense implies anything". Yet, following implications is not sufficient for arriving at a solution to a problem - the solution statement(s) must imply back at the original equation (or system of equations). In other words, we need an equivalence between the original problem's conditions and its solution set.

For example, the equation [math]\sqrt{13-4x} = 2-x[/math] implies that [math]x \in \{-3, 3\}[/math]. However, only [math]-3[/math] is a solution to the original equation. Thus, [math](\sqrt{13-4x} = 2-x) \iff (x \in \{-3\})[/math] provides a solution, while [math]\sqrt{13-4x} = 2-x \Rightarrow x \in \{-3,3\}[/math] does not. If you move to an equation which is being implied rather than one which is equivalent, you are only guaranteed to arrive at a _superset_ of the solution set to the original problem.

Specifically, if the original conditions (as in OP's problem) are contradictory, then you may, by virtue of "vacuous truth", imply its (non-existent) solutions to belong to _any_ arbitrary set, but the actual solution set will be a subset of that set - namely, the empty set [math]\emptyset[/math] (which is the only subset guaranteed to be found in any arbitrary set).

>>8013779
Again, it depends upon phrasing.

I agree that it is more correct to say there are no solutions, per your phrasing, which would be more standard.

I was only saying that every triangle that satisfies OP's condition has area 42, which is also true. It is correct to assert "Every triangle with hypotenuse 10 and altitude 6 has area 42."

>>8013786
>I agree that it is more correct to say there are no solutions, per your phrasing, which would be more standard.

A solution set must strictly satisfy the original problem, otherwise it's not a solution set (see the example surd equation shown above). I guess your high school math teacher wouldn't be too enthusiastic if you came to him at the beginning of the semester and told him that you already "solved" all the excercises he might come up with during the semester, because you know for a fact that the solution set of any of them is some subset of the reals (in the worst case, that subset being [math]\emptyset[/math]). You might troll him that way, but I strongly suspect he wouldn't accept that. Providing some superset of the solution set of a problem is not a solution.

By virtue of vacuous truth, you can say that [math](x \neq x) \Rightarrow (x = 42)[/math], and this is true. But [math]42[/math] is not a solution to [math]x \neq x[/math], because the reverse implication (i.e. [math](x \neq x) \Leftarrow (x = 42)[/math]) is false, as [math]42[/math] is not a number which doesn't equal itself. If you want to provide a solution set, it must be equivalent to the original equation - thus, [math](x \neq x) \iff (x \in \emptyset)[/math] states that there are no solutions, because the set of numbers not equal to themselves is the empty set.

>>8013847
I agree it's not a solution. I'm only saying that every triangle with hypotenuse 10 and altitude 6 has area 42.

This is crazy.

THIS IS CRaZY.

Area of a triangle is height times base divided by two.

Just because the triangle is turned on its side the formula still holds.

1/2 x 6 x 10 = 30

Oh well, time to return to pottery on this ceramics discussion board.

>>8013874
>I'm only saying that every triangle with hypotenuse 10 and altitude 6 has area 42

Not really. You misrepresented OP's problem conditions (which require that the triangle be a right one and the height equal to 6 be towards the hypotenuse, these details you ommitted), thus your statement is false.

>>8013922
See above.

>>8011307
>why dont you show us the solution

Sure, here: the area is [math]A[/math] such that [math]A \in \emptyset[/math]. Enjoy.

>>8013947
The statement is true. This is unequivocal.

>>8014048
No, the statement "every triangle with hypotenuse 10 and altitude 6 has area 42" is false. Triangles with hypotenuse 10 and height 6 (whithout any furhter restrictions, as opposed to OP's problem which demands a right triangle and the height of 6 to be towards the hypotenuse) are possible to construct and the area of none of them equals 42 (let alone the set [math]\{42\}[/math] be a superset of all sets of such areas).

>>8014122
... Euclidean triangles of course. You can't change the domain of discourse arbitrarily; that's basically re-defining terms.

>>8014130
You are saying "every triangle..." without restricting the domain to non-Euclidian triangles either. Thus, "ever triangle" includes Euclidian triangles. Therefore your statement "every triangle with hypotenuse 10 and altitude 6 has area 42" is false, because a counterexample can easily be provided.

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>>8011185

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A plausible triangle's area is equal to bh/2.
The fuck's with Euros/dingdongs and their obsession with shitty geometry problems, anyway?

0.5 (10)(11.66...)(sin(30.96...))=30

if we go don't go full autism one could say that the area is (10*6)/2=30.
however you should have realized the triangle in question cannot be a right triangle.
the height of a right triangle with the hypothenuse equal to 10 can only be smaller or equal to 5.
so the triangle in question doesn't actually exists and the area itself has no meaning.

The "answer" is i*sqrt (136) right?

>>8014169
Unless I'm mistaken, this is simple trigonometry.

10*6/2 dummies

>>8014186
>sqrt (136)
Ugh. How about [math]2\sqrt{34}[/math]?

>>8014194
Dunno what you'd call "complex trigonometry" if that shit's "simple."
Actually, looking at it, BD is perpendicular to A1C because BD is perpendicular to AC, right? Do you have to prove that A1C and AC interesect with BD at the same angle, or is it just a given that AA1 is perpendicular to the bottom and top faces?

>>8014199
>being a right triangle was irrelevant information
Well fine, I'll just take my trigonometry elsewhere.

>>8014180
Since when is strictly following the given conditions "going full autism"? Did you ever say to yourself during an exam "nah, I can't be arsed to go full autism on this and will instead just conveniently ignore some arbitrary parts of the problem specification, hopefully nobody notices anyway"?

>>8014284
>[X] FILE DELETED

Lol, I saw you pic earlier and there was nothing offensive I noticed about it. What gives?

>>8014288
That's what most high school geometry courses entail, yes.

>>8014362
Entail what, ignoring random parts of the given problems?

>>8014141
Listen. Suppose you're taking the SAT. The question displays a right triangle with the sides of the non-hypotenuse 3 and 4. It asks you what is the length of the hypotenuse. Possible answers:

A) 4
B) 5
C) 6
D) Cannot be determined from information given

Are you seriously going to select (D) and go "hurr I can't answer that because it depends on the curvature of the Riemmanian manifold". Because (D) is obviously the wrong answer, and you would get the question wrong.

>>8014359
Nothing, I just didn't mean to post it.

>>8014381
Entail taking information that is explicitly stated
>it's a right triangle
>it has a hypotenuse of length 10
>it has a height of length 6
And demonstrating that you know how to find the area of a triangle in which the base and altitude are given, (bh)/2.
They don't give you a shitty little ruler with these problems and ask you to calculate the actual unit length of the figure. They don't care about your super dank measuring skills. They care about you recognizing how you'd actually solve that problem, given an ACTUAL triangle.

It's a high school math problem. Get the fuck over yourself.

>>8014277
>square prism
That means the vertical edges are perpendicular to the horizontal edges.

24

>>8014386
Have you even read what you're replying to? The statement "every triangle with hypotenuse 10 and altitude 6 has area 42" is trivially false, because you can provide a trivial counterexample (such as a regular Euclidian triangle with an area of 30, which was mentioned in this thread countless times by confused anons who ommited to read the "right triangle" requirement.

The statement "every RIGHT triangle with a hypotenuse of 10 and an altitude TO ITS HYPOTENUSE equal to 6 has an area of 42" is vacuously true, because such triangles don't exist. But the statement "every triangle with a hypotenuse of 10 and an altitude equal to 6 has an area of 42" is false, because there exist triangles as described which have an area unequal to 42. That's all, and curvatures of Riemannian manifolds don't have anything to do with it.

>>8014411
But the height of 6 has to be the height to the hypotenuse, and if the latter is 10, then the triangle can't be a right triangle. Get the fuck over yourself.

>>8014423
Oh, then yeah, it is easy.
c1c = a1a. take the inverse tangent for a1a/ab and c1c/cd to get the angles of the two prisms with the horizontal, then subtract both from pi radians.
I don't get how you can use trig for the last one though.

>>8014439
>then the triangle can't be a right triangle
bh/2 doesn't only apply to right triangles, you dingbat.

Use the triangle postulate to find out the lengths of the sides geniuses.

>>8014461
Draw a line parallell to AD from B to a point on CD.

>>8014464
does it apply to triangles that dont exist

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new question

>>8014502
Yes, as does everything.

So every right triangle with hypotenuse 10 and altitude 6 has altitude [math] \leq [/math] 5.

>>8011185
a^2+b^2=100
36+x^2=a^2
(10-x)^2+36=b^2
136-20x+x^2=b^2
36+x^2+136-20x+x^2=100
36-10x+x^2=0
x1=5+i(sqrt11)
x2=5-i(sqrt11)
(10-x1)=5-i(sqrt11)
(10-x2)=5+i(sqrt11)
A1=(15+i3sqrt11)+(15-i3sqrt11)=30
A2=(15-i3sqrt11)+(15+i3sqrt11)=30
The area is 30.

>>8014553
If A1 is the area of one of the two smaller triangles, shouldn't it be 6*x1 / 2? Why are you adding things to find the area?

You should wind up with two possible areas: a complex number and its conjugate.

(Solving your way, the area must be a complex number because it is an impossible triangle.)

>>8014169
(i) A1C is perpendicular to BD because it lies within the plane of AC and AA1, both of which are perpendicular to BD.

(ii) Because |AC|2 = |AD|2 + |DC|2, |AC| = 4. Therefore, because |AE|/|AD| = |AD|/|AC|, |AE| = 1. Consequently the angle between A1E and the plane B1D1DB is arctan(1/√3) = 30º.

Likewise, because |EC|/|DC| = |DC|/|AC|, |EC| = 3, and the angle between C1E and the plane B1D1DB is arctan(3/√3) = 60º.

Therefore the angle between the two planes A1BD and BC1D is 30º + 60º = 90º.

(iii) Let BD and AC be the y and x axes (respectively) of a cartesian coordinate system. Then the vector AD = (|AE|, |ED|, 0) = (1,√3,0). Since ABD is isosceles, then |BE| = |ED|, and BC1 = (|EC|, |BE|, |CC1|) = (3,√3,√3).

Then (1,√3,0)·(3,√3,√3) = AD·BC1 = |AD||BC1|cosθ = 2√15 cosθ, where θ is the angle between AD and BC1. Therefore θ = arccos(√3/√5) ≈ 39.23º.

>>8014800
Congrats, you passed the pre-entry exam and got into college.

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>>8014800
>>8014553

Autists.

>>8014832
The first guy you replied to was answering a different question.

>>8014842
Oh, my bad.

>>8014849
Also you're wrong

>>8014851
I can't be wrong, I was arrogant about my proof.
Joke's on you.

On a hunch, shouldn't the hypothenuse be irrational?

>>8011185
what's worse than a problem you can't solve
thanks anon

>>8011226
yeah I almost got duped until I read some of the replies and realised that I need to brush up on geometry

>>8014862
>On a hunch, shouldn't the hypothenuse be irrational?
I think you mean imaginary. There's nothing wrong with an irrational hypotenuse, just look at the right triangle 1 by 1 by sqrt(2)

It's OP. Here is the answer.

Evil Geometry Problem https://youtu.be/GbQQcrwYHt0

>>8011185

it's 30.

>>8017804
Then construct a triangle exactly as described and show how its area is "30".

All you people who keep ignoring the requirement for the triangle to be a right triangle need either glasses, better glasses, or a brain.

>>8011185
The geometry is invalid.

any similar problems to this one that are actually solvable?

>>8017522
Holy shit how did I not see that the first time.

>>8019409
That's why OP posted this. Typical bait thread like many others, with the problem either being phrased somwhat ambiguously, or with a high chance that many if not most will miss some part of it (like is the case here in particular).

>>8019371
What do you mean by "similar"? If the conditions are a contradiction, then it obviously isn't solvable.

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Area=1/2 base times height.
Height is 6. Base is 10. 0.5*6*10=30
How about you leave your basement and attend a 6th grade maths class?

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>>8020160
>How about you leave your basement and attend a 6th grade maths class?
OK, if you're so well-educated: what's the sum of the lengths of the three sides?

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am i doing it right

>>8020160
A right triangle can't have a height perpendicular to its hypotenuse larger than the length of the hypotenuse divided by 2 (as has been pointed out a countless number of times by those who actually bothered to read and understand the conditions of the problem). Morons.

>>8011226
Do you mean 10^2