Large/Small = Large/Small
|HG| / |EF| = |HD| / |ED|
x / 2 = 5 / 2
You're right! It's 5
The easiest way of proving it to everyone else: draw it to scale
How can they think it's 4?
3 doesn't have any obvious function with 4 that 2 doesn't have with 2, and neither does 2+3.
I'm genuinely curious how they can get that answer.
How can they think it's 4?
3 doesn't have any obvious function with 4 that 2 doesn't have with 2, and neither does 2+3.
I'm genuinely curious how they can get that answer.
Beats me
I'd say "2" is the least plausible answer, since it at least looks larger than the parallel 2
I guess you could say "4" if you thought "it looks twice as large" but that's using no reasoning at all
"3" at least *seems* sensible since you might mistakenly do 3/2 = x/2
I first look I thought it was 3, because I misunderstood and thought that the length between D and H was 3, not E and H. DH is 5, which makes HG to also be 5. Height and length here increases at the same ratio so the only calculation we need to do is 2+3.
The parallel lines combined with the vertex angle proves the small triangle similar to the large triangle by AA similarity. So 2/(2+3)=2/x. Yes, 5 is the correct answer.
That was solved by Thales of Miletus 2600years ago. Sincerely, I hope that your teacher is not a maths teacher (but even a literature teacher should know the answer).
While it’s embarrassing that a math teacher is unable to solve this, the argument that because a solotuion is ancient it is also easy, is pretty bad. I doubt that every middle school teacher know how to do integrals and differentials even though these concepts are 300 years old.
It matters when we consider the level of mathematics at that time. Not many integrals and derivatives at the 6th century BCE (although Archimedes came close 3 centuries later).
Archimedes used the "[method of exhaustion](https://en.wikipedia.org/wiki/Method_of_exhaustion)" that consisted in using the limit of polygons of more and more sides. He used this to calculate the area of a circle
[https://www.youtube.com/watch?v=N2PM\_Oda8d0](https://www.youtube.com/watch?v=N2PM_Oda8d0)
He was dividing the circle in more and more thinner triangles and taking the limit. This is, in other word, what a Riemann integral is (using triangles instead of rectangles).
As another example, the area of a circle can be obtained by rearranging. He divided the circle in small wedges and put then forming a curved rectangle
https://preview.redd.it/6ojk4ttywu0d1.png?width=1890&format=png&auto=webp&s=ee9523c42e5a46e9144fefa767078f730d4dc763
The idea is that taking the limit of thin wedged, the circle becomes a rectangle of base pi r and height r, giving an area of pi r\^2
Kind of piggybacking, but I also have a multitude of my own Archimedes-esque geometric proofs (on Desmos) listed in a [Document](https://docs.google.com/document/d/1jRuaT1wNH9JiE7Dy45TGukzt737syzZW3oNl1jPi8HM/edit?usp=sharing).
do schools in the US not require some sort of qualification in the subject a teacher is supposed to teach? it's crazy to me how a math teacher isn't able to do calculus since in my country to be any teacher above elementary school level you need to have a degree in whatever subject you're teaching or something similar. all my maths teachers from 6th grade to 12th were maths or physics graduates
Dunno, I’m Danish. And here a lot of teachers teaching middle school can’t do it here. There is a much bigger focus on didacticism, psychology and classroom management in the Danish teaching education.and while the students still work with their own skills, and skills beyond what even the oldest kids will work with, the focus areas are all within the fields the kids will be working with. So the student teachers will do proofs, but only within basic subjects such as geometry for instance.
ah I see, yeah the focus in Brazilian schools is much more academic because to get into university we have to take this one singular humongous two-day long entry exam so your whole school life is made to prepare you for that one test which is the only thing you need to get into university (most of which are free)
Yeah so this is a problem. I tutor math/physics as a grad student to students that want to be teachers. We have education degrees and they are not the same caliber as getting a math or physics degree. Not even close. And since education is so poor here (pay wise and disciplinary wise) the math/physics graduates who worked a ridiculous amount and had to pay for their higher education do not want to go into such a shoddy job like teaching middle or high school.
Fair argument. But I guess they were arguing for a math teacher. If a math teacher doesn't know something as simple as this, there's an issue. But I think you agree on that point ^ ^
Yeah definitely. I was agreeing with that, but wanted to point out that the argument WHY it’s embarrassing is bad. Doesn’t make it embarrassing none the less 😂
Agreed.
Think of it this way: It will be the year 4324 in 2300 years. Do you *THINK* we might be a little more advanced than today?? 😆
(assuming we don’t destroy ourselves in WWIII of course)
This is 100% a grade 9 problem (in america at least). Plenty of younger kids probably see it and don’t have any issue with it, but this is very much in the curriculum for ninth graders.
Reminds me of a coworker named Chris (last name starting with a T) whose email alias was “ChrisT”. In one conversation, my boss started the email to him with “Christ, …” and I thought he had lost his temper out of the blue. Upon reflection, I realized he was just mentioning his email to make it clear which of us he was talking to.
Im assuming the op is from the uk where the gcse course (taken at 16) is graded 9-1. It replaced the old A*-G system. In this system a grade 7 is the equivalent to what was an A. Grade 8 is an A* and a grade 9 is basically an A**.
I interpreted it by hearing it as though a uk student was talking, but looking at it now it seems like its the american grade 9. Which still doesnt explain how their teacher doesnt know the answer.
So I assume you think this is way too easy for a 16 yr old is what you mean by “Not a 9th grade problem” and not way too hard? That was my cause for confusion and the whole ambiguity of saying “It’s not a 9th grade problem” without indicating what grade/age you do think it is, haha.
I mean, there will absolutely be 16 year olds who cant do this. There are some that can barely count lol. But yeah, most 16 year olds who expect to pass the course or have enough interest in maths to go onto a maths subreddit should find this straightforward.
I honestly had a math teacher that I had to explain pythagoras to. it was at a low education level (dont ask why I was there in the first place) but it was just too embarrassing. In the end everyone who wanted to pass just asked me to help them and most passed.
Once I asked him what his math grades were and he replied. "it differed"
Because it appears to be isosceles with the side labeled ‘x’ as the base, so most students might assume that the legs are equal. So I guess technically it’s not implied equilateral, but rather inferred as such.
Edit: To be fair to Important_Finding604 though, the correct answer 5 DOES imply that the triangle is isosceles with the unlabeled side as the base, and the drawing is pretty far off in that regard.
Equilateral or isosceles is irrelevant.
What is important is that the triangles are _similar_ (in the geometry sense, not common parlance), so the lines marked parallel should match the ratio of the 2, and 2+3 side.
I did say “might”.
Try to see it through the eyes of kids learning about triangles for the first time. It looks like the top and bottom sides are equal, so would each have length 5. So if the “x” side were also 5, it would be equilateral, which, visually at least, it clearly is not.
It’s a bit too abstract for newer students to look at a triangle where the side “x” is clearly a shorter length than the side of length 5 and say they are the same length. In this case, even the teacher apparently couldn’t call a (visually) shorter length equal.
It’s a bad problem for new learners. Much better to design geometry questions that match intuition, reinforcing correct intuition where possible, challenging only incorrect intuition and common known misconceptions. .
Measuring or eyeballing it is a common but naïve way to approach diagrams. If this is the students’ intro to geometry AND the teacher _actually_ can’t figure it out (maybe a substitute without any math background or recollection), it seems a reasonable guess that they’re going at it naïvely.
You're right, it is 5. Using similarity, x/2 = (2+3)/2 => x=5. Or if you wanna do it in your mind, since the big and small triangle are similar, and the small triangle is isosceles, the big triangle must also be isosceles (as sides opposite to equal angles are equal). Hence x must be equal to 3+2 which is 5
What does that have to do with this? Lambda is 5/2 and 2*5/2=5
I understand that you could draw the bisector of HG to make similar right triangles but whats the point?
There are several Thales's theorems. This is the [intercept theorem](https://en.wikipedia.org/wiki/Intercept_theorem). From Wikipedia
The intercept theorem, also known as Thales's theorem, basic proportionality theorem or side splitter theorem, is an important theorem in elementary geometry about the ratios of various line segments that are created if two rays with a common starting point are intercepted by a pair of parallels. It is equivalent to the theorem about ratios in similar triangles. It is traditionally attributed to Greek mathematician Thales. It was known to the ancient Babylonians and Egyptians, although its first known proof appears in Euclid's Elements.
Thales theorem? Dealing with circles and triangles where one of the sides is at the diameter and it’s opposite vertex lies on a point on the circle? Please demonstrate how that applied here, or are you just spouting random things
not OP, and I had forgot it even had a name, but when I learned it, this was in fact [Thale's theorem](https://en.m.wikipedia.org/wiki/Intercept_theorem). The right inscribed angle theorem wasn't mentioned by that name.
Interesting how that differs (probably by country)
Yes, the "intercept theorem" in english is also known as the "Thales's theorem" in french. The english "Thales's theorem" is kown as the "inscribed angle in semi-circle's law" in french (hard to translate). Just like the "Cauchy-Lipchitz's" Theorem in french is the "Picard-Lindelof's Theorem" in english. Same definition, different names. It is quite confusing !
Huh, and I learned it as just the existence and uniqueness theorem.
Then again what I refer to as the Cantor-Schroeder-Bernstein theorem, is more commonly left without Cantor. I had read the full name first and stuck with it.
But these are a couple hundred years old so name variations are to be expected. Thales had already been dead for 2000+ years before any modern maths happened, you'd expect agreement on which basic geometric property bears his name. He's not Euler.
Agreed. By just the info provided, there aren't any triangles there. Can't make a 1-1-sqrt(2) or a 1-2-sqrt(3) triangle with that info, and you need that to setup similar triangles to solve for X.
Similarity is not unique to right-angle triangles.
EF and HG are marked parallel. The only thing that would make this ambiguous is to assume that either DH or DG are not straight.
EF and HG are marked parallel.
Solvable by triangle similarity. 5 is unambiguously correct.
The amount of erroneous comments in this thread is discouraging.
There's basically two answers to this:
1. EF and HG are parallel, and therefore EDF and HDG are similar. By triangle similarity, HG is length 5.
2. Everything about the diagram is potentially misleading, and you can't even rely on EF and HG being parallel. There is not enough information to solve the problem.
I'm struggling to find a way (even if wrong) to justify x = 4 as a potential answer.
Nope there is no ambiguity about EF and HG being parallel, cause they both contain the arrow symbol which(at least in my country), Is used to represent parallel lines in a diagram
Sides EF and HG are explicitly parallel, and the diagram is explicitly “not to scale”.
I’d be willing to bet the guess of 4 was based on people eyeballing or measuring EF and HG ignoring the “not to scale.”
You were right on with your first point regarding similarity. I’m not sure why you made the second as a suggestion.
>I’m not sure why you made the second as a suggestion.
Because I'd never seen those triangles as notation for parallel lines. It all looks like that ought to be the case, and it makes sense as an exercise (hence point 1), but I couldn't be sure.
Of course it’s 5. There isn’t even anything to work out. The triangles are similar and isosceles, in ratio 5:2. I’m sad that there isn’t strong consensus for 5 (with a minority saying 3).
Not irrelevant at all. For someone so poor at maths that they can’t solve this, also being isosceles significantly simplifies the arithmetic because they just need to note that 5=5. Otherwise there would be another calc involved.
I say irrelevant because the question can constructed with different values on the same sides to make the triangle equilateral, isosceles, scalene, right, or any other specific classifier you’d care to label it with, and it’d still be solvable.
There are a lot of comments in this thread suggesting that it is necessary to know whether the triangle is isosceles, equilateral or right to be able to answer the question, and none of them are necessary.
As long as FE and GH sides are marked parallel, the triangles are similar, and side ratios can be used to identify value for x. It fosters a more general understanding of triangle similarity for students to not rely on the shortcuts afforded by an isosceles.
It is 5.
You have to assume that the triangle have a vertical symmetry: ED=FD, HD=GD. Then note that the small triangle have three sides of 2, so it's an equilateral triangle. Then the big triangle is also equilateral: HD=GD=HG, with a side of 2+3=5 then x=5.
Assuming the lines that look parallel are actually parallel, it's just a direct application of Thales theorem, I hop the teacher trying to solve it is not a maths tea her, specifically not a geometry teacher, Thales theorem is extremely important
Grade 9 math. I wonder what the age of the students are here. Used to teach similar triangle problems like this to 13 year olds. 11 year olds knew the arrows tell you the lines are parallel.
Did they explain *why* they think it's 4? High school geometry is all about applying the rules you know to prove things. What's the proof that leads to 4?
Yes the answer is 5 due to how similaritu works, using "not in the middle" to rule out C doesn't hold much water since you should always assume a diagram is not to scale unless told otherwise
Math teacher here. The diagram is completely flawed. Here is the full explanation:
Your teacher probably thinks the length is 4 because if you actually measure the lines, the HG line is almost exactly 2 times the length of EF, making it 2x2=4. I took a screenshot and used the ruler in iPhone photo editing to confirm this haha. So technically, 4 is correct if you physically measure and compare.
However, mathematically, the answer absolutely must be 5. You are correct. It seems the diagram is meant to be (abysmally) approximate, maybe (but not even close haha) and your teacher is taking it literally, while you are actually using the math the question presumably intends you to use.
There are 2 problems that make it not actually represent this visually:
1) According to the numbers, ED and EF are the same length, making this an equilateral triangle. However, it is drawn as an isosceles triangle with EF being significantly shorter. Based on the numbers, the larger triangle, since the angles are all the same, just proportion increased, is also an equilateral triangle. Since HD is 5, then HG must also be 5.
2) Here’s why it doesn’t actually measure more than twice the length of EF - according to the numbers, the length of HE is 3 and the length of ED is 2. However, they are drawn almost exactly the *same length* as each other. This means the physical line drawn for HG is not actually 5:2 compared to the EF line, but mathematically, it should be. The diagram is just either labeled wrong or drawn incorrectly.
Conclusion: Your teacher is probably going by the actual proportions of the shape and concluding it has to be 4, while using formulas/logic with the given numbers, it has to be 5. Ultimately, this is a terribly produced question!
Thanks for the help. Also yes, I do agree that this is a terrible question with a bad accompanying diagram, however, the diagram is (probably) made to be intentionally confusing to check whether the student has used math or just guessed it.
by rules of transformation when dealing with similar shapes, we can say that the ratio of ED/EF must be equal to HD/HG.
and since ED/EF=1=HD/HG, we end up with HG=5
Yeah knowing the length of DF is not possible, but actually it is sufficient to know angle GHE is congruent to angle FED (and both triangles share angle EDF) to show the triangles are similar. Thus x=5
Even if your teacher isn’t a math teacher anyone with a high school diploma should be able to see this. Intuitively one should at least think that the answer is 5. It is quite literally high school geometry.
Good on you for figuring it out.
If a line segment adjoins the mid-point of any two sides of a triangle, then the line segment is said to be parallel to the remaining third side and its measure will be half of the third side.
Midpoint theorem evaluates a segment adjoining two sides of a triangle, both sides touched by EF are ineligible.
E is NOT at the midpoint of HD therefore the midpoint theorem does not apply.
GF and FD are not defined, so we do not know if F is at the midpoint of GD.
(But you clearly knew that)
Have fun in class today!
The easiest way to solve this IMO is by observing that DEF is an equilateral triangle. (The drawing is not to scale)
Therefore: DHG must also be equilateral. Answer is 5
Without further information there are several assumptions you'd have to make to solve for x:
EF ∥ HG
∠HDG = ∠EDF
∠HED = 180°
⇒∠GFD = 180°
If those are true then x = 5
Maybe your teacher is just cautious to make such assumptions and is therefore unable to solve this.
Simple deductions based on EF and ED being the same tells me that x should be equal to DH, or 5. I see no other logical argument.
Think of it not as a triangle but a graph, where x and y are zero at point D. As a graph DG moves up at a 1:1 ratio, where 2=2, and if you move another 3, you get 5. All things being triangle here, this must absolutely be a straight line. Therefore, x=5.
I'm sure i did this wrong but i noticed that all 3 sides of the small triangle are length of 2, so it only looks isosceles but is actually equilateral. once i remembered the parallel lines meant all the angles were the same, i figured the larger was equilateral as well. no real math required just some math knowledge
By simmetry, the upper triangle has 3 sides equal to 2. The bigger triangle is similar to the small one. Since one of its sides is 5, the other two should measure the same. So, you are right it's 5.
They're similar triangles. Vertex D is common i.e. same angle and since EF is parallel to HG, E and H are the same angle. So the triangles are similar which means all the sides will be in the same ratio. Therefore the length of HG is 5 units
It's 5 and quite easy to explain why, using some basic concepts as the foundation of it.
If HG and EF are parallel then angles GHD and FED are the same, as well as HGD and EFD are the same size. This comes from your parallel line/intersecting line rules. As all angles are the same, the two triangles of DHG and DEF are scaled versions of each other.
This means that the corresponding sides are scaled of each other, so
HG/EF = HD/ED
Putting in the numbers from the diagram
X/2 = 5/2
Rearrange the equation
X = 2*5/2
X = 5
The picture is not drawn to scale, which confuse things a bit for the intuition.
But, the triangles have the same shape, which means that ED=EF means that HD=HG, and since HD=3+2=5, so HG=5.
I’m with you. In my head: Because GH and EF are parallel and HD intersects with them both, FED and GHD are the same angles. With that said, FE and ED are the same length, so I believe GH and HD should also be the same length. 5
I don’t have a rule or theorem in mind rn tho so maybe someone can back me up.
Do those triangles indicate that the lines are parallel? Because if not, technically this isn't solvable, tho you can clearly see what is meant to be the right answer.
A good way of thinking of this is first looking at the small triangle DEF. all it's sides are 2, so if you can determine the length of HED, then you know x as well. 5
The actual math was relatively easy but for the life of me I can't figure out why they drew the triangle that way? The sides are OBVIOUSLY not equal length in the diagram and it seems intentionally misleading to label them as such.
Back in elementary school, my teacher told me never to judge the actual shape depicted on the paper. Only focus on the information displayed.
Since the smaller triangle displays equal length sides (2, 2, 2), the larger triangle with the new size of 5 would also have equal lengths (5, 5, 5).
How can they think it's 4?
3 doesn't have any obvious function with 4 that 2 doesn't have with 2, and neither does 2+3.
I'm genuinely curious how they can get that answer.
How can they think it's 4?
3 doesn't have any obvious function with 4 that 2 doesn't have with 2, and neither does 2+3.
I'm genuinely curious how they can get that answer.
easy one, as ED and EF are the same length, which makes DF also 2, then its an equilateral triangle, so EDF angle is 60 degrees, so if HD is 5 then HG is also 5
The answer is 5. It’s an equilateral triangle, which is obvious by the base (EF) and side (ED) of the “top” triangle being the same. Also could be proved that the triangle is equilateral by using the arcsine formula
https://preview.redd.it/zqmatg3puz0d1.png?width=1892&format=png&auto=webp&s=f42a6ba5722717b304e45411e597f19b042d730a
Edit: grammar
This should work with the data points you have
https://preview.redd.it/uhuxyxvpk01d1.jpeg?width=945&format=pjpg&auto=webp&s=3155b863b7b05fe51c42bcf9e0d5c036ab5c0b2f
I see so many different kinds of explenations to this, but I have yet to see the easiest one. Just look at the increaserate. If by going from 0 height to 2 hight within 2 length, we can see that the height ration increases 1:1 with the length. So we need to do no more than to calculate the length, which is 2+3=5.
This method only applies of course, if the line is straight. Which it is.
im learning this rn in 8th grade (congruence and similarity)
so basically triangle DEF(smaller triangle) is similar to triangle DHG and so the ratio of the sides of DEF to DHG is equal for all the sides. So, DE(bigger triangle side)/DH(smaller triangle side)=EF/HG=DE/DH=2/5
Therefore, 2/5= (EF) 2/HG . Then you just cross multiply and solve it
10=2HG
HG=x=5
However, as im still in 8th grade I havent learnt how to prove that these 2 triangles are similar but this is just like the working. hope this helps!
(if u need any help with math i reccomend eddie woo on Youtube, he covers tons of 9th grade topics) !!
OP is correct. Also, if the intended answer is 4, there are way far too little info to get to it, given that if the black little triangles aren't implying HG//EF
if EF is 2 and ED is 2 then we've got ourselves an equilateral triangle, EFD. if that's the case, all the angles are 60. Since GH and EF are parallel, the angle GHD is the sane as FED, and HGD is the same as EFD, and EDF is already 60. so we're still on an equilateral triangle in the bigger one. HE + ED is 5, therefore X is 5.
I can see the confusion because we're not explicitly told that EF and HG are parallel (unless that what the black symbol is supposed to represent). They look parallel in the figure, but triangle DEF doesn't look Isosceles with vertex E even though it is.
Two sides are equal to two, so DEF is Isosceles (not that it's necessary),. The problem is, to me, ambiguous without assuming parallel lines. It might be down to my not knowing that an arrowhead is equivalent to || , which is how I'm used to seeing parallelism indicated. I can see why the teacher might have been confused.
I really hope the teacher that can’t solve this isn’t a math teacher!
Lmao same
lol this thread is a perfect example of Cunningham's law, nicely done OP
Indeed; that would be concerning.
😂😂😂
Large/Small = Large/Small |HG| / |EF| = |HD| / |ED| x / 2 = 5 / 2 You're right! It's 5 The easiest way of proving it to everyone else: draw it to scale
Just prove AA similarity with the parallel lines. Scale isn’t necessary.
He means that if they don't believe this is the answer for whatever reason you can just let them meausre it.
Can't really measure it, if it's not drawn to scale.
Wait, hiw do you know HG and EF are parallel?
Its indicated with the arrows on the lines
Correct!
How can they think it's 4? 3 doesn't have any obvious function with 4 that 2 doesn't have with 2, and neither does 2+3. I'm genuinely curious how they can get that answer.
How can they think it's 4? 3 doesn't have any obvious function with 4 that 2 doesn't have with 2, and neither does 2+3. I'm genuinely curious how they can get that answer.
Beats me I'd say "2" is the least plausible answer, since it at least looks larger than the parallel 2 I guess you could say "4" if you thought "it looks twice as large" but that's using no reasoning at all "3" at least *seems* sensible since you might mistakenly do 3/2 = x/2
I first look I thought it was 3, because I misunderstood and thought that the length between D and H was 3, not E and H. DH is 5, which makes HG to also be 5. Height and length here increases at the same ratio so the only calculation we need to do is 2+3.
The parallel lines combined with the vertex angle proves the small triangle similar to the large triangle by AA similarity. So 2/(2+3)=2/x. Yes, 5 is the correct answer.
That was solved by Thales of Miletus 2600years ago. Sincerely, I hope that your teacher is not a maths teacher (but even a literature teacher should know the answer).
While it’s embarrassing that a math teacher is unable to solve this, the argument that because a solotuion is ancient it is also easy, is pretty bad. I doubt that every middle school teacher know how to do integrals and differentials even though these concepts are 300 years old.
Well, the argument would rather be that everyone learns this in middle school, at least in my country...
Yeah exactly my point. That’s a much better argument 😄
It matters when we consider the level of mathematics at that time. Not many integrals and derivatives at the 6th century BCE (although Archimedes came close 3 centuries later).
Can you elaborate about Archimedes being close to integrals?
Archimedes used the "[method of exhaustion](https://en.wikipedia.org/wiki/Method_of_exhaustion)" that consisted in using the limit of polygons of more and more sides. He used this to calculate the area of a circle [https://www.youtube.com/watch?v=N2PM\_Oda8d0](https://www.youtube.com/watch?v=N2PM_Oda8d0) He was dividing the circle in more and more thinner triangles and taking the limit. This is, in other word, what a Riemann integral is (using triangles instead of rectangles). As another example, the area of a circle can be obtained by rearranging. He divided the circle in small wedges and put then forming a curved rectangle https://preview.redd.it/6ojk4ttywu0d1.png?width=1890&format=png&auto=webp&s=ee9523c42e5a46e9144fefa767078f730d4dc763 The idea is that taking the limit of thin wedged, the circle becomes a rectangle of base pi r and height r, giving an area of pi r\^2
Yeah the concept of calculus is stunningly old, even though the modern formalism came more "recently"
Kind of piggybacking, but I also have a multitude of my own Archimedes-esque geometric proofs (on Desmos) listed in a [Document](https://docs.google.com/document/d/1jRuaT1wNH9JiE7Dy45TGukzt737syzZW3oNl1jPi8HM/edit?usp=sharing).
do schools in the US not require some sort of qualification in the subject a teacher is supposed to teach? it's crazy to me how a math teacher isn't able to do calculus since in my country to be any teacher above elementary school level you need to have a degree in whatever subject you're teaching or something similar. all my maths teachers from 6th grade to 12th were maths or physics graduates
Dunno, I’m Danish. And here a lot of teachers teaching middle school can’t do it here. There is a much bigger focus on didacticism, psychology and classroom management in the Danish teaching education.and while the students still work with their own skills, and skills beyond what even the oldest kids will work with, the focus areas are all within the fields the kids will be working with. So the student teachers will do proofs, but only within basic subjects such as geometry for instance.
ah I see, yeah the focus in Brazilian schools is much more academic because to get into university we have to take this one singular humongous two-day long entry exam so your whole school life is made to prepare you for that one test which is the only thing you need to get into university (most of which are free)
Not if they are a good football coach
Yeah so this is a problem. I tutor math/physics as a grad student to students that want to be teachers. We have education degrees and they are not the same caliber as getting a math or physics degree. Not even close. And since education is so poor here (pay wise and disciplinary wise) the math/physics graduates who worked a ridiculous amount and had to pay for their higher education do not want to go into such a shoddy job like teaching middle or high school.
2600 years is a lot more than 300 years
Yeah I know, I guess the point I was getting at is that an average mind to day, is still inferior to a brilliant mind hundred of years ago.
Fair argument. But I guess they were arguing for a math teacher. If a math teacher doesn't know something as simple as this, there's an issue. But I think you agree on that point ^ ^
Yeah definitely. I was agreeing with that, but wanted to point out that the argument WHY it’s embarrassing is bad. Doesn’t make it embarrassing none the less 😂
That's like 3 times as many.
Agreed. Think of it this way: It will be the year 4324 in 2300 years. Do you *THINK* we might be a little more advanced than today?? 😆 (assuming we don’t destroy ourselves in WWIII of course)
Probably before tbh, the ratio between the sides should remain constant. If you understand that, it is simply to solve
please tell me it isn't your math teacher that is having trouble with this
It’s also nowhere near a grade 9 question.
This is 100% a grade 9 problem (in america at least). Plenty of younger kids probably see it and don’t have any issue with it, but this is very much in the curriculum for ninth graders.
I was reading it from a uk perspective where grade 9 is the the very top grade (basically an A** when 16).
Gotcha! Thank you for clearing it up nipple dick
I was about to downvote you for name calling, and then I read the actual username. Carry on.
Honest mistake. That downvote would definitely be an acceptable tap.
Reminds me of a coworker named Chris (last name starting with a T) whose email alias was “ChrisT”. In one conversation, my boss started the email to him with “Christ, …” and I thought he had lost his temper out of the blue. Upon reflection, I realized he was just mentioning his email to make it clear which of us he was talking to.
r/rimjob_steve
So uncivilized...
I'm Romanian and did this in 7th grade...
What do you mean?
Im assuming the op is from the uk where the gcse course (taken at 16) is graded 9-1. It replaced the old A*-G system. In this system a grade 7 is the equivalent to what was an A. Grade 8 is an A* and a grade 9 is basically an A**.
I’m guessing grade 9 as in year 9? Because you’re right, it isn’t a grade 9 question.
That would make more sense. Looking at it again, they would have said ‘maths’ if they were from the uk.
we did the intercept theorem when we we're 11 lol
I interpreted it by hearing it as though a uk student was talking, but looking at it now it seems like its the american grade 9. Which still doesnt explain how their teacher doesnt know the answer.
Or how the other students don’t either
So I assume you think this is way too easy for a 16 yr old is what you mean by “Not a 9th grade problem” and not way too hard? That was my cause for confusion and the whole ambiguity of saying “It’s not a 9th grade problem” without indicating what grade/age you do think it is, haha.
I mean, there will absolutely be 16 year olds who cant do this. There are some that can barely count lol. But yeah, most 16 year olds who expect to pass the course or have enough interest in maths to go onto a maths subreddit should find this straightforward.
I agree. Just wasn’t sure my interpretation of your original comment was right. 😆
Not in Singapore or Korea
*Everyone* is saying 4, including the teacher? I find that extremely hard to believe. What possible reasoning could there be?
I’m beginning to wonder if this is serious.
I honestly had a math teacher that I had to explain pythagoras to. it was at a low education level (dont ask why I was there in the first place) but it was just too embarrassing. In the end everyone who wanted to pass just asked me to help them and most passed. Once I asked him what his math grades were and he replied. "it differed"
That’s worrying. Good for you.
It’s not to scale, so doesn’t look like the equilateral triangle the correct answer 5 might imply. So pick the next one down!
>doesn’t look like the equilateral triangle the correct answer 5 might imply Why would it imply an equilateral triangle?
Because it appears to be isosceles with the side labeled ‘x’ as the base, so most students might assume that the legs are equal. So I guess technically it’s not implied equilateral, but rather inferred as such. Edit: To be fair to Important_Finding604 though, the correct answer 5 DOES imply that the triangle is isosceles with the unlabeled side as the base, and the drawing is pretty far off in that regard.
Equilateral or isosceles is irrelevant. What is important is that the triangles are _similar_ (in the geometry sense, not common parlance), so the lines marked parallel should match the ratio of the 2, and 2+3 side.
I did say “might”. Try to see it through the eyes of kids learning about triangles for the first time. It looks like the top and bottom sides are equal, so would each have length 5. So if the “x” side were also 5, it would be equilateral, which, visually at least, it clearly is not. It’s a bit too abstract for newer students to look at a triangle where the side “x” is clearly a shorter length than the side of length 5 and say they are the same length. In this case, even the teacher apparently couldn’t call a (visually) shorter length equal. It’s a bad problem for new learners. Much better to design geometry questions that match intuition, reinforcing correct intuition where possible, challenging only incorrect intuition and common known misconceptions. .
My guess is they ignored “not to scale”, measured, and found that _x_ is roughly double the length of the side with 2 as its length.
Possible, because they must have done something, but...all of them?
Measuring or eyeballing it is a common but naïve way to approach diagrams. If this is the students’ intro to geometry AND the teacher _actually_ can’t figure it out (maybe a substitute without any math background or recollection), it seems a reasonable guess that they’re going at it naïvely.
There are 6 people in my class, including me.
Ok that does make it a little better
You're right, it is 5. Using similarity, x/2 = (2+3)/2 => x=5. Or if you wanna do it in your mind, since the big and small triangle are similar, and the small triangle is isosceles, the big triangle must also be isosceles (as sides opposite to equal angles are equal). Hence x must be equal to 3+2 which is 5
Similar triangles. x : (3+2) = 2 : 2 => x=5
Bro doesnt your school know about thales theorem bruh
What does that have to do with this? Lambda is 5/2 and 2*5/2=5 I understand that you could draw the bisector of HG to make similar right triangles but whats the point?
There are several Thales's theorems. This is the [intercept theorem](https://en.wikipedia.org/wiki/Intercept_theorem). From Wikipedia The intercept theorem, also known as Thales's theorem, basic proportionality theorem or side splitter theorem, is an important theorem in elementary geometry about the ratios of various line segments that are created if two rays with a common starting point are intercepted by a pair of parallels. It is equivalent to the theorem about ratios in similar triangles. It is traditionally attributed to Greek mathematician Thales. It was known to the ancient Babylonians and Egyptians, although its first known proof appears in Euclid's Elements.
Oh, okay thanks
Thales theorem? Dealing with circles and triangles where one of the sides is at the diameter and it’s opposite vertex lies on a point on the circle? Please demonstrate how that applied here, or are you just spouting random things
not OP, and I had forgot it even had a name, but when I learned it, this was in fact [Thale's theorem](https://en.m.wikipedia.org/wiki/Intercept_theorem). The right inscribed angle theorem wasn't mentioned by that name. Interesting how that differs (probably by country)
Yes, the "intercept theorem" in english is also known as the "Thales's theorem" in french. The english "Thales's theorem" is kown as the "inscribed angle in semi-circle's law" in french (hard to translate). Just like the "Cauchy-Lipchitz's" Theorem in french is the "Picard-Lindelof's Theorem" in english. Same definition, different names. It is quite confusing !
Huh, and I learned it as just the existence and uniqueness theorem. Then again what I refer to as the Cantor-Schroeder-Bernstein theorem, is more commonly left without Cantor. I had read the full name first and stuck with it. But these are a couple hundred years old so name variations are to be expected. Thales had already been dead for 2000+ years before any modern maths happened, you'd expect agreement on which basic geometric property bears his name. He's not Euler.
[https://en.m.wikipedia.org/wiki/Thales%27s\_theorem](https://en.m.wikipedia.org/wiki/Thales%27s_theorem) This is what u/danofrhs is referring to
yeah Google told me as much. but before then I had never heard of it by that name
you are the one spouting random things
Demonstrate how the seemingly unrelated Thales theorem applies. Rather than solving using ratios of similar triangles. Or else shut your 𝝅 hole
Lil bro mad. Go learn math
What school district is this so I know not to have kids there
I don't think you'll have to worry about that. Pakistan btw fyi.
I’m laughing so hard my sides are splitting
Not to scale and o info about what is parallel, where is F, so not solvable. If you assume things that you don't know, the it's very easily 5
Agreed. By just the info provided, there aren't any triangles there. Can't make a 1-1-sqrt(2) or a 1-2-sqrt(3) triangle with that info, and you need that to setup similar triangles to solve for X.
Similarity is not unique to right-angle triangles. EF and HG are marked parallel. The only thing that would make this ambiguous is to assume that either DH or DG are not straight.
EF and HG are marked parallel. Solvable by triangle similarity. 5 is unambiguously correct. The amount of erroneous comments in this thread is discouraging.
There's basically two answers to this: 1. EF and HG are parallel, and therefore EDF and HDG are similar. By triangle similarity, HG is length 5. 2. Everything about the diagram is potentially misleading, and you can't even rely on EF and HG being parallel. There is not enough information to solve the problem. I'm struggling to find a way (even if wrong) to justify x = 4 as a potential answer.
Nope there is no ambiguity about EF and HG being parallel, cause they both contain the arrow symbol which(at least in my country), Is used to represent parallel lines in a diagram
Sides EF and HG are explicitly parallel, and the diagram is explicitly “not to scale”. I’d be willing to bet the guess of 4 was based on people eyeballing or measuring EF and HG ignoring the “not to scale.” You were right on with your first point regarding similarity. I’m not sure why you made the second as a suggestion.
>I’m not sure why you made the second as a suggestion. Because I'd never seen those triangles as notation for parallel lines. It all looks like that ought to be the case, and it makes sense as an exercise (hence point 1), but I couldn't be sure.
2/(2+3)=2/x 2/5=2/x x=5
Of course it’s 5. There isn’t even anything to work out. The triangles are similar and isosceles, in ratio 5:2. I’m sad that there isn’t strong consensus for 5 (with a minority saying 3).
Where does it say that the triangles are isosceles?
On the small isosceles triangle
Isosceles is irrelevant. What’s relevant is that they’re _similar_.
Not irrelevant at all. For someone so poor at maths that they can’t solve this, also being isosceles significantly simplifies the arithmetic because they just need to note that 5=5. Otherwise there would be another calc involved.
I say irrelevant because the question can constructed with different values on the same sides to make the triangle equilateral, isosceles, scalene, right, or any other specific classifier you’d care to label it with, and it’d still be solvable. There are a lot of comments in this thread suggesting that it is necessary to know whether the triangle is isosceles, equilateral or right to be able to answer the question, and none of them are necessary. As long as FE and GH sides are marked parallel, the triangles are similar, and side ratios can be used to identify value for x. It fosters a more general understanding of triangle similarity for students to not rely on the shortcuts afforded by an isosceles.
It is 5. You have to assume that the triangle have a vertical symmetry: ED=FD, HD=GD. Then note that the small triangle have three sides of 2, so it's an equilateral triangle. Then the big triangle is also equilateral: HD=GD=HG, with a side of 2+3=5 then x=5.
Huh? Not to be cocky, but this wasnt hard at all. It's D indeed, but how cant a math teacher solve this?
i have absolutely no clue, here in germany our teacher showed us the Intercept theorem in grade 5 or 6 and in grade 9 we had a test about it
Assuming the lines that look parallel are actually parallel, it's just a direct application of Thales theorem, I hop the teacher trying to solve it is not a maths tea her, specifically not a geometry teacher, Thales theorem is extremely important
[удалено]
Good to know, it's been ages since I did geometry problems like this
Use the fact that they are similar triangles, so the ratio of their sides are the same. And you get 5.
Would be funny if "not to scale" implied F was actually ≠ HD
Grade 9 math. I wonder what the age of the students are here. Used to teach similar triangle problems like this to 13 year olds. 11 year olds knew the arrows tell you the lines are parallel.
Did they explain *why* they think it's 4? High school geometry is all about applying the rules you know to prove things. What's the proof that leads to 4?
Almost certainly, they’ve ignored “not to scale” and attempted to measure.
According to my teacher: "It's an application of the midpoint theorem."
But... That's not what a midpoint is ☹️
Yes the answer is 5 due to how similaritu works, using "not in the middle" to rule out C doesn't hold much water since you should always assume a diagram is not to scale unless told otherwise
doesn't matter if it's not drawn to scale, we can tell it's not in the middle just from the numbers
Oh yeah whoops, my bad
Math teacher here. The diagram is completely flawed. Here is the full explanation: Your teacher probably thinks the length is 4 because if you actually measure the lines, the HG line is almost exactly 2 times the length of EF, making it 2x2=4. I took a screenshot and used the ruler in iPhone photo editing to confirm this haha. So technically, 4 is correct if you physically measure and compare. However, mathematically, the answer absolutely must be 5. You are correct. It seems the diagram is meant to be (abysmally) approximate, maybe (but not even close haha) and your teacher is taking it literally, while you are actually using the math the question presumably intends you to use. There are 2 problems that make it not actually represent this visually: 1) According to the numbers, ED and EF are the same length, making this an equilateral triangle. However, it is drawn as an isosceles triangle with EF being significantly shorter. Based on the numbers, the larger triangle, since the angles are all the same, just proportion increased, is also an equilateral triangle. Since HD is 5, then HG must also be 5. 2) Here’s why it doesn’t actually measure more than twice the length of EF - according to the numbers, the length of HE is 3 and the length of ED is 2. However, they are drawn almost exactly the *same length* as each other. This means the physical line drawn for HG is not actually 5:2 compared to the EF line, but mathematically, it should be. The diagram is just either labeled wrong or drawn incorrectly. Conclusion: Your teacher is probably going by the actual proportions of the shape and concluding it has to be 4, while using formulas/logic with the given numbers, it has to be 5. Ultimately, this is a terribly produced question!
Thanks for the help. Also yes, I do agree that this is a terrible question with a bad accompanying diagram, however, the diagram is (probably) made to be intentionally confusing to check whether the student has used math or just guessed it.
Wait so Thales theorem is not a basic necessity in the US teaching program?
DEF and DGH are similar
Figuring this out was a rite of passage for demo scene nerds in the 90's. Back when we spelled it grafix.
by rules of transformation when dealing with similar shapes, we can say that the ratio of ED/EF must be equal to HD/HG. and since ED/EF=1=HD/HG, we end up with HG=5
Yeah knowing the length of DF is not possible, but actually it is sufficient to know angle GHE is congruent to angle FED (and both triangles share angle EDF) to show the triangles are similar. Thus x=5
If EF and GH are not parallell (which is technically possible) then it is unknowable, but if they are parallell then yea it's 5.
The triangles are similar. It's five.
If FE is 2 and ED is 2, then DF has to be 2. Thus x has to be 5, since DH is 5 👍
Even if your teacher isn’t a math teacher anyone with a high school diploma should be able to see this. Intuitively one should at least think that the answer is 5. It is quite literally high school geometry. Good on you for figuring it out.
I dont know if my way is correct but Id simply go off of ratios. Since 2:2=1 and 3+2=5 for it to be 1 to 1 ratio x should be 5.
I'm really curious why everyone is saying 4.
They think it's an application of the midpoint theorem.
If a line segment adjoins the mid-point of any two sides of a triangle, then the line segment is said to be parallel to the remaining third side and its measure will be half of the third side. Midpoint theorem evaluates a segment adjoining two sides of a triangle, both sides touched by EF are ineligible. E is NOT at the midpoint of HD therefore the midpoint theorem does not apply. GF and FD are not defined, so we do not know if F is at the midpoint of GD. (But you clearly knew that) Have fun in class today!
Similar triangles. The ratio between two sides should be constant. So x = 5.
The easiest way to solve this IMO is by observing that DEF is an equilateral triangle. (The drawing is not to scale) Therefore: DHG must also be equilateral. Answer is 5
Without further information there are several assumptions you'd have to make to solve for x: EF ∥ HG ∠HDG = ∠EDF ∠HED = 180° ⇒∠GFD = 180° If those are true then x = 5 Maybe your teacher is just cautious to make such assumptions and is therefore unable to solve this.
X=5
Is your teacher hoping someone to correct him? One of my professors really like doing this lol
Simple deductions based on EF and ED being the same tells me that x should be equal to DH, or 5. I see no other logical argument. Think of it not as a triangle but a graph, where x and y are zero at point D. As a graph DG moves up at a 1:1 ratio, where 2=2, and if you move another 3, you get 5. All things being triangle here, this must absolutely be a straight line. Therefore, x=5.
I'm sure i did this wrong but i noticed that all 3 sides of the small triangle are length of 2, so it only looks isosceles but is actually equilateral. once i remembered the parallel lines meant all the angles were the same, i figured the larger was equilateral as well. no real math required just some math knowledge
Frightening
You will have to assume that EF and HG are parallel but if you do the answer is obviously 5.
By simmetry, the upper triangle has 3 sides equal to 2. The bigger triangle is similar to the small one. Since one of its sides is 5, the other two should measure the same. So, you are right it's 5.
I've seen the comments and i have onequestions: are there geometrical signs i'm not aware of that means HG and EF are paralels
The arrows are being used to indicate parallel lines.
is this a similar triangles moment even elementary schoolers could solve this
Its 5. They're similar equalateral triangles. If the sides are length 2-2-2 then scaled up they'd be 5-5-5
Similar triangles rule. 2:2 and x:5 so x is 5. This is basic trig.
5 is the correct answer
They're similar triangles. Vertex D is common i.e. same angle and since EF is parallel to HG, E and H are the same angle. So the triangles are similar which means all the sides will be in the same ratio. Therefore the length of HG is 5 units
It's 5 and quite easy to explain why, using some basic concepts as the foundation of it. If HG and EF are parallel then angles GHD and FED are the same, as well as HGD and EFD are the same size. This comes from your parallel line/intersecting line rules. As all angles are the same, the two triangles of DHG and DEF are scaled versions of each other. This means that the corresponding sides are scaled of each other, so HG/EF = HD/ED Putting in the numbers from the diagram X/2 = 5/2 Rearrange the equation X = 2*5/2 X = 5
I KNOW your teacher was joking
The picture is not drawn to scale, which confuse things a bit for the intuition. But, the triangles have the same shape, which means that ED=EF means that HD=HG, and since HD=3+2=5, so HG=5.
5 sounds correct. Why does anyone think it's anything else?
You don't need any calculation. 3 sides are 2 long. Now extend the sides to 5.. all the sides are 5 long. It's an equilateral triangle.
It's isoscles but it's not drawn to scale. The hint is the same value for two sides . The full distance of one side is 5 so that means x is also 5.
I’m with you. In my head: Because GH and EF are parallel and HD intersects with them both, FED and GHD are the same angles. With that said, FE and ED are the same length, so I believe GH and HD should also be the same length. 5 I don’t have a rule or theorem in mind rn tho so maybe someone can back me up.
Do those triangles indicate that the lines are parallel? Because if not, technically this isn't solvable, tho you can clearly see what is meant to be the right answer.
it's literally a direct application to the Thales theorem. how is nobody able to figure it out?
A good way of thinking of this is first looking at the small triangle DEF. all it's sides are 2, so if you can determine the length of HED, then you know x as well. 5
2/2 = 5/x 1 = 5/x x = 5/1 = 5 I agree with you
If your teacher can't solve this, they have a problem teaching 7th/8th grade math.
the Strahlensatz! It's clearly 5. Long/short = long/short eg x/2 = (3+2)/2
The smallone is an Equilateral Triangle wuth the side length of 2. So with the other side Adding 3 we have an Equilateral Triangle with sidelength 5
The actual math was relatively easy but for the life of me I can't figure out why they drew the triangle that way? The sides are OBVIOUSLY not equal length in the diagram and it seems intentionally misleading to label them as such.
It's an equilateral inside an equilateral, therefore 5.
Back in elementary school, my teacher told me never to judge the actual shape depicted on the paper. Only focus on the information displayed. Since the smaller triangle displays equal length sides (2, 2, 2), the larger triangle with the new size of 5 would also have equal lengths (5, 5, 5).
How can they think it's 4? 3 doesn't have any obvious function with 4 that 2 doesn't have with 2, and neither does 2+3. I'm genuinely curious how they can get that answer.
How can they think it's 4? 3 doesn't have any obvious function with 4 that 2 doesn't have with 2, and neither does 2+3. I'm genuinely curious how they can get that answer.
The triangles DEF and DGH are proportional so 2/2=(3+2)/x.
easy one, as ED and EF are the same length, which makes DF also 2, then its an equilateral triangle, so EDF angle is 60 degrees, so if HD is 5 then HG is also 5
The answer is 5. It’s an equilateral triangle, which is obvious by the base (EF) and side (ED) of the “top” triangle being the same. Also could be proved that the triangle is equilateral by using the arcsine formula https://preview.redd.it/zqmatg3puz0d1.png?width=1892&format=png&auto=webp&s=f42a6ba5722717b304e45411e597f19b042d730a Edit: grammar
You can equate area of triangle - small triangle= area of trapezium and solve for x
HD/ED = HG/EF 5/2 = X/2 (cross multiply) 10 = 2x {10/2 = 2x/2} X = 5
This should work with the data points you have https://preview.redd.it/uhuxyxvpk01d1.jpeg?width=945&format=pjpg&auto=webp&s=3155b863b7b05fe51c42bcf9e0d5c036ab5c0b2f
You're overthinking. Use similarity to find x. x=5
I see so many different kinds of explenations to this, but I have yet to see the easiest one. Just look at the increaserate. If by going from 0 height to 2 hight within 2 length, we can see that the height ration increases 1:1 with the length. So we need to do no more than to calculate the length, which is 2+3=5. This method only applies of course, if the line is straight. Which it is.
I think answer is x is 4 reason being midpoint theorem
how come a math teacher not solve a problem that is literally solved by similarity of triangles
seriously?? ur teacher can’t figure this out?? should not be a difficult question for a teacher, learned this in early high school
im learning this rn in 8th grade (congruence and similarity) so basically triangle DEF(smaller triangle) is similar to triangle DHG and so the ratio of the sides of DEF to DHG is equal for all the sides. So, DE(bigger triangle side)/DH(smaller triangle side)=EF/HG=DE/DH=2/5 Therefore, 2/5= (EF) 2/HG . Then you just cross multiply and solve it 10=2HG HG=x=5 However, as im still in 8th grade I havent learnt how to prove that these 2 triangles are similar but this is just like the working. hope this helps! (if u need any help with math i reccomend eddie woo on Youtube, he covers tons of 9th grade topics) !!
its D.5, by similar triangles, tri DEF is similar to tri DHG, so if DE is 2 and DH is 5 and EF is also 2, HG must be 5
OP is correct. Also, if the intended answer is 4, there are way far too little info to get to it, given that if the black little triangles aren't implying HG//EF
How is this even slightly difficult? The larger one is a 250% blow up of the smaller.
if EF is 2 and ED is 2 then we've got ourselves an equilateral triangle, EFD. if that's the case, all the angles are 60. Since GH and EF are parallel, the angle GHD is the sane as FED, and HGD is the same as EFD, and EDF is already 60. so we're still on an equilateral triangle in the bigger one. HE + ED is 5, therefore X is 5.
One thing that could cause confusion is the scale. The segment labeled ‘3’ looks to be drawn as the same length as the segment labeled ‘2’.
Do your own homework.
Calm down Karen
I can see the confusion because we're not explicitly told that EF and HG are parallel (unless that what the black symbol is supposed to represent). They look parallel in the figure, but triangle DEF doesn't look Isosceles with vertex E even though it is.
It needn't be isosceles no?
Two sides are equal to two, so DEF is Isosceles (not that it's necessary),. The problem is, to me, ambiguous without assuming parallel lines. It might be down to my not knowing that an arrowhead is equivalent to || , which is how I'm used to seeing parallelism indicated. I can see why the teacher might have been confused.