"99 Percent" Miss This. What Is The Length?
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- Опубліковано 15 жов 2024
- It may not be exactly 99 percent, but many people will get the incorrect answer. It's a great teaching opportunity. Learn how to solve this problem! (There are multiple ways, I am presenting one valid method).
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Great! Thanks for the shout out. My only thought at the end was that people might be left thinking "so why did it give the right answer before?" It might be nice to point out the coincidence of the original height of 4.
Thanks for the feedback! I will think about this, but I haven't come up with a good explanation really! In the original problem, I did notice the answer was 8 = 5 + 3, so I did try to do the "shortcut" way of drawing 3-4-5 triangles. But I didn't have a convincing proof of why the horizontal lines were coincident and also parallel to the top line, so I figured it was a coincidence. I then suspected someone would find a valid proof. So I was surprised when so many people merely used the incorrect method without proper justification! I suspect someone will email me a good reason.
@@MindYourDecisions You can see why it must work if we build up the shape differently. Start with a 4x5 rectangle and build a 3-4-5 triangle on the right side and another to the bottom edge to match this shape. Then remove the bottom and right edges of the rectangle. We've now recreated this shape, save for proving that the angle on the bottom right is 90°. But that's a simple proof. That angle is the sum of two angles we created. the right side is the 4-5 angle from a 3-4-5 right triangle. The left side was the complement of the 4-5 angle of a 3-4-5 right triangle.
We now have enough information to conclude that the original shape and our constructed shape must be identical. If we try this again with the 3x5 rectangle as our starting point, we end up failing to prove that the bottom right corner is a right angle, since it would be the sum of a 3-4-5 right triangle's 4-5 angle and the complement it's 3-5 angle. Which cannot be 90°.
@@MindYourDecisions I don't think there's a good proof for it other than that you can attach a 345 triangle to the right of the 5 line by extending the 4 line of the smaller 345 triangle by 3, so that you always get a 4 line on the right side, now if you say that the left line is 4 it just aligns which it will not do for any other number
Not true. The two "3" sides on the left side of the problem with the 90 degree angle mean that the remaining side of the short triangle is zero, hence, the correct answer to the problem is the square root of 41.
@@MindYourDecisions Well, I have found the condition for the lengths. Assume we want to find the height (in the first video 4, here 3) so that the "wrong" solution is right. Then the height = (the length of 4 from video)*(the length of 5 from video)/(square root of (the legth of 3)^2 + (the length of 4)^2). You can put any numbers instead of 3, 4 and 5 and find the height. It is all based on similarity of triangles. I can try to write more proper solution once I am back home.
Lets assume you have height of 3 and legths of 3,4 and want to find something instead of 5. According to my solution the missing length is sqrt(3^2+4^2)*3/4 = 3,75. So in this video, if the last length is 3,75 instead of 5, the incorrect method will lead to the same solution.
In fact, I have found the condition that the incorrect method produces the same result as the correct method ;-)
Always take that "Diagram not to scale" warning seriously
Gotta learn that the hard way
But scale isn't the same as proportions.
I always thought it just meant you can’t use a ruler 🤦♂️
@@kubush Yeah original plot is very misleading.
If it's not scale, it's false, he changes the problem after asking the question? And I thought these were honest questions?
The error is 99% of people who tried this, assumed that you drew it correctly in the first place.
@@Mausoc I don't think you understood the error, then... The problem is not the answer but the used method.
The "99% error" is assuming that a line equals 4 when there's no reason to, as you find the same answer no matter the actual length of the lines
Ah, therein lies the lesson. It's okay to assume, but verify your assumptions are correct.
He changed a number in the problem and got a “wrong” answer lol I wasted my time watching this
I just saw that not to scale was acceptable, so my answer was also not to scale.
@@willhanlon5863 Did you even watch the video? People were using a method that just happened to turn out correct in that specific case. As shown that when you change the length of any line, the method doesn't work anymore. It's like when a kid sees x^2 for the first time and the first example says, 2^2 = 2 and then assumes that 3^2 = 6. Theres no proof given at any stage that moving said line would connect to the 2nd right angle, it's an assumption people made and the answer just happened to fit.
Ah yes, the "not to scale" diagram. Another one of those things teachers like to laugh about in the break room after everybody failed the test.
It’s kind of a dirty trick tbh
@@bobbytrap2554 not really, it tells you that it's not to scale and it's expected that you actually "solve" the problem instead of just using a ruler.
Yes, but this test gives a good insight into maths. It's like hooping a basketball, we're never as accurate as we believe we are.
@bobbytrap surely not using a ruler is why you fail this question. if you were to measure it on the page you would see that the lengths were totally off and that would definitely prompt you to work out the answer using the numbers.
@@ryansmith9138 Yeah. My problem was I have no grid paper, ruler within reach and I wasn't going to wait ten minutes to open up an art program and try to make a grid in it and then draw in it just because someone else can't be bothered for their own UA-cam video and just recycled the diagram from a different problem.
I did find these though:
www.ginifab.com/feeds/angle_measurement/
www.mathsisfun.com/algebra/trig-solving-triangles.html
FOLLOW-UP COMMENT: (BTW, I DO LOVE YOUR VIDEOS.) I wanted to see why we got two different answers - I obtained 8 (in two different ways); you obtained sqrt(71). While your *method* is correct, your answer of sqrt(71) is for the *revised* problem where the left most vertical line segment is of length 3. With the *original* problem, that segment had length 4 --> so using your logic, your answer for the *original* problem would be x^2 = (sqrt(80))^2 - 4^2 = 64 --> x = 8, which was my answer (along with a few others.) Perhaps you could add a comment at the end of your video, since the length of interest was labelled x in both cases.
I think a useful way of looking at this problem is by transforming it into a simpler problem that most people would recognize. In this case you can make the shape convex by flipping over the first right angle lines thus creating a quadrilateral. From there most people would realize the method for calculating the side.
As many have said the question itself is somewhat misleading, but because I work as a 3d draftsperson I can promise those questioning the drawing that sketches from customers (boss's) can be much worse. If you work with only known information instead of implied information you are much more likely to give the customer what they want.
Oh, that's easy. It's 12 1/4" - I measured it on the screen.
Whit what
@@ashirbadbehera4129 Whit a whruler.
My screen is smaller than that, so?
@@jlnbroadcast aouko
Your ruler must be wrong. My ruler measures it ~6 1/2.
When the test says "Solve for X" and then purposely shows the wrong diagram, they might as well just show a picture of a honey baked ham.
The diagram is not wrong. The diagram shows measurements you're supposed to use (lines with dimensions) and lines that are meant to be perpendicular (the symbol). It was never meant to be a tool to measure everything. It does serve its purpose properly. As it's somewhat my field (construction, architecture and the like), I can say that the constraints are set properly and there is one solution that does work universally. The issue was that many people assumed proportions were proper and they tried to find a solution that was not following the constraints given.
LOL
Plus, I was thinking this is solvable simply in your head, and not needing a square root table or SINE, TANGENT, COTANGENT tables.... This is not math, this is DECEPTION on part of lamo MIND YOUR DECISIONS>
@@yousurf374 all you need is Pythagoras rule and a bit of logic. You don't need to collapse the square root to know if you're right
@@europpe if you left the answer as squarer root of X, as the answer, you would be marked UNSOLVED... as the basis is in real numbers. I solved the original one, where he says proof method would be wrong, or less correct. IT was correct, regardless of his MORE correct method, as given... It was a whole number solution.
Now, my issue was that he did not state answer can be yielded as want for square root of a number, or that calculator could be used, with a square root function. Regardless of being able to get there without a calculator.. I get that part, but he did not fully solve this second iteration. My Advanced statistics professor would have given him a zero for the solution and even no partial for his solution/math method... as he did not fully solve. He was a harsh professor. I feared a question answer on a final exam as possibly being wrong, that I had erased the entire thing, as he marked number correct vs. number of questions asked... his final had only 8 questions, so I tried to increase my final score by removing it.... but later found out I was actually correct and should have left the math and final answer.. I had doubts that got the better of me... he caused me ANXIETY!!!! LOL.
@@yousurf374 no. If you solved the question using the method that was shown as incorrect in this video then there was a fundamental mistake in your assumptions. The fact it actually leads to the correct number is a coincidence caused by an unlucky choice of variables.
I took a different approach at it (which might seem more difficult)
once you find that the hypotnuse for the first triangle is length 5, you can notice that it creates an isoceles triangle with the base as the hypotnuse for the bigger triangle (if that makes sense), next you find the angle which is 90 + sin^-1(3/5) (i know, its unnecesarily complicated) and using the cosine rule you get a length of 80 which gives the correct x value
Its nice seeing a different method, I was pleasantly surprised!
+1, I comes this method to mind at first, though it is a little brutal not elegant
THE DIAGRAM IS ALWAYS WRONG.
EVEN WHEN IT'S RIGHT, IT'S WRONG.
Academy, Khan
Yap
This
Stop shouting your post in all caps.
As a crosscheck, you could consider this which doesn't consider the construction from bottom left horizontally to the junction of the 4 and 5 lines.
1) Start by adding your own construction line from bottom left to top right. Now, as before, slide the '4' length along to the top right vertex creating an overall triangle from the original "z" shape.
2) You now have two triangles based on the same base, (the current overall hypotenuse). They both have a side of length 4. They also both have a right angle in their upper left vertex. These two triangles have "Right Angle + Hypotenuse + 1 Other Side" congruency. The other two matching sides must therefore be the same length.
3) Just as you say, in sliding the line up in step 1) we created a side of length 3 + 5 =8. This is the equivalent to the side X in the other congruent triangle. Length X therefore is 8.
It's even simpler if you use the original value of 4 on the left hand vertical length.
Move the diagonal 4 to the other end of the diagonal 3 (as Presh did) and add the 3 to the 5 to get a side of 8.
You now have two congruent triangles (side of 4, right angle and a shared hypotenuse), so the top length is 8 too!!
Correct answer: The title should be ' 'Here's a deliberately misleading problem see if you can work out why'.
Bingo ... you nailed it!
This is the kind of BS that really weak teacher's use to "appear" smart to their students.
@@leahcasey2678 Not really, this teaches critical thinking. Like if you see "diagram not to scale" why would you assume the enclosure forms a perfect rectangle?
Heck even if the diagram looked super super close to being the same length, how do you know there isn't a .1 degree difference in the angles which would throw off your answer (without working out the geometry first)
A student shouldn't feel "tricked" if they get this wrong, they should realize they made assumptions that weren't true. Cause you know doing that in the real world is what gets you in deep poo poo.
@@leahcasey2678 This illustrates an important point of mathematics; you can't trust that something is so just because you assume it is so. There are many cases where mathematicians have demonstrated hundreds of thousands of cases where their theorem is correct and they cannot find a case where it is not, but they cannot prove it using pure mathematical logic and thus it is as "good" as any random statement.
@Mike L @Vesa E but we are to assume the diagram is on a flat two dimensional surface?
@@myclamish If this is a teaching moment, then they have to make it clear from the start the lesson is about just that. Because it looks like a geometry lesson. Had it been a trig lesson, maybe the student would have tried more trigonometry-centered solutions. You're right, we shouldn't assume things in mathematics, and it's an important lesson. But the teacher is resonsible for pointing that out. A "gotcha" math problem teaches nothing and only serves to frustrate a student.
Thanks for leaving “drawing not to scale” out of the thumbnail
Even if that note was there I would have noticed the different lengths of the two "3" lines before I read the note.
@@stevenvanhulle7242 No you wouldn't have. 1:23 shows they were drawn exactly the same which was the point.
@@brandonfrancey5592 Except that, in the thumbnail, the leftmost side is given as 3, whereas at the start of this video, it's given as 4...
@@brandonfrancey5592 Uhm, yes. It was obvious really. When the angle is 90°, the opposing side cannot be the same length as one of the adjacent sides. You see that at a glance
@@stevenvanhulle7242 Even when I know not to trust anything that's not specified, I was still saying 'that is not a triangle' and 'that is not a parallel line' over and over again because I kept thinking they were.
didn't think about rearranging the lines like that! You solved this so much more elegantly than I. I wrote out a system of equations then spent half an hour solving them 😆
I think that's where the illustrated approach really helps, a lot of people (including myself) think of the algebraic approach first, without thinking about possibility of rearrangements.
you still solved it !
You must've used the 3:5 ratio, right??
The scale of the figure is the most deceiving part of geometry. Most people use wrong procedure because they seem the right way in the given scale.
Apparent scales are completely irrelevant - you should ALWAYS work with actual numbers, variables and geometric relationships.
If you said that something was twice as big as something else in a Geometry test just because it looked twice as big then you would get 0 marks on that problem.
@@Peter_1986 There are some problems with that. First it's just how the human mind works. We see something and immediatly make connections. Second is that even if the scale is wrong it's generally just smaller or larger, the relations between each number stays the same. And third, if the scale is wrong to this degree then why even bother with a diagram? May as well train people to draw out a figure from the numbers of the text. Personally I think the diagram should always give the right idea and what a person should do is to find a way to prove it mathematically.
@@shadowcloud1994
Yes, a diagram should preferably resemble its intended values.
For example, most people would get very emotional if they saw an image of an angle that claimed to be 90° but actually looked like a small acute angle.
Those diagrams are pretty annoying, and can cause some confusion.
Perfect example when drawing scales don't matter
@@Peter_1986 I agree there's no point of drawing a diagram if it isn't intuitively correct. Your example was spot on. The diagrams should clearly indicate that the side is not 3.
“Triangles are not always right (although they are never wrong)” - Khan Academy
Hehe I like that
that 'right' means the right angle, right?
@@user-matlee2477 right
@@RummanNaser I didn't know until I watch the video that "right" (American) translates to "Right-angled" (English).
I guess you never stop learning.
diagram not to scalene triangle ~not khan academy
As long as the two sides of length 4 are equal the answer will be the sum of the other two sides. Take the drawing from 2:52 and slide the side of length 4 down like you did at 1:30t o form a rt triangle whose sides are 4 and 8. If you draw the remaining two sides of length 4 and x, you have a sort of diamond that can be split down the middle, into similar triangles: the two sides of length 4 match and the right angles match, and so the side of length x matches the side of length 8.
"diagram not to scale" is a cheap way to make a problem way more difficult than it would be in a real life situation
Actually when we sketch things in the field, we don't do anything to scale, but solely rely on dimensions and shapes...we nice it
up when we get back to the office. That's the way it is in the real world of survey and engineering.
Is this a Steven Universe reference?
@@Wormhole798 yes, and presenting a problem for a student should be "niced up". you sketch when you already have the skills to unsketch it, not beforehand
@@beevee7078 i dont know what that is, so no
Yep, stuff like this is why I lost interest in mathematics, after taking part in competitions featuring problems just like this one. Instead of being in any way clever, far too many people rely on cheap shots like this that would get them laughed out of the room in a real world situation.
More direct proof: the 3 and 4 at right angles form two sides of a rectangle. Complete the rectangle. The other length 3 side is collinear with the 5 because 90 plus 90 = 180. Erase the original two sides to obtain a quad with sides 3, 4, 8 and x and right angles between 4 and 8 and between 3 and x. Divide the quad into two RA triangles. Apply Pythagoras: 3 squared plus x squared = 4 squared plus 8 squared. So x squared = 71. Solved.
Your method is more efficient 👍🏻
"Diagram not to scale" literally means that the distances writetn in it are not equal to the actual drawing.
IT DOESN"T mean "angles are not correct in the drawing".
No it doesn’t. It is ‘not to scale’ I.e not a scalar drawing. If the angles were the same it would be the same shape just scaled down or up. I see your point but a ‘not to scale’ diagram doesn’t need the same angles or you could just work out the enlargement and figure out the final length.
Drawing the image out of scale with line segments longer than they actually are is a dirty trick.
Agree .
It's so disgusting! That shape is so freaking weird since.It is "not to scale"
As someone with a PhD in physics, I'd give this video a bit of a thumbs down. Every problem has a context and it is important to convey that clearly. The English language and the tools that we have developed to diagram problems always go along with a set of assumptions because as humans we tend to be imprecise in our communication. For instance, I am assuming that the sides are straight even though the author does not mention this. And it would be standard to assume that the author intends the left side to be assumed to be vertical and the top to be horizontal. The words "Diagram not to scale" is not as clear as "The bottom 2 points are not necessarily on a horizontal axis". Or, sometimes one would draw a dashed line slightly above the bottom left point and towards and meeting the bottom right point to clearly indicate that the 2 points might be assumed to be different. Otherwise, it is trickey and not math or science.
It's not a dirty trick. If you don't assess the validity of the information you think is available, you will reliably miss the solution. Whether malice from someone or not.
@Jan Lenz Yes, it is important to learn not to assume. To be fair, this is the first "MindYourDecisions" video that I have watched - so I don't know how common videos like this are here. But, in my experience many people who teach these topics tend to significantly overplay problems with purposeful deceptions and I've seen hundreds of students get turned away from the wonderful study of science and math. They fall for these repeated deceptions and then think they can't can't handle the subject and drop courses. If the diagram made it clear that the 2 bottom points were not necessarily at the same vertical position then you still would have had a challenging little problem to many people. They would have gotten stuck, but then pondered the options and developed theirs skills a bit while trying to solve it. Nature should be a marvel to learn - and it is not a box of tricks.
Whenever something is not drawn on scale, never assume a "horizontal" line is horizontal.
it kinda has to be since there are 90° marks
@@nathanyam2310 So, because there are three right angles you deduce that the lower two vertices are at the same height? That's so... odd.
@@stevenvanhulle7242 well no that isn't true.
@@stevenvanhulle7242 look at the angle between the side lengths that are 3 units long, that angle can be any angle since it is not defined. So then those two lower vertices can be at different heights but you can't really determine where they are located based on the graphic.
Horizontal is meaningless anyway, it's just to describe a line to talk about it. You could also give it a name like 'a'.
Thanks for drawing it in such a purposely misleading way.
*Really helpful*
It is very interesting for a person from a completely different country to develop the exact same process as me. Even without translating, the video was understood. Math hooray!
Yeah same
how are u typing english then
@@doraelog690 I used a translator after writing down Korean.
@@TheHorrorsPersistButSoDoI I don't quite understand what you're saying. What is Collatz concepture?
@@TheHorrorsPersistButSoDoI Oh, I see. Thank you for your kind explanation of the Collatz conjecture. I hope someone who can answer your question will come!
I gazed and contemplated for 15-20 minutes at the initial question of this video. Thankfully I realized the fact that the easiest way to solve this one is to imagine the hypotenuse of the 4 and (3+5) right triangle and then relate it to the hypotenuse of the right triangle with legs 4 and x.
Using Pythagoras (GOUGU 😊) Theorem,
Initial case :
(x^2) + 16 = 80
x = 8
Thumbnail case :
(x^2) + 9 = 80
x = √71
At the age of 57 having not done much maths for 30+ years i did get the right answer!!!.....Yeah me!!!!....
yay !
gg
Snap.....it’s something to do with being 57, 😝
No offense sir but this is a 6th grade problem...
@@ilovehomies in which country
First, find the length from bottom left point to upper right. That's the hypotenuse of a triangle whose sided are 4 and 8. That's Sqrt(80). Using that number as the length of the hypotenuse of the upper left triangle and the measure of the side given (3 or 4), you can find x. If the left side is 3, the answer is sqrt(71). If the left side is 4, answer is 8.
I did this all mentally.
Pausing at 3:53: The length of X is nothing because the shape you've drawn cannot exist. A right triangle with a hypotenuse of length 3 and another side with length 3 is impossible. One of the sides has been labeled with an incorrect length.
exactly
You’re right but that’s not the point of the problem
I'll assume you meant 2:53 because I can't figure out house to pause a video 5 seconds after it finishes.
But I still don't see a right triangle with two sides of length 3 that you describe. If you're talking about the two 3 lines on the left side at 2:53 then nothing in the diagram says those two lines form a right triangle.
I even sketched it up in Fusion 360 and it worked fine and gave me the correct answers for both versions of the problem.
I did this:
Let's say that the angle between the sides of length 4(or 3) and 3 is "a"
So
3.cos(a)+4.cos(a+90)+5.cos(a)=4(or 3)
Once you have a then:
X = 3sin(a)+4sin(a+90)+5sin(a)
"once you have a" :D
@@Samurook yes...is easy with this identify: ua-cam.com/video/G5LEOplnBxg/v-deo.html
Well I chose to solve this for a non-euclidean space since that wasn't specified, so the answer is 9.
That's like saying "well I choose to focus my exercise on my butt supporting muscles, which is why I sit on the couch all day". In this case you gain points for being a smart alec, but you don't actually improve your mathematics skills, which is the point of the exercise here.
in most of the non-euclidean spaces, the 3/4/5 rule is wrong. So in my opinion, if you truly want to solve it this way, you have to describe the properties of your space that give an answer equal to 9 in this problem. good luck
@@DinoDiniProductions Its more like saying "you give a purposely misleading question so I give a ridiculous answer" - and thats something I am very fond of. The whole video has the ring of "being a smart alec" to it, he should be able to take his own medicine.
I did it in my head by just using the thumbnail lol
What’s that
I simply drew the diagram and added a rectangle around the sides length3 and 4. It then become obvious that you have two Pythagorean sums to do with the first having right angle sides of 4 and 8 (3+5). With the 4 3 4 5 version you then have two identical triangles so the answer is 8.
>trying to solve with trigonometry
>cant
>"ok, ill just move these two lines"
>answer becomes obvious
but don't give it .
I just added 5 + 3 = 8 and then we get a right triangle then use Pythagorean theorem and find x
I didn't move the lines. I just saw, that the "4" line segment can be divided in 3:5 ratio
We should all thank him for teaching us to pay more attention to the diagram not to scale
No, we should all resent him for giving a deceptive problem. That goes beyond "is not drawn to scale."
@@robertveith6383 "deceptive" assuming that the height equals 3 is the reason why everyone needed this lesson
@@robertveith6383 it was a mistake lol. Everyone makes mistakes, he clarified it.
The problem with an initial length of 4 was more interesting because, when answered correctly, 8 contained a hidden trap. It was also necessary to prove that the length of the vertical segment from the top of the lower right angle is 4. But she had another beautiful solution. We will complement the right angle with sides 3 and 4 to the rectangle and break the resulting structure into two equal right triangles (according to a common hypotenuse and legs 4). In one of them, the other leg is equal to 8. Therefore, the desired segment is also equal to 8.
It was the first version of the problem that I proposed at the Ivano-Frankivsk Regional Olympiad in Mathematics for 8th grade students in 2020. With many correct answers, only one student gave a complete solution (with a justification of length 4 for a vertical segment). Others fell into the laid trap, or did not solve the problem at all, because they had not yet learned the Pythagorean theorem. The solution I proposed was not seen by any of the participants.
I got 8 as well, i'm annoyed because I paused it as soon as the video started, solved it then played it to check my answer just to see him change one of the values. It was so frustrating but I was relatively confident in my answer due to how round of a number the answer was
8.
Extend side 5 by 3 and connect with corner between 3 and vertical 4. The length of the connection is also 4, so x is equal 5+ 3 which is 8.
In the original task, moving two straight lines (4 units and 3 units) in parallel results in a kite. The two short sides of it are 4 units long, one of the long ones is 8 units long. Consequently, the other long side must also be 8 units long.
I would draw an extra line from the 4-4 vertex to the 8-8 vertex and call it two right triangles just because more people can understand right triangles. But yeah, obvious solution is obvious.
The way I thought to do it was to start, again, with the 3-4-5 triangle. Given all sides are known, you can employ trigonometric functions to find all angles inside it. Lets call the left one L and the right one R.
The given diagonals, lengths given as 3 and 5, are parallel as they're shown to be 90 degrees from each other. And they both intersect with lines (the leftmost line originally a 4 turned to 3 in the modified example, and the line of distance x we want to find) that we know are orthogonal. Quick geometry shows us that the bottom left angle and the top right angle sum to 90 given they're parallel and orthogonal counterparts. Lets call the bottom left angle B and the top right is 90-B.
If we turn the image into a trapezoid, and we know the too left angle and the two angles formed by the 3-4-5 triangle, and recognizing the relationship B and 90-B, we can use the angle sum property of a quadrilateral to determine the value for B.
Given B we can find the angle 90-B and then draw a line parallel to x at the bottom. We know the angle between X and the right most line and can inflect 5 into the x direction. And we can inflect the hypotenuse of the 3-4-5 triangle's side (length of 5) using the angle found as 180 - (90-B) - 90 - R into the x direction and sum the two to find X
8 or sqrt(71) depending of the length of the first edge (sometimes is 4, sometimes is 3 in your video).
If it is 4 then the right answer is 8
@@ilyshi Thank you!
I got x=8 for the first diagram with a height of 4. There are two congruent 345 triangles, so x = one hypotenuse + one short side = 5+3 = 8. Making the far left side 3 means making the rightmost angle of the polygon more acute and lengthening x, so square root of 71 makes sense.
Whenever there is a "Diagram not to scale", it means that the lengths arent drawn to scale. The actually shape configuration should be the same, same angles and side proportions. Cap', this is cap.
Wrong.
When you can't even figure out the wrong method...
hahha! pure comment bro!
Haha
Haven't watched either video yet, found the solution!!!!!!
1: Label each point 'A' through 'E' starting at the top left and going CCW.
2: Connect points B and E with a line.
3: Draw a 4X8 rectangle (Length Wise).
4: Divide it into two rectangles: 4X3 (Left) and 4X5 (Right).
5: Label each point ''F' through 'K' starting at the top left and going CCW.
6: Connect points F and I with a line.
7: You should be able to figure out the rest from here, but if not, continue reading.
8: Find length of line 'FI': (4^2)+(8^2) = 16+64 = 80; FI = sqrt(80)
9: Fl = BE = sqrt(80)
10: (x^2)+('AB'^2) = 'BE'^2 = (sqrt(80))^2 = 80
11: x^2 = 80 - ('AB'^2) ; x = sqrt(80 - ('AB'^2))
** {AB=3, x=sqrt(71)} {AB=4, x=8} **
Another variant - finish triangle, with sides x, 8+a and 3+b. From little triangle, a2-b2=-16. From little and big triangle similarity, a/b=(3+b)/(8+a)==8a-3b=16. And for finish for the big triangle x2+(b+3)2=(a+8)2, calculate and keep in mind previously results get x2=71
This is an intentional ace breaker. In college, an instructor tried to bust everyone with a similar an ace breaker. I used a scrap of paper to make a "ruler" to draw the correct scale, showing the actual solution was very different from the implied solution. I drew the correct scale object next to the incorrect version in the test. In so doing, I was accused of cheating on the test! After appealing to the academic review board, the problem was thrown out and grades were adjusted. The instructor, after further investigation, had used other ace breakers on previous tests maliciously to kill students chances for 4.0 grades. His excuse...no one is perfect, so no one should get perfect grades. He was dismissed at the end of the year.
And everyone clapped
Teachers like that seem to miss the point of teaching. It is to share knowledge and take pride in the success of the pupil. It is not a freaking competition. I had a few of those teachers, especially in math class. Just miserable people bro :) They don't deserve the privilege of teaching.
Cheers to you for applying to the review board and getting rid of the teacher. Your resolve and determination saved a lot of future students.
I was the guy who got those problems and invalidated the "but this isn't fair because I got sucked into doing something wrong because I'm less smart" people's whining...
The point of the exercise is to learn to validate how robust your assumptions about your methodology is. Otherwise you're approaching the people who YOLO their understanding of math and think they're getting ripped off buying a 1/3lb hamburger ;)
@@thomas.m5223 Yup, teachers like that forgot the point of learning.
"Hey, you're gonna learn math! You're either here to learn or to use the degree for a job position!"
>So I made this test where problems are as annoying to actually understand as possible, and no diagrams are actually to scale just to mess with you. This is meant to truly let you get your hands in there and show all the math you've learned! Also, question 5 is written in Mandarin.
Cool Story Bro.
The original image is clearly intended to trick us. So of course everyone would assume the second triangle has the same height. This one is disappointing.
no it isn't
This video (without the previous) just ended up in my list, so I tried it, but took a slightly different path (probably already shown in other comments): the missing part of the triangle can be seen as half of a rectangle, so you end up with a trapezium with opposing right angles. This can be divided in two triangles with right angle, both with the hypothenuse of sqrt(80), giving the same results as in the video.
I think this is the simplest solution
You can also overlap the 4 with the 5, and you know if you follow a straight line over the 3, the length will be 5, that will make the undefined line sqrt(50)+4. Since the triangles overlap, you will have to use pytagoras. 1^2+2^2=5. and subtract sqrt(5) from( sqrt(50)+4) and you`ll have the answer.
You could also criticise how the incorrect proof is based on the (unproven) assumption that the line between the points (the bottom two at 2:08) will be parallel to x.
The diagram is not to scale.
The diagram is wrong and meant to mislead you like all mathematics.
The diagram clearly states everything you need to know; all necessary side lengths and all necessary angles are known. Assuming anything that isn't explicitly stated is your own fault.
@@kjerins that's like giving a student a blank paper and then marking them on knowing whether it is white or not. Just another 3rd finger question.
My first approach : x = 9 I have used the 3-4-5- rule in a couple of right triangles
But proven wrong when I watched the video further. Good trick
Extend length 4 to left top corner, equilateral triangle of side 3,
Now consider length of big triangle formed, length will be 4+3=7, other is 5, and is a right angle triangle of length 7, 5, and hypotenuse x.
X= root square of 7 plus squire of 5.
49+25= 74
X=root 74=8.6
for left hand line length = 4, the answer is 8
for left hand line length = 3, the answer is (squrt of 71)
It is simple, the construction of the rest of 3,4 & 5 is defined.
"The "diagram not to scale" makes this a trick question, I think.
ua-cam.com/video/yu8XiGCjhcY/v-deo.html
in mathematics and geometry, no diagram is ever assumed to be to scale unless stated otherwise. the only things that are assumed to be correct are the given angles and values as well as any explicitly named properties, such as "construct the line to be a tangent of a circle" and such.
diagrams are typically drawn such that all relevant information is presented clearly, and that sometimes means drawing things not to scale.
the primary reason for this is because when a mathematician constructs a diagram to solve a problem, they may not know the exact dimensions ahead of time, for example because it could be that the dimensions are precisely the thing that the mathematician is trying to work out, in this situation it makes no sense to expect the diagram to already be to scale, it makes a whole lot more sense to draw the diagram such that it is easy to write in all relevant information so it becomes easy to overview.
No, in geometry problems, it is standard to not give a to scale picture unless otherwise mentioned.
actually saying 'not to scale' means use geometry to prove it . (assume nothing)
In an exam whilst mathematicians calculate all the data if it were accurately drawn to scale candidates would just get the ruler into play.
Real world example of that figure our country's flag 🇳🇵😂
The precise way to draw the Nepalese flag is actually written in the constitution and it’s far more complex than this. Numberphile has a video on it
Only Nepal has unique flag in the whole world!🤩 Other countries flag is rectangular in shape.
Yeah
At 2:18 the two triangles are similar (all angles are congruent ) so their are in proportion to each other
•1,5 + 2,5 =4
•3/5 =1,5/2,5
Small triangle sides length : 3 , 1,5 , x
Big triangle sides length : 5 , 2,5 , y
We can find x,y through Pythagorean theorem
then x + y is the blue line
Then Pythagorean theorem again 2:55
That’s it we found upper side!!!
Good problem. I eventually got this right, but I only got a decimal approximation because I used the sine function a lot. Always nice to have many ways to get to the solution.
I was grinning the moment i saw the wrong answer...it was exactly what I thought..
The left side is NOT 3, it is 4! (given by your original question) This gives x = 8
This massively annoyed me too. The length changes to 3 and then never goes back to 4.
I first thought that to solve it, we first join the bottom most vertices and form a line segment of 5 units
Then draw a perpendicular from there and then it forms a rectangle so that perpendicular becomes 4 units and the second part of x becomes 3 by Pythagoras theorem so answer becomes 5+3=8
I also got the answer 8 as I tried to form as many right triangles as I could ultimately I got an equation y²+8y+9= x. Then I got the value of y which is 3 and got final answer 8.
The second he mentioned the "wrong method" I knew it would be the exact way I solved it in the thumbnail
Same
I guess the "99 percent" in the title is "not to scale" either... :P
These problems are trickier when you mess with the dimensions. That tends not to happen in real life. Most people would (I suggest) have got the correct answer if you hadn't gone out of your way to deceive them. More like psychology than maths.
But had the length of the side been 3.9 instead many people would falsely assume the bottom hypotenuse to be parallel with the top line because it would look really close to being parallel, and then the answer would be wrong.
The issue with thinking you can rely on the visual representation is that the brain always tricks you visually. It tries to see balance where there isn't by giving different weight to different shapes and lines. Typography is a good example - individual letter forms do not line up at the geometric top or bottom of each in a word or sentence. Yet the brain (and eye) sees them as aligned. People got the correct answer because the question was badly conceived. They essentially made a guess based on assumption. And the proof it was badly conceived is this "3" alteration.
@@bro748 Nice
Thinking x=8. If you finish out the 3 x 4 box at the bottom by extending the 5 line down 3 more then a right angle 4 long back to the other 3 line. That's a 3 x 4 box with 90 degree corners. If you draw a line from the bottom left apex where the initial vertical 4 line meets the 3 line up to the top right corner. (x and 5 meet) At that point you have two identical right angle triangles with the hypotenuse being common. The top left one is 4, x and the hypotenuse and the bottom right is 5+3extension, 4 at right angles, and the same hypotenuse. Since the triangles are identical the x = 8 which is the sum of the 5 and 3 extension.
I'm a complex man. I see a math problem, I click
There are no imaginary parts, so a better definition is "real man" 😁
it took me 10s from graphic to know its 9
or am i wrong?
Poo in the loo 👳🏿💩🐮🇮🇳
I almost think this video should replace the previous one... It might also be worth mentioning how easy it is to make false assumptions by relying too much on the way the diagram looks.
Why is this comment 2 weeks old while the video was just released
How?
@@detaqwer Probably Patreon supporter. They have early access to videos.
@@detaqwer Presh gave a link to this video three weeks ago in the comments to the "previous video": ua-cam.com/video/amLDMD8KPzQ/v-deo.html. He invited us (or at least Michael Timothy, whose idea it was) to comment on it.
My father often told me something one of his math teacher said... "a good mathematician can think right on a wrong diagram!"
I checked the heights before I did it the "wrong way".
If cosA=4/5, then the height on the left would need to be equal to 5cosA (which is 4) for the hypotenuse of the bottom 3-4-5 triangle to be parallel to x.
The non-vertical length '3' and the length '5' are parallel - the length '4' is perpendicular to both. Slide the length '4', keeping it parallel to itself, such that the end joining the '3' goes along the entirety of this latter length, effectively 'shrinking' it to zero and 'growing' the '5' to now '8'. Join the two non-right-angled vertices of the resultant quadrilateral with a straight line. You now have two right-angled triangles sharing the same hypotenuse and with remaining sides 3, x and 4, 8. Therefore 3^2 + x^2 = 4^2 + 8^2 or x^2 = 71.
Scale and proportion isn't the same thing! You can have an accurate diagram that is not to scale but has correct proportions.
the english language phrase "not to scale" indicates that two or more parts of the diagram do not use the same scale; thus, the proportions on the diagram are wrong.
@@sirmoonslosthismind In the English language, "scale" and "proportion" have specific definitions and aren't the same thing. "Not to scale" doesn't necessarily mean the proportions are wrong. 🤦♂️
There's nothing wrong with the previous proof as long as the picture is to scale with the measurements. In fact, changing the 4 to a 3 in the diagram not only makes it go from a diagram to scale to NOT a diagram to scale, but making it to scale would completely change the shape.
Unfortunately, with geometry diagrams like this, the only things you can be certain of are specified lengths or angles, denoted right angles, and denoted "equivalent" angles/lengths.
@@becomingsentient1208 This is only true because the people who make the diagrams are doing it that way so that you *have* to use a specific mathematical method to solve it. However, it makes more sense that whoever was making the diagram could have just as easily made it to scale. There are plenty of tools that people use to make these diagrams and making it to scale is in many cases easier than not.
@@lelouchlamperouge8286 By the same token, the problems weren't written so that everything could be solved with a ruler and protractor. That's learning how to measure, not how to solve the problem in front of you.
Edit for typo
@@becomingsentient1208 More than one way to skin a rabbit. Obviously the only reason they make it not to scale is for that purpose of testing math skills. I'm more interested in how practical it is. Once you take Calculus, you'll be finding the area of things until you don't even understand what the thing is.
Oh wow, I solved it in my head from the thumbnail and made the exact mistake he predicted ( essentially )...
Because the problem intentionally misleads people to using the wrong method, for whatever reason. To prove a point, I guess, but then this isn't a math problem. It's a logic problem featuring math
If you extend the diagonal length 4 to the top left this length completes the vertical 4 and diagonal 3 as a RA triangle and gives a value of square root of 7. So x is the square root of the sum of the squares of 5 and 6.64 which is square root of 69.08
Errr.... With the "point-and-click " YT page shows a different height of the vertical side as 4 thus with you changing the dimensions of the shape; 100% will fail to find ANY answer that's correct. As there isn't one. Unless, of course, we use all 3 dimensions......
EVEN 4 if you want.....
"Sometime geometry problem are trickier than they look."
Me, who always gets barely passing grade, but solved this in correct way from thumbnail: **visible confusion**
Yea, I remember in high school my geometry teacher said “students who usually do really good in math wind up being back at geometry while the opposite is true...this is because geometry uses common sense.”
@@fosterwalrus8413 So I’m a math genius? I failed at geometry so many times and only managed to pass because my teacher decided to let me cheat because that was more than her paygrade
@@fosterwalrus8413 I attended a state math competition in 5th grade and placed 5th. It was divided into four categories, and I got 4/10 in the geometry part where you had to measure objects (without a ruler) and get within 30% of the specified unit. Everything else I scored almost perfectly on. 39/40 on the exam, the largest part of the competition, 10/10 on a mental math exact test, and 8/10 on a mental math estimation test. Geometry is probably the only reason I didn't place 1st or 2nd in the state.
AAAA, REDDIT MOMENT, ama right????????????????????????? HAAAAAAAHAHAHAHAAAAAAAAAAAAAAAAAAAAAAAAAAA
Too many times I’ve doubted my answer in a maths exam because my teacher drew the acute angle that I had to find as an obtuse angle just for the fun of it.
I'm an online physics tutor, and one of my Wyzant students once asked me a question that made ME doubt myself and do it a few times "just to make sure," because it had a coefficient of friction of 1.4!
I spent like 3 hours trying to solve this because my teacher wanted me to have a try (this was 2 months ago, and i did get it right) … then this month i found you channel
I'm more amazed it's not 100% miss, given that the thumbnail's different to the actual problem. Downvoted accordingly.
*Dislike
@@furioustester4056 Way to avoid the substance of my comment and the difference between an easy-peasy visual solution and a difficult square root. Well done.
The image very misleading to be honest LoL
Yes, and when trigonometry is used it ignores the misleading nature of the image.
yup when its to scale the whole image changes
@@codibanks2315 The scale doesn't matter one bit, trigonometry doesn't need scale to find the solution, it just needs enough data.
@@martintaper7997 my brain is small don't confuse me xD
Length of the lines originally were 4, 3, 4, 5 when the problem was shown. Eventually the lengths became 3, 3, 4, 5. (!)
I was wondering that too. When I solved for the original 4,3,4,5 figure using trig I came up with x = 8.
ikr! So annoying
Right answer is x=8. Originally length of the lines: 4,3,4,5.
If you watch the video he states why he changed the length of that side. It is to help him show that people were using an incorrect proof. The goal, he points out, is to learn something, not get a correct answer through incorrect reasoning. If you originally came up with an answer of 8 by reasoning it correctly, then congrats, you did it right!
I solved this using similar triangles and Pythagorean theorem only, without moving the lines as Presh did in the video.
First, connect the bottom left corner (where the two lines of length 3 are connected) to the top right corner (where the end of the line whose length we want to calculate and the end of the line of length 5 meet).
Then, two right triangles are made, which are similar, cause they have the same angles. We can easily see that the scale factor is 3/5. If we set the base of each triangle as x and 4-x, we can find that x=3/2 from the scale factor.
So knowing the two sides of the right triangles, using the Pythagorean theorem, we can find that the hypotenuses of the triangles are 3*sqrt(5)/2 and 5*sqrt(5)/2 for the small and big triangles respectively.
Finally, we can do the Pythagorean theorem one final time to find the length of the line we want to find. One side is 3, the other side is the sum of the hypotenuses we found in the previous step, so 4*sqrt(5), so the third side turns out to be sqrt(71).
As soon as you mentioned "incorrect proof", the thought that the 3/4 hypotenuse is not necessarily parallel to x popped into my head.
atn (3/4) + atn (3/4) ≠ 90
But sir in the previous problem we could easily prove that the angle is equal to 180°. And then we can solve it easily by the other method
Damn, that triangle even changes it side lengths in the video from start to end.
The impossible right angle marked on there is just unethical
I get 8 as the value of x. At 1:29 you show another 3-4-5 right triangle but you invert the 3 and the 4. This means that the 3 side is on top and must be added to the 5 that you get from the lower 3-4-5- triangle. Then, 5 + 3 = 8 instead of the 5 +4 that you show in the video. I recreated this puzzle using SketchUp and also got 8 as the value of x.
9 = 5 + 4 using just the 3,4, 5, rule and seeing the two triangles that must be a 4 and a 5 along the X axis - takes about 10 seconds once you see it!
but thats wrong
@@joyishankujurkujurjoykujur2737 don't think so - the triangle at the bottom has edges of 3 and 4 around right angle - hence the hypotenuse must be 5, then draw a vertical lie from the line on the far right straigh up so it joins line X at a right angle - then its hypotenuse muse be 5 and its height is 4 - so the other side is 3 - and 3 + 5 generally equals 8!
Please watch the video again 😅
@@joyishankujurkujurjoykujur2737 I see what they have done in their proof - but to someone who did technical drawing for 5 years - it's not a very good (precise) way to depict your starting puzzle! Ciao guys - I normally find the questions on this excellent channel don't carry this sort of sub-optimal initial framing!
I worked it out from the thumbnail and got an answer of 9, you had missed the "Not drawn to scale" comment on there. Would have been helpful, kinda felt a bit cheated there, ...
I looked at the thumbnail and got sqrt(59).
@@ZenoDovahkiin Got the exact same answer; worked it out on thumbnail as well :(
Before viewing the video, I created the drawing, starting with the side of length 4 on the left and the side on the right with length 5, cannot be 5, but should be longer, around 6.2 if "x" is supposed to be a straight line with a 90 degree angle on the left. So, diagram not to scale is the beginning and end of the problem. So let's see what the video says.
I solved this problem using vector addition. I got an answer of 8.63 , almost the same as your answer.
At first it was 4 .. not 3
My ans was wrong this time..
8..by drawing 2 triangle..
right, the left-most side (vertical) is 4 at time 0:17 i the video, while later on it's 4 (after 1:59). :)
Me too.
You are right.
I tried for like 20 minutes, I got no where but I'm at least proud to say I didn't fall for the easy trap
For the original solve i also thought that the incorrect way might be incorrect because of the incorrect scales. But you can easily shift it into a Kite and then it is easily solved by symmetry. Not so easy on with 3 instead of four because you have to actually use pythagoras but still pretty straight forward. Nice Puzzles.
nah, this is how it's really done:
5+3=8
x2=16
and 4x2=8
then you subtract that 8 from 16 which equals 8
and then you add the 4+3+5 together which equals 12
and we all know 12 is made up of 4+2+6
so we add a decimal point to the 8 from earlier
and add the 426 to that
and we get 8.426
which is the square root of 71.
So, Presh got it wrong. He forgot to revert the original value of the left vertical back to 4. so x is not sqrt 71.
True, but it's solved in exactly the same way. x is 8 in the original problem; the "hypotenuse" is still sqrt(80), but you subtract 4^2 from sqrt(80)^2 rather than subtracting 3^2.
But, you can just read off the answer from the diagram with essentially no computation at all if you just "complete the rectangle" for the triangle with legs of length 3 and 4, and then notice that the resulting shape is a kite that has equal outer angles (90-degrees), equal-length short sides (4), and one long side's length effectively "given" (5+3). The two halves of the kite share a bisecting/central hypotenuse (which happens to be sqrt(80) in length, but you don't even need to bother computing that). Long story, short: equal outer angles and equal short edges, and one long edge provided, means the other long edge must be the same length: 8.
@@jamesedwards6173 I agree!
I agree, 8 seems correct. Not sure why is is saying people came up with the answer 9 and the incorrect proof. 9 seems wrong.
In your answer you make a wrong assumption by saying “you NOTICE that the resulting shape is a kite”. The problem is: if you had the left side of length 3.99, you probably wouldn’t notice that your end-shape doesn’t build a kite any longer, which is why you came to the same conclusion “8”, which would be wrong.
@@dmitrikonnov922 Except that x=8 is the correct answer. (What he's talking about from 2:12 to 2:30 is essentially what I was saying, but without the rotation, which is unnecessary---other than maybe making it a bit easier to visualize?? 2:36 shows a short and long side of the [half-]kite, with a right angle between, and the shared bisection/hypotenuse of the two sides of the kite. If you keep this construction in mind, then flip back to the initial diagram of the problem, you can clearly see that it's impossible for x to be anything but 8, because of the shared hypotenuse, the equal-length short edges given as 4, the long-edge length of 8 that's now obvious, and the equal [right] angles between short & long edges. You can't have any other value for x, given these facts. ... That being said, yes, if those short edges weren't both of length 4, you'd have to perform the more generalized solution approach outlined in the video. But, they are both of length 4, so you don't, and x=8 can very nearly be read directly off the diagram.)
Well I understand people not liking trick questions I think there a good way to see if you truly grasp what your learning. Especially in mathematics they are a perfect way of showing math doesn’t lie. As the old saying goes ,” Small mistakes crash rockets”.
This man is taking high school maths to the next level. Thanks for challenging the idle mind!
I solve it in a different way and the "scale" was not considered because I designed the shape to paper :)
So, what I did was to extend the size 3 line that created an isosceles triangle of 4 length so it was easy to fine the total length of the 3-lenght line.
After that, and from the same point, I have extended a new line to the right-bottom corner in which I now know the two sides, and it's easy to find the third (the rest of the "X" line that will be the first side and the 4 which is the second).
So the total "X" would be, 4 (as the side of the isosceles triangle)+ the number that I would find.
I'm pretty sure this is a correct approach too, but I'm really bored to make the calculations to compare it with your approach :)
In the beginning, when you change the value of a dimension (such as the 4 to 3), you must also change the diagram of the dimension, or else you are making an incorrect premise, so to speak. The diagram becomes impossible. This IS geometry after all.
I always thought that 'not to scale' means that you cant measure it with a ruler without knowing the ratio. But when the lines lengths proportions are incorrect, then I think there should be another type of notice...
If you had draw this diagram to scale in the first place, people's reasoning would be quite different: therefore an invalid question that you posed.
That's what you get for trying to eyeball a geometry problem instead of actually working it out logically
@@colehanna4040 As a qualified educator, I would never put up this type of question without being super specific, the whole question was vague and ambiguous and most students would look at the diagram expecting some sort of reality.
After this question, how we can trust our teachers? how can we trust Presh Tawalker?
Again, misdirection/ question! Deception!
They are trying to have you use logic instead of going by appearances which, as the saying goes, can be deceptive, and are not to be relied upon.
@@transklutz Wish I did that when I picked a wife. Her circumference is expanding faster than the universe. I can't even weigh her anymore. I just shoot a laser next to her and see how much the light bends.