Since the answer mustn’t depend on the exact values of x and r, there is a very fast solution: Assume r is very, very small. This trivially makes the blue rectangle a square with a side of length 5.
I’ve used that trick a few times. “Oh it works for all x? Let me just set it to an awfully convenient number and solve that instead”, never thought to take it to the limit though! That’s a grade above
It's a pity when i was at university i struggled a lot on passing math exams and i didn't appreciate the little things like this. Now after a few years i appreciate it
It does not look like a tangent (Try extending the line past the contact, does it look tangent anymore?) Assuming things like this from the diagram, instead of relying on the problem statement is not a good idea. You couldve also argued that the quarter circle ‘looks’ like it has the same radius, but you would not have got full points for your work.
Another great video. This problem is a wonderful opportunity to use the Power of a Point Theorem. This says that if a line through point P intersects a circle and we measure the near and far distances from P to the circle along that line, the product of the two measurements is the same...no matter which line it was! In this case, P is the bottom left corner of the rectangle, and the circle I care about is the one with radius r. Measuring horizontally, we see that the "near distance" to the circle is x, while the "far distance" to the circle is x+2r. The product of these measurements is x(x+2r). Measuring along the red segment, the "near distance" is 5cm, and so is the "far distance". Their product is 25cm^2. By PoP, x(x+2r)=25cm^2.
also you can use circle properties. idk the name of the theorem, but it states that tangent^2 = secant * secant's external segment. this problem is a unique case when the secant (the rectangle's lower side) passes through the circle's center.
My way of solving: 5 is a leg of a 30 - 60 - 90 triangle which hypotenuse is 2 r and the other leg r. So r is 5 over sqrt 3. The rectangle has a side which is 3r long and the other which is as long as r. The area is 5 over sqrt 3 multiplied by 3 times 5 over sqrt 3 which is equal to 75 over 3 which is 25.
I've tried this question before but I made the assumption that the radius of semicircle and quater circle are same. It's amazing to see how that would not affect the answer at all, we still get the same result even if the radius was not same.
@@notryangosling2011 To be fair, if the radius of the quarter circle changes, the value of 5 should change as well. So it actually does depend on the radius.
@@Marcus-qh3qy yes, by Changing the radius of quater circle length of tangent must change if the radius of semicircle is fixed. But in this case radius of both the circles are not given and the only certain value that we know is the length of tangent (i.e. 5 cm). So if length of quatercircle changes, it would change the radius of semicircle, instead of affecting the length of tangent. length of the tangent still remains and have to remain 5 cm as it is defined by the question. And that is what's truly amazing about it, by keeping the length of tangent same, if we change the radius of one circle then the radius of other circle would change in such a way that the area of rectangle remains same.
Hes assuming the line is tangent to the semi-circle. You could argue it looks like its a tangent, but its not stated, so it could be sightly off. So the answer should be approximately 25. So the answer should be ≈25 not =25.
I literally screamed in excitement when you said that tangents of circles are perpendicular to the radius that touches at the same point. I completely forgot about that when I was trying to solve it on my own 😂
I tried it by considering the radius of the semicircle and the quater circle to be same and got the same answer. However when you consider x and r , I was really curious to see where i went wrong. Amazing sum. Thanks for explaining.
I haven't been into math for years, but I really appreciate how elegant this solution is. Your enthusiasm is really infectious. You're a great teacher.
I loved math in high school. I thought I wanted to be a mathematician cause I loved stuff like this. Ended up in accounting and slowly got away from math like this. This video reignited that love. That's so cool. I love this. Thank you!
This was the coolest problem i have ever seen. At first i thought that this was impossible. But today i learnes that when the radius meetss at the r tangent point it creates right angle
Its amazing how hard and difficult problems, once split into easier and simpler steps crumble and just becomes solvable and clear, almost obvious. Exciting indeed
I got very close with a triangle calculator, and having forgotten every concept used in this video. I eyeballed the angle, and it looked close enough to 90 degrees, so I went with it. The angles of a triangle have to add up to 180, so I eyeballed the one opposite the 5cm side of the triangle, and came up with 60 degrees. I popped both that and the 5cm side into a right angle triangle calculator, and got 5.7735 for the longest side of the triangle - the one touching the rectangle. That looked to me like about 2/3rds of the long side of the rectangle, and twice the size of the short side, so I divided it by two (to get 2.88675), multiplied that by 3 (to get 8.66025), then multiplied those together and got 24.9999766875, which is close enough to 25 for me.
I have followed the same line of thinking as you have. With the same result.Maths have always been my weakness. Guaranteed I would not pass 5th class elementary school maths tests.
Sadly, this wouldn't get me any points in an math exam, because they graded not just the result, but how i got there. If i can solve it in my head and just wrote the answer, best i could hope for is that they let me redo the test in front of them/explain verbally how i got there. But most teachers wouldn't even give you that chance because it means A LOT of extra work when you just could write a basic way of solving down.
I think a more interesting question would be to find the base (x + 2r) and height (x) separately, which I initially thought you had to do to find the area, but this is still a cool solution.
@@jcnot9712 You can't determine x in this problem. That is the beauty of this problem. When R approaches to zero X approaches 5 and the area remains constant that is 25. Similarly when r approaches infinity x tends to some value such that the area of rectangle remains 25. by the equation.
I’d like to add another way of solving this that I found more natural: Draw the right triangle as shown and since 5 is a base of the triangle, you can use the 5,12,13 Pythagorean triple With this triple you can find that r is 12 since it is the base and that x + r is 13 Then we find that x = 1 and add up x + r + r (12*2 + 1) to get 25
There's another way to solve this that's much easier, which is to shrink the half-circle down to radius 0, which leaves us with a 5 by 5 square. We can do this because the result must be independant of the actual values of r, since it's possible to draw the problem with all restraints given in multiple ways.
Huh? What do you mean? For each value of '5', there are infinite pairs (x,r) satisfying the situation in the problem. There is the case with x=r, '5'² = x²+2xr = x²+2x² = 3x² So, x = '5' √3/3.
That's not really how this problem works. There is no single value of x or r; x and r can be anything as long as they satisfy the relation "25 = x^2 + 2xr". x = r is already a valid possibility. In other words, if you assume x = r, you'll still get the same solution.
Another method is using tangent-secant relationships: We know the relationship between the tangent and secant in the problem is 5^2= x(x+r+r) This equals 25 = x^2+ 2rx which is also the area of the rectangle.
Man, if only my teachers in school had this kind of optimism, the reason I started to hate maths was because it seemed like they hated it too much, on top of that we had physical abuse as punishment so yeah , that’s tough. Maybe ill start again, math does sound magical to me.
I literally forgot the tangential line is always perpendicular to the radius 😂 it's amazing how you can solve the problem with such limited information. you're a genius, bruh
A meta-solution: if you know that the problem can be solved without any more information (i.e. without knowing the r/x ratio), then you may as well set r=0, which yields 25 immediately.
@@seant2432the 5 becomes x as the radius goes to 0, the hypotenuse becomes the 5cm side. 2 sides of the “triangle” are now 5cm and one side is 0cm. 5*5 = 25
Wow! I heard somewhere that people solve some questions just by putting 0 for something which seems unrealistic for example, area of triangle or a length of a line. But I always doubted how it was possible. Now with your comment I am getting it to some extent that If there is a lack of information and we know that the problem can be solved without such information, we can simply use 0 in such situations. Impressive! Please tell me more about it.
Yeah this doesn't work, if you substitute 0 you get 25 = 0, meaning you need to back up and try again. Just because 25 is "all that is left" doesn't mean you can take it out of context of the equation he constructed and call it an answer
Didn't expect I would make an audible sound when I saw the solution. And I'm not the type of guy who's the best at maths, to put it lightly Love your vids!
You don't need algebra to solve this problem. The key is to realize that the problem is under-constrained meaning the relative sizes of the circles can change. If you assume the right circle has radius zero, then the rectangle becomes a square and the side length must be 5cm making the area 25cm^2
The fact the line just shows one point of intersection means that it is a tangent point. If it wasn't, it would show two points. That's part of the unspoken rules between the ones that make exercises and the ones that try to solve exercises.
@@samueldeandrade8535 the unspoken rule in my home country is that the diagrams on the exam paper are not drawn to scale. That's why I raise my question
Looking at the problem, since the radius of the circle doesn't matter, we can assume that r get arbitrarily close to zero, letting the rectangle be a 5x5 square.
I was going to ask how you knew the triangle was a right triangle, but I decided to rewatch the video to see if you in fact did explain it, and...you did! Great precise teaching. The way the video is so succinct and easy to follow really helps you learn!
With channels like this kids have no excuses today. Just watching the videos for fun should help when you inevitably see similar problems in class and on tests.
youre not wrong but this a very basic problem, i was doing trig substitution integration at 17, i dont think people 40 years ago were. My a-level course for maths consisted of stuff that was being done by undergrads 40 years ago. That annoys me, because i found my a-levels easy as fuck, so to think i could've had a piss easy degree instead of one that makes me want to jump off a cliff....
@@wavingbuddy3535What does you gloating about trig substitutions have to the OP? There are kids doing that at 10. All he said was that videos like this are helpful; you went on a tangent about how advanced you were, and why you want an easier degree, which says something about your work ethic.
I am a Biolog Student But love these videos, it checks your ability to breakdown complex problems into simpler and not panic all the time. Great Videos❤❤❤
Just like a challenging puzzle, but once you utilize the concept of the tangent line at a point on the circle, everything begins to fall into place effortlessly.
One thing that I perceive as an error in the problem. The 5 cm line looks to be tangent to the circle in the same way that the radius of the 1/4 circle appears to be the same as that of the 1/2 circle. On the picture of the problem, it does not state that the line is tangent to the 1/2 circle, so maybe if one was able to zoom in on the picture, could not the intersects the circle and not tangent? It is only when you state that the 5 cm line is tangent in the video is it that we can take that to be the actual case. You could have also said that the x = r. Either one of those statements were not in the original problem.
No. This would be unreasonable. The fact it shows just one point of intersection means it is a tangent point. It also doesn't say the curve lines are part of circles. We assume they are because that's reasonable. More, we don't actually need to assume anything. If the intersection point is not a tangent point, then the solution of the problem would be "the value of the area is indetermined (in this case). There is not enough information". But this would be the analysis of one case, one interpretation of the figure.
0:11 "... and then this red line with the length of 5 that goes from the vertex of this rectangle, tangent to the semicircle." Only if don't think that Andy's narration forms part of the question would you think it was an assumption.
So, taking r = quarter-circle radius (also side a) R = half-circle radius side b = r + 2R To find: ab as the radius to the point of tangency is perpendicular to the tangent, the tangent line and the radius make two legs of a right triangle who's hypotenuse by construction is r + R. Thus you get: R² + 5² = (R + r)² R² + 25 = R² + r² + 2Rr Subtracting R² from both sides: 25 = r² + 2Rr taking r common: 25 = r(r + R) But as a = r and b = R + r, 25 = ab Now, ab is the area of the rectangle, which is 25 square units.
I am not sure why I was haunted by this video's recommendation for weeks, but I am glad I watched it. It's nice to know... I expected some derivative or integral calculation instead, whereas the curved line is cosinus, and its x has to be found... where the distance from the starting point of the 5 cm has to fit - and then put it into an equation.
I know you don’t have anything saying that x=r, but if you assume that x=r then you would have: 5^2 + x^2 = 4x^2 25=3x^2 X=sqrt(25/3) And the area would be given by x*(3x) or 3x^2 So we have 3*(sqrt(25/3))^2 That would be 25
Out of all the math problems u posted, funnily enough this was the first one I've solved and answered completely by myself. I've been a math geek way back in elementary who joins in tons of math contests. But because of the pandemic, I lost my interest for the talents which I was born with. But because of people like you, I found the fun in math again, and realized I always loved it. Keep doing these kinds of videos!
I saw the thumbnail and calculated it and found 25. Watched the video for confirmation. I’m a high school student and it s a fairly easy question, but your explanation is really sick! Can’t deny I feel good solving this ahaha
Back in the early 80’s in high school, we would all have written “Undefined” as the answer and moved on. There are no indications other than the word “Rectangle” in the question to indicate anything is actually right angles. (Except the tangent right angle of course.) It was an easy way to get half a mark with a technically correct answer in response to bad diagram construction.
Before I watch the video, I'm going to say that the rectangle has a length of 1r and a width of 3r. The 'r' meaning the radius of the semicircle. The line that intersects with the semicircle is tangent, so it forms a right triangle with a hypotenuse with a length 2r and height of r. ---------
Oh wow that was fun! I was looking at the problem before realizing the tangency and thinking more information was needed. A few seconds into the vid cleared that up lol
You actually can assume that x=r, because if you draw the semicircle as a full circle the radius of the semicircle reaches the top of the rectangle, which means that the short side of the rectangle is r, and the short side is equal to the radius of the quarter circle. r=x
The figure is misleading, and what you're doing isn't proof. All that we have obtained is a quadratic equation that ties x and r, it's: x^2+2rx=25. In other words, r=(25-x^2)/2
You can also easily derive the values of x and R. X^2 +2X -25 = 0 Solving that equation for X gives you X = sqrt(26) - 1 You can plug that X value back into the equation for the area to see that R = 1. So the two circles are not the same!
I did it a very different way which I thought might be interesting to share: sec(θ) = 2 r tan(θ) = 5 r² + tan²(θ) = sec²(θ) r² + 25 = 4 r² 3 r² = 25 Area = 3 r² = 25
Both x and r can vary between 0 and 5. When x = 5, r = 0. When r = 5, x = 5*((sqrt(2)-1), approx. 2,08. If x = r, => x = r = 5/sqrt(3), approx. 2,89. Pretty exiting stuff.
I always hated problems like this. We're taught not to assume that the x and r radii (as he labeled them) are the same because there's no proof in the diagram. At the same time, we then are forced to make the assumption that the 5 cm line is a tangent of which there's also no proof. You can't have it both ways.
The red line being tangent to the semicircle is a given not an assumption. The diagram depicts it touching the semicircle at a single point. If it wasn't a tangent to the semicircle, then it would either intersect the semicircle at 2 points, or not touch the semicircle at all.
shot in the dark buffer space buffer space buffer space buffer space based on the fact that the 5cm line is tangent with the circle and thus perpendicular to it's radius, r, so 5² + r² = 4r² (4r² from squaring 2r, from the center of the semicircle to the left-hand edge of the rectangle) so 5² = 3r² rectangle has an area of 3r² (r•3r), thus the area is... 25 cm² edit: so. the exact logic used in the video, except for the unnecessary assumption that x = r on my part. whoops. basically replaces 2xr and x² with 2r² and r², respectively
Couldn't you assume the area is the same for any r, set r=0, and then just solve for x^2 where x=5? Something something not to scale. Also, this reminds me of the napkin ring problem, but in 2D.
Though I really like this and it leads to some good intuition, I did think of an issue: you can't assume the area will always be the same. In this case it is, so if you find the area for r = 0, you've found it for all valid r, but I can't think of how you'd jump to your initial assumption unless you already figured it out the other way.
@@atrus3823 The way I reason it: You're solving for the area. The radius of the right semicircle isn't given. There is an infinite range of semicircles with an infinite range of radii that have a tangent line that length through that corner (easily provable by the fact that If you draw a line from the corner to any given point the given distance away to the bottom right, you could then construct a semicircle tangent to that line flush with the horizontal line in the problem and then construct a quarter circle and then a rectangle from there). So given you're solving for an area of an infinite range of possible shapes, all the shapes must have the same area, unless we weren't given enough information to solve the problem in the first place. So a semicircle of infinitely small radius must result in the same area. So you can do the simple solve by treating r=0. If all the questions have the same answer, solve the easiest question.
@@atrus3823 I guess you have to assume that there is a solution, ie a number and not an expression that depends on x or r. Since we cant assume any relationship x/r between x and r this would have to be the same for any r arbitrarily close to 0, and also r=0 (?). Ie x/r arbitrarily large. Also see the comment by stirdix. I’m really impressed by this setting r=0 idea!
Think r radius getting small until it equals to zero. Then 5cm line's corner create with middle of quarter circle getting small and it lands on rectangle base. So we can assume X+r+r = 5 Because r≈0 X = 5 So 5x5 rectangle area = 25 Sorry for bad English. I hope you understand.
Sorry if this had been answer before but, how do we know that the angle is a right angle? Better said, how do we know it is a tangent line?, Thank you in advance
We know that the red line is tangent to the semicircle because Andy states that it is when he's explaining the question. It is therefore a given of the question. If it wasn't a tangent to the semicircle, then it would either intersect the semicircle at 2 points, or not touch the semicircle at all. Neither of these are the case in the diagram. Apply the Tangents to a Circle Theorem, and you've proven the right angle.
It's cool that you can solve this but I'm so dumb I can't even think of why or how this would translate into building or any type of work, Pythagorean helps by squareing off fences to building
How exciting
Isn’t that cool?
*video ends*
You click on the comments section
You see this comment and open it
You read the replies
Love watching these. Hated working as an engineer, but I always had a love for math and this takes me back to solving fun puzzles like these in school
Aren’t you a mukbanger
@@riprider5626 always thought they were Chinesers
This comment makes me think that you might like a degree in mathematics, if it were feasible for you.
@@michaelbujaki2462man does makbangs for a living now
I'm looking to possibly become an engineer and have no idea what it's like, why was it so bad for you?
My wife doesn’t understand why I love these videos
I love that you love these videos. Thank you!
my subconscious , which knows I'm not good at math, does not understand why I watch these type of videos .
@bikesboardsbeats which begs the question--why DO you love your wife ?
@@edwardmacnab354 well since the variable wife intersects the tangent line of the circle at a right angle we can use a^2*b^2=c^2 to solve.
"How exciting!"
I have not done math for 3 years since my studies are in different fields, but this was truly exciting. Nice video!
Since the answer mustn’t depend on the exact values of x and r, there is a very fast solution: Assume r is very, very small. This trivially makes the blue rectangle a square with a side of length 5.
This is thinking outside the box! Or actually thinking inside the box?
Ha, how elegant.
Limits are cool 😎
i dont get it
I’ve used that trick a few times. “Oh it works for all x? Let me just set it to an awfully convenient number and solve that instead”, never thought to take it to the limit though! That’s a grade above
It's a pity when i was at university i struggled a lot on passing math exams and i didn't appreciate the little things like this.
Now after a few years i appreciate it
If these videos were around when I was in college I would have done so much better.
this guy solved the problem before i even knew what was happening
lmao same
x3
I missed that the 5cm line was tangent to the semi-circle. Cool problem. I have enjoyed. Thanks once again!
thats the first thing i noticed but couldnt figure out what to do with the perpendicular
wdym you ‘noticed’? Theres no way you couldve figured that without listening to the problem stament. He explicitly says that it is tangent
@@juv7026 it looked like atangent in the figure :) i didnt really listen to him thats why i couldnt solve the rest by myself
@@juv7026yes he could. It was the most obvious part of the whole problem.
It does not look like a tangent (Try extending the line past the contact, does it look tangent anymore?)
Assuming things like this from the diagram, instead of relying on the problem statement is not a good idea. You couldve also argued that the quarter circle ‘looks’ like it has the same radius, but you would not have got full points for your work.
you know a problem is even more exciting than usual when he hit us with the "isn't that cool?"
Another great video.
This problem is a wonderful opportunity to use the Power of a Point Theorem. This says that if a line through point P intersects a circle and we measure the near and far distances from P to the circle along that line, the product of the two measurements is the same...no matter which line it was!
In this case, P is the bottom left corner of the rectangle, and the circle I care about is the one with radius r.
Measuring horizontally, we see that the "near distance" to the circle is x, while the "far distance" to the circle is x+2r. The product of these measurements is x(x+2r).
Measuring along the red segment, the "near distance" is 5cm, and so is the "far distance". Their product is 25cm^2.
By PoP, x(x+2r)=25cm^2.
This was sleek.
Never could have thought the process would be this simple
I mean I looked at it and was like calculus for sure
They made it simple.
also you can use circle properties. idk the name of the theorem, but it states that tangent^2 = secant * secant's external segment. this problem is a unique case when the secant (the rectangle's lower side) passes through the circle's center.
It is called the tangent-secant theorem
en.wikipedia.org/wiki/Tangent%E2%80%93secant_theorem
Way easier, solved that way in matter of seconds
Your enthusiasm when saying "isn't that cool? :)" is really lovely
yeah its not the usual depressed "how exciting"
My way of solving: 5 is a leg of a 30 - 60 - 90 triangle which hypotenuse is 2 r and the other leg r. So r is 5 over sqrt 3. The rectangle has a side which is 3r long and the other which is as long as r. The area is 5 over sqrt 3 multiplied by 3 times 5 over sqrt 3 which is equal to 75 over 3 which is 25.
You assumed the two circle portions share the same radius, but they don't necessarily.
I've tried this question before but I made the assumption that the radius of semicircle and quater circle are same.
It's amazing to see how that would not affect the answer at all, we still get the same result even if the radius was not same.
Same, i labeled the radius on quarter circle and semi circle as x. Still got 25.
@@tcjgaming9813 that's what is amazing about this question that the area of rectangle does not depend on radius of the quater circle
@@notryangosling2011 To be fair, if the radius of the quarter circle changes, the value of 5 should change as well. So it actually does depend on the radius.
@@Marcus-qh3qy yes, by Changing the radius of quater circle length of tangent must change if the radius of semicircle is fixed.
But in this case radius of both the circles are not given and the only certain value that we know is the length of tangent (i.e. 5 cm).
So if length of quatercircle changes, it would change the radius of semicircle, instead of affecting the length of tangent. length of the tangent still remains and have to remain 5 cm as it is defined by the question.
And that is what's truly amazing about it, by keeping the length of tangent same, if we change the radius of one circle then the radius of other circle would change in such a way that the area of rectangle remains same.
Hes assuming the line is tangent to the semi-circle. You could argue it looks like its a tangent, but its not stated, so it could be sightly off. So the answer should be approximately 25.
So the answer should be ≈25 not =25.
I’ve been looking for channels like this for ages because I *love* math! Just hearing one of the math terms just makes me smile!
I literally screamed in excitement when you said that tangents of circles are perpendicular to the radius that touches at the same point. I completely forgot about that when I was trying to solve it on my own 😂
I tried it by considering the radius of the semicircle and the quater circle to be same and got the same answer.
However when you consider x and r , I was really curious to see where i went wrong. Amazing sum. Thanks for explaining.
The values of x and r are not determined. And they can be the same.
Try drawing the same image, keeping x equal but r larger or smaller. You'll see where your assumption went wrong and why it still holds true.
By actually applying Power of point Theorem you can end on the same result directly.
Exactly! And that's the superior way. You may be strange, Sam. But you are smart.
I haven't been into math for years, but I really appreciate how elegant this solution is. Your enthusiasm is really infectious. You're a great teacher.
I loved math in high school. I thought I wanted to be a mathematician cause I loved stuff like this. Ended up in accounting and slowly got away from math like this. This video reignited that love. That's so cool. I love this. Thank you!
This was the coolest problem i have ever seen. At first i thought that this was impossible. But today i learnes that when the radius meetss at the r tangent point it creates right angle
Dang you bring my math vibes back to prime bro. I wanna try solving some math problem
Its amazing how hard and difficult problems, once split into easier and simpler steps crumble and just becomes solvable and clear, almost obvious. Exciting indeed
I got very close with a triangle calculator, and having forgotten every concept used in this video.
I eyeballed the angle, and it looked close enough to 90 degrees, so I went with it. The angles of a triangle have to add up to 180, so I eyeballed the one opposite the 5cm side of the triangle, and came up with 60 degrees. I popped both that and the 5cm side into a right angle triangle calculator, and got 5.7735 for the longest side of the triangle - the one touching the rectangle. That looked to me like about 2/3rds of the long side of the rectangle, and twice the size of the short side, so I divided it by two (to get 2.88675), multiplied that by 3 (to get 8.66025), then multiplied those together and got 24.9999766875, which is close enough to 25 for me.
I have followed the same line of thinking as you have. With the same result.Maths have always been my weakness. Guaranteed I would not pass 5th class elementary school maths tests.
Sadly, this wouldn't get me any points in an math exam, because they graded not just the result, but how i got there.
If i can solve it in my head and just wrote the answer, best i could hope for is that they let me redo the test in front of them/explain verbally how i got there.
But most teachers wouldn't even give you that chance because it means A LOT of extra work when you just could write a basic way of solving down.
And eyeballing or even meassuring the drawing normally didn't go because they say that the drawing is not up to scale / it's not precise enought.
I think a more interesting question would be to find the base (x + 2r) and height (x) separately, which I initially thought you had to do to find the area, but this is still a cool solution.
But he already did at the end. x(x+2r) = x2 +2xr
@@bladeoflucatiel so what’s x?
@@jcnot9712 You can't determine x in this problem. That is the beauty of this problem. When R approaches to zero X approaches 5 and the area remains constant that is 25. Similarly when r approaches infinity x tends to some value such that the area of rectangle remains 25. by the equation.
@@siddhantyadav27you can find X and R. 5 12 13 are pythagorus triplets so its so simple to find them
@@siddhantyadav27 that’s such a cool visual. Thanks for the explanation.
I’d like to add another way of solving this that I found more natural:
Draw the right triangle as shown and since 5 is a base of the triangle, you can use the 5,12,13 Pythagorean triple
With this triple you can find that r is 12 since it is the base and that
x + r is 13
Then we find that x = 1 and add up x + r + r (12*2 + 1) to get 25
There's another way to solve this that's much easier, which is to shrink the half-circle down to radius 0, which leaves us with a 5 by 5 square. We can do this because the result must be independant of the actual values of r, since it's possible to draw the problem with all restraints given in multiple ways.
Next step maybe would be to find the value instead of the '5', that makes x=r
Huh? What do you mean? For each value of '5', there are infinite pairs (x,r) satisfying the situation in the problem. There is the case with x=r,
'5'² = x²+2xr
= x²+2x²
= 3x²
So, x = '5' √3/3.
@@samueldeandrade8535yeah but this not the solution buddy
@@alexfoley9103 about what are you talking about? What did I say is a or the solution?
That's not really how this problem works. There is no single value of x or r; x and r can be anything as long as they satisfy the relation "25 = x^2 + 2xr".
x = r is already a valid possibility. In other words, if you assume x = r, you'll still get the same solution.
@@alexfoley9103 gave up? Great.
As one of the only people I know who likes math, your videos make me very happy.
How exciting
Another method is using tangent-secant relationships:
We know the relationship between the tangent and secant in the problem is 5^2= x(x+r+r)
This equals 25 = x^2+ 2rx which is also the area of the rectangle.
Man, if only my teachers in school had this kind of optimism, the reason I started to hate maths was because it seemed like they hated it too much, on top of that we had physical abuse as punishment so yeah , that’s tough. Maybe ill start again, math does sound magical to me.
as soon as I wrote out the Pythagorean for the tangent line I was like "wait a minute" surprisingly simple
I literally forgot the tangential line is always perpendicular to the radius 😂 it's amazing how you can solve the problem with such limited information. you're a genius, bruh
Yeah, that was the key to this one.
Such an exciting problem visual, looked difficult at first but after making that extra construction it obly took 2 mins
That caught me by surprise! I did made the assumption the circles were the same, but seeing how they don't need to be made my day!
A meta-solution: if you know that the problem can be solved without any more information (i.e. without knowing the r/x ratio), then you may as well set r=0, which yields 25 immediately.
Instinctively, that feels like it's impossible. I believe you. You sound very confident, but I dont' understand
@@seant2432the 5 becomes x as the radius goes to 0, the hypotenuse becomes the 5cm side. 2 sides of the “triangle” are now 5cm and one side is 0cm. 5*5 = 25
Wow! I heard somewhere that people solve some questions just by putting 0 for something which seems unrealistic for example, area of triangle or a length of a line.
But I always doubted how it was possible.
Now with your comment I am getting it to some extent that If there is a lack of information and we know that the problem can be solved without such information, we can simply use 0 in such situations.
Impressive! Please tell me more about it.
How does setting r=0 give you 25 "immediately"? You get Area=x^2 and you still have to solve it the same way.
Yeah this doesn't work, if you substitute 0 you get 25 = 0, meaning you need to back up and try again. Just because 25 is "all that is left" doesn't mean you can take it out of context of the equation he constructed and call it an answer
Didn't expect I would make an audible sound when I saw the solution. And I'm not the type of guy who's the best at maths, to put it lightly
Love your vids!
You don't need algebra to solve this problem. The key is to realize that the problem is under-constrained meaning the relative sizes of the circles can change. If you assume the right circle has radius zero, then the rectangle becomes a square and the side length must be 5cm making the area 25cm^2
If the question didn't mention the line to be a tangent line, how am I supposed to prove that?
I don't believe anything could be concluded if the lines wasn't tangent.
The fact the line just shows one point of intersection means that it is a tangent point. If it wasn't, it would show two points. That's part of the unspoken rules between the ones that make exercises and the ones that try to solve exercises.
@@samueldeandrade8535 the unspoken rule in my home country is that the diagrams on the exam paper are not drawn to scale. That's why I raise my question
@@yucalvin3205 what you said now has nothing to do with your question. You are questioning the tangency of a point. Scales are something else.
0:11
Holy s**t. I was looking at this for so long and just said ‘nope’. Wow I love this channel!!!
This is one of those problems with a simple solution, but you have to know exactly what you're doing to find it. Well done.
Looking at the problem, since the radius of the circle doesn't matter, we can assume that r get arbitrarily close to zero, letting the rectangle be a 5x5 square.
I was going to ask how you knew the triangle was a right triangle, but I decided to rewatch the video to see if you in fact did explain it, and...you did! Great precise teaching. The way the video is so succinct and easy to follow really helps you learn!
With channels like this kids have no excuses today. Just watching the videos for fun should help when you inevitably see similar problems in class and on tests.
youre not wrong but this a very basic problem, i was doing trig substitution integration at 17, i dont think people 40 years ago were. My a-level course for maths consisted of stuff that was being done by undergrads 40 years ago. That annoys me, because i found my a-levels easy as fuck, so to think i could've had a piss easy degree instead of one that makes me want to jump off a cliff....
@@wavingbuddy3535What does you gloating about trig substitutions have to the OP? There are kids doing that at 10. All he said was that videos like this are helpful; you went on a tangent about how advanced you were, and why you want an easier degree, which says something about your work ethic.
No its just your generations lack of care and lability
@@wavingbuddy3535a level maths is of no value tbf, further maths on the other hand
"... kids have no excuses today". How funny it is that we normalized the "duty to study". And by funny I mean terrible, of course.
Your voice turns into Schwarzenegger at around 2:08 for a second, did you do that on purpose? :D
I am a Biolog Student But love these videos, it checks your ability to breakdown complex problems into simpler and not panic all the time. Great Videos❤❤❤
That's a really neat solution!
Thank you, I thought so too. I was excited to share it!
Just like a challenging puzzle, but once you utilize the concept of the tangent line at a point on the circle, everything begins to fall into place effortlessly.
One thing that I perceive as an error in the problem. The 5 cm line looks to be tangent to the circle in the same way that the radius of the 1/4 circle appears to be the same as that of the 1/2 circle. On the picture of the problem, it does not state that the line is tangent to the 1/2 circle, so maybe if one was able to zoom in on the picture, could not the intersects the circle and not tangent? It is only when you state that the 5 cm line is tangent in the video is it that we can take that to be the actual case. You could have also said that the x = r. Either one of those statements were not in the original problem.
No. This would be unreasonable. The fact it shows just one point of intersection means it is a tangent point. It also doesn't say the curve lines are part of circles. We assume they are because that's reasonable. More, we don't actually need to assume anything. If the intersection point is not a tangent point, then the solution of the problem would be "the value of the area is indetermined (in this case). There is not enough information". But this would be the analysis of one case, one interpretation of the figure.
0:11 "... and then this red line with the length of 5 that goes from the vertex of this rectangle, tangent to the semicircle."
Only if don't think that Andy's narration forms part of the question would you think it was an assumption.
So, taking
r = quarter-circle radius (also side a)
R = half-circle radius
side b = r + 2R
To find: ab
as the radius to the point of tangency is perpendicular to the tangent, the tangent line and the radius make two legs of a right triangle who's hypotenuse by construction is r + R.
Thus you get:
R² + 5² = (R + r)²
R² + 25 = R² + r² + 2Rr
Subtracting R² from both sides:
25 = r² + 2Rr
taking r common:
25 = r(r + R)
But as a = r and b = R + r,
25 = ab
Now, ab is the area of the rectangle, which is 25 square units.
I got soooo lost. Not because you didn't explain well, it's my thick skull. I'll watch again till I get it
I am not sure why I was haunted by this video's recommendation for weeks, but I am glad I watched it. It's nice to know... I expected some derivative or integral calculation instead, whereas the curved line is cosinus, and its x has to be found... where the distance from the starting point of the 5 cm has to fit - and then put it into an equation.
I was a huge STEM kid but moved into Law when I was at Uni. Watching these vids make me miss STEM 😂
I know you don’t have anything saying that x=r, but if you assume that x=r then you would have:
5^2 + x^2 = 4x^2
25=3x^2
X=sqrt(25/3)
And the area would be given by x*(3x) or 3x^2
So we have 3*(sqrt(25/3))^2
That would be 25
you'll probably lose marks in an exam if you assume that even if you get the correct answer
@@AlviStudiesagree with you
Yeah. You can assume x=r. But in a test you would have to justify that.
Out of all the math problems u posted, funnily enough this was the first one I've solved and answered completely by myself.
I've been a math geek way back in elementary who joins in tons of math contests. But because of the pandemic, I lost my interest for the talents which I was born with.
But because of people like you, I found the fun in math again, and realized I always loved it. Keep doing these kinds of videos!
I saw the thumbnail and calculated it and found 25. Watched the video for confirmation. I’m a high school student and it s a fairly easy question, but your explanation is really sick! Can’t deny I feel good solving this ahaha
By using tangent theorem: sq 5= x.(x+2r) = area of the rectangle
Area= 25 sq units
I was fully prepared for this video to somehow use sine and tangent lines, I guess I gotta start doing some actual math problems
can watch these all day
Back in the early 80’s in high school, we would all have written “Undefined” as the answer and moved on. There are no indications other than the word “Rectangle” in the question to indicate anything is actually right angles. (Except the tangent right angle of course.) It was an easy way to get half a mark with a technically correct answer in response to bad diagram construction.
Before I watch the video, I'm going to say that the rectangle has a length of 1r and a width of 3r. The 'r' meaning the radius of the semicircle.
The line that intersects with the semicircle is tangent, so it forms a right triangle with a hypotenuse with a length 2r and height of r.
---------
A reverse problem would look even more sophisticated and exciting.
seemingly difficult yet clean and simple, nice
Oh wow that was fun! I was looking at the problem before realizing the tangency and thinking more information was needed. A few seconds into the vid cleared that up lol
It pleases me greatly how many of these puzzles are solved by drawing a right triangle.
Didn’t we were supposed to assume the red line was a tangent line…
You actually can assume that x=r, because if you draw the semicircle as a full circle the radius of the semicircle reaches the top of the rectangle, which means that the short side of the rectangle is r, and the short side is equal to the radius of the quarter circle. r=x
The figure is misleading, and what you're doing isn't proof. All that we have obtained is a quadratic equation that ties x and r, it's: x^2+2rx=25. In other words, r=(25-x^2)/2
that was honestly so cool. its one of those answers that sneaks up on you!
I like this. Seems hard at a glance but really intuitive when you sit down with it
You can also easily derive the values of x and R.
X^2 +2X -25 = 0
Solving that equation for X gives you X = sqrt(26) - 1
You can plug that X value back into the equation for the area to see that R = 1.
So the two circles are not the same!
How did you obtain your quadratic equation in X?
Would there be a way to solve this with no evidence whatsoever that (r, 5cm) is a right angle?
this is genius, I never thought to think of it like this, I feel stupid now.
We can also solve by using tangent secant theoram. It'll take seconds to solve.
What do you mean?
Fun for the sake of fun. A lot of us lose this being adults in a rat race. Thank you.
I thought it will be harder when we just imagine it. It turns out, it is easier when we try to breakdown every part of it
You can do this problem with tangent to the circle theorem too:
5² = x(x+2r) = x² + 2xr = Area
I did it a very different way which I thought might be interesting to share:
sec(θ) = 2 r
tan(θ) = 5
r² + tan²(θ) = sec²(θ)
r² + 25 = 4 r²
3 r² = 25
Area = 3 r² = 25
omfg that’s wild. I knew what the steps would be, but didn’t feel like writing it down, so I was thrown for a loop when the equations were the same
I miss this. I studied engineering and now working as a Project Manager is cool but I won't ever need to use these skills...
Both x and r can vary between 0 and 5.
When x = 5, r = 0.
When r = 5, x = 5*((sqrt(2)-1), approx. 2,08.
If x = r, => x = r = 5/sqrt(3), approx. 2,89.
Pretty exiting stuff.
x = 2.5
r = 3.75
Oh this gave me such joy when that solution presented itself!
I always hated problems like this. We're taught not to assume that the x and r radii (as he labeled them) are the same because there's no proof in the diagram. At the same time, we then are forced to make the assumption that the 5 cm line is a tangent of which there's also no proof. You can't have it both ways.
You don’t have to assume, because he said it was.
The red line being tangent to the semicircle is a given not an assumption.
The diagram depicts it touching the semicircle at a single point. If it wasn't a tangent to the semicircle, then it would either intersect the semicircle at 2 points, or not touch the semicircle at all.
It’s crazy how simple that was
He actually said "isn't that cool" after solving this one so you KNOW it's an absolute BANGER
shot in the dark
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based on the fact that the 5cm line is tangent with the circle and thus perpendicular to it's radius, r, so 5² + r² = 4r² (4r² from squaring 2r, from the center of the semicircle to the left-hand edge of the rectangle) so 5² = 3r²
rectangle has an area of 3r² (r•3r), thus the area is...
25 cm²
edit: so. the exact logic used in the video, except for the unnecessary assumption that x = r on my part. whoops. basically replaces 2xr and x² with 2r² and r², respectively
Couldn't you assume the area is the same for any r, set r=0, and then just solve for x^2 where x=5? Something something not to scale. Also, this reminds me of the napkin ring problem, but in 2D.
This is very clever!
Though I really like this and it leads to some good intuition, I did think of an issue: you can't assume the area will always be the same. In this case it is, so if you find the area for r = 0, you've found it for all valid r, but I can't think of how you'd jump to your initial assumption unless you already figured it out the other way.
@@atrus3823 The way I reason it: You're solving for the area. The radius of the right semicircle isn't given. There is an infinite range of semicircles with an infinite range of radii that have a tangent line that length through that corner (easily provable by the fact that If you draw a line from the corner to any given point the given distance away to the bottom right, you could then construct a semicircle tangent to that line flush with the horizontal line in the problem and then construct a quarter circle and then a rectangle from there). So given you're solving for an area of an infinite range of possible shapes, all the shapes must have the same area, unless we weren't given enough information to solve the problem in the first place. So a semicircle of infinitely small radius must result in the same area. So you can do the simple solve by treating r=0. If all the questions have the same answer, solve the easiest question.
@@atrus3823 I guess you have to assume that there is a solution, ie a number and not an expression that depends on x or r. Since we cant assume any relationship x/r between x and r this would have to be the same for any r arbitrarily close to 0, and also r=0 (?). Ie x/r arbitrarily large.
Also see the comment by stirdix. I’m really impressed by this setting r=0 idea!
So basically for anything: when life gives you circles, make triangles.
wow i just found the area of a quarter circle with the radius of 5 and subtracted the triangle to find the area of half the overlap and just double it
Think r radius getting small until it equals to zero. Then 5cm line's corner create with middle of quarter circle getting small and it lands on rectangle base.
So we can assume
X+r+r = 5
Because r≈0
X = 5
So 5x5 rectangle area = 25
Sorry for bad English. I hope you understand.
I swear if this dude was my math teacher I would be willing to get extra homework
Sorry if this had been answer before but, how do we know that the angle is a right angle? Better said, how do we know it is a tangent line?, Thank you in advance
We know that the red line is tangent to the semicircle because Andy states that it is when he's explaining the question.
It is therefore a given of the question.
If it wasn't a tangent to the semicircle, then it would either intersect the semicircle at 2 points, or not touch the semicircle at all. Neither of these are the case in the diagram.
Apply the Tangents to a Circle Theorem, and you've proven the right angle.
Not strict but fastest solution: let r=0, then x=5 and area =x*x=5*5=25.
Or just apply secant - tangent theorem, that's is 5² is equal to (x)(x +2r) which is the width × height
And boom here's your area in one line
I think I’ve found a new way to stay sane. No, I’m not genius. Or even good at math. But I love getting lost in it.
interesting part is this: the answer shows that the radii of both circles are equal
It's cool that you can solve this but I'm so dumb I can't even think of why or how this would translate into building or any type of work, Pythagorean helps by squareing off fences to building
Tested this in Fusion 360 to see if the area was 25 cm^2, and this is indeed true. How Exciting.
The line could have not been tangential to the circumference
bro makes me open youtube instead of scrolling instagram at 2 am