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
I always sucked at math. The numbers always kept getting jumbled and I could not memorize the formulas if my life depended on them. Your videos on the other hand are delightful. You really know how to make math short, concise, and pretty fun. Love these vids!
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.
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
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.
The solution only holds for any size half-circle if you adjust accordingly the length of the 5cm line tangent to the circle. On a bigger or smaller half circle, the 5cm line would not be be tangent to the circle. And you wouldn't get the square triangle needed for the solution.
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.
Oh i really love this one. Cause i was going everywhere and everythink started making sense and then, you showed the solution, bypassing ALL my steps by just... using math. Love it.
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!
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 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.
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 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❤❤❤
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.
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!
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 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!
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
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
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
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
This is one of the few geometry puzzles that I find absolutely incredible. Usually, you use concepts to measure certain sides to use to find an area but somehow the equation of the triangle also leads us for the area of the rectangle. Coming up with a puzzle like this is a different level of genius. I would love to see a more in depth explanation of this. What is the relationship of the hypothenuse of the triangle to the area of the rectangle? Why does squaring it gives us the formula for the area? I don't think it is just a happy coincidence.
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
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.
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.
Well, an easy way that you can use in your head: knowing the radius of the left circle is square root of (25+r^2) - r, which is the short side of the rectangle, and the long side of the rectangle is square root of (25+r^2) + r, the area will be the multiplication, that is 25 +r^2-r^2 which is 25.
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 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 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.
As there's no restriction to the size of each semi-circle, the choice of the angle at which the 5cm segment is inclined is arbitrary. We can thus choose 45 degrees to make the problem much simpler. The Radius of the circle on the right then becomes 5cm, forcing the radius of the circle on the left to become sqrt(50) - 5. We get the horizontal length of our rectangle with sqrt(50) - 5 + 2*5 = sqrt(50)+5 The height of the rectangle is simply equal to the radius of the left circle so the area of the rectangle is obtained with (sqrt(50) - 5) * (sqrt(50) + 5) = 50-25 = 25
I size of half circle does not matter, then make it very big or very small. Now the 5cm line would not be tangent to the circle. The solution only holds for any size half-circle if you adjust accordingly the length of the 5cm line tangent to the circle. On a bigger or smaller half circle, the 5cm line would not be be tangent to the circle. And you wouldn't get the square triangle needed for the solution.
The solution only holds for any size half-circle if you adjust accordingly the length of the 5cm line tangent to the circle. On a bigger or smaller half circle, the 5cm line would not be be tangent to the circle. And you wouldn't get the square triangle needed for the solution.
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
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
Andy, I agree that this if a fun. The solution only holds for any size half-circle if you adjust accordingly the length of the 5cm line tangent to the circle. On a bigger or smaller half circle, the 5cm line would not be be tangent to the circle. And you wouldn't get the square triangle needed for the solution. Just saying…
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.
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.
Got the same answer even though I assumed the radius of the quarter circle was the same as the radius of the half circle. This makes me want to make an animation that shows x getting larger and smaller and how the dimensions of the rectangle change, but the area stays the same.
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.
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.
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.
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.
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.
![Felix "Unfair" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/Oswujxm2Ag0/mqdefault.jpg)
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?
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
I have not done math for 3 years since my studies are in different fields, but this was truly exciting. Nice video!
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!"
this guy solved the problem before i even knew what was happening
lmao same
x3
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.
you know a problem is even more exciting than usual when he hit us with the "isn't that cool?"
I always sucked at math. The numbers always kept getting jumbled and I could not memorize the formulas if my life depended on them. Your videos on the other hand are delightful. You really know how to make math short, concise, and pretty fun. Love these vids!
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.
Your voice turns into Schwarzenegger at around 2:08 for a second, did you do that on purpose? :D
Your enthusiasm when saying "isn't that cool? :)" is really lovely
yeah its not the usual depressed "how exciting"
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
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.
The solution only holds for any size half-circle if you adjust accordingly the length of the 5cm line tangent to the circle. On a bigger or smaller half circle, the 5cm line would not be be tangent to the circle. And you wouldn't get the square triangle needed for the solution.
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.
He is in awe
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!
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
Oh i really love this one. Cause i was going everywhere and everythink started making sense and then, you showed the solution, bypassing ALL my steps by just... using math. Love it.
Thank you. You're educating and helping me recover so much lost knowledge.
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!
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 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.
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!
It’s crazy how simple that was
Ah - this one is not only exciting, but also cool!
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❤❤❤
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.
Holy s**t. I was looking at this for so long and just said ‘nope’. Wow I love this channel!!!
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!
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.
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!
These problems actually remind me of the art of problem solving intro to geometry book.
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 😂
Dang you bring my math vibes back to prime bro. I wanna try solving some math problem
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!
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
as soon as I wrote out the Pythagorean for the tangent line I was like "wait a minute" surprisingly simple
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
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.
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
I love your enthusiasm for math
This is one of the few geometry puzzles that I find absolutely incredible. Usually, you use concepts to measure certain sides to use to find an area but somehow the equation of the triangle also leads us for the area of the rectangle. Coming up with a puzzle like this is a different level of genius.
I would love to see a more in depth explanation of this. What is the relationship of the hypothenuse of the triangle to the area of the rectangle? Why does squaring it gives us the formula for the area? I don't think it is just a happy coincidence.
That's a really neat solution!
Thank you, I thought so too. I was excited to share it!
As one of the only people I know who likes math, your videos make me very happy.
How exciting
Such an exciting problem visual, looked difficult at first but after making that extra construction it obly took 2 mins
I like these because looking at them makes you think differently than you're used to. Math isn't hard, you just have to learn it.
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.
can watch these all day
Oh this gave me such joy when that solution presented itself!
that was honestly so cool. its one of those answers that sneaks up on you!
A reverse problem would look even more sophisticated and exciting.
It pleases me greatly how many of these puzzles are solved by drawing a right triangle.
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
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.
I swear if this dude was my math teacher I would be willing to get extra homework
He actually said "isn't that cool" after solving this one so you KNOW it's an absolute BANGER
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 was a huge STEM kid but moved into Law when I was at Uni. Watching these vids make me miss STEM 😂
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
What an elegant answer!
I like this. Seems hard at a glance but really intuitive when you sit down with it
So basically for anything: when life gives you circles, make triangles.
The way you found The answer makes me so happy 😊
Well, an easy way that you can use in your head: knowing the radius of the left circle is square root of (25+r^2) - r, which is the short side of the rectangle, and the long side of the rectangle is square root of (25+r^2) + r, the area will be the multiplication, that is 25 +r^2-r^2 which is 25.
seemingly difficult yet clean and simple, nice
I got soooo lost. Not because you didn't explain well, it's my thick skull. I'll watch again till I get it
What an elegant solution BRAVO!!!!
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 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.
Love your videos! Keep up the great work.
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.
As there's no restriction to the size of each semi-circle, the choice of the angle at which the 5cm segment is inclined is arbitrary. We can thus choose 45 degrees to make the problem much simpler. The Radius of the circle on the right then becomes 5cm, forcing the radius of the circle on the left to become
sqrt(50) - 5.
We get the horizontal length of our rectangle with
sqrt(50) - 5 + 2*5 = sqrt(50)+5
The height of the rectangle is simply equal to the radius of the left circle so the area of the rectangle is obtained with
(sqrt(50) - 5) * (sqrt(50) + 5) = 50-25 = 25
I size of half circle does not matter, then make it very big or very small. Now the 5cm line would not be tangent to the circle. The solution only holds for any size half-circle if you adjust accordingly the length of the 5cm line tangent to the circle. On a bigger or smaller half circle, the 5cm line would not be be tangent to the circle. And you wouldn't get the square triangle needed for the solution.
The solution only holds for any size half-circle if you adjust accordingly the length of the 5cm line tangent to the circle. On a bigger or smaller half circle, the 5cm line would not be be tangent to the circle. And you wouldn't get the square triangle needed for the solution.
MAGIC! And it sounds so soothing as well
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
Fun for the sake of fun. A lot of us lose this being adults in a rat race. Thank you.
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.
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Andy, I agree that this if a fun. The solution only holds for any size half-circle if you adjust accordingly the length of the 5cm line tangent to the circle. On a bigger or smaller half circle, the 5cm line would not be be tangent to the circle. And you wouldn't get the square triangle needed for the solution. Just saying…
This is a very cool problem, good job!
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.
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.
This is the definition of trust the process
this is genius, I never thought to think of it like this, I feel stupid now.
I was fully prepared for this video to somehow use sine and tangent lines, I guess I gotta start doing some actual math problems
it was so exciting that he had to add something to the regular end of the video
Got the same answer even though I assumed the radius of the quarter circle was the same as the radius of the half circle. This makes me want to make an animation that shows x getting larger and smaller and how the dimensions of the rectangle change, but the area stays the same.
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.
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.
Didn’t we were supposed to assume the red line was a tangent line…
Very nice. Very succinct explanation, too. Well done!
Brilliant. Love your videos. 👏👏👏
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.
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.
this is one of the coolest 30 seconds problem!
I really enjoy this kind of videos 😅
Lets gooooooo!!!!! I got this one on my own! Time to watch the video! :D
Accidentally assumed both circles had the same radii but luckily I did every other step the same way and it worked out!
bro makes me open youtube instead of scrolling instagram at 2 am
Freaken awesome problems! Beautiful
Oh, neat. So this gives a way to construct a square with the same area as a rectangle, by drawing two circles and constucting a tangent line.
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.
I have to say. This makes me want to read more math.
DID NOT EXPECT TO BE THIS SIMPLE bravo