I’m a mechanical engineer and haven’t been in school for nearly 30 years. I still watch these to see how neat and tidy most of these problems end. It is satisfying to put a box around a whole number. Maybe because real life is never that clean. I’m afraid I might start saying “how exciting” when a meeting ends or when we solve a problem at work.
🎉Area is 36.🎉 Let radius of small circle is r. Using same technique as used in previous problem, half the base of triangle is √3r. Cut triangle vertically in half, and using tan60°, height of triangle is 3r. Same is the height of rhombus.. and hence diameter of circle is also 3r. So radius of circle is 3/2(r) Therefore area is 9/4(πr²) =9/4(16) =36🎉🎉🎉🎉
You don't need to calculate using the angles at all. If you look at the diagram at 1:20 of the video, the 30-60-90 triangle shows that the top point of the equilateral triangle is 2 times the radius of the circle from the center. Given that the center of the circle is 1 radius base the base of the triangle, the triangle therefore has a height of 3 radii. And since there's no numbers involved it means that in an equilateral triangle, the height of the triangle will always be 3 times the radius of an inscribed circle.
@andymath … Love your channel! Aren’t there an infinite number of top circles that are tangent to the semicircle and have a point on that long diagonal? Why can’t we shift the top circle left or right, making its area bigger or smaller? I think, at 2:38, you have to assume that the center of the top circle lies on the radius of the semicircle that bisects the long diagonal. If that radius bisects the line, then it is a 90 degree angle at the point of tangency. Otherwise, it might not be, and the area of the circle could be any nonzero number between 0 and 8.
Or you have to assume that the radius of the top left circle that intersects the long diagonal at the point of tangency is on the same line as the radius of the semicircle. That isn’t necessarily true. If you moved the top circle left or right, those radii would rest on different lines.
Finally back, huh. I've been waiting for this vid with anticipation since you do great insights and generalize the idea of the end result a pot of the time, which is awesome from my point of view
36 the height of the equilateral triangle is 1.5 times the height of the inscribed circle, the diameter of the bigger circle is the height of the triangle, so the radius of the big circle is 1.5 times the radius of the little one area=pi(1'5r)^2 area=2.25*pir^2 pir^2=16 so area is 2.25*16=36 also, where you been? we missed you
Answer to the next puzzle (will it be tomorrow?) Left circle radius = √(16/π) The height of the equilateral triangle = 3√(16/π) = the diameter of the right circle The area of the right circle = 36
Wasn't sure how to approach this until I saw you mention height of the equilateral triangle, then it clicked. I stopped reading your comment and worked it out myself. Thanks.
I did it differently: r of yellow circle (which I will just call r) is (R-x)/2, where R is the radius of the big circle and x is the distance from O to the blue chord line. This means A = π(R-x)²/4. Consider the right triangle with sides x, R, and chord/2. The angle between x and R is 30º because the big angle on the circumference is 90º and the portion of it which is spanned by the angle of the equilateral triangle is 60º. This means x/R=sin30º, so x=R/2. A = π(R-x)²/4 = π(R-(R/2))²/4 = π(R/2)²/4 = πR²/16 Consider a right triangle inside the equilateral triangle made up of hypotenuse d, the distance from the centre of the equilateral triangle to its vertex, r of the red circle, and R/2. d= R/√3, (the distance from the centre of an equilateral triangle to its vertex is (1/√3) times its side length). r of the red circle = √(8/π) So (R/√3) = (R/2)² + (√(8/π))² R²/3 = R/4 + 8/π R²/12 = 8/π R²= 96/π A = πR²/16 = π(96/π)/16 = 6.
2:39 How do you know that the radius of the semi-circle would pass through the center of the small circle? What if it doesn't pass through it in a straight line?
I was also hoping for a better explanation of this intersection. It should be noted that the small circle is presumed to be tangent to the semicircle at such a point that allows the small circle to be as large as possible, which means it has to be perfectly in the middle of that circle segment, meaning it's symmetrical, and that's why the cross-intersection between the chord and the large radius are all at exact right-angles.
It's because the circles are not independent from each other. The length of a side of the triangle, which is also the radius of the semicircle, depends on the length of the radius of the larger circle. The length of the radius of the smaller circle in turn depends on the length of the radius of the semicircle.
@jamessanchez3032 ahh right I think I understand. I was confused because I was just assuming r was the symbol for all radius' rather than a value for that specific circles radius
My answer for tomorrow's question is: 36 square units. In my calculations, i got the answer by finding the radius of the small circle, multiplying it by 3/2, then squaring it and multiplying it by pi to get the area of the big circle. Idk tho, i could be wrong
jsut middle school student trying i dont know whetehr its correct or not, the answer for next question is 31, here the work - - pir^2 = 16, r = 4/rootpi, draw the line that start from middle of top trangle, and the degree will be 30 since all the angle are same and 60 60.2 = 30 , so sin30/4/rootpi= sin90/x , so x = 4.5 we now have right trangle so a^2 +(4/rootpi)^2 = 4.5^2, a = 4.32, and since a is equal to radius of big circle, now you plugi into formula and got = 31.00627
Ok, hol'up right there man. First of all I'll start by saying the area is actually exactly 36 NOT approximately 31, secondly you didn't need the sine law at all here, 30-60-90 triangles are well known so you could've just used the fact that the hypotenuse of that triangle was gonna be 2r (exactly as Andy did in this video's puzzle, and lastly the "a" variable you named is NOT the radius of the big circle at all, it's just the length of a tangential line to the circle. Now, as for how I reached my answer. First do literally everything Andy did at the start of the video till you get the 2r length. Then extend that same line to the base of the equilateral triangle, you get an extra length of r to the height of the equilateral triangle. Hence the height of the equilateral triangle is 3r. Since the yellow shape is a rhombus, we know each 2 opposite sides are parallel, the circle touches the rhombus at the ceiling and the base at exactly 90deg angles. So the height of the rhombus is 2R (assuming R is the big circle's radius). The height of the equilateral triangle is exactly the the same as the height of the rhombus. So 3r = 2R, R = 3r/2. Hence the area = π*(3r/2)² = π*9r²/4 = 9/4 * πr² = 9/4 * 16 = 36
@@Grizzly01-vr4pn I tried keeping up with his steps to understand where he went wrong, and I think it was only when he assumed that the variable he named "a" (which is r√3 according to normal calculations without substituting for r) is the radius of the big circle. The other steps are right, just approximated early on.
@@catalinagonzalezoliva2738 Nah, it's a geometry & Pythagoras application. Trigonometry just stemed from it. Watch khan academy's geometry playlist and you'll find out all about it, you won't be disappointed trust me. Also it's steming from the idea of similar triangles.
Btw 30-60-90 triangles are taught mostly in, like, 7th or 8th grade (not sure which it is) in most education systems I know, just go back to your old textbooks if you're older
It has to. A chord like this forming a 60 degree angle will be the big radius long, and that point is the big radius from the center because it's the big radius.
It may be a little clearer if you go back to the start and label the 8 circle radius as x maybe. Then x is just a unit of length. Carry that through to the end and then the general formula for area of a circle will contain r but the radius of the ? circle will be a length containing x. A different length to the 8 circle radius, but still related because they will both be in terms of x.
In this case, “r” isn’t a variable. It’s a constant. It wasn’t directly calculated, but you can treat is like any other number in your formulas. He could have done the calculation, but the benefit of formulas is you can leave constants alone until you need to use them for a final calculation to keep things neat. r = 1.59578… and to avoid writing that down twelve times and adding rounding errors, he left it as r.
At 4:20 you claim Pi-r^2=8 to satisfy that expression in the new formula 3Pi-r^2/4... but... those two "r"s refer to the radii of two different circles and you didn't prove that the two circles have the same radius.
He's measuring the second circle in terms of the first circle's radius, so the length of the smaller circle's radius is √3/2 times the length of the first circle's.
i really like it when a complicated puzzle ends with the answer being an integer
My answer to tomorrow's puzzle is 36
@@adityagoyal7110 You mean “the next puzzle”? which may not be tomorrow’s puzzle.
@@adityagoyal7110 I also got that.
longest time for a haircut ever
Incognito. 😊😊
The best haircuts take 3 days.
Lol, btw is answer of tomorrow's puzzle is 36 units² ¿
@@adityagoyal7110 Since the radius is 3/2 the length of the other one, the area is 9/4 the area of the other one. 16 * 9/4 = 36. :D
THE KING IS BACK
We were getting worried… glad you’re back, how exciting!
I’m a mechanical engineer and haven’t been in school for nearly 30 years. I still watch these to see how neat and tidy most of these problems end. It is satisfying to put a box around a whole number. Maybe because real life is never that clean. I’m afraid I might start saying “how exciting” when a meeting ends or when we solve a problem at work.
I already have!
Do it!
🎉Area is 36.🎉
Let radius of small circle is r.
Using same technique as used in previous problem, half the base of triangle is √3r.
Cut triangle vertically in half, and using tan60°, height of triangle is 3r.
Same is the height of rhombus.. and hence diameter of circle is also 3r.
So radius of circle is 3/2(r)
Therefore area is 9/4(πr²)
=9/4(16)
=36🎉🎉🎉🎉
You don't need to calculate using the angles at all. If you look at the diagram at 1:20 of the video, the 30-60-90 triangle shows that the top point of the equilateral triangle is 2 times the radius of the circle from the center. Given that the center of the circle is 1 radius base the base of the triangle, the triangle therefore has a height of 3 radii. And since there's no numbers involved it means that in an equilateral triangle, the height of the triangle will always be 3 times the radius of an inscribed circle.
2:07 moving the radius made the funny shape
I love the duck down exit! Glad you’re ok!
Glad you are back! Will we get the catch up videos as Christmas Day presents? How exciting!
Nice haircut too!
Love your channel, BTW!!!!!
I have never been more concerned about the well being of a TH-cam content creator than I was these last few days. How exciting!
He back
🎉🎉🎉🎉
Glory be. You're back ! Welcome, and thanks now we can continue with our Christmas season.
That was really very neat, had the same method too. Glad you've found some time, welcome back.
Welcome back Andy
I really missed you 😊
May not have been able to do one every every day but it’s nice how you still do it
@andymath … Love your channel! Aren’t there an infinite number of top circles that are tangent to the semicircle and have a point on that long diagonal? Why can’t we shift the top circle left or right, making its area bigger or smaller? I think, at 2:38, you have to assume that the center of the top circle lies on the radius of the semicircle that bisects the long diagonal. If that radius bisects the line, then it is a 90 degree angle at the point of tangency. Otherwise, it might not be, and the area of the circle could be any nonzero number between 0 and 8.
Or you have to assume that the radius of the top left circle that intersects the long diagonal at the point of tangency is on the same line as the radius of the semicircle. That isn’t necessarily true. If you moved the top circle left or right, those radii would rest on different lines.
Finally back, huh.
I've been waiting for this vid with anticipation since you do great insights and generalize the idea of the end result a pot of the time, which is awesome from my point of view
I feel like the next problem is going to be the same from the first problem if I’m looking at it correctly. Nice hair cut.
Oh thank God you're not dead! I only just found your channel a few days ago, would have been really frustrating.
And he got a haircut. How exciting!
I liked his wild hair 😭 but he is still hot with this style. I'm so happy he puts out this content. His enthusiasm makes me enthusiastic.
The king is back again+ the first day i solve a problem
36
the height of the equilateral triangle is 1.5 times the height of the inscribed circle, the diameter of the bigger circle is the height of the triangle, so the radius of the big circle is 1.5 times the radius of the little one
area=pi(1'5r)^2
area=2.25*pir^2
pir^2=16
so area is 2.25*16=36
also, where you been? we missed you
This one was very exciting!
I solved this one on my own, and I put a box around the answer. I was so excited to put that box around it.
How exciting!
Great problem! Uses many different geometric theorems all together.
Answer to the next puzzle (will it be tomorrow?)
Left circle radius = √(16/π)
The height of the equilateral triangle = 3√(16/π) = the diameter of the right circle
The area of the right circle = 36
Wasn't sure how to approach this until I saw you mention height of the equilateral triangle, then it clicked. I stopped reading your comment and worked it out myself. Thanks.
@ It’s a fun puzzle. Good job 👍 and happy holidays 😊
@@cyruschang1904 You too.
He just went for a haircut. Thats why hes late
Your haircut looks nice🎉
I did it differently:
r of yellow circle (which I will just call r) is (R-x)/2, where R is the radius of the big circle and x is the distance from O to the blue chord line.
This means A = π(R-x)²/4.
Consider the right triangle with sides x, R, and chord/2. The angle between x and R is 30º because the big angle on the circumference is 90º and the portion of it which is spanned by the angle of the equilateral triangle is 60º. This means x/R=sin30º, so x=R/2.
A = π(R-x)²/4 = π(R-(R/2))²/4 = π(R/2)²/4 = πR²/16
Consider a right triangle inside the equilateral triangle made up of hypotenuse d, the distance from the centre of the equilateral triangle to its vertex, r of the red circle, and R/2.
d= R/√3, (the distance from the centre of an equilateral triangle to its vertex is (1/√3) times its side length).
r of the red circle = √(8/π)
So (R/√3) = (R/2)² + (√(8/π))²
R²/3 = R/4 + 8/π
R²/12 = 8/π
R²= 96/π
A = πR²/16 = π(96/π)/16 = 6.
You're amazing
Yeaaa boii he's backkk
how exciting!
36 is the area of circle inside rhombus
I got the same answer!
@@Waterhiyouhey how exciting
How exciting.
Nice haircut!!!
2:39 How do you know that the radius of the semi-circle would pass through the center of the small circle? What if it doesn't pass through it in a straight line?
The centers of tangent circles are collinear.
I was also hoping for a better explanation of this intersection. It should be noted that the small circle is presumed to be tangent to the semicircle at such a point that allows the small circle to be as large as possible, which means it has to be perfectly in the middle of that circle segment, meaning it's symmetrical, and that's why the cross-intersection between the chord and the large radius are all at exact right-angles.
Guess who's back 🎶🎵 back again
Aggvent calender day 16- the area of the circle inscribed in the rhombus is 36 square units
welcome backkk
Amazing
Tu t'laisses pousstache ???
The area of the red circle looks to be about 24, put a box around that.
This one was hard. Nice one!
This problem is unsolvable - you cannot prove that the 2 circle centers and the tangent Point are on the same line. :)
The only thing I'm not understanding is that the final step he uses Pi R² equals 8 to solve for a completely different circle...
But it doesn't say that point O is the center of the diameter. Can't assume that the side of the equilateral triangle is a radius of the semicircle.
The center of a circle is traditionally designated as O.
I'm very confused by the fact that you can use the area of one circle with a different radius to work out the area of another circle.
It's because the circles are not independent from each other. The length of a side of the triangle, which is also the radius of the semicircle, depends on the length of the radius of the larger circle. The length of the radius of the smaller circle in turn depends on the length of the radius of the semicircle.
@jamessanchez3032 ahh right I think I understand. I was confused because I was just assuming r was the symbol for all radius' rather than a value for that specific circles radius
man looks so different with a beard 😭
My answer for tomorrow's question is:
36 square units. In my calculations, i got the answer by finding the radius of the small circle, multiplying it by 3/2, then squaring it and multiplying it by pi to get the area of the big circle. Idk tho, i could be wrong
❤
jsut middle school student trying i dont know whetehr its correct or not, the answer for next question is 31, here the work - - pir^2 = 16, r = 4/rootpi, draw the line that start from middle of top trangle, and the degree will be 30 since all the angle are same and 60 60.2 = 30 , so sin30/4/rootpi= sin90/x , so x = 4.5 we now have right trangle so a^2 +(4/rootpi)^2 = 4.5^2, a = 4.32, and since a is equal to radius of big circle, now you plugi into formula and got = 31.00627
Ok, hol'up right there man.
First of all I'll start by saying the area is actually exactly 36 NOT approximately 31, secondly you didn't need the sine law at all here, 30-60-90 triangles are well known so you could've just used the fact that the hypotenuse of that triangle was gonna be 2r (exactly as Andy did in this video's puzzle, and lastly the "a" variable you named is NOT the radius of the big circle at all, it's just the length of a tangential line to the circle.
Now, as for how I reached my answer.
First do literally everything Andy did at the start of the video till you get the 2r length.
Then extend that same line to the base of the equilateral triangle, you get an extra length of r to the height of the equilateral triangle.
Hence the height of the equilateral triangle is 3r.
Since the yellow shape is a rhombus, we know each 2 opposite sides are parallel, the circle touches the rhombus at the ceiling and the base at exactly 90deg angles.
So the height of the rhombus is 2R (assuming R is the big circle's radius).
The height of the equilateral triangle is exactly the the same as the height of the rhombus.
So 3r = 2R, R = 3r/2.
Hence the area = π*(3r/2)² = π*9r²/4 = 9/4 * πr² = 9/4 * 16 = 36
Not quite.
Your answer should be an integer.
@@Grizzly01-vr4pn I tried keeping up with his steps to understand where he went wrong, and I think it was only when he assumed that the variable he named "a" (which is r√3 according to normal calculations without substituting for r) is the radius of the big circle. The other steps are right, just approximated early on.
@@Z-eng0 thanks
@@Z-eng0 thanks man
Don’t worry I watch at nights
what do you use to edit all these problems
Was beginning to get genuinely concerned for your welfare.
Welcome back.
And a haircut, too. HOW EXCITING
I know nothing about 30-60-90 triangles 😭😭😭
It's an application of trigonometry and pythagoras
@@catalinagonzalezoliva2738 Nah, it's a geometry & Pythagoras application.
Trigonometry just stemed from it.
Watch khan academy's geometry playlist and you'll find out all about it, you won't be disappointed trust me.
Also it's steming from the idea of similar triangles.
Btw 30-60-90 triangles are taught mostly in, like, 7th or 8th grade (not sure which it is) in most education systems I know, just go back to your old textbooks if you're older
Andy please give some tips for (IOQM) preparation
How do you know the equilateral triangle has one vertex on the center of the semi circle?
It has to. A chord like this forming a 60 degree angle will be the big radius long, and that point is the big radius from the center because it's the big radius.
Why did the orange circle's diameter have to be root 3 r? I guess I missed the explanation
1:30 to 2:15, he explains it as well as anyone could
Wait the background changed
Are you moving or something and that's why you didn't post videos for a while
what is he cooking again
Bro ur like a week behind please pre record these lmao
Andy how's the answer not 16?
I verify that it is 16 and not 6 in my calculations
Maybe I will try again
Turns out I forgot to divide the diameter of the unknown circle by 2
how can u sub in the pi r squared value of one circle for another one
It may be a little clearer if you go back to the start and label the 8 circle radius as x maybe. Then x is just a unit of length. Carry that through to the end and then the general formula for area of a circle will contain r but the radius of the ? circle will be a length containing x. A different length to the 8 circle radius, but still related because they will both be in terms of x.
In this case, “r” isn’t a variable. It’s a constant. It wasn’t directly calculated, but you can treat is like any other number in your formulas. He could have done the calculation, but the benefit of formulas is you can leave constants alone until you need to use them for a final calculation to keep things neat. r = 1.59578… and to avoid writing that down twelve times and adding rounding errors, he left it as r.
Nice
I have solved it myself but ❤
Your new haircut made you look like Mrbeast lol
New question's answer is 36 sq. Units
Yipppe
Day 16: 36 sq units
i got 48 but i will double check!
i assumed that all of the rhombus sides where the same length! i understand where 36 comes from now!
At 4:20 you claim Pi-r^2=8 to satisfy that expression in the new formula 3Pi-r^2/4... but... those two "r"s refer to the radii of two different circles and you didn't prove that the two circles have the same radius.
He's measuring the second circle in terms of the first circle's radius, so the length of the smaller circle's radius is √3/2 times the length of the first circle's.
Next problem: 36.
40 second 😆✌️
Hi
6
first
second
One thousand four hundred thirty ninth
Second
You're late.