🤓Actually it is one of the most imp theorem of tangents that is it is always perpendicular to the point of contact to the radius that is the angle will be always 90 and it also have its proof 🤓
I was like "one side is 4, the other is a little less, let's say 16, and if we are buying tiles or something, just get 20 to give us a margin of error"
this is the first one i was able to solve myself, and i did it instantly, i knew that the distance through the circles was pretty much the only thing we knew, so when i drew it out in my head i recognized the diagonol as a hypotenuse, made the triangle and knew the answer. very proud
Same, except I did it by drawing an equilateral triangle side 2cm and then using 1/2*a*b*SinC to find the area, where I then rearranged 1/2*base*height to find the height, added 2 to the height to find the total rectangle height, and then multiplied it all by 4. Got 14.9282.
Same, I did the entire thing mentally while laying down on the bed lol. Height of an equilateral triangle is (root3)/2 times the side. Connect all the centres of the circles and the answer becomes obvious.
@@itsjustme9691 because 14.928 is not the exact answer. It's 14.9282032303... and keeps going on forever. This is what you would call a irrational number. Whenever a root of a number results in an irrational number you should leave the root in the final answer in order to get the exact correct answer
This channel doesn't just teach me math. It also teaches me to approach problems in ways that I'd never consider on my own. I think this is fantastic. I never made it to Algebra 2 in high school before quitting. I did somehow place into Algebra 3 at my local community college and got out with an A, an A that I lost a lot of sleep trying to earn. I love math, but I hate school. Resources like this channel are great. Thank you.
I know absolutely nothing about maths beyond the basics but this is the 4th video of yours that I've watched and I must really be learning a lot from you as I actually figured this one out for the most part, so thank you 👍🏾🙏🏾
Question: at 0:42 when you drew the two red lines, how do you know the hypotenuse of this triangle is straight and intersects at the tangent point between the two circles? I was going to solve it like that initially but wasn’t sure if it was correct to make this assumption (the assumption being that the hypotenuse is 2cm)
Well, consider the fact that those two circles meet at a point; that means that you can draw a straight line between their centers. Since their radii are equal, we know that that distance is 2 (1cm (the first radius) + 1cm (the second radius)) Hope that helps?
2 circles touching means there is a straight line between the circles centres, and the line distance is always r#1 + r#2 or in our case, since the circles are the same size its r+r or 2r. And since we know the circles are 1cm, it's just 2.
Sees thumbnail "Oh, this is gonna be tricky~" Sees video duration "It's not even 2 minutes?!?!" Sees the solution "Dang, I never thought of triangle in there"
my solution involved seeing that the centers of the circles formed an equilateral triangle, then cutting that in half to make a 30/60/90 right triangle with two known side lengths. yours was a lot less complicated!
My attempt: The width of the rectangle is equal to the sum of the diameters of the top two circles. Radius of 1 means diameter of 2, 2*2=4, so the width is 4. The height, meanwhile, is equal to the radius of the top two circles, plus the height of the triangle formed by the centers of the three circles, plus the radius of the bottom circle. That triangle is an equilateral triangle with side length 2cm. Thus, its height is 2sin(60), or sqrt(3). The radius of all three circles is 1, so that's 1+sqrt(3)+1, or 2+sqrt(3). Area is width times height, so the area of the rectangle is 4(2+sqrt(3)), or 8+4sqrt(3)
@@shreyshah110 Draw a line between any two of the circles' centers. Because the circles are all tangent to one another, the length of the line will be equal to the sum of the two circles' radii. And because all the circles have the same radius of 1cm, that line will always be 2cm long, regardless of which circles you use. Thus, the triangle of centers is equilateral
When adding a unit, this is actually a multiplication. So the result is not 8 + 4 * SQRT(3) cm² but either 8 cm² + 4 * SQRT(3) cm² or alternativly [8 + 4 * SQRT(3)] cm² . Otherwise you would add a scalar to an area, which is rather awkward.
Had a similar question. At the same time stamp, how can we know that it's a 90° triangle? It feels like he skips a step here. Don't we need to prove it's a right triangle first? Or that the proposed hypotenuse is actually a straight line? Studied languages and lit at the uni, but math has been slowly growing on me over the years.
@@kathorseesfirst of all, the proposed hypotenuse is a straight line because it's just the shortest distance between the centres and it goes through the tangent (he deliberately created this line to be straight so that he could use Pythagoras on the resulting triangle). Second of all, I see your point here, the proof that the triangle is a right angled triangle is: if you were to draw a straight line between the top 2 circles' centres which goes through the tangent, then draw a straight line between the top left and bottom circles (similar to the first line he drew), you would get a an equilateral triangle connecting each of the circle centres, we know it's equilateral because each side is the length of 2 radii (2cm) If you half this equilateral triangle you get the right angled triangle he has in the video, and we know it's a right angled triangled because whenever an equilateral triangle is cut down the middle, it turns into a right angled triangle. Sorry if the explanation is bad, it's hard to explain without drawing it for you
It’s like I know how to do it, (find the area of the whole rectangle without the circles in it, then find the area of one of the circles, multiply it by three, and subtract that from the area of the rectangle without the circle, then you get the area of the little spaces in between everything) but I don’t want to do the math lol (Idk if my train of thought was way off tho lol) On further thinking I see it’s not a square lol,
So what im getting is, you find and classify the know, assign variables to the unknown, and form a right triangle to solve for the variables and maybe even a system of equations if nessesary, and you've got yourself a solution!
The three centers make an equilateral triangle with side = 2cm. It has 60 degrees angles. So the height of the triangle is h = (side * squareroot(3) / 2) = squareroot(3). So the height of the rectangle is (from top to bottom) = 1 + h + 1 = 2 + squareroot(3). So the area of the rectangle is: 4 * (2 + squareroot(3)).
Well, that was a spectacular failure for me. I got almost everything, minus some (no doubt crucial) difference in how to find X on the triangle, and I got 14.34 cm²... And on second attempt, 12.56. I suppose call it half marks for right idea.
I haven't been in a class since 2011 and that was end level computer science so it's been even longer since I had to look at this stuff... yet I'm super engaged with this channel content lol
I don't know the precise measurements, and I haven't watched the video, but my immediate gut instinct was to just call it 4x4. But it's not, the radius of the circles overlaps slightly. Maybe by a quarter. So it's nearer to 4x3.75 than to 4x4
That's what I did too. If I ever had to solve this in real life, like maybe we gotta fit these cylinders into a box, 4*3.75 is a valid solution, even though it's not the mathematically correct one.
Yay! Although I squared off the vertical radius of the lower circle from the tangent and used to give height = 1 + (sin60) + (sin60) + 1 Area = 4 * (2 + 2(sin60)) Same beans
@ThrowTop some people might say "oh, my bad" or "didn't get it but I do now", but the really cool cats just delete their comments, because that's very mature.
0:51 you don’t actually have to do Pythagorean Theorem to find X here. You can actually use the 30-60-90 right-triangle rule. This can be proven because if you draw 2 more radii (radius in plural) to mimic the right triangle and create an equilateral triangle (we know this is true because each side length of the equilateral triangle would equal to 2) and splitting it again just gives you the 30-60-90 right triangle. Back to the 30-60-90 right-triangle rule after proving to you all that you can use the 30-60-90 right-triangle rule instead of the Pythagorean Theorem to find x, we know that in a 30-60-90 right-triangle that opposite = x (whatever number you are using), adjacent = x times sq root of 3, and the hypotenuse = 2x. In this case, we can see here that 1 is the opposite, 2 is the hypotenuse, x is the adjacent, and that we are dealing with a 30-60-90 right triangle. Therefore, we can plug in the values we are using for the right triangle we are dealing with here (opposite = 1, adjacent = x, hypotenuse = 2). 1 times 1 = 1, 1 Times Square root of 3 = 1 (sq root of 3), and 1 times 2 = 2. Thus, the solution for x is simply sq root of 3. In conclusion, you don’t need Pythagorean Theorem to figure the right triangle shown in 0:51, you can instead use the 30-60-90 right-triangle rule. Note: Not all right-triangles are 30-60-90 right-triangles.
Since the sides are all the same length, I would think the answer is 8? since the top one is 4 meaning width, and the height of the rectangle is the same? Im bad at math but to me that makes sense NVM I misunderstood, still I'd think the answer is 16 because 4 * 4
@@truebye Huh? That is the only 'rectangle' I see, even though a rectangle is longer in one measurement right? But theres no rectangle at all in the picture
I keep seeing this come up in random places and know I can solve it, but I definitely didn't have the fastest solution. Puts my mind at ease knowing how simple it really is...
nice channel, problem is I hardly ever see the whole videos I simply take the problem from the cover and skip directly to the end to see if my results are right 😂
This is one of those peoblems, that are easy to approximate, without doing the weird shit. If we pretend that the cube is 4*4 than the area is 16. We know that it aint. And theres something like a .25-.5 overlap. I'll go with 0.25, so the height is 3.75. so ill say the area is in the ballpark of 15(or 17 at max). Now, ill watch the video. I dont think ill be off by more than 0.5 Edit: The final solution is 14.92
If you make a triangle by drawing lines from the centre of all circles you get a triangle with three equally big angles (60). the angle 60 provides that the "x" that you made has to be the standard angle ( sqrt3 / 2 ) x 2 which also gives square root of three plus 2 additional cm from the original. Then its just 4cm times (sqrt3 + 2vm) = 4(sqrt3) + 8 cm
Should be idk… close to 14.8 or so in a range of + or - 0.12? Edit jumped ahead to find the answer figuring it’d be some shenanigans which I seem to be right on, however if you plug it into a calculator I wasn’t far off :>
before you explain it let me answer it. for the length, just add up the diameter of the circle which means 4, for the height, just do the Pythagorean calculation from the root of 3, which is 1.73 and add 2, 1 from the radius. then 1 from the top which is clearly 1 last thing you just need do 4 x 5.73 = 22.92
Wasnt hard. The width is 4cm as itsbthe length of 2 circles with a radius of 1cm. The height is a bit more difficult. The height is 2cm (radius of bottom circle plus radius of one of the other circle) plus the height of the triangle formed by the 3 center points of the circles. The length of all 3 sides is 2 cm and the height of and equilateral traingle with s=2 is sqrt 3. So the height of the rectangle is (2+sqrt3)cm The area is 4cm x ((2+sqrt3)cm) Which equals (8+4sqrt3)cm^2
I'm embarrassed. I thought the rectangle was a square. And no, I didn't get 16cm^2 for the answer. I decided to find the more difficult side and square it. How exciting
I noticed that the center of each circle would form an equilateral triangle of side length 2cm. From there I remembered that you can split the 60° angle in two to get a 30-60-90 triangle, and that the lengths are 1,2,√3 .
Before I watch it: Notice the side length of the square is 4 times the radius, which is one. Therefore the area is 4² = 16cm² EDIT: I was foolish and thought that it was a square without proof
Didn't played the video, the solution can be, find the height of the triangle inside formed by joining centres. Add 2cms to it, it's the height of the rectangle. And 4cm is the width of the rectangle.
My HS trig teacher would take points off for the error in the final expression. Either put brackets around the math (everything before the units) or add the units after the “4”. Yes, he was that pedantic.
What if you make a triangle going from the top of the top left circle (or top right) to the bottom of the bottom circle to the right angle between both points at the bottom? You know the distance of the hypothenus (4x radius) and the base (2x radius) then Pythagorean Theorem and you got the height of the rectangle
I didn't watch this video before commenting we can see that top ones show that one of the x is 4 that side that we know is 100% bigger than x, but not much, so second x is 3
One thing I'd like to mention is that you do need to prove the triangle is right triangle. Then, the solution is complete. Otherwise, you can't assume it's a right triangle in the first place.
If it was a square and it said it was without a visual representation even, just 16 cm squared minus 3 pi because the area of a circle with radius 1 is always pi units, to get 6.58 cm squared
How come he didn't apply the same logic of the 1cm radius to the bottom circle? It looks equal to the other 2. How do you know to not assume the same for that circle?
I might be dumb but could you just plus the circls diameters to get the length and width then calculate area Idk maby im wrong or you wanted to go the long way round. I didnt watch the video
If there's one thing I have learnt, it's that there is always a right angled triangle whenever the balls touch.
so true, it felt like bro just summoned a right angled triangle there.
i love it when balls touch
the right angle is stored in the balls
My grandma used to tell me this every night at bedtime 🥲
🤓Actually it is one of the most imp theorem of tangents that is it is always perpendicular to the point of contact to the radius that is the angle will be always 90 and it also have its proof 🤓
Remember: if you're doing anything geometry related, the Pythagorian Theorum will save your ass.
I fully agree
Pythagoras - 2500 years later we still remember you and praise your work.
That's winning at life.
@@GustavSvard well, Pythagoras didn’t actually come up with the theorem himself, there are records of it in use far before he was even born
Al-kashi >>>>>>Pythagoras
@@MathsAbde-q2e🤓
My attempt: look at it for about 8 seconds, "ehhh about 15".
Turns out it's 14.93
Oh wow you're such a genius
@@jmckendry84 thanks man
Found the engineer.
I was like "one side is 4, the other is a little less, let's say 16, and if we are buying tiles or something, just get 20 to give us a margin of error"
14.928
The problem solving (not just the math) in your content is next level
I'm sure he just had a lot of time to think about the question. If you have enough time, you will eventually solve every problem
fr, real problem solving skill is if you know how to solve it at first sight @@gdmathguy
@@gdmathguyplot twist : he did it in one take
But that's an extremely basic math problem
Everyone who has worked with triangles in high school and is a little skilled in math is capable of solving this
The second reason why I watch till the end is to hear "How exciting"😂
Math being the first one
& putting a box around the answer
And ‘plug it in’
@@TravellingMark5454💀💀
this is the first one i was able to solve myself, and i did it instantly, i knew that the distance through the circles was pretty much the only thing we knew, so when i drew it out in my head i recognized the diagonol as a hypotenuse, made the triangle and knew the answer. very proud
I'm proud of you.
Same, I have had practice with other similiar problems. Got this one too
Same, except I did it by drawing an equilateral triangle side 2cm and then using 1/2*a*b*SinC to find the area, where I then rearranged 1/2*base*height to find the height, added 2 to the height to find the total rectangle height, and then multiplied it all by 4. Got 14.9282.
Same, I did the entire thing mentally while laying down on the bed lol. Height of an equilateral triangle is (root3)/2 times the side. Connect all the centres of the circles and the answer becomes obvious.
Bro thats class 8th question 😂
for those wondering the area without the root is 14.928 sq. cm
But why does he leave the root
Easier and less calculation to be done@@itsjustme9691
@@itsjustme9691 because 14.928 is not the exact answer. It's 14.9282032303... and keeps going on forever. This is what you would call a irrational number. Whenever a root of a number results in an irrational number you should leave the root in the final answer in order to get the exact correct answer
@@itsjustme9691because the actual answer not rounded is like 14,928203230275...
So 8+4√3 just sound and looks better and is also correct
@@itsjustme9691 so we get The exact value
This channel doesn't just teach me math. It also teaches me to approach problems in ways that I'd never consider on my own. I think this is fantastic. I never made it to Algebra 2 in high school before quitting. I did somehow place into Algebra 3 at my local community college and got out with an A, an A that I lost a lot of sleep trying to earn. I love math, but I hate school. Resources like this channel are great. Thank you.
I know absolutely nothing about maths beyond the basics but this is the 4th video of yours that I've watched and I must really be learning a lot from you as I actually figured this one out for the most part, so thank you 👍🏾🙏🏾
Question: at 0:42 when you drew the two red lines, how do you know the hypotenuse of this triangle is straight and intersects at the tangent point between the two circles? I was going to solve it like that initially but wasn’t sure if it was correct to make this assumption (the assumption being that the hypotenuse is 2cm)
Well, consider the fact that those two circles meet at a point; that means that you can draw a straight line between their centers.
Since their radii are equal, we know that that distance is 2 (1cm (the first radius) + 1cm (the second radius))
Hope that helps?
2 circles touching means there is a straight line between the circles centres, and the line distance is always r#1 + r#2 or in our case, since the circles are the same size its r+r or 2r. And since we know the circles are 1cm, it's just 2.
I think you forgot to subtract the area of the circles from the square
The thumbnail made me think this was a Terry A Davis video 💀
Sees thumbnail
"Oh, this is gonna be tricky~"
Sees video duration
"It's not even 2 minutes?!?!"
Sees the solution
"Dang, I never thought of triangle in there"
took me 4 mins to see it on my own
Oh come on, this one was not hard...
Have you seen the animation vs math of Alan Becker?
my solution involved seeing that the centers of the circles formed an equilateral triangle, then cutting that in half to make a 30/60/90 right triangle with two known side lengths. yours was a lot less complicated!
I don't know how he got to a right angle without your step.
My attempt:
The width of the rectangle is equal to the sum of the diameters of the top two circles. Radius of 1 means diameter of 2, 2*2=4, so the width is 4.
The height, meanwhile, is equal to the radius of the top two circles, plus the height of the triangle formed by the centers of the three circles, plus the radius of the bottom circle.
That triangle is an equilateral triangle with side length 2cm. Thus, its height is 2sin(60), or sqrt(3). The radius of all three circles is 1, so that's 1+sqrt(3)+1, or 2+sqrt(3).
Area is width times height, so the area of the rectangle is 4(2+sqrt(3)), or 8+4sqrt(3)
how do you know that its an equilateral triangle?
i got it my bad
@@shreyshah110 Draw a line between any two of the circles' centers. Because the circles are all tangent to one another, the length of the line will be equal to the sum of the two circles' radii. And because all the circles have the same radius of 1cm, that line will always be 2cm long, regardless of which circles you use. Thus, the triangle of centers is equilateral
Damn exactly how i did it and it took just a few secs lmao
Which is ~14.93 cm^2
When adding a unit, this is actually a multiplication. So the result is not 8 + 4 * SQRT(3) cm² but either 8 cm² + 4 * SQRT(3) cm² or alternativly [8 + 4 * SQRT(3)] cm² . Otherwise you would add a scalar to an area, which is rather awkward.
Very straight to the point
Very straight forward
Very nice video
👍
At 0:46 how are you sure the red diagonal line intersects the point where the 2 circles touch?
The shortest path between the centres of 2 circles touching always goes through the point at which they touch
@@elijahknox4421 aaaah yes of course. Thanks buddy
Had a similar question. At the same time stamp, how can we know that it's a 90° triangle? It feels like he skips a step here. Don't we need to prove it's a right triangle first? Or that the proposed hypotenuse is actually a straight line? Studied languages and lit at the uni, but math has been slowly growing on me over the years.
@@kathorseesfirst of all, the proposed hypotenuse is a straight line because it's just the shortest distance between the centres and it goes through the tangent (he deliberately created this line to be straight so that he could use Pythagoras on the resulting triangle).
Second of all, I see your point here, the proof that the triangle is a right angled triangle is: if you were to draw a straight line between the top 2 circles' centres which goes through the tangent, then draw a straight line between the top left and bottom circles (similar to the first line he drew), you would get a an equilateral triangle connecting each of the circle centres, we know it's equilateral because each side is the length of 2 radii (2cm)
If you half this equilateral triangle you get the right angled triangle he has in the video, and we know it's a right angled triangled because whenever an equilateral triangle is cut down the middle, it turns into a right angled triangle.
Sorry if the explanation is bad, it's hard to explain without drawing it for you
It’s like I know how to do it, (find the area of the whole rectangle without the circles in it, then find the area of one of the circles, multiply it by three, and subtract that from the area of the rectangle without the circle, then you get the area of the little spaces in between everything) but I don’t want to do the math lol
(Idk if my train of thought was way off tho lol)
On further thinking I see it’s not a square lol,
Not me jumping to conclusions and saying 16
Using a calculator and rounding up to two decimal places, the non-equation answer is about 14.93cm^2
So what im getting is, you find and classify the know, assign variables to the unknown, and form a right triangle to solve for the variables and maybe even a system of equations if nessesary, and you've got yourself a solution!
It’s easy, the area is what it is and there’s not much you can do about it.
My attempt :
L=4r
W = L
W= 4r
So area of rectangle is 4r²
And r = 1cm so the answer is 16cm²
Me too!
The three centers make an equilateral triangle with side = 2cm. It has 60 degrees angles.
So the height of the triangle is h = (side * squareroot(3) / 2) = squareroot(3).
So the height of the rectangle is (from top to bottom) = 1 + h + 1 = 2 + squareroot(3).
So the area of the rectangle is: 4 * (2 + squareroot(3)).
Well, that was a spectacular failure for me. I got almost everything, minus some (no doubt crucial) difference in how to find X on the triangle, and I got 14.34 cm²... And on second attempt, 12.56.
I suppose call it half marks for right idea.
I haven't been in a class since 2011 and that was end level computer science so it's been even longer since I had to look at this stuff... yet I'm super engaged with this channel content lol
The only problem i was able to solve on my own
oh my GOD YOU LOOK LIKE A YOUNG TERRY DAVIS
Ey I got this one pog
Nice I did it in my head and it ended up being similar to what he did. I thought of sin60 to solve.
I didn't even realize this was a rectangle even though it says it in the thumbnail
As i iitian this problem quite eassy. But your some problem are quite good
It’s 16 I calculated properly using Pi, and some calculas concepts that actually helped
i appreciate that these videos get to the point and explain the process. i want to study math again and this is waking my brain up
The solution is simple
But I’m too dumb to think of that
Love the “how exciting” tag as the end of all your videos. Lol
I have done this question before solution 🎉
When I saw a right angled triangle I mouthed out Pythagoras like he was making a cameo in the Avengers
I don't know the precise measurements, and I haven't watched the video, but my immediate gut instinct was to just call it 4x4.
But it's not, the radius of the circles overlaps slightly. Maybe by a quarter. So it's nearer to 4x3.75 than to 4x4
That's what I did too. If I ever had to solve this in real life, like maybe we gotta fit these cylinders into a box, 4*3.75 is a valid solution, even though it's not the mathematically correct one.
Yay! Although I squared off the vertical radius of the lower circle from the tangent and used to give height = 1 + (sin60) + (sin60) + 1
Area = 4 * (2 + 2(sin60))
Same beans
Finally, one i can solve
1:31
@@ThrowTop Seriously? Check the timestamp.
@@ThrowTop My timestamp was 1:31, Check what he says in the video at 1:31.
You might just get it if you think really, really hard.
@ThrowTop some people might say "oh, my bad" or "didn't get it but I do now", but the really cool cats just delete their comments, because that's very mature.
From watching a bunch of your other videos, the tools learned helped me solve this while paused at the start. I can't stress how excited I am
I learned a lot by this problem solving based method.
Yet kinda make me angry the area not being a natural number, not exciting
i see the reason why it has to be written that way but i agree its really annoying. its like 14.928 rounded
0:51 you don’t actually have to do Pythagorean Theorem to find X here. You can actually use the 30-60-90 right-triangle rule.
This can be proven because if you draw 2 more radii (radius in plural) to mimic the right triangle and create an equilateral triangle (we know this is true because each side length of the equilateral triangle would equal to 2) and splitting it again just gives you the 30-60-90 right triangle.
Back to the 30-60-90 right-triangle rule after proving to you all that you can use the 30-60-90 right-triangle rule instead of the Pythagorean Theorem to find x, we know that in a 30-60-90 right-triangle that opposite = x (whatever number you are using), adjacent = x times sq root of 3, and the hypotenuse = 2x. In this case, we can see here that 1 is the opposite, 2 is the hypotenuse, x is the adjacent, and that we are dealing with a 30-60-90 right triangle. Therefore, we can plug in the values we are using for the right triangle we are dealing with here (opposite = 1, adjacent = x, hypotenuse = 2). 1 times 1 = 1, 1 Times Square root of 3 = 1 (sq root of 3), and 1 times 2 = 2. Thus, the solution for x is simply sq root of 3.
In conclusion, you don’t need Pythagorean Theorem to figure the right triangle shown in 0:51, you can instead use the 30-60-90 right-triangle rule.
Note: Not all right-triangles are 30-60-90 right-triangles.
Since the sides are all the same length, I would think the answer is 8? since the top one is 4 meaning width, and the height of the rectangle is the same? Im bad at math but to me that makes sense
NVM I misunderstood, still I'd think the answer is 16 because 4 * 4
Its not the square
@@truebye Huh? That is the only 'rectangle' I see, even though a rectangle is longer in one measurement right? But theres no rectangle at all in the picture
thats not easy?
Even a child can solve this question maybe 15years old
How exciting... :)
Engineers getting an aneurysm looking at the final answer.
There is an implicit triangle which measures 4 by 4 by 4
I got Africa as my answer. I'm not good at shapes
Nah bro my answer was Antigua and Barbuda because I'm terrible at math ye
I keep seeing this come up in random places and know I can solve it, but I definitely didn't have the fastest solution. Puts my mind at ease knowing how simple it really is...
4cm across, 2cm + sqr(3)/2 * 2cm tall. Multiply those. Done.
nice channel, problem is I hardly ever see the whole videos I simply take the problem from the cover and skip directly to the end to see if my results are right 😂
Huh. Similar answer, but built an equilateral triangle in the center and did height as 2*sin(60degrees) = 2 (sqroot(3)/2) = sqroot(3)
This is one of those peoblems, that are easy to approximate, without doing the weird shit.
If we pretend that the cube is 4*4 than the area is 16. We know that it aint. And theres something like a .25-.5 overlap. I'll go with 0.25, so the height is 3.75. so ill say the area is in the ballpark of 15(or 17 at max). Now, ill watch the video. I dont think ill be off by more than 0.5
Edit:
The final solution is 14.92
I solved this mentally using Pythagorean theorem on that
radius + radius = 2
c² = 2²
a² = 1
b = √(4-1)
b = √3
height = 2 + √3
width = 4
there
If you make a triangle by drawing lines from the centre of all circles you get a triangle with three equally big angles (60). the angle 60 provides that the "x" that you made has to be the standard angle ( sqrt3 / 2 ) x 2 which also gives square root of three plus 2 additional cm from the original. Then its just 4cm times (sqrt3 + 2vm) = 4(sqrt3) + 8 cm
Should be idk… close to 14.8 or so in a range of + or - 0.12?
Edit jumped ahead to find the answer figuring it’d be some shenanigans which I seem to be right on, however if you plug it into a calculator I wasn’t far off :>
before you explain it let me answer it. for the length, just add up the diameter of the circle which means 4, for the height, just do the Pythagorean calculation from the root of 3, which is 1.73 and add 2, 1 from the radius. then 1 from the top which is clearly 1
last thing you just need do 4 x 5.73 = 22.92
Me: the fuck am I supposed to get the hight?
Op: *draws triangle
Me: *Thermonuclear Facepalm
Wasnt hard. The width is 4cm as itsbthe length of 2 circles with a radius of 1cm. The height is a bit more difficult. The height is 2cm (radius of bottom circle plus radius of one of the other circle) plus the height of the triangle formed by the 3 center points of the circles. The length of all 3 sides is 2 cm and the height of and equilateral traingle with s=2 is sqrt 3. So the height of the rectangle is (2+sqrt3)cm
The area is 4cm x ((2+sqrt3)cm)
Which equals (8+4sqrt3)cm^2
15.50 +-1.00 cm, next question. Why do math guys have to complicate things, when approximate numbers are more than enough?(im joking)
I'm embarrassed. I thought the rectangle was a square. And no, I didn't get 16cm^2 for the answer. I decided to find the more difficult side and square it.
How exciting
Turn on of the circles in the corner into squares and then multiply by four for each quadrant
I noticed that the center of each circle would form an equilateral triangle of side length 2cm. From there I remembered that you can split the 60° angle in two to get a 30-60-90 triangle, and that the lengths are 1,2,√3 .
Before I watch it: Notice the side length of the square is 4 times the radius, which is one.
Therefore the area is 4² = 16cm²
EDIT: I was foolish and thought that it was a square without proof
Before watching the video, i got something around 13.27
Didn't played the video, the solution can be, find the height of the triangle inside formed by joining centres. Add 2cms to it, it's the height of the rectangle. And 4cm is the width of the rectangle.
My HS trig teacher would take points off for the error in the final expression. Either put brackets around the math (everything before the units) or add the units after the “4”. Yes, he was that pedantic.
I thinks 16
I did this on my own before watching and got 13.86 cm. Let’s see if I’m right.
(I messed up the units😭😭😭)
Call me stupid but... That's a Square, isn't it?
Nevermind, i just saw the video.
I solved it similarly. Though I used the height of a equilateral triangle which is L * sqrt(3) / 2. where L was 2, so that height is just sqrt(3)
Quite simple! Long (horiz) side is 4cm. Vertical = 1cm + 1cm + height of triangle where all sides are 2cm
So, when two circles touch we can form a right angled triangle,with sides r1, r1+r2 & √(r2² + 2r1r2). Nice intuition
Why didn't anybody tell me it ISN'T a square???
I was so confident when I saw the side length was 4 so the area must be 16
What if you make a triangle going from the top of the top left circle (or top right) to the bottom of the bottom circle to the right angle between both points at the bottom?
You know the distance of the hypothenus (4x radius) and the base (2x radius) then Pythagorean Theorem and you got the height of the rectangle
I solved it by connecting all the 3 centres, they makes an equilateral triangle, I have one side, know the area formula and boom...
I didn't watch this video before commenting
we can see that top ones show that one of the x is 4
that side that we know is 100% bigger than x, but not much, so second x is 3
Bro always does some shit that makes me say oh shit! Then I realize I'm getting hyped about math and I'm a fucking nerd.
One thing I'd like to mention is that you do need to prove the triangle is right triangle. Then, the solution is complete. Otherwise, you can't assume it's a right triangle in the first place.
If it was a square and it said it was without a visual representation even, just 16 cm squared minus 3 pi because the area of a circle with radius 1 is always pi units, to get 6.58 cm squared
I knew it had something to do with Pitágoras but didn't think of that, f me
The only thing i hate about this is that he didnt finish the equation, so the area is 14.9282032303 cm²
How come he didn't apply the same logic of the 1cm radius to the bottom circle? It looks equal to the other 2. How do you know to not assume the same for that circle?
Cool video. I learned something quickly and easily
I might be dumb but could you just plus the circls diameters to get the length and width then calculate area
Idk maby im wrong or you wanted to go the long way round. I didnt watch the video
Height of same handed triangle is a⅓
what is wrong with just saying that the area is 4 x 3.732 cm or 14.928cm sq
Yay I solved this one! ...sort of I did look up the 90 60 30 triangle to do it. I should have thought to use Pythagoras.
im dumb, especially in math, is 8 + 4✓3cm² the answer?, it looks like it needs to be solved, help
Coool
14.93, leaving the square root in the answer seems rather pretentious
Why couldnt you have eliminated the square root and put it in a simple form, then aXb the rectangle?
One can also use trig to solve this... cos = adj/hyp. cos 30° = √3 etc...
How does a man look like a frat in his early 20s but sound like my middle aged highschool teacher?