2:13 before squaring both sides you must demand the right side to be greater or equal to 0 2:49 try to find the x using Viet's theorem or something else before using Discriminant to avoid making minor mistakes
Funny that when using calculus to solve this you get that the area of the square is equal to (20-2(10sin(pi/4)))^2 which equals (20-2(10sqrt(2)/2))^2 or (20-2(5sqrt(2))^2 which equals 34.3146
Oh that's clever, I did it a little different. I defined the right circle to be (x-10)^2+(y-10)^2=10^2, so a circle with radius 10 that touches the x- and y-axes, so the square will be divided evenly in two by the y-axis. The top right corner of the square would then be at point (s/2,s), where s is the side length, and would have to sit on the line y=2x. I substituted 2x for y and solved to get (x-2)(x-10)=0. Obviously the 10 solution was nonsense so x=2, and the side length is 4 meters.
So. I looked at the preview and said to myself it's 100% gonna be 4, skipped through whole video, it's 4. Math is nice but isn't always needed with a good eye, I guess.
![[LIVE] BABYMONSTER(베이비몬스터) - DRIP | 두시탈출 컬투쇼](http://i.ytimg.com/vi/UG05pxLGwzc/mqdefault.jpg)
I never thought I would spend 3 minutes learning how to workout the area of the square from two circles. I loved the randomness of youtube sometimes!
Glad you liked it!
You know, I think this is why measurement devices were invented instead of using complicated mathematics.
Same here 😂
It's actually easier if you substitute y = (10 - x/2) into the first equation (y^2 = 20x - x^2). No mucking about with roots if you do it that way.
That is very clever
By symmetry the green line is just 10-x/2, and the quadratic is way easier to solve from there
You can find y directly because it is ((20-x)/2) you can just put that into the Pythagorean theorem
2:13 before squaring both sides you must demand the right side to be greater or equal to 0
2:49 try to find the x using Viet's theorem or something else before using Discriminant to avoid making minor mistakes
Funny that when using calculus to solve this you get that the area of the square is equal to (20-2(10sin(pi/4)))^2 which equals (20-2(10sqrt(2)/2))^2 or (20-2(5sqrt(2))^2 which equals 34.3146
why would you use calculus?
Oh that's clever, I did it a little different. I defined the right circle to be (x-10)^2+(y-10)^2=10^2, so a circle with radius 10 that touches the x- and y-axes, so the square will be divided evenly in two by the y-axis. The top right corner of the square would then be at point (s/2,s), where s is the side length, and would have to sit on the line y=2x. I substituted 2x for y and solved to get (x-2)(x-10)=0. Obviously the 10 solution was nonsense so x=2, and the side length is 4 meters.
A clever solution as well
really fun problem!
So. I looked at the preview and said to myself it's 100% gonna be 4, skipped through whole video, it's 4. Math is nice but isn't always needed with a good eye, I guess.
I did it in my head and got 0....
(10-x)^2 + (10-0.5x)^2 = 100
(10)^2 (10)^2={100+100}=200 10^20 10^2^10 2^5^2^2^5 1^1^12^1 2^1 (x ➖ 2x+1) .
hey man, i found an easier solution, I just uploaded it so you can see it's a LOT easier
the solution:
th-cam.com/video/FyWjh4R3QPg/w-d-xo.htmlsi=MSK3MSluNnO1gaUU
th-cam.com/video/FyWjh4R3QPg/w-d-xo.htmlsi=MSK3MSluNnO1gaUU
Nice one! I'll check it out