if you hold both side of the line and join them together, the line will be half its lenght going down. 80 x ½ = 40 for the lowest part of the cable to be 10m off the ground the poles are ZERO meter apart. 50-40=10. any distance between the poles and the cable would be higher off the ground
I'm in Finnish middle school (8th grade). I like maths a lot. I didn't understand much of this, but it's great to watch your videos. I always learn something new!
If the problem depends on the relative size of the two squares, then it is not solvable. If it does not, then the solution is the same for any relative size of the two squares, so pick the particular case where the two squares are the same size. Their bottom sides meet at the circle's center (symmetry), so the surface is 2 * the area of one square which diagonal is the radius of the circle.
Diameter = 16. Radius = 8. If when we draw a square in the one right quarter circle touching the quarter circles perimeter, the distance from the rightmost extreme of the quarter circle to the side of the square is z, which can be found by symmetry, then the side of the larger square is, a = 8 - z.
Alternative solution: Imagine both the squares are same, so they meet in the center of the circle. So, Diagonal is the radias of the circle. According to Pythagoras, square's diagonal = x√2. So, radius = 8 = x√2. So, x=8/√2. Thus, 2x^2 = 64
are you fucking kidding me? you are a disgrace.. an imbecile. remove yourself from the maths community at once. the answer was 16e+303. absolutely mad of you to think it was whatever that RUBBISH product was insinuating. return to the academy and refresh and refine your techniques. sincerely, mathemetician and astrophysicist, 'Young Nigga Bonin' 88'.
Solved it in 10 fking seconds but my indian ahh thought he will give an answer in whole number and answer i am getting is wrong !!! I should stop doubting myself so much i am 18 now and need to be more confident
I'm too lazy to do the first step, which is to find the radius of the blue circles in terms of pi. Hence, the radius of the blue circles is k. Let the radius of the big circle be y and the radius of the small circle be x. 2x+2k=2y, so y-x=k The centers of the small circle have a vertical displacement of 2k and a horizontal displacement of something. This something is the absolute value of y-x, which is k. Distance between the two: rt(k² + (2k)²) = krt5 x + y = krt5 2x = krt5-k x = ½(krt5-k) y = x+k = ½(krt5+k) ½(pir² + pir²) = ½pi(x² + y²) = ½pi(½)²((krt5-k)² + (krt5+k)²) = ⅛pi(k²)((rt5-1)²+(rt5+1)²) = ⅛pi(k²)(5-2rt5+1+5+2rt5+1) = ⅛pik²(12) = ½(3)pik² ½pik²=36. k² = 72/pi = ½(3)pi(72/pi) = 36(3) = 108
You lost me at exactly 0:01 when the question started. 😂😂😂 Joking. I lost it on the last part about the small triangle being 1.5 something or another than the larger triangle. I'm gonna have to watch again. Thank you!