User:Nikhilrd

i love math!!

So for Question #20 Amc 10b, we can see that we can create a square from the edges of each diameter because each semicircle is congruent. From there, we realize that because the diagonals are 4, the side lengths of the square are 4/sqrt(2). We realize that we can draw a right triangle connecting the center of the circle to the edge of the sphere and the center of the diameter of one of the semicircles. With this information, we realize that we must set one of the lengths as half of the length of the side of the square so we get 4/2sqrt(2). using Pythagorean theorem, we get 2+x^2 = 4(the radius squared of the sphere which happens to be the hypotenuse). we solve for x and we get sqrt2 which is the radius of each semicircle. The circumference of each is pi*sqrt(2). we multiply that by 4 and get 4sqrt2*pi and that equals sqrt(32)pi so we get A, 32. ~KEVIN_LIU