These came off the Rocksprings math contest, which was, for the most part, outstandingly easy. I scored a 252 on their scale. But with the 2 questions that the answer key had wrong, I scored a 268. Not bad, eh?
Anyways,
31) What is the area of the circle formed when a plane passes 6-m from the center of a sphere with radius 7-m?
This one seems to involve some geometric info that I am not aware of yet.
49) A circle has an area of 20pi. It is divided into 10 equal pie sections and seperated. What is the periphery of one section?
Dont know what periphery means in this case.
56) What is the number of diagonals for a regular polygon with 16 sides?
Um... Forgot the diagonal formula. Answer is 104, but dunno why.
I was wondering if anybody else here has taken any of these weird, 3rd party tests that are given out in rural places? And if so, do you know who the maker is? It wasnt a useless test, I must say. I like that some of the questions are easy enough so that freshmen wont get discouraged, but hard enough to challenge a practiced mathie like anyone who posts on this site. But... I'm pretty sure that Sam would score a 360 and Brad 300+, had they taken it too.
31) What is the area of the circle formed when a plane passes 6-m from the center of a sphere with radius 7-m?
If the plane intersected right at the center, the circle would have the same radius as the sphere. If it intersected at one point (tangent to the sphere), it would form a point circle (degenerate). Anywhere inbetween should scale in a linear fashion between 0 and 7. So 6 m away from the center would intersect a radius of 7-6 = 1 ... area is pi. Is that what the answer was?
49) From dictionary.com, periphery seems to be the circumference of something. So I think this question is asking the arc length of each section.
area = 20pi, radius = sqrt(20) diameter = 4sqrt(5), circumference = 4 pi sqrt(5).
divide that circumference by 10 and you get 2*pi*sqrt(5) / 5 or approximately 2.8
52) Think of a polygon with n vertices. To make a diagonal, one vertex can connect with any other vertex except the 2 ones adjacent to it and itself. So each vertex makes (n-3) diagonals... n vertices times n-3 diagonals gives n(n-3), but that counts each diagonal twice(once from one endpt and once from the other) so the formula is n(n-3)/2
I figured out #31 in health class today. I was being dumb. The plane cuts into the sphere 6 inches away from the center. Since they told us that the radius of the sphere was 7, we can make a right triangle, with the unknown radius of the cross section being one leg, the 6 inches being the other leg, and the 7 inch radius being the hypotenuse. C^2 - A^2 = B^2, so the radius of the new circle is sqrt(13). The area is therefore 13pi square inches.
On 49, the correct answer is 4sqrt(5) + 2pisqrt(5)/5. Unfortunately, I dont know where they got that 2nd part.
And the diagonal question makes alot more sense now, thanks!
Oh, and one more thing: be sure that before you use a program that finds you the distance between a point and a line that the point is not already on the line. For some reason, the formula doesnt work in that case. Try it with this for example (-4,1) and line x + 2y = -2.
I surmise, since polygon has perimeter and circle or other self-enclosed curvature has circumference, a pie section that has both lines and curves has a hybrid of perimeter and circumference -- periphery.
I should have known it wouldn't scale linearly.
