This Blog exists for the collective benefit of all geometry students. While the posts are specific to Mr. Chamberlain's class, any and all "geometricians" are welcome. The more specific your question (including your own attempts to answer it) the better.
hi number 19 on pg 348. the wording on this problem confused me. and i dunno how to do it. and number number 4 on the self test- i dont even know what its asking. thank you and good night.
For #19, you need to visualize a beach ball (sphere) being "cut" by a large flat piece of cardboard (plane). The outer part of the beach ball "touches" the cardbard to form a circle. If the cardboard is 5 cm from the center, imagine the chords that are being created. Then, picture a specific (random) chord and radii that meet at a right angle. You should see them as legs of a right triangle, do you?????
For 349#4, you can use the beachball again with a concentric basketball on the inside. Then cut both with the cardboard plane thru the common center. What do the intersections with the plane look like (Ans: two concentric circles).
i know we did #9 in class, (like megan said...) but same deal. I got a different answer. And I've checked my work against my notes and i dont know but something's off. I got 5sqrt5 instead of 10sqrt5. the radius is 15. a^2+b^2=c^2. 10^2+x^2=15^2. 100+x^2=225. 225-100=x^2=125. sqrtx= sqrt125=sqrt25+sqrt5=5sqrt5. i still don't see where I'm off
i know we were supposed to do # 17 in class (and we didn't get to it) and i have absolutely no idea how to do it. I want to drop a perpendicular but then there's two variables. and then i wanted to draw ok but that doesn't help much. it gives that ok is 10 but i just don't know :(
Circle O has a radius of 10cm A chord XY exists that is 8cm in length How far is the chord from center O?
Draw a radius that perp bisects chord XY. Draw another radius to X. You now have a right triangle with a leg length 4 and a hypotenuse (OX) of length 10, yes?
Enter Pythagorus, stage left. 10^2 - 4^2 = 86
So the distance of the chord from center O is sqrt(86) or 2sqrt(21), yes?
yeah and 19. no idea even after your explanation. i still don't see it. and 21 is a What the H*ll? ok. on #5 (selftest) i'm not understanding how to find that length of the chord with just the radius and visaversa. those are 4 my theses. i think i'll start my own religion: the anticirclests.
I don't know how else to explain #19 except to say that there is a right triangle with legs of 5cm and 12cm inside the sphere that is connecting the center of the sphere to the plane and the circle that is "cut" by the plane. The 5 is given as the distance of the plane from the center, and the 12 is the radius of the circle cut by the plane, leaving the hypotenuse (radius of the sphere) for you to calculate... 5^2+12^2=169, yes?
Draw OB as a perp bisector of chord AC. Draw radius OA. That will give you a right triangle with leg AB 6cm and hypotenuse OA 10cm. Angle-BOA has a sine (opposite/hyp of AB/OA, right?) of 6/10 or .60. The inverse sine (see pg 311) of .60 is appx = 37deg. That is half of angle AOC, which is the central angle for chord AC. So arc-AC measures 74deg, ca-peesh?
hi number 19 on pg 348.
ReplyDeletethe wording on this problem confused me. and i dunno how to do it.
and number number 4 on the self test- i dont even know what its asking.
thank you and good night.
Dear Deprived,
ReplyDeleteFor #19, you need to visualize a beach ball (sphere) being "cut" by a large flat piece of cardboard (plane). The outer part of the beach ball "touches" the cardbard to form a circle. If the cardboard is 5 cm from the center, imagine the chords that are being created. Then, picture a specific (random) chord and radii that meet at a right angle. You should see them as legs of a right triangle, do you?????
For 349#4, you can use the beachball again with a concentric basketball on the inside. Then cut both with the cardboard plane thru the common center. What do the intersections with the plane look like (Ans: two concentric circles).
I think that we did # 11 in class, but when I tried it on my own, I got a different answer than the book. I used ratios of 9: x = 12: 10 and got 7.5.
ReplyDeletei know we did #9 in class, (like megan said...) but same deal. I got a different answer. And I've checked my work against my notes and i dont know but something's off. I got 5sqrt5 instead of 10sqrt5. the radius is 15. a^2+b^2=c^2. 10^2+x^2=15^2. 100+x^2=225. 225-100=x^2=125. sqrtx= sqrt125=sqrt25+sqrt5=5sqrt5. i still don't see where I'm off
ReplyDeletenever mind. I got it now :)
ReplyDeletei know we were supposed to do # 17 in class (and we didn't get to it) and i have absolutely no idea how to do it. I want to drop a perpendicular but then there's two variables. and then i wanted to draw ok but that doesn't help much. it gives that ok is 10 but i just don't know :(
ReplyDeletefor 347#11,
ReplyDeleteCircle O has a radius of 10cm
A chord XY exists that is 8cm in length
How far is the chord from center O?
Draw a radius that perp bisects chord XY. Draw another radius to X. You now have a right triangle with a leg length 4 and a hypotenuse (OX) of length 10, yes?
Enter Pythagorus, stage left. 10^2 - 4^2 = 86
So the distance of the chord from center O is sqrt(86) or 2sqrt(21), yes?
yeah and 19. no idea even after your explanation. i still don't see it. and 21 is a What the H*ll? ok. on #5 (selftest) i'm not understanding how to find that length of the chord with just the radius and visaversa. those are 4 my theses. i think i'll start my own religion: the anticirclests.
ReplyDeletefor 348#17
ReplyDeleteYou gotsk to know your 30-60-90 relationships... ARRGGGHH!!
You can draw a perp bisector from O to a point P on chord JK. Since arc-JK is 120deg, angl-POJ must be 60deg.
Since OJ=10, OP must=5 (30-60-90 rules) and therefore PJ=5*sgrt(3). PJ is half of JK, so JK=10*sqrt(3)... ca-peesh??
yeah. i got it now. hehe
ReplyDeleteI don't know how else to explain #19 except to say that there is a right triangle with legs of 5cm and 12cm inside the sphere that is connecting the center of the sphere to the plane and the circle that is "cut" by the plane. The 5 is given as the distance of the plane from the center, and the 12 is the radius of the circle cut by the plane, leaving the hypotenuse (radius of the sphere) for you to calculate... 5^2+12^2=169, yes?
ReplyDeletesqrt(169)=13, so the diameter is 26cm, yes?
#21 is trig... grab a pencil...
ReplyDeleteWe have a chord 12cm in a circle of radius 10.
Call it Circle O with chord AC.
Draw OB as a perp bisector of chord AC. Draw radius OA. That will give you a right triangle with leg AB 6cm and hypotenuse OA 10cm. Angle-BOA has a sine (opposite/hyp of AB/OA, right?) of 6/10 or .60. The inverse sine (see pg 311) of .60 is appx = 37deg. That is half of angle AOC, which is the central angle for chord AC. So arc-AC measures 74deg, ca-peesh?