Author Topic: A Hard(For me) math problem I need help with, Circles and what not :) (Read 8071 times)

DAaaMan64 · « **on:** November 01, 2007, 01:39:42 PM »

Quote
A circle of unknown radius has a chord of length d whose midpoint is 1 foot from the circle. Find the radius, r, of the circle [in inches].

I have been told that I can find it because of an isosceles triangle I can create, but I have yet to see way to create an isosceles triangle that is isosceles in all instances!

Thank you!!!

Athrun Zala · « **Reply #1 on:** November 01, 2007, 02:50:14 PM »

um, for any given chord of a circle, the triangle AOB (with A and B being the chord's edges, and O the circle's centre) is ALWAYS an isosceles, because OM (M being the AB midpoint) is perpendicular to AB...

hope that helps

DAaaMan64 · « **Reply #2 on:** November 01, 2007, 07:22:25 PM »

Hm... Maybe I don't understand that information's application in this case. Can you elaborate? I only know the distance from the midpoint of the chord to the circle, the outer part of the circle on the center. Thank you!

Kairon · « **Reply #3 on:** November 01, 2007, 07:46:28 PM »

The outer part of the circle, 1 ft from the midpoint to the outer rim of the circle? Hmm... *opens up paint*

Is this trig, or geometry, btw?

GoldenPhoenix · « **Reply #4 on:** November 01, 2007, 07:47:49 PM »

I hate math, glad I don't need to do anything but simple mathematics now. Even this problem makes my head hurt without even thinking about a solution!

Kairon · « **Reply #5 on:** November 01, 2007, 07:53:53 PM »

I can solve this with trig I bet but pure geometry is... uh.... that was 7-8 years ago... hmmm

scratch that, screw trig

wait... this may be all about trig... *looks up chords on wikipedia*

DAaaMan64 · « **Reply #6 on:** November 01, 2007, 08:03:10 PM »

no tell me any solution. I am simply doing extra credit.

But as I am but a humble crap-popsicle at math, I cannot figure it out without some guidance.

I have gotten different answers everywhere I turn and my head asplode soon. Thank you Kairon if you have anything

Kairon · « **Reply #7 on:** November 01, 2007, 08:05:27 PM »

What type of math class is this, may I ask?

ShyGuy · « **Reply #8 on:** November 01, 2007, 08:09:13 PM »

Is this some SCC class?

DAaaMan64 · « **Reply #9 on:** November 01, 2007, 08:21:17 PM »

lol yes it is ShyGuy, how did you know?

Well my math is Pre-Calc, but this is math level wide. I have tried many different theories of my own, I at least need direction...

Kairon · « **Reply #10 on:** November 01, 2007, 08:29:30 PM »

Gah... I'm at a loss. Is that even enough information? unknown radius, unknown chord length... that could be any sort of set-up couldn't it?

DAaaMan64 · « **Reply #11 on:** November 01, 2007, 08:34:27 PM »

ya I know, that is my problem, I believe that ultimately it is simply going to be a function of r and not fulling simplified..... Thank you for trying.

GoldenPhoenix · « **Reply #12 on:** November 01, 2007, 08:35:25 PM »

I think the answer is .1235467890 inches.

DAaaMan64 · « **Reply #13 on:** November 01, 2007, 08:53:44 PM »

COULD BE

ShyGuy · « **Reply #14 on:** November 01, 2007, 09:08:18 PM »

Here's an idea. divide known line into 12, see how long one unit is, then physically measure the unknown line, convert it to the one unit scale and round to the nearest number.

NinGurl69 *huggles · « **Reply #15 on:** November 01, 2007, 09:27:54 PM »

ANSUR:

r = [ (d^2) / 96 ] + 6 ~~ (inches)

where `d` must be 24in or greater.

if chord `d` is 24in and the midpoint is 12in to the closest point on the circle (perpendicular to chord), then the chord is basically the diameter of the circle and the 12in ends up being the radius. r=12in (d=24in) is the smallest the circle can be.

ShyGuy · « **Reply #16 on:** November 01, 2007, 09:32:15 PM »

Just don't ask KDR for help, rumor has it he doesn't help people with their homework.

Kairon · « **Reply #17 on:** November 01, 2007, 11:03:12 PM »

Quote
Originally posted by: Professional 666
ANSUR:

r = [ (d^2) / 96 ] + 6 ~~ (inches)

where `d` must be 24in or greater.

if chord `d` is 24in and the midpoint is 12in to the closest point on the circle (perpendicular to chord), then the chord is basically the diameter of the circle and the 12in ends up being the radius. r=12in (d=24in) is the smallest the circle can be.

Yeah, good catch that the chord cannot be smaller than 24 in. I was actually starting to wonder whether this was a trick question...

How did you calculate r though for chords... huh? confused? how did... blargh!

edit

AH. Darn. It's so SIMPLE TOO!!! GGGLLLLAAARRGHH

I now officially hate the pythagorean theorem.

NinGurl69 *huggles · « **Reply #18 on:** November 01, 2007, 11:26:43 PM »

let r = L + 12in

so L = r - 12in

by the pythagorean theorem, L^2 + (d/2)^2 = r^2

substitute, so (r-12)^2 + (d/2)^2 = r^2

Solve for `r`

Shift Key · « **Reply #19 on:** November 01, 2007, 11:54:45 PM »

Reminds me of an idea that me and my friends devised.

We need to build a time machine to go back to the time of Pythagoras and beat the sh*t out of him, for inventing such a theorem.

Yes, swear filters. I'm keeping that there because that was how we felt about the matter.

Crimm · « **Reply #20 on:** November 02, 2007, 06:00:38 AM »

Number is the ruler of forms and ideas and the cause of gods and demons.

DAaaMan64 · « **Reply #21 on:** November 02, 2007, 07:41:54 AM »

Professional 666, thank you! Thank you very much, this may get me the grade in this class I need! I will just need to do a write up on the problem now...

Thank you Kairon and Athrun Zala for your help too.

ShyGuy, you still never answered how you knew this was an SCC problem. How did you know?

Kairon · « **Reply #22 on:** November 02, 2007, 08:02:21 AM »

What is SCC?

Nick DiMola · « **Reply #23 on:** November 02, 2007, 08:05:46 AM »

Quote
Originally posted by: Kairon
What is SCC?

According to DAaaMan64's profile he lives in Spokane, I'm guessing it means Spokane Community College. Could be wrong though.

DAaaMan64 · « **Reply #24 on:** November 02, 2007, 08:14:47 AM »

Bingo

ShyGuy · « **Reply #25 on:** November 02, 2007, 08:41:58 AM »

How did I know?

I'm Batman.

bustin98 · « **Reply #26 on:** November 02, 2007, 08:57:07 AM »

How funny if ShyGuy is the teacher or someone else connected to the class.

"Honest Mr. Alling, I didn't go to the Internet for help. Nintendo World Report? Nope, never heard of it..."

DAaaMan64 · « **Reply #27 on:** November 02, 2007, 10:57:45 AM »

Personally I don't think that is that big of deal. If I ask for a formula from some people, thats the same as going to a website to find a formula written by some people. Also, I didn't exactly go with what Proffessional 666 gave me. I varied it a little. Plus had to explain it in good detail for credit.

NinGurl69 *huggles · « **Reply #28 on:** November 02, 2007, 11:08:11 AM »

Took me a good 10min.

DAaaMan64 · « **Reply #29 on:** November 02, 2007, 12:06:43 PM »

lol they give us 2 weeks :p

Caliban · « **Reply #30 on:** November 02, 2007, 04:13:42 PM »

Quote
Originally posted by: Shift Key
Reminds me of an idea that me and my friends devised.

We need to build a time machine to go back to the time of Pythagoras and beat the sh*t out of him, for inventing such a theorem.

Don't you dare to ever try that, I need that theorem for my CNC Applications course.

ThePerm · « **Reply #31 on:** November 02, 2007, 10:06:04 PM »

KDR doesnt help with homework

DAaaMan64 · « **Reply #32 on:** November 06, 2007, 06:44:20 AM »

Alright so I turned this in. Got my grade back, I got 1/10!! Wow, I guess I didn't understand something! I know it is only extra credit, but I will take what I can get. I don't know what I did wrong... WOW I pretty much turned in a version of what Professional 666 gave me...

Maybe ShyGuy told on me. ha ha

ShyGuy · « **Reply #33 on:** November 06, 2007, 07:23:51 AM »

That's not all. Mr. Cash is keeping an eye you for inventory shrinkage.

Kairon · « **Reply #34 on:** November 06, 2007, 07:45:02 AM »

OUCH! Did uh...is there uh... any more extra credit? Because Pro's solution should be correct, if a little sparsely explained. Did your teacher want you to write out a more detailed explanation or something?

Oh, and I think that Shyguy DID tell on you...

DAaaMan64 · « **Reply #35 on:** November 06, 2007, 08:02:23 AM »

Mr. Cash no longer threatens me. I do not work there.

There is more extra credit and I will do as much as I can Kairon.

Ya I don't understand it either. Considering I testing it again the idea that the smallest possible radius is 12 inchs. Hmm...

But seriously ShyGuy, I still don't know who you are, but whatever... haha Your just a good guesser

stevey · « **Reply #36 on:** November 06, 2007, 09:26:29 AM »

Pro is 100% right in his math.

Did you show your work clearly? Drew a circle, shown the right triangle, ect?

odifiend · « **Reply #37 on:** November 06, 2007, 09:27:32 AM »

Is there a numerical answer?
I got r = sqrt(144+d^2/4) inches.
That problem is not hard unless you are supposed to get a number. Radius is your hypotenuse, the foot is your short side and d/2 is your long side. I don't think pro's logic is right or I don't get it anyway...

stevey · « **Reply #38 on:** November 06, 2007, 09:56:50 AM »

no it's r=d²/96+6

work

GoldenPhoenix · « **Reply #39 on:** November 06, 2007, 10:01:19 AM »

See, I KNEW my answer was right!

Athrun Zala · « **Reply #40 on:** November 06, 2007, 03:49:14 PM »

Quote
Originally posted by: DAaaMan64
Alright so I turned this in. Got my grade back, I got 1/10!! Wow, I guess I didn't understand something! I know it is only extra credit, but I will take what I can get. I don't know what I did wrong... WOW I pretty much turned in a version of what Professional 666 gave me...

Maybe ShyGuy told on me. ha ha

pretty much? Pro¿s solution is the answer... what did you turn in exactly?

DAaaMan64 · « **Reply #41 on:** November 07, 2007, 03:47:48 AM »

Ya I think you guys are right. I'll go in today and find out why.

News:

Author Topic: A Hard(For me) math problem I need help with, Circles and what not :) (Read 8071 times)