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Math Riddle: How much string would you add....
Posted on 7/25/14 at 6:05 pm
Posted on 7/25/14 at 6:05 pm
Let's say the earth was perfectly round and smooth, and you ran a string all the way around the equator - 25,000 miles. And then your wife comes along and wants the string to sit exactly 1 foot off the ground - all the way around.
Question: how much new string do you add to the original string to make that happen?
Here's the cryptic answer (shhhhh! don't reveal it!): LINK
Question: how much new string do you add to the original string to make that happen?
Here's the cryptic answer (shhhhh! don't reveal it!): LINK
This post was edited on 7/25/14 at 6:11 pm
16
Posted on 7/25/14 at 6:06 pm to tigerpawl
Do you even circumference, bro?
Posted on 7/25/14 at 6:07 pm to tigerpawl
wut
Posted on 7/25/14 at 6:07 pm to tigerpawl
Can't I just put all of it in a big container and put the container one foot off the ground.
Posted on 7/25/14 at 6:08 pm to tigerpawl
I'd tell her to get her arse back in the kitchen and make me dinner.
Posted on 7/25/14 at 6:08 pm to tigerpawl
I'm not married. Next question.
Posted on 7/25/14 at 6:09 pm to tigerpawl
Is this actually a riddle?
Posted on 7/25/14 at 6:12 pm to tigerpawl
C= 25000 = 2*PI*R ==> R=25000/(2*PI)
C' = 2*PI*(25000/(2*PI)+1)
Calculate the radius, add 1, recalculate Circumference. Subtract C'-C
Oops. Correction as noted: Convert 1' increment to 1/5280 miles.
C' = 2*PI*(25000/(2*PI)+1)
Calculate the radius, add 1, recalculate Circumference. Subtract C'-C
Oops. Correction as noted: Convert 1' increment to 1/5280 miles.
This post was edited on 7/25/14 at 6:22 pm
Posted on 7/25/14 at 6:14 pm to tigerpawl
The surprising thing is that the answer is independent of the size of the sphere.
Posted on 7/25/14 at 6:16 pm to tigerpawl
Approximately 25,004' ?
Posted on 7/25/14 at 6:20 pm to tigerpawl
6.28 feet
Posted on 7/25/14 at 7:05 pm to tigerpawl
Surprisingly simple:
The circumference of a circle = 2pi * r (radius)
Therefore, adding 1 foot to the circumference means this is now 2pi * (r+1) = (2pi*r) + 2pi
The extra amount of string required is therefore 2pi times the additional height, in this case 2pi feet.
The circumference of a circle = 2pi * r (radius)
Therefore, adding 1 foot to the circumference means this is now 2pi * (r+1) = (2pi*r) + 2pi
The extra amount of string required is therefore 2pi times the additional height, in this case 2pi feet.
Posted on 7/25/14 at 8:04 pm to tigerpawl
1. 2pi x r = C
2. 2pi x (r + 1ft) = C'
3. (2pi x r) + (2pi x 1ft) = C'
4. C + 2pi = C'
5. C' - C = 2pi
Assume all units are in feet...
2. 2pi x (r + 1ft) = C'
3. (2pi x r) + (2pi x 1ft) = C'
4. C + 2pi = C'
5. C' - C = 2pi
Assume all units are in feet...
This post was edited on 7/25/14 at 8:12 pm
Posted on 7/25/14 at 8:18 pm to tigerpawl
The square root of 25001 squared minus 25000 squared....or some stupid shite like that.
Posted on 7/25/14 at 8:23 pm to tigerpawl
This is really easy.
Posted on 7/26/14 at 8:59 am to tigerpawl
This thread is both hilarious and sad.
Posted on 7/26/14 at 9:17 am to tigerpawl
That's just like a woman. I just ran a MFing string around the whole damn earth for you, and of course...it's not good enough. Now you want it to be 1 foot off the ground. Nothing is ever good enough for you.
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