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 Post subject: Can you solve this?
PostPosted: April 17th, 2007, 10:10 pm 
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The hardest problem on a sheet I had for geometry, Mike and I worked it out..well he showed me why most of the stuff was wha it was, but in the end I got it so .. now it's your turn.

How many square feet does Betsy have to graze?


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PostPosted: April 17th, 2007, 10:10 pm 
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PostPosted: April 17th, 2007, 10:28 pm 
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She can't graze because Sev killed her. (Look down so you know this isn't spam)

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But I will print this out and try to figure it out tomorrow. It's late here.

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PostPosted: April 17th, 2007, 10:43 pm 
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i remember doing problems like this... i hated them.

well 100ft rope, she can basically cut the bottom half of circle, on the right side it is a 25ft before hitting wall so turning she can cut another 75ft going straight down the fence. on the left it's 80ft so she can cut another 20ft ar the side. case close.

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PostPosted: April 17th, 2007, 11:12 pm 
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23875.25?

I worked this out on Illustrator... so yeah XD.

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PostPosted: April 18th, 2007, 2:27 pm 
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I make a scale drawing, then did the calculations ... totaling = 21,902.69ft^2

^ looking now @ agent waffles, I did much the same; 1/2 main circle, 1/4 small circle on left side ..
But instead of subtracting(for the 3 top circle, I added: 1 triangle + 2 arks)

Scanning pic .. hopefully posting

EDIT: Redid, using an even more basic calculation then waffles = 22,001.1ft^2

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 Post subject: Re: Can you solve this?
PostPosted: April 20th, 2007, 12:53 am 
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I got 21589.18 sq.ft. Here's my calculations:

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I used this diagram. First, it's obvious that C (look in diagram for area with letter c in it), is 5000pi, and that d is 100pi. Figure B is somewhat simple, because instead of trying to figure about A and B together, I split them apart. So since the central angle of sector B is 45 degrees, it's 1/8 of a full circle, making it's area 1250pi. the last part is stupid because of the square root of 2 making things complicated. Using geometric theorems (vert. angles, alt. interior, etc) it can be proved that the central angle of sector A is also 45 degrees, making it's area (1406.25pi-625pi * root2) When you add them all together you get 21590.178, rounded up is 21590.18.

Of course, logically the the cow can also graze in the small triangle next to the barn, adding 312.5 square units. But whatever, that's a technicality.

Hope I got it right. ;O

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 Post subject: Register and login to get these in-post ads to disappear
PostPosted: April 20th, 2007, 12:53 am 
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PostPosted: April 20th, 2007, 8:27 pm 
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Yuck, geometry. Give me algebra 2. Love that stuff. Qudratic functions, conic sections, and polynomial fractions ftw! :lol:


Off topic: Anyone want to do another problem (alg2)? :lol: (I had alot of trouble with this one :oops: Probly won't be too hard to you guys though, seeing how easily you did the geometry one, lol.)

It is three separate fractions. Parenthesis group the (numerator)/(denominator) of each.

Simplify:
(m^2+n^2)/(m^2-n^2) + (m)/(n-m) + (n)/(m+n)

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Spoiler for Nietszche Quote:

But what did such a Teuton afterwards look like when he had been "improved" and led into a monastery? Like a caricature of a human being, like an abortion: he had become a "sinner," he was in a cage, one had imprisoned him behind nothing but sheer terrifying concepts... There he lay now, sick, miserable, filled with ill-will towards himself; full of hatred for the impulses toward life, full of suspicion of all that was still strong and happy. In short, a "Christian"... - Twighlight of the Idols
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Last edited by ryan1 on April 20th, 2007, 9:04 pm, edited 1 time in total.

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PostPosted: April 20th, 2007, 8:43 pm 
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I got 21,064.9^2 feet, using a scale model made on paint.


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