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NeoPets Riddles (Lenny Conundrums) and Answers Here

 
 
andrewkatz
 
  1  
Reply Wed 4 Oct, 2006 08:05 pm
This can be solved using a^2+b^2=c^2.
No need to find any areas,
0 Replies
 
Lykandhed
 
  1  
Reply Wed 4 Oct, 2006 08:13 pm
to use the pythagorean theorem, we first need to know what the 2 of the side lengths (a, b, c) are. The shape formed is a quadrilateral, where 2 of the sides are equal to less than the 400m length of the larger square's side.
0 Replies
 
mommytbe
 
  1  
Reply Wed 4 Oct, 2006 08:15 pm
Care to elaborate?? I have been out of school so long, I can't even begin to solve that math problem. Will you help out a mommy_tbe who has been out of school way to long? I would appreciate a little help. Thanks so much!
~T.

I sure wish my daughter was here instead of college, she is really a math wiz!
0 Replies
 
dawnfur
 
  1  
Reply Wed 4 Oct, 2006 08:19 pm
Hi, I've never posted on here before, but is dividing the shape up into two right angled triangles and one rectangle the right ting to do? Or am I off track? Question
0 Replies
 
dawnfur
 
  1  
Reply Wed 4 Oct, 2006 08:20 pm
*thing
0 Replies
 
owlette
 
  1  
Reply Wed 4 Oct, 2006 08:25 pm
LC186 image

http://img.photobucket.com/albums/v27/gemma58/LC186_meriacres.gif
0 Replies
 
adamonroy
 
  1  
Reply Wed 4 Oct, 2006 08:28 pm
mommytbe sende me your neo user
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Amoora
 
  1  
Reply Wed 4 Oct, 2006 08:29 pm
Confuzzle
Very confused I got an answer I just think its not a big number so it must be wrong. Confused
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Amoora
 
  1  
Reply Wed 4 Oct, 2006 08:30 pm
Mommytb I understand how you feel I've been out of school for a few years. I have always been very bad in math this is like reading something in french I have no clue Laughing
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mommytbe
 
  1  
Reply Wed 4 Oct, 2006 08:35 pm
adamonroy it is simple, mommy_tbe. thanks so much, I will look for a neomail. I appreciate the help.
0 Replies
 
debezzzz
 
  1  
Reply Wed 4 Oct, 2006 08:55 pm
bad at math
I have been out of school along time, my kids are gone for the rest of the week, I could use some help. you can neomail me, same user name,.please and thank you
0 Replies
 
owlette
 
  1  
Reply Wed 4 Oct, 2006 09:02 pm
I've used a very simple method (not formulae) to calculate, hopefully it is right....

amazing what a pen paper and ruler can do Razz
0 Replies
 
adamonroy
 
  1  
Reply Wed 4 Oct, 2006 09:07 pm
thanks owlette
i will try it, because is SOOOOO difficult to day for me. i only get the distace from the down left corner to the center.. is
282.84m and the area of each piece is 22857.14m2 thats all that i have bye now.. good luck to every body
0 Replies
 
mommytbe
 
  1  
Reply Wed 4 Oct, 2006 09:07 pm
care to share with your friends owlette? or you could neomail at mommy_tbe?
0 Replies
 
owlette
 
  1  
Reply Wed 4 Oct, 2006 09:14 pm
Accuracy still has to be confirmed but:

Basically all I did was measure the "units" of each red line. Expressed that as a percentage of total and worked out actual...

For example
top red line is 8 "units", side of square is 9.3 "units" thus 8/9.3 = 0.860
Multiply this by 400 metre (known length of side of square) and the top red line equals 344.086 metres
0 Replies
 
judesalb
 
  1  
Reply Wed 4 Oct, 2006 09:16 pm
Just a little note to those here. Answers have always been posted at this site. It is not neopets and no one monitors us. Feel free to share with those that are mathematically challenged.
0 Replies
 
theturtlehermit
 
  1  
Reply Wed 4 Oct, 2006 09:31 pm
first off...

we got 882 meters.

we will post how in a seperate message.
0 Replies
 
adamonroy
 
  1  
Reply Wed 4 Oct, 2006 09:39 pm
with your method i get 934.8 m but may be is not the exactly number T.T
0 Replies
 
theturtlehermit
 
  1  
Reply Wed 4 Oct, 2006 09:40 pm
heres how:

the length of D1 of the kite (the 1/7th in the corner) is 200*sqrt(2), and its area is (400^2)/7.

so we multiply the area by two and divde by 200*sqrt(2) to find the length of D2 (the short diagonal of the kite). D2 is the hypotenuse of a 45-45-90 right triangle, both sides are a part of the perimeter we need. so we will divid the length of D2 by sqrt(2) and then multiply by 2.

this is the first part of the answer, it is 1600/7.

then we can see that the length from the corner to D2 is a part of a 45-45-90 triangle with one of the two sides that we just found as a hypotenuse, and so 800/7/sqrt(2) is equal to the base of yet another right triangle that we will need, and the other side of it is equal to D1- 800/7/sqrt(2). we can then use pyagorthus to find the length of the long side of the kite that is included (which comes out to be 200*sqrt(58)/7) and we add that into our perimiter figure.

Now turning our attention to the last two sides we need, it can be seen through a bit of trig that we do not feel like going into that the angle at the base of that triangle is 60 degrees, making that triangle equalateral, with the inside edge of the kite, and making the final answer.

(600*sqrt(58)/7)+(1600/7) which is evaluated as 881.351980503, and rounds up to 882 meters.
0 Replies
 
adamonroy
 
  1  
Reply Wed 4 Oct, 2006 09:50 pm
OMG thanks
0 Replies
 
 

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