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A square park of an area 4225 sq. m has to be fenced. The fencing will require the use of a wire that can enclose the park 4 times and each circuit of the wire would be 5 % greater than the perimeter of the park. What length of the wire is needed for this?
 

 Oct 3, 2019
 #1
avatar+104969 
+2

sqrt (4225)  =   65  = side of the square

 

So....the perimeter of the park  =  4 * 65  =  260 ft

 

So......the length of wire needed  is

 

First perimeter + Second perimeter + Third perimeter + Fourth perimeter  = 

 

260 + 260(1.05) + 260(1.05)^2  + 260(1.05)^3  ≈  1120.6 ft

 

 

cool cool cool

 Oct 3, 2019
 #2
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0

wait CPhill the answer you provided didn't match the multiple-choice, here are the four answer available. "1092,  1082.31,  1094.22, and 1089.21" did you do something wrong or is there something wrong with the answer choices?

Guest Oct 4, 2019
 #3
avatar+104969 
+3

Mmmm....don't know......I was assuming that each new "wrap" of wire would create a "new" perimeter

 

I see that hectictar is answering......she probably has the correct interpretation!!!!

 

 

cool cool cool

CPhill  Oct 4, 2019
 #4
avatar+8810 
+1

length of one side of park   =   √[ 4225 sq m ]  =   65 m

 

perimeter of park   =   4 * 65 m   =   260 m

 

length of each circuit  =   5% greater than 260 m   =   105% * 260 m   =   273 m

 

total length needed  =  4 * 273 m   =   1092 m

 

smileylaugh

 Oct 4, 2019
 #6
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Yeah that's what I thought too thanks for answering smiley

Guest Oct 4, 2019
 #7
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Yeah that's what I thought too thanks for answering smiley

Guest Oct 4, 2019
 #5
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+1

I think they have used 5% simple (non-compound) increase:

 

[260 x 4] + [5% x 260 x 4] = 1040 + 52 =1092 m.

 Oct 4, 2019

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