Question 881304: Please help me solve this question:
You are given two pieces of wire, each 20 cm long. One piece is to be bent to form the biggest possible rectangle and the other piece to form a golden rectangle.Determine the length and width of each rectangle.Use two methods in each case.
Note:Golden rectangle have length y cm and width x cm,where y > x,is such that y/x = (x+y)/y = golden ratio that is 1.618
Answer by dkppathak(439)   (Show Source):
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You are given two pieces of wire, each 20 cm long. One piece is to be bent to form the biggest possible rectangle and the other piece to form a golden rectangle.Determine the length and width of each rectangle.Use two methods in each case.
Note:Golden rectangle have length y cm and width x cm,where y > x,is such that y/x = (x+y)/y = golden ratio that is 1.618
let length is Y and width is X
as per given first condition
perimeter is 20 for both case
p =2(l+b)
20= 2(Y +X)
10= Y +x if we have pair as 1,9 area is 9 2,8 area 16 3,7 area 21 4,6 area is 24 5,5 area is 25 ( but it will be square ) therefor
maximum area will be 24 cm^2
so we can say biggest possible rectangle is 6,4
for golden ratio Y>x and Y +X/Y =1.6
therefor for second pair will be (1,9) (2,8( (3,7)(4,6) only
by which we are getting golden ratio Y+X/Y= 10/6 =1.66
so golden ratio shows the side should be 6 and 4 only which shows golden ratio as length =6 cm and width =4 cm
Answer
biggest rectangle will be l=6 cm and width =4 cm
golden rectangle will be also 6 and 4 only
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