Question 164862
let p1 = perimeter of one square.
let p2 = perimeter of the other square.
let a1 = area of one square.
let a2 = area of the other square.
let s1 = side of one square.
let s2 = side of the other square.
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p1 = p2 + 16
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perimeter of a square equals 4 times one side.
p = 4*s
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4s1 = 4s2 + 16 (first equation) *****
s1 = (4s2 + 16)/4
s1 = s2 + 4
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a1 = 4a2 - 14
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area of a square equals one side times the other side (one side squared).
a = s^2
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s1^2 = 4(s2^2) - 44 (second equation) *****
since s1 = s2+4, this equation becomes
(s2+4)^2 = 4(s2^2) - 44
to make calculating less confusing, let x = s2
equation becomes:
(x+4)^2 = 4x^2 - 44
this becomes:
x^2 + 8x + 16 = 4x^2 - 44
subtract the equation on the left from both sides of the equation and it becomes
0 = 4x^2 - 44 - x^2 - 8x - 16
this becomes
0 = 3x^2 - 8x - 60
which is the same as
3x^2 - 8x - 60 = 0
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that equation factors out to be
(3x+10)*(x-6) = 0
factors are:
x = 6
or
x = -10/3
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since the side of the square can't be negative, the only answer that looks promising is x = 6.
since x = s2, that means that ...
s2 = 6.
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first equation is reproduced here for ease of reference.
4s1 = 4s2 + 16 (first equation) *****
since s2 = 6, this equation becomes
4s1 = 4*6 + 16 = 24 + 16 = 40
s1 = 10.
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second equation is reproduced here for ease of reference.
s1^2 = 4(s2^2) - 44 (second equation) *****
since s1 = 10, and s2 = 6, this equation becomes
10^2 = 4*(6^2) - 44
100 = 4*36 - 44
100 = 144 - 44
100 = 100
s1 and s2 yield correct values for the equations.
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answer is:
s1 = 10 = side of one square.
s2 = 6 = side of the other square.
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