Question 164862: The perimeter of one square exceeds that of another by 16 and its area is 44 less than 4 times the area of the other. What is the length of the side of each side of each square?
Found 2 solutions by gonzo, padmameesala:
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! 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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Answer by padmameesala(42) (Show Source):
You can put this solution on YOUR website! Let x be the length of the side of the square1,
p be the perimeter of the square 1
and a be the area of the square 1.
Then p=4x and a=x^2
Let y be the length of the side of the square2,
q be the perimeter of the square2
and b be the area of the square2.
Then q=4y and b=y^2
Given that the perimeter of one square exceeds that of another by 16 and its area is 44 less than 4 times the area of the other
that is p=q+16 and a=4b-44
that is 4x=4y+16 and x^2=4y^2-44
Now solving these two equations for x and y we get
plug x=y+4 in x^2=4y^2-44
that gives (y+4)^2=4y^2-44
y^2+8y+16=4y^2-44
3y^2-8y-60=0
(3y+10)(y-6)=0
that is y=-10/3 or y=6
since y is the length of the side of the square2 it cant be negative so -10/3 is discarded
hence y=6
therefore x=y+4=6+4=10
Thus length of the side of square1 is 4units and length of the side of the square2 is 10units.
My suggestion is not to worry that the answer is too length try to go through it twice or thrice so that you will understand it more explicitly
I hope you will enjoy it.
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