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In any square, the relationship between the area A and the perimeter P
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05 Jan 2023, 00:12
05 Jan 2023, 00:12
Expert Reply
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Question Stats:
87% (01:32) correct
12% (01:49) wrong based on 16 sessions
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In any square, the relationship between the area A and the perimeter P is \(A =nP^2\), where n is a constant. What is the value of \(n^2\)?
A. 1/4
B. 1/8
C. 1/32
D. 1/64
E. 1/256
A. 1/4
B. 1/8
C. 1/32
D. 1/64
E. 1/256
In any square, the relationship between the area A and the perimeter P
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30 Jan 2023, 10:55
30 Jan 2023, 10:55
1
Expert Reply
The area of a square is given by the formula A = s^2, where s is the length of the square's side. The perimeter is given by P = 4s. Plugging these values into the equation given in the question, you can solve for n
s^2=n * (4s)^2
s^2=16s^2n
n=s^2/16s^2=1/16
n^2=(1/16)^2=1/256
s^2=n * (4s)^2
s^2=16s^2n
n=s^2/16s^2=1/16
n^2=(1/16)^2=1/256
Re: In any square, the relationship between the area A and the perimeter P
[#permalink]
27 Dec 2023, 20:17
27 Dec 2023, 20:17
If it helps to have something more concrete, you could just make up a square and plug those values in. For example, use square with length 1, and see where that gets you.
