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Manager
Joined: 10 Dec 2020
Posts: 50
Given Kudos: 955
GRE 1: Q166 V162
Julius has a piece of string that is 24 inches long. He is stretching
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29 Nov 2021, 14:54
29 Nov 2021, 14:54
2
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Question Stats:
70% (01:56) correct
29% (01:33) wrong based on 24 sessions
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Julius has a piece of string that is 24 inches long. He is stretching the string around pegs on a pegboard to create different shapes.
a). Quantity A is greater.
b). Quantity B is greater.
c). The two quantities are equal.
d). The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The maximum area of a regular hexagon created by the string |
The maximum area of a square created by the string |
a). Quantity A is greater.
b). Quantity B is greater.
c). The two quantities are equal.
d). The relationship cannot be determined from the information given.
Re: Julius has a piece of string that is 24 inches long. He is stretching
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30 Nov 2021, 01:06
30 Nov 2021, 01:06
Do we have perhaps an answer explanation to this problem? I'm not sure how to determine the max area of the hexagon. Thanks.
Manager
Joined: 10 Dec 2020
Posts: 50
Given Kudos: 955
GRE 1: Q166 V162
Julius has a piece of string that is 24 inches long. He is stretching
[#permalink]
30 Nov 2021, 01:22
30 Nov 2021, 01:22
2
Sorry, don't have the official explanation handy.
My thoughts: The word "strech" may create an ambiguity - as in how long a string is supposed to be streched to create the max. area? "Wrap" can be a better alternative. Solution --
Regular Hexagon ----> 6 eq. triangles (with side 'a')
perimeter = 6 * a
area = 6 * (√3/4 * a^2)
ref: https://www.math-only-math.com/perimete ... xagon.html
quantity A: 41.57
Perimeter = 6*a = 24
a = 4
Therefore, Area = 6 * (√3/4 * 4^2) = 24√3 = 41.57
quantity A: 36
Perimeter = 4*s = 24
s = 6
Therefore, Area = s^2 = 36
a). Quantity A is greater.
My thoughts: The word "strech" may create an ambiguity - as in how long a string is supposed to be streched to create the max. area? "Wrap" can be a better alternative. Solution --
Regular Hexagon ----> 6 eq. triangles (with side 'a')
perimeter = 6 * a
area = 6 * (√3/4 * a^2)
ref: https://www.math-only-math.com/perimete ... xagon.html
quantity A: 41.57
Perimeter = 6*a = 24
a = 4
Therefore, Area = 6 * (√3/4 * 4^2) = 24√3 = 41.57
quantity A: 36
Perimeter = 4*s = 24
s = 6
Therefore, Area = s^2 = 36
a). Quantity A is greater.
elkk wrote:
Do we have perhaps an answer explanation to this problem? I'm not sure how to determine the max area of the hexagon. Thanks.
