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The perimeter of the shaded region in the rectangle
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17 Dec 2020, 09:51
17 Dec 2020, 09:51
Expert Reply
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Question Stats:
29% (01:11) correct
70% (01:06) wrong based on 31 sessions
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Quantity A |
Quantity B |
The perimeter of the shaded region in the rectangle |
\(2 \sqrt{2} +2\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Re: The perimeter of the shaded region in the rectangle
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17 Dec 2020, 19:46
17 Dec 2020, 19:46
D. we need to know if the triangle is isosceles or that the white areas are equal sized or that the top of the triangle intercepts the line directly in the middle or that the two sides of that line are equal length.
Assuming we know one of the above is true, calculate the hypotenuse of the triangle which is sqrt(2) we have two of them and the bottom length so 2*sqrt(2)+2
But again we can't make any assumption for GRE problems that they don't explicitly tell us.
Assuming we know one of the above is true, calculate the hypotenuse of the triangle which is sqrt(2) we have two of them and the bottom length so 2*sqrt(2)+2
But again we can't make any assumption for GRE problems that they don't explicitly tell us.
Re: The perimeter of the shaded region in the rectangle
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28 Feb 2021, 00:23
28 Feb 2021, 00:23
Solution:
Although it looks like the triangle divides the rectangle into two equal part on both side but, it is possible that it does not.
So for eg. one of the sides of the triangle could be 1 the other being 2, using the pythagoras theorem we'll get the hypotenuse as 5
Perimeter=1+2+\(\sqrt{5}\)=3+\(\sqrt{5}\)
Qty A>Qty B.
Alternatively, the upper portion could be of same length i.e. 1 & 1
In this case the perimeter=2+2\(\sqrt{2}\)
Qty A=Qty B
Thus IMO D
Hope this helps!
Although it looks like the triangle divides the rectangle into two equal part on both side but, it is possible that it does not.
So for eg. one of the sides of the triangle could be 1 the other being 2, using the pythagoras theorem we'll get the hypotenuse as 5
Perimeter=1+2+\(\sqrt{5}\)=3+\(\sqrt{5}\)
Qty A>Qty B.
Alternatively, the upper portion could be of same length i.e. 1 & 1
In this case the perimeter=2+2\(\sqrt{2}\)
Qty A=Qty B
Thus IMO D
Hope this helps!
