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(AB)^2 + (BD)^2
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25 Sep 2019, 16:00
25 Sep 2019, 16:00
Expert Reply
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Question Stats:
26% (00:15) correct
73% (00:34) wrong based on 41 sessions
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Attachment:
#greprepclub ABCD is a rectangle..jpg [ 16.99 KiB | Viewed 3292 times ]
ABCD is a rectangle.
Quantity A |
Quantity B |
\((AB)^2 + (BD)^2\) |
\(AD\) |
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: (AB)^2 + (BD)^2
[#permalink]
26 Sep 2019, 21:30
26 Sep 2019, 21:30
1
We have a rectangle.
By Pythagoras theorem, Diagonal \(AD^2 = AB^2 + BD^2\)
Therefore, we can replace qty A with \(AD^2\)
The main idea the question is trying to test is that side AB and BD could both be an integer number or a fractional number.
In case \(AB\) and \(BD\) are fractional number their square will be less and hence \(AD\) would be bigger than \(AD^2\). However if \(AB\) and \(BD\) are integer number \(AD^2\) will be bigger than \(AD\). Also keep in mind that \(AB\) and \(BD\) cannot be \(0\) or \(-ve\) number.
By Pythagoras theorem, Diagonal \(AD^2 = AB^2 + BD^2\)
Therefore, we can replace qty A with \(AD^2\)
The main idea the question is trying to test is that side AB and BD could both be an integer number or a fractional number.
In case \(AB\) and \(BD\) are fractional number their square will be less and hence \(AD\) would be bigger than \(AD^2\). However if \(AB\) and \(BD\) are integer number \(AD^2\) will be bigger than \(AD\). Also keep in mind that \(AB\) and \(BD\) cannot be \(0\) or \(-ve\) number.
