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(BC)^2+(BA)^2
[#permalink]
31 Aug 2020, 10:06
31 Aug 2020, 10:06
Expert Reply
00:00
Question Stats:
70% (00:49) correct
29% (01:12) wrong based on 47 sessions
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Attachment:
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Quantity A |
Quantity B |
\((BC)^2+(BA)^2 \) |
\((BD)^2\) |
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: (BC)^2+(BA)^2
[#permalink]
02 Sep 2020, 12:47
02 Sep 2020, 12:47
1
Carcass wrote:
Quantity A |
Quantity B |
\((BC)^2+(BA)^2 \) |
\((BD)^2\) |
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.
First notice that \(BC\) and \(BA\) are legs of the right triangle \(ABC\). Therefore:
\((BC)^2+(BA)^2 = (CA)^2\) from the pythagorean theorem.
So now, we're comparing both diagonals in the trapezoid.
Let \(CA\) be a fixed length.
Since we should never trust the dimensions of the figure (unless given information, such as the right angle), the length of \(BD\) can be anything.
You can move \(D\) and extend it far to the right, making the diagonal \(BD\) very large.
Or, you can move point \(D\) diagonally along the dashed line as close as you can to line \(CA\), making \(BD\) small.
Since we can change the length of \(BD\), we can make it smaller, equal, or larger than \(CA\).
Therefore, the answer is D
gmatclubot
Re: (BC)^2+(BA)^2 [#permalink]
02 Sep 2020, 12:47
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