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M10-34

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Bunuel
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M10-34 [#permalink] New post   16 Sep 2014, 00:43
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93% (00:31) correct 7% (00:21) wrong based on 46 sessions
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If the diagonal of a rectangle is 13, what is the area of the rectangle?


(1) The length of the rectangle is 12.

(2) The width of the rectangle is 5.
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D
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Bunuel
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Re M10-34 [#permalink] New post   16 Sep 2014, 00:43
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Statement (1) by itself is sufficient. Knowing the diagonal and the length we can find the width.

Statement (2) by itself is sufficient. Knowing the diagonal and the width we can find the length.


Answer: D
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yogi02
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Re: M10-34 [#permalink] New post   18 Sep 2019, 13:01
Bunuel, Cant we simply get it from diagonal alone. Any rectangle with diagonal 13 will have width and length = 12 and 5 in any order. As per Pythagoras triplet 13-12-5
What is even a need of given 2 statements
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Re: M10-34 [#permalink] New post   18 Sep 2019, 20:27
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yogi02 wrote:
Bunuel, Cant we simply get it from diagonal alone. Any rectangle with diagonal 13 will have width and length = 12 and 5 in any order. As per Pythagoras triplet 13-12-5
What is even a need of given 2 statements


You assume with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 13 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 5:12:13. Or in other words: if \(a^2+b^2=13^2\) DOES NOT mean that \(a=5\) and \(b=12\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=13^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=5\) and \(b=12\).

For example: \(a=1\) and \(b=\sqrt{168}\) or \(a=2\) and \(b=\sqrt{165}\) ...
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Re: M10-34 [#permalink] New post   11 Oct 2019, 23:57
Bunuel wrote:
yogi02 wrote:
Bunuel, Cant we simply get it from diagonal alone. Any rectangle with diagonal 13 will have width and length = 12 and 5 in any order. As per Pythagoras triplet 13-12-5
What is even a need of given 2 statements


You assume with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 13 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 5:12:13. Or in other words: if \(a^2+b^2=13^2\) DOES NOT mean that \(a=5\) and \(b=12\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=13^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=5\) and \(b=12\).

For example: \(a=1\) and \(b=\sqrt{168}\) or \(a=2\) and \(b=\sqrt{165}\) ...


Bunuel
Although extending this discussion makes no sense but again I hope we can find the area using: d1xd2/2. d1=d2 (rectangle and hence congruent).

Just a question for concept clarity.

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Re: M10-34 [#permalink]
11 Oct 2019, 23:57
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Bunuel
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