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Bunuel
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M10-34 16 Sep 2014, 00:43
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5% (low)

Question Stats:

94% (00:31) correct 6% (00:21) wrong based on 47 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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M10-34 16 Sep 2014, 00:43
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Official Solution:


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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M10-34 18 Sep 2019, 13:01
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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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Bunuel
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M10-34 18 Sep 2019, 20:27
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yogi02
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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M10-34 11 Oct 2019, 23:57
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Bunuel
yogi02
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.

Regards
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