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04 Feb 2012, 17:12
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
55% (01:53) correct
45%
(01:39)
wrong
based on 896
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Given that ABCD is a rectangle, is the area of triangle ABE > 25?
(Note: Figure above is not drawn to scale).
(1) AB = 6
(2) AE = 10
Rectangle.PNG [ 2.86 KiB | Viewed 78814 times ]
How come the answer is B and not C? Can someone please explain?
PS: I tried the jpeg and bitmap format to attach the picture, but it says these two formats are not supported. Therefore attached the .pdf.
(Note: Figure above is not drawn to scale).
(1) AB = 6
(2) AE = 10
Attachment:
Rectangle.PNG [ 2.86 KiB | Viewed 78814 times ]
PS: I tried the jpeg and bitmap format to attach the picture, but it says these two formats are not supported. Therefore attached the .pdf.
04 Feb 2012, 17:38
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Given that ABCD is a rectangle, is the area of triangle ABE > 25? (Note: Figure above is not drawn to scale).
Rectangle.PNG [ 2.86 KiB | Viewed 78343 times ]
\(Area=\frac{1}{2}*AB*BE\)
(1) AB = 6 --> clearly insufficient: BE can be 1 or 100.
(2) AE = 10 --> now, you should know one important property: for a given length of the hypotenuse a right triangle has the largest area when it's isosceles, so for our case area of ABE will be maximized when AB=BE. So, let's try what is the largest area of a right isosceles triangle with hypotenuse equal to 10. Finding legs: \(x^2+x^2=10^2\) (where x=AB=BE) --> \(x=\sqrt{50}\) --> \(area_{max}=\frac{1}{2}\sqrt{50}^2=25\). Since it's the maximum area of ABE then the actual area cannot be more than 25. Sufficient.
Answer: B.
Attachment:
Rectangle.PNG [ 2.86 KiB | Viewed 78343 times ]
(1) AB = 6 --> clearly insufficient: BE can be 1 or 100.
(2) AE = 10 --> now, you should know one important property: for a given length of the hypotenuse a right triangle has the largest area when it's isosceles, so for our case area of ABE will be maximized when AB=BE. So, let's try what is the largest area of a right isosceles triangle with hypotenuse equal to 10. Finding legs: \(x^2+x^2=10^2\) (where x=AB=BE) --> \(x=\sqrt{50}\) --> \(area_{max}=\frac{1}{2}\sqrt{50}^2=25\). Since it's the maximum area of ABE then the actual area cannot be more than 25. Sufficient.
Answer: B.
04 Feb 2012, 17:41
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nglekel
Hi, and welcome to GMAT Club.
Unfortunately your reasoning is nor correct.
You assume with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 10 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 6:8:10. Or in other words: if \(a^2+b^2=10^2\) DOES NOT mean that \(a=6\) and \(b=8\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=10^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=6\) and \(b=8\).
For example: \(a=1\) and \(b=\sqrt{99}\) or \(a=2\) and \(b=\sqrt{96}\) or \(a=4\) and \(b=\sqrt{84}\) ...
Hope it's clear.
