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Given that ABCD is a rectangle, is the area of triangle ABE>

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Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink] New post 04 Feb 2012, 16:12
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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).
Attachment:
Rectangle.PNG
Rectangle.PNG [ 2.86 KiB | Viewed 5833 times ]

(1) AB = 6
(2) AE = 10

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.
[Reveal] Spoiler: OA

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Re: Area > 25? [#permalink] New post 04 Feb 2012, 16:31
Hi,

B states that AE = 10 and triangle ABE is a right triangle. So it makes it a special case "side-based" right triangle where one of the lengths of the sides form ratios of whole numbers, such as 3 : 4 : 5.

Side AE = 10, which means that side AB = 6 and side BE = 8 (ratio 6:8:10 = ratio 3:4:5). Now knowing the sides, you can easily calculate the area which equals 24 < 25.
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Re: Area > 25? [#permalink] New post 04 Feb 2012, 16:38
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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).
Attachment:
Rectangle.PNG
Rectangle.PNG [ 2.86 KiB | Viewed 5805 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.
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Re: Area > 25? [#permalink] New post 04 Feb 2012, 16:41
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Hi,

B states that AE = 10 and triangle ABE is a right triangle. So it makes it a special case "side-based" right triangle where one of the lengths of the sides form ratios of whole numbers, such as 3 : 4 : 5.

Side AE = 10, which means that side AB = 6 and side BE = 8 (ratio 6:8:10 = ratio 3:4:5). Now knowing the sides, you can easily calculate the area which equals 24 < 25.


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.
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Re: Area > 25? [#permalink] New post 09 Mar 2012, 23:33
Bunuel wrote:
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.

This is what's so great about the forum. One's faulty assumptions get checked in time. In this case, I had also fallen into the trap of thinking that since hypotenuse is 10 the other sides are 8 and 6. As Bunuel points out, that's clearly the wrong way to think about this.

And knowing the isosceles-right triangle property certainly helps!
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Re: Area > 25? [#permalink] New post 21 Nov 2012, 06:55
Bunuel wrote:
Given that ABCD is a rectangle, is the area of triangle ABE > 25? (Note: Figure above is not drawn to scale).
Attachment:
Rectangle.PNG
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: the 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 can not be more than 25. Sufficient.

Answer: B.

Well an isosceles triangle has maximum area given a hypothenuse. The hypothenuse doesn't seem to be given here, side AB can be as long or as short as you want, thereby making the area larger or smaller than 25.

edit: sorry didnt read the question correctly, i somehow read that BE was given as 10.
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Re: Area > 25? [#permalink] New post 19 Oct 2013, 04:44
Bunuel wrote:
Given that ABCD is a rectangle, is the area of triangle ABE > 25? (Note: Figure above is not drawn to scale).
Attachment:
Rectangle.PNG
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: the 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 can not be more than 25. Sufficient.

Answer: B.


Hey Bunuel,

The property you mentioned only stands true when the hypotenuse is fixed and that is the reason it cannot be applied to the option A. Else, the answer would have been D.

Thought I should clarify for the people reading the post.
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Re: Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink] New post 06 Dec 2013, 09:05
My first thought was to try and determine the possible lengths of side AB and AE. Is this possible with only one given side measurement?
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Re: Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink] New post 06 Dec 2013, 10:04
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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).
Attachment:
The attachment Rectangle.PNG is no longer available

(1) AB = 6
(2) AE = 10

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.


F.S 1 is clearly Insufficient.

Another approach for F.S 2 :

We know that a^2+c^2 = 10^2 \to a^2+c^2 = 100

Also, area of \triangle ABE - \frac{1}{2}*a*c

Is\frac{1}{2}*a*c>25 \to Is a*c>50 \to 2*a*c>100?

Is 2*a*c>a^2+c^2 \to Is (a-c)^2<0. Of-course, the answer is NO.Sufficient.
Attachments

img.png
img.png [ 3.58 KiB | Viewed 2453 times ]


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Re: Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink] New post 01 Jan 2014, 04:52
Quote:
(1) AB = 6 --> clearly insufficient: BE can be 1 or 100.
(2) AE = 10 --> now, you should know one important property: the 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 can not be more than 25. Sufficient.


Hi Bunnel,
Why cant the reasoning that a right triangle has greatest area when it is isosceles be applied to the first statement as well. Which says AB = 6, hence assuming BE = 6 we would get the area = 1/2*6*6 = 18 < 25
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Re: Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink] New post 01 Jan 2014, 05:03
Expert's post
Rohan_Kanungo wrote:
Quote:
(1) AB = 6 --> clearly insufficient: BE can be 1 or 100.
(2) AE = 10 --> now, you should know one important property: the 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 can not be more than 25. Sufficient.


Hi Bunnel,
Why cant the reasoning that a right triangle has greatest area when it is isosceles be applied to the first statement as well. Which says AB = 6, hence assuming BE = 6 we would get the area = 1/2*6*6 = 18 < 25


The property says: for a given length of the hypotenuse a right triangle has the largest area when it's isosceles. Thus you cannot apply it to the first statement.
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Re: Given that ABCD is a rectangle, is the area of triangle ABE>   [#permalink] 01 Jan 2014, 05:03
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