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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.
Given that ABCD is a rectangle, is the area of triangle ABE > 25? (Note: Figure above is not drawn to scale).
Attachment:
Rectangle.PNG [ 2.86 KiB | Viewed 11422 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.
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 \(b=\sqrt{84}\) ...
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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"There is no alternative to hard work. If you don't do it now, you'll probably have to do it later. If you didn't need it now, you probably did it earlier. But there is no escaping it."
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.
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.
Re: Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink]
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
Re: Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink]
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. _________________
Re: Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink]
14 Mar 2015, 09:20
Bunuel, have you encountered real gmat questions testing this concept: for a given length of the hypotenuse a right triangle has the largest area when it's isosceles ? _________________
If my post was helpful, press Kudos. If not, then just press Kudos !!!
Re: Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink]
14 Mar 2015, 11:18
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Hi Ergenekon,
This is a rarer concept (from the realm of Multi-Shape Geometry), but the GMAT has been known to test it.
The broader issue is more about comparing squares and rectangles though.
For example, compare the areas of this square and rectangles....
10x10 9x11 8x12
Areas: (10)(10) = 100 (9)(11) = 99 (8)(12) = 96
By increasing one side and decreasing the other by an "equivalent amount", the area decreases.
When it does appear on the GMAT, it's often themed around 'percentage change' in side lengths (re: length is 10% greater, width is 10% less), but the pattern is still the same.
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