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Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink] 
04 Feb 2012, 16:12
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Question Stats:
58% (01:53) correct 42% (01:39) wrong based on 680 sessions
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Given that ABCD is a rectangle, is the area of trilateral ABE > 25?
(Note: Figure above is non drawn to scale).
Attachment:
Rectangle.PNG [ 2.86 KiB | Viewed 65083 times ]
(1) AB = 6
(2) AE = 10
How come through the suffice is B and non C? Fanny someone please explain?
PS: I tried the jpeg and electronic image formatting to attach to the picture, but it says these two formats are not supported. Therefore attached the .pdf.
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Re: Area > 25? [#permalink] 04 Feb 2012, 16:38
Given that ABCD is a rectangle, is the area of triangle ABE > 25? (Notation: Figure above is not drawn to scale).
Attachment:
Rectangle.PNG [ 2.86 KiB | Viewed 64969 times ]
\(Area=\frac{1}{2}*AB*BE\)
(1) AB = 6 --> clearly lean: BE can comprise 1 or 100.
(2) AE = 10 --> at once, you should know one important dimension: for a given duration of the hypotenuse a right triangle has the largest area when it's isosceles, and then for our case area of ABE will be maximized when Group AB=Represent. So, let's try what is the largest area of a right isosceles triangle with hypotenuse equal to 10. Determination legs: \(x^2+x^2=10^2\) (where x=AB=Make up) --> \(x=\sqrt{50}\) --> \(area_{max}=\frac{1}{2}\sqrt{50}^2=25\). Since it's the maximum area of ABE then the actual area cannot be more than 25. Sufficient.
Response: B.
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Re: Surface area > 25? [#permalink] 04 Feb 2012, 16:31
Hi,
B states that AE = 10 and triangle ABE is a in good order Triangulum. Then it makes IT a specialized case "side-based" right triangle where one of the lengths of the sides form ratios of whole numbers, such equally 3 : 4 : 5.
Root 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 well calculate the area which equals 24 < 25.
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Re: Area > 25? [#permalink] 04 Feb 2012, 16:41
nglekel wrote:
Hi,
B states that AE = 10 and triangle ABE is a precise triangle. So it makes it a special case "root-supported" right triangle where one of the lengths of the sides form ratios of whole numbers game, such A 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 easy calculate the area which equals 24 < 25.
Howdy, and welcome to GMAT Club.
Unfortunately your reasoning is nor correct.
You assume with nobelium ground for IT that the lengths of the sides are integers. Knowing that hypotenuse equals to 10 DOES Non entail that the sides of the right Triangulum necessarily must be in the ratio of Pythagorean triple - 6:8:10. Or in former words: if \(a^2+b^2=10^2\) DOES NOT mean that \(a=6\) and \(b=8\), certainly this is nonpareil of the possibilities but definitely not the only one. As a matter of fact \(a^2+b^2=10^2\) has infinitely many solutions for \(a\) and \(b\) and only when one of them is \(a=6\) and \(b=8\).
For example: \(a=1\) and \(b=\sqrt{99}\) operating theatre \(a=2\) and \(b=\sqrt{96}\) or \(a=4\) and \(b=\sqrt{84}\) ...
Bob Hope it's take in.
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Re: Area > 25? [#permalink] 09 Mar 2012, 23:33
Bunuel wrote:
You assume with no solid ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 10 DOES Non mean that the sides of the right triangle needfully must be in the ratio of Pythagorean triple - 6:8:10. Or in other words: if \(a^2+b^2=10^2\) DOES Non mean that \(a=6\) and \(b=8\), certainly this is one of the possibilities but definitely not the only single. In fact \(a^2+b^2=10^2\) has infinitely many solutions for \(a\) and \(b\) and exclusively same 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 defective assumptions get checked in time. In this case, I had also fallen into the trap of reasoning that since hypotenuse is 10 the other sides are 8 and 6. Eastern Samoa Bunuel points retired, that's clearly the wrong way to think approximately this.
And informed the isosceles-right triangle property certainly helps!
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Atomic number 75: Area > 25? [#permalink] 21 Nov 2012, 06:55
Luis Bunuel wrote:
Given that ABCD is a rectangle, is the arena of triangle ABE > 25? (Tone: Figure above is non drawn to scale).
Attachment:
Rectangle.PNG
\(Orbit=\frac{1}{2}*Type AB*Personify\)
(1) AB = 6 --> intelligibly scant: Glucinium seat be 1 or 100.
(2) AE = 10 --> now, you should know one important property: the right-angled triangle has the largest area when it's isosceles, so for our case area of ABE will represent maximized when AB=Embody. So, let's hear 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=Abdominal muscle=BE) --> \(x=\sqrt{50}\) --> \(area_{max}=\frac{1}{2}\sqrt{50}^2=25\). Since IT's the utmost orbit of ABE then the actual area can not be many than 25. Sufficient.
Answer: B.
Well an isosceles Triangle has maximum area conferred a hypothenuse. The hypothenuse doesn't seem to live given here, side AB can be As long operating room as short as you want, thereby making the arena big Beaver State smaller than 25.
delete: disconsolate didnt read the question correctly, i somehow read that Glucinium was granted as 10.
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Re: Area > 25? [#permalink] 19 Oct 2013, 04:44
Bunuel wrote:
Given that ABCD is a rectangle, is the area of triangle ABE > 25? (Billet: Figure above is not drawn to scale).
Attachment:
Rectangle.PNG
\(Area=\frac{1}{2}*AB*Live\)
(1) AB = 6 --> clearly insufficient: Represent can embody 1 or 100.
(2) AE = 10 --> now, you should know one important property: the right-angled triangle has the largest expanse when it's isosceles, so for our case domain of ABE will be maximized when AB=Glucinium. So, let's try what is the largest area of a right isosceles triangle with hypotenuse equal to 10. Determination 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 supreme area of ABE then the actual country can non be more than 25. Comfortable.
Answer: B.
Hey Bunuel,
The property you mentioned exclusively stands true when the hypotenuse is fixed and that is the reason IT cannot be practical to the option A. Else, the answer would have been D.
Thinking I should clarify for the masses Reading the post.
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Re: Given that ABCD is a rectangle, is the arena of triangle ABE> [#permalink] 06 Dec 2013, 09:05
My first thought was to adjudicate and make up one's mind the possible lengths of side Av and AE. Is this possible with sole one given side measurement?
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Re: Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink] 06 Dec 2013, 10:04
enigma123 wrote:
Given that ABCD is a rectangle, is the country of Triangle ABE > 25?
(Banker's bill: Figure above is not drawn to scale).
Attachment:
The affixation Rectangle.PNG is atomic number 102 longer in stock
(1) Abdominal muscle = 6
(2) AE = 10
How come the answer is B and not C? Can mortal please excuse?
PS: I tried the jpeg and bitmap format to attach the picture, just 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, orbit 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 OrdinalAmple.
Attachments
img.png [ 3.58 KiB | Viewed 60250 times ]
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Re: Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink] 01 Jan 2014, 04:52
Quote:
(1) AB = 6 --> distinctly insufficient: BE can live 1 or 100.
(2) AE = 10 --> now, you should know one cardinal belongings: the reactionary triangle has the largest domain when it's isosceles, so for our cause area of ABE will be maximized when Av=BE. So, let's try what is the largest area of a right isosceles triangle with hypotenuse equal to 10. Determination legs: x^2+x^2=10^2 (where x=AB=BE) --> x=\sqrt{50} --> area_{soap}=\frac{1}{2}\sqrt{50}^2=25. Since IT's the maximum region of ABE then the actual area can buoy not equal more than 25. Decent.
Hi Bunnel,
Why cant the reasoning that a right-wing Triangle has superlative sphere when it is isosceles be practical to the first statement as well. Which says Ab = 6, therefore assuming Glucinium = 6 we would mystify 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] 01 January 2014, 05:03
Rohan_Kanungo wrote:
Inverted comma:
(1) Abdominal muscle = 6 --> clearly insufficient: Glucinium can be 1 or 100.
(2) AE = 10 --> immediately, you should know one important place: the right-hand triangle has the largest area when it's isosceles, so for our showcase area of ABE bequeath be maximized when AB=Represent. Indeed, 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=Bachelor of Arts=BE) --> x=\sqrt{50} --> area_{grievous bodily harm}=\frac{1}{2}\sqrt{50}^2=25. Since IT's the maximum area of ABE then the actual area can non be more than 25. Sufficient.
Hi Bunnel,
Why cant the thinking that a right triangle has sterling field when it is isosceles be practical to the for the first time instruction as well. Which says BA = 6, hence assuming BE = 6 we would get the arena = 1/2*6*6 = 18 < 25
The property says: for a given length of the hypotenuse a right-angled triangle has the largest area when it's isosceles. Thus you cannot apply it to the first assertion.
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Ray: Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink] 14 Spoil 2015, 09:20
Bunuel, have you encountered actual gmat questions testing this concept: for a given length of the hypotenuse a right-angled triangle has the largest sphere when it's isosceles ?
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Re: Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink] 14 Mar 2015, 11:18
Hi Ergenekon,
This is a rarer concept (from the realm of Multi-Shape Geometry), simply the GMAT has been known to test it.
The broader effect is more near comparing squares and rectangles though.
For representative, compare the areas of this square and rectangles....
10x10
9x11
8x12
Areas:
(10)(10) = 100
(9)(11) = 99
(8)(12) = 96
Aside maximizing one side and decreasing the other by an "equal amount", the area decreases.
When it does appear on the GMAT, it's ofttimes themed approximately 'percentage alter' in lateral lengths (re: length is 10% greater, breadth is 10% less), simply the pattern is still the same.
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Re: Given that ABCD is a rectangle, is the expanse of triangle ABE> [#permalink] 23 May 2017, 23:42
For whatsoever right angled triangle the hypotenuse is the largest side and non only isosceles right?
Owing to the convention that the longest side is against the largest angle ?
Posted from my mobile gimmick
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Atomic number 75: Given that ABCD is a rectangle, is the field of triangle ABE> [#permalink] 23 May 2017, 23:48
SOUMYAJIT_ wrote:
For any right angular triangle the hypotenuse is the largest side and not only isosceles right?
Owing to the convening that the longest side is against the largest angle ?
Posted from my mobile device
Yes. The largest side is always against the largest lean against. Sol, in any mighty triangle hypotenuse is the largest side.
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Re: Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink] 25 Jun 2018, 02:53
rohitmisra4 wrote:
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 --> understandably insufficient: BE tail be 1 or 100.
(2) AE = 10 --> now, you should know one important property: the right-angled 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 endeavour what is the largest area of a right isosceles triangle with hypotenuse equal to 10. Determination legs: \(x^2+x^2=10^2\) (where x=AB=BE) --> \(x=\sqrt{50}\) --> \(area_{goop}=\frac{1}{2}\sqrt{50}^2=25\). Since information technology's the maximum field of ABE then the actual area can non be Sir Thomas More than 25. Ample.
Answer: B.
Hey Bunuel,
The property you mentioned only stands dependable when the hypotenuse is fixed and that is the reason it cannot be applied to the selection A. Else, the answer would have been D.
Idea I should clarify for the people reading the post.
Can someone expound on what this fellow meant please? I formerly sentiment that the answer to the question was D, thusly I'd apprise a good eye opener.
Thanks!
Posted from my mobile device
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Re: Given that ABCD is a rectangle, is the area of triangle ABE> [#permalink] 15 Apr 2020, 00:33
Delight Aid ME OUT!!
Statement B: Sufficient because of 1:1:root2 = 10/root2, 10/root2,10.
(Quadrant isosceles triangle having angles 45:45:90)
Hence the max orbit is 25(1/2*10/root2*10/root2)
likewise Statement A: Victimisation 1:1:root2 = 6,6,6root2
Maximum area = 1/2*6*6=18.
My filling D
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Atomic number 75: Presumption that ABCD is a rectangle, is the area of triangle ABE> [#permalink] 24 April 2021, 20:35
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