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If the shaded area is one half the area of the triangle ABC

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If the shaded area is one half the area of the triangle ABC [#permalink] New post 15 Jul 2010, 07:57
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

59% (03:01) correct 41% (01:31) wrong based on 79 sessions
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d1.JPG
d1.JPG [ 5.29 KiB | Viewed 2524 times ]


If the shaded area is one half the area of the triangle ABC and angle ABC is a right angle, then the length of the line segment AD is:

A. \frac{1}{2}*w
B. \frac{1}{2}*(w+x)
C. \sqrt{2x^2+z^2}
D. \sqrt{w^2-3y^2}
E. \sqrt{y^2+z^2}
[Reveal] Spoiler: OA
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Re: PS: Shaded area is half the triangle [#permalink] New post 15 Jul 2010, 08:30
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Nusa84 wrote:
Hi guys,

I have some trouble with this exercise, the theory is clear, but I see too much information and I dont know how to proceed. Any ideas would be helpful, thanks!

Attachment:
d1.JPG


If the shaded area is one half the area of the triangle ABC and angle ABC is a right angle, then the length of the line segment AD is:

A.(1/2)*w
B.(1/2)*(w+x)
C.\sqrt{2x^2+z^2}
D.\sqrt{w^2-3y^2}
E.\sqrt{y^2+z^2}


Property of median: Each median divides the triangle into two smaller triangles which have the same area.

So, AD is median and thus y=z, (we could derive this even not knowing the above property. Given: area_{ABC}=\frac{x(y+z)}{2}=2*area_{ABD}=2*\frac{xy}{2} --> y=z).

Next: AD is hypotenuse in right triangle ABD and thus AD=\sqrt{x^2+y^2}=\sqrt{x^2+4y^2-3y^2}=\sqrt{x^2+(2y)^2-3y^2}=\sqrt{w^2-3y^2} (as 2y=y+z=BC and AC^2=w^2=AB^2+BC^2=x^2+(2y)^2, so w^2=x^2+(2y)^2).

Answer: D.
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Re: PS: Shaded area is half the triangle [#permalink] New post 15 Jul 2010, 08:44
Bunuel wrote:
Nusa84 wrote:
Hi guys,

I have some trouble with this exercise, the theory is clear, but I see too much information and I dont know how to proceed. Any ideas would be helpful, thanks!

Attachment:
d1.JPG


If the shaded area is one half the area of the triangle ABC and angle ABC is a right angle, then the length of the line segment AD is:

A.(1/2)*w
B.(1/2)*(w+x)
C.\sqrt{2x^2+z^2}
D.\sqrt{w^2-3y^2}
E.\sqrt{y^2+z^2}


Property of median: Each median divides the triangle into two smaller triangles which have the same area.

So, AD is median and thus y=z, (we could derive this even not knowing the above property. Given: area_{ABC}=\frac{x(y+z)}{2}=2*area_{ABD}=2*\frac{xy}{2} --> y=z).

Next: AD is hypotenuse in right triangle ABD and thus AD=\sqrt{x^2+y^2}=\sqrt{x^2+4y^2-3y^2}=\sqrt{x^2+(2y)^2-3y^2}=\sqrt{w^2-3y^2} (as 2y=y+z=BC and AC^2=w^2=AB^2+BC^2=x^2+(2y)^2, so w^2=x^2+(2y)^2).

Answer: D.


Quite a tough one, very good explanation, thanks
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Re: PS: Shaded area is half the triangle [#permalink] New post 15 Jul 2010, 08:58
I understood everything up until the \sqrt{x^2+4y^2-3y^2} part. how was this derived?
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Re: PS: Shaded area is half the triangle [#permalink] New post 15 Jul 2010, 11:04
Bunuel wrote:
azule45 wrote:
I understood everything up until the \sqrt{x^2+4y^2-3y^2} part. how was this derived?


y^2=4y^2-3y^2.



i'm sorry, i should have been more specific. i wasn't talking about substituting 4y^2-3y^2 for y^2, but rather where did 4y^2-3y^2 come from? i can't seem to figure this out. everything else makes perfect sense, but this.
much appreciated and sorry if this should be obvious.
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Re: PS: Shaded area is half the triangle [#permalink] New post 15 Jul 2010, 11:37
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azule45 wrote:
Bunuel wrote:
azule45 wrote:
I understood everything up until the \sqrt{x^2+4y^2-3y^2} part. how was this derived?


y^2=4y^2-3y^2.



i'm sorry, i should have been more specific. i wasn't talking about substituting 4y^2-3y^2 for y^2, but rather where did 4y^2-3y^2 come from? i can't seem to figure this out. everything else makes perfect sense, but this.
much appreciated and sorry if this should be obvious.


This is just an algebraic operation in order to pair x^2 and 4y^2, (added they give w^2).

Question is AD=? Well it's simple to get that AD=\sqrt{x^2+y^2}, but this option is not listed and thus we need to transform it to get the option which is listed (or the other way around we need to transform the options to get this one).

Now, options A and B does not make any sense and we can get rid of them.

Option C: \sqrt{2x^2+z^2}=\sqrt{2x^2+y^2} as y=z hence it's out too as \sqrt{2x^2+y^2}\neq{\sqrt{x^2+y^2}}.

Option D: \sqrt{w^2-3y^2}, w is hypotenuse hence it's equal to w^2=x^2+(y+z)^2=x^2+(2y)^2=x^2+4y^2, so \sqrt{w^2-3y^2}=\sqrt{x^2+4y^2-3y^2}=\sqrt{x^2+y^2}=AD.

Hope it's clear.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: PS: Shaded area is half the triangle [#permalink] New post 16 Jul 2010, 07:09
ahhh haa, Bunuel. so you worked the problem using the answer choices. got it. smart move. much appreciated for your help.
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Re: If the shaded area is one half the area of the triangle ABC [#permalink] New post 06 Mar 2014, 01:36
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Re: If the shaded area is one half the area of the triangle ABC [#permalink] New post 02 Apr 2014, 06:24
Wouldnt it be easier to assume the triangle to a 3,4,5 triple? AD would thus be sqrt(13) (z=y=2) just plug in the values starting option C. Takes less than a minute. (better than writing the lengthy equations)

Am I missing something here?
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Re: If the shaded area is one half the area of the triangle ABC [#permalink] New post 05 Apr 2014, 14:26
pikwik wrote:
Wouldnt it be easier to assume the triangle to a 3,4,5 triple? AD would thus be sqrt(13) (z=y=2) just plug in the values starting option C. Takes less than a minute. (better than writing the lengthy equations)

Am I missing something here?


Yes, that is in fact preferred approach
See below:

OK, so let's say that triangle ABC is 3,4,5. Therefore X=3, W=5 and z+y = 4. Now, smaller triangl to have 1/2 area of larger, then Y=2 and X=2 as well. Now replacing this value in answer choices we can see that only D fits the bill.

Therefore OA is D.

Hope this helps

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Re: If the shaded area is one half the area of the triangle ABC   [#permalink] 05 Apr 2014, 14:26
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