Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 30 May 2010
Posts: 29
Schools: YALE SOM

If the shaded area is one half the area of the triangle ABC [#permalink]
Show Tags
15 Jul 2010, 08:57
7
This post was BOOKMARKED
Question Stats:
66% (03:03) correct
34% (02:02) wrong based on 209 sessions
HideShow timer Statistics
Attachment:
d1.JPG [ 5.29 KiB  Viewed 5498 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. \(\sqrt{w^23y^2}\) E. \(\sqrt{y^2+z^2}\)
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Sep 2009
Posts: 39626

Re: PS: Shaded area is half the triangle [#permalink]
Show Tags
15 Jul 2010, 09:30
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.\(\sqrt{w^23y^2}\) 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 \(AD=\sqrt{x^2+y^2}=\sqrt{x^2+4y^23y^2}=\sqrt{x^2+(2y)^23y^2}=\sqrt{w^23y^2}\) (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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 30 May 2010
Posts: 29
Schools: YALE SOM

Re: PS: Shaded area is half the triangle [#permalink]
Show Tags
15 Jul 2010, 09:44
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.\(\sqrt{w^23y^2}\) 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 \(AD=\sqrt{x^2+y^2}=\sqrt{x^2+4y^23y^2}=\sqrt{x^2+(2y)^23y^2}=\sqrt{w^23y^2}\) (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



Manager
Joined: 19 Jul 2009
Posts: 52
Location: baltimore, md
Schools: kellogg, booth, stern, ann arbor

Re: PS: Shaded area is half the triangle [#permalink]
Show Tags
15 Jul 2010, 09:58
I understood everything up until the \sqrt{x^2+4y^23y^2} part. how was this derived?
_________________
Paaaaayyy Meeeee!!!!!



Manager
Joined: 19 Jul 2009
Posts: 52
Location: baltimore, md
Schools: kellogg, booth, stern, ann arbor

Re: PS: Shaded area is half the triangle [#permalink]
Show Tags
15 Jul 2010, 12:04
Bunuel wrote: azule45 wrote: I understood everything up until the \sqrt{x^2+4y^23y^2} part. how was this derived? \(y^2=4y^23y^2\). i'm sorry, i should have been more specific. i wasn't talking about substituting \(4y^23y^2\) for \(y^2\), but rather where did \(4y^23y^2\) 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.
_________________
Paaaaayyy Meeeee!!!!!



Math Expert
Joined: 02 Sep 2009
Posts: 39626

Re: PS: Shaded area is half the triangle [#permalink]
Show Tags
15 Jul 2010, 12:37
azule45 wrote: Bunuel wrote: azule45 wrote: I understood everything up until the \sqrt{x^2+4y^23y^2} part. how was this derived? \(y^2=4y^23y^2\). i'm sorry, i should have been more specific. i wasn't talking about substituting \(4y^23y^2\) for \(y^2\), but rather where did \(4y^23y^2\) 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: \(\sqrt{w^23y^2}\), \(w\) is hypotenuse hence it's equal to \(w^2=x^2+(y+z)^2=x^2+(2y)^2=x^2+4y^2\), so \(\sqrt{w^23y^2}=\sqrt{x^2+4y^23y^2}=\sqrt{x^2+y^2}=AD\). Hope it's clear.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 19 Jul 2009
Posts: 52
Location: baltimore, md
Schools: kellogg, booth, stern, ann arbor

Re: PS: Shaded area is half the triangle [#permalink]
Show Tags
16 Jul 2010, 08:09
ahhh haa, Bunuel. so you worked the problem using the answer choices. got it. smart move. much appreciated for your help.
_________________
Paaaaayyy Meeeee!!!!!



Math Expert
Joined: 02 Sep 2009
Posts: 39626

Re: If the shaded area is one half the area of the triangle ABC [#permalink]
Show Tags
06 Mar 2014, 02:36



Current Student
Joined: 25 Mar 2014
Posts: 11
WE: Operations (Investment Banking)

Re: If the shaded area is one half the area of the triangle ABC [#permalink]
Show Tags
02 Apr 2014, 07:24
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?



Current Student
Joined: 06 Sep 2013
Posts: 1997
Concentration: Finance

Re: If the shaded area is one half the area of the triangle ABC [#permalink]
Show Tags
05 Apr 2014, 15:26
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 Peace J



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15939

Re: If the shaded area is one half the area of the triangle ABC [#permalink]
Show Tags
09 Jul 2015, 22:38
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



CEO
Joined: 17 Jul 2014
Posts: 2524
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: If the shaded area is one half the area of the triangle ABC [#permalink]
Show Tags
27 Nov 2015, 16:29
I tried 345 and 6810 still didn't get to the answer:
y=z=1.5 or y=z=3 y^2/z^2 = 2.25 or y^2/z^2 = 9
x=4 or x=8 x^2 = 16 x^2 = 64
AD^2 = 2.25+16 or 9+64 AD is sqrt(18.25) or sqrt(73)
neither of the answer choices yields any results even close to this.
I doubt such a question would be on the actual gmat...



Intern
Joined: 26 Aug 2016
Posts: 10

Re: If the shaded area is one half the area of the triangle ABC [#permalink]
Show Tags
10 Mar 2017, 06:26
I am trying to use the formula Area/area=s^2/s^2. Is it possible to use this formula for this problem?



Chat Moderator
Joined: 04 Aug 2016
Posts: 474
Location: India
Concentration: Leadership, Strategy
GPA: 4
WE: Engineering (Telecommunications)

Re: If the shaded area is one half the area of the triangle ABC [#permalink]
Show Tags
11 Mar 2017, 03:26
iqahmed83 wrote: I am trying to use the formula Area/area=s^2/s^2. Is it possible to use this formula for this problem? If I understood your query correctly, then you are trying to use the gist that if two similar triangles have corresponding sides in proportion, then their areas will be squared. Except from the first part of the sentence from which we get z=y, I don't think their is much information left to deduce to use above corollary.



Senior Manager
Joined: 12 Nov 2016
Posts: 365

Re: If the shaded area is one half the area of the triangle ABC [#permalink]
Show Tags
14 Mar 2017, 23:21
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. 12∗w12∗w B. 12∗(w+x)12∗(w+x) C. 2x2+z2−−−−−−−√2x2+z2 D. w2−3y2−−−−−−−√w2−3y2 E. y2+z2−−−−−−√
This question asks us to identify the length of line segment AD; therefore, we must assess the area of triangle ACB, which subsumes triangle ADB. We can plug in hypothetical values in order to derive the answer. For example, assume that we have an isosceles triangle with the ratio 45 : 45 : 90 ( we can make this assumption because we are not limited to a 30 : 60 : 90 triangle although such a triangle could also be used as a demonstration and to derive the answer). The sides of this triangle, considering the ratio of sides in a 45 : 45 : 90 triangle (1 : 1 : √2), are 2 : 2 : √2. We then calculate the area of the triangle (1/2 bh) which equals 2. We can now plug in for the other variables in the triangle: w = 2√2, z= 1 (this is true because if triangle ADB is half the area of triangle ACB then Z and Y must be midpoints of CD and DB) y= 1, x= 2. We can then use the Pythagorean theorem to calculate line segment AD. Whatever this value is, which in this hypothetical scenario happens to be √5, it must equate to the value of the correct equation. Now that we have values for all the variables, we can simply plug in values into each equation and eliminate answer choices until we arrive at a value that equals are hypothetical value of AD.




Re: If the shaded area is one half the area of the triangle ABC
[#permalink]
14 Mar 2017, 23:21







