Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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:

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 \(w^2=x^2+(2y)^2\)).

Re: PS: Shaded area is half the triangle [#permalink]

Show Tags

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:

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 \(w^2=x^2+(2y)^2\)).

Re: PS: Shaded area is half the triangle [#permalink]

Show Tags

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.
_________________

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 \(\sqrt{w^2-3y^2}=\sqrt{x^2+4y^2-3y^2}=\sqrt{x^2+y^2}=AD\).

Re: If the shaded area is one half the area of the triangle ABC [#permalink]

Show Tags

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)

Re: If the shaded area is one half the area of the triangle ABC [#permalink]

Show Tags

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.

Re: If the shaded area is one half the area of the triangle ABC [#permalink]

Show Tags

09 Jul 2015, 21:38

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.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...