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.

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:

Re: PS: Shaded area is half the triangle [#permalink]
15 Jul 2010, 08:30

3

This post received KUDOS

Expert's post

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]
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]
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. _________________

Re: PS: Shaded area is half the triangle [#permalink]
15 Jul 2010, 11:37

1

This post received KUDOS

Expert's post

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.

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

Peace J

gmatclubot

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