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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]

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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]

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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]

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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]

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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]

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09 Jul 2015, 21:38

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