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# PS: Shaded area is half the triangle

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PS: Shaded area is half the triangle [#permalink]  15 Jul 2010, 07:57
00:00

Difficulty:

5% (low)

Question Stats:

80% (03:06) correct 20% (01:09) wrong based on 5 sessions
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 [ 5.29 KiB | Viewed 1418 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.(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}
[Reveal] Spoiler: OA
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Re: PS: Shaded area is half the triangle [#permalink]  15 Jul 2010, 08:30
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:

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).

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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:

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).

Quite a tough one, very good explanation, thanks
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Re: PS: Shaded area is half the triangle [#permalink]  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]  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]  15 Jul 2010, 11:37
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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.

Hope it's clear.
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Re: PS: Shaded area is half the triangle [#permalink]  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: PS: Shaded area is half the triangle   [#permalink] 16 Jul 2010, 07:09
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