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# If the shaded area is one half the area of the triangle ABC

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Intern
Joined: 30 May 2010
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If the shaded area is one half the area of the triangle ABC [#permalink]  15 Jul 2010, 07:57
00:00

Difficulty:

55% (hard)

Question Stats:

62% (03:09) correct 38% (01:39) wrong based on 95 sessions
Attachment:

d1.JPG [ 5.29 KiB | Viewed 2979 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. $$\sqrt{w^2-3y^2}$$
E. $$\sqrt{y^2+z^2}$$
[Reveal] Spoiler: OA
Math Expert
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Re: PS: Shaded area is half the triangle [#permalink]  15 Jul 2010, 08:30
3
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:

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
Manager
Joined: 19 Jul 2009
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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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Manager
Joined: 19 Jul 2009
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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
1
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 $$\sqrt{w^2-3y^2}=\sqrt{x^2+4y^2-3y^2}=\sqrt{x^2+y^2}=AD$$.

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: If the shaded area is one half the area of the triangle ABC [#permalink]  06 Mar 2014, 01:36
Expert's post
Bumping for review and further discussion.

GEOMETRY: Shaded Region Problems!
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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)

Am I missing something here?
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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
Re: If the shaded area is one half the area of the triangle ABC   [#permalink] 05 Apr 2014, 14:26
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# If the shaded area is one half the area of the triangle ABC

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