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

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

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15 Jul 2010, 07:57
1
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Difficulty:

65% (hard)

Question Stats:

64% (03:00) correct 36% (02:51) wrong based on 256 sessions

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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. $$\sqrt{y^2+z^2}$$
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Joined: 02 Sep 2009
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15 Jul 2010, 08:30
4
1
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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Joined: 30 May 2010
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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:

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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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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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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15 Jul 2010, 11:37
2
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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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]

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06 Mar 2014, 01:36
Bumping for review and further discussion.

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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)

Am I missing something here?
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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.

Therefore OA is D.

Hope this helps

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

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27 Nov 2015, 15:29
I tried 3-4-5 and 6-8-10 still didn't get to the answer:

y=z=1.5
or y=z=3
y^2/z^2 = 2.25 or y^2/z^2 = 9

x=4 or x=8
x^2 = 16
x^2 = 64

neither of the answer choices yields any results even close to this.

I doubt such a question would be on the actual gmat...
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Re: If the shaded area is one half the area of the triangle ABC  [#permalink]

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10 Mar 2017, 05:26
I am trying to use the formula Area/area=s^2/s^2. Is it possible to use this formula for this problem?
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Re: If the shaded area is one half the area of the triangle ABC  [#permalink]

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11 Mar 2017, 02:26
iqahmed83 wrote:
I am trying to use the formula Area/area=s^2/s^2. Is it possible to use this formula for this problem?

If I understood your query correctly, then you are trying to use the gist that if two similar triangles have corresponding sides in proportion, then their areas will be squared.

Except from the first part of the sentence from which we get z=y, I don't think their is much information left to deduce to use above corollary.
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Re: If the shaded area is one half the area of the triangle ABC  [#permalink]

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14 Mar 2017, 22:21
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. 12∗w12∗w
B. 12∗(w+x)12∗(w+x)
C. 2x2+z2−−−−−−−√2x2+z2
D. w2−3y2−−−−−−−√w2−3y2
E. y2+z2−−−−−−√

This question asks us to identify the length of line segment AD; therefore, we must assess the area of triangle ACB, which subsumes triangle ADB. We can plug in hypothetical values in order to derive the answer. For example, assume that we have an isosceles triangle with the ratio 45 : 45 : 90 ( we can make this assumption because we are not limited to a 30 : 60 : 90 triangle although such a triangle could also be used as a demonstration and to derive the answer). The sides of this triangle, considering the ratio of sides in a 45 : 45 : 90 triangle (1 : 1 : √2), are 2 : 2 : √2. We then calculate the area of the triangle (1/2 bh) which equals 2. We can now plug in for the other variables in the triangle: w = 2√2, z= 1 (this is true because if triangle ADB is half the area of triangle ACB then Z and Y must be midpoints of CD and DB) y= 1, x= 2. We can then use the Pythagorean theorem to calculate line segment AD. Whatever this value is, which in this hypothetical scenario happens to be √5, it must equate to the value of the correct equation. Now that we have values for all the variables, we can simply plug in values into each equation and eliminate answer choices until we arrive at a value that equals are hypothetical value of AD.
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Re: If the shaded area is one half the area of the triangle ABC  [#permalink]

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05 May 2018, 03:09
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Re: If the shaded area is one half the area of the triangle ABC &nbs [#permalink] 05 May 2018, 03:09
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