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OG 79

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massi2884 Senior | Next Rank: 100 Posts Default Avatar
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OG 79

Post Thu May 10, 2012 10:54 am
In the figure above, is the area of triangular region ABC equal to the area of triangular region DBA ?

(1) AC^2 = 2(AD)^2

(2) -ABC is isosceles

ttp://postimage.org/" target="_blank">

OA C

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Post Thu Jan 24, 2013 5:47 am
Ron,

Thanks so much for the thorough explanation. Your comprehensive understanding of this test and your willingness to openly share that understanding has been instrumental to my comprehension of ways to approach problems I would have never thought of on my own. It is really appreciated.

Thanks
-Bp

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Post Sat Jan 26, 2013 8:50 am
bpolley00 wrote:
Ron,

Thanks so much for the thorough explanation. Your comprehensive understanding of this test and your willingness to openly share that understanding has been instrumental to my comprehension of ways to approach problems I would have never thought of on my own. It is really appreciated.

Thanks
-Bp
you're welcome, thanks for reading.

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Post Thu Jan 24, 2013 5:47 am
Ron,

Thanks so much for the thorough explanation. Your comprehensive understanding of this test and your willingness to openly share that understanding has been instrumental to my comprehension of ways to approach problems I would have never thought of on my own. It is really appreciated.

Thanks
-Bp

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Post Sat Jan 26, 2013 8:50 am
bpolley00 wrote:
Ron,

Thanks so much for the thorough explanation. Your comprehensive understanding of this test and your willingness to openly share that understanding has been instrumental to my comprehension of ways to approach problems I would have never thought of on my own. It is really appreciated.

Thanks
-Bp
you're welcome, thanks for reading.

_________________
Ron has been teaching various standardized tests for 20 years.

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Pueden hacerle preguntas a Ron en castellano
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mcdesty Master | Next Rank: 500 Posts
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Post Wed Jul 09, 2014 10:29 am
See Img
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Post Thu Jan 24, 2013 3:21 am
bpolley00 wrote:
Also with it not being A, I am assuming we are assuming the photo is not drawn to scale? Is that correct? Thanks!
-BP
that is correct. as it seems you've already noticed, once you have both statements, it turns out that AD is much shorter than any other side in the picture -- definitely not the way it looks.

this shouldn't surprise you, though. in fact, NO data sufficiency diagram will ever be "to scale" at the start of the problem!
that may sound like a strange claim, but think about it for a second -- at the beginning of a data sufficiency problem, there are always unknowns, or quantities that can vary, in the diagram. (for instance, in this diagram, there are various pairs of sides that can exist in any imaginable proportion until you restrict them with one or the other of the numbered statements.)
clearly there is no "accurate" depiction of something with many possible appearances, so all data-sufficiency diagrams are "not to scale" by default.

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Post Thu Jan 24, 2013 3:17 am
bpolley00 wrote:
Can someone in more detail touch on being able to put those two together?
most likely, the notation is at least partly to blame; a lot of people forget that you can type "√" by pressing alt+v (if you have a mac) or, failing that, by just googling "square root sign" and then copying and pasting.

the first statement says that (AC)^2 = 2(AD)^2.
if you take the square root of both sides, you get AC = (√2)AD.

the second statement implies that AC = BC, since those are the only two sides of triangle ABC that can actually be the same. (the hypotenuse of a right triangle can't equal one of the legs.) so, each of AC and BC is equal to (√2)AD.

the only side you're missing now (as far as the sides necessary to find the triangles' areas) is AB.

three ways to go from here:

1/
remember that it's data sufficiency. because triangle ABC is a 45º-45º-90º triangle (isosceles right triangle), you know that it has to have some constant proportions (even if you don't remember what they are). so, AB is some constant multiple of AC or BC, so it's AD times something.
if that's the case, then the area of triangle ABD is (base)(height)/2 = (some # times AD)(AD)/2 = some number times (AD)^2.
the area of triangle ABC is (base)(height)/2 = (AD√2)(AD√2)/2 = AD^2. that's either definitely equal or definitely not equal to the area of ABD (it can't be both), so, sufficient.

2/
using the pythagorean theorem, actually find the value of AB:
(AB)^2 = (AD√2)^2 + (AD√2)^2
this gives AB = 2AD.
so, then, the area of triangle ABD is (base)(height)/2 = (2AD)(AD)/2 = (AD)^2.
this is the same as the area of triangle ABC (calculated above), so you get "yes" = sufficient.

3/
if you have the proportions for 45º-45º-90º triangles memorized, you can also find AB that way.
the hypotenuse of such a triangle is √2 times either of the legs, so AB = (√2)(AD√2) = 2AD.
same as #2 from here.

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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

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Post Wed Jan 23, 2013 5:39 pm
Can someone in more detail touch on being able to put those two together? I am having a tough time conceptually understanding. Also with it not being A, I am assuming we are assuming the photo is not drawn to scale? Is that correct? Thanks!
-BP

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