Property of a triangle within a triangle
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In above triangle,
assume ce is perpendicular to ab
Now, add all conditions of statement 2 of the problem.
ae=7
be=9
Angle bce = Angle cae
Scenario 1:
Angle bce = Angle cae = 30
In this case, x = 14/sqrt(3) and y = 18
Scenario 2:
Angle bce = Angle cae = 60
In this case, x = 14 and y = 18/sqrt(3)
So, even after adding 1 more assumption that line ce is perpendicular to ab, we can't solve the question by using this statement alone.
Hope I am correct. Let me know if there is a problem with the approach.
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St1: X = 5
Y can be any length from 12 <= Y <= 20 (using triangle property that 3rd side must be less than sum of other 2 sides and greater than difference between other 2 sides)
Hence Insufficient
St2: a = b
Triangles CBD and ABC are similar as:
An. C is common
An CBA = An CAB
and since 2 angles in the traingles are equal, the third angle (An CDB and An CBA) must also be equal.
AC / BC = AB / BD = BC / CD or y / 16 = 9 / y
y^2 = 16 * 9 = 144
y = 12
Sufficient
Hence B
Y can be any length from 12 <= Y <= 20 (using triangle property that 3rd side must be less than sum of other 2 sides and greater than difference between other 2 sides)
Hence Insufficient
St2: a = b
Triangles CBD and ABC are similar as:
An. C is common
An CBA = An CAB
and since 2 angles in the traingles are equal, the third angle (An CDB and An CBA) must also be equal.
AC / BC = AB / BD = BC / CD or y / 16 = 9 / y
y^2 = 16 * 9 = 144
y = 12
Sufficient
Hence B
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srcc25anu wrote:St1: X = 5
Y can be any length from 12 <= Y <= 20 (using triangle property that 3rd side must be less than sum of other 2 sides and greater than difference between other 2 sides)
Hence Insufficient
St2: a = b
Triangles CBD and ABC are similar as:
An. C is common
An CBA = An CAB
and since 2 angles in the traingles are equal, the third angle (An CDB and An CBA) must also be equal.
AC / BC = AB / BD = BC / CD or y / 16 = 9 / y
y^2 = 16 * 9 = 144
y = 12
Sufficient
Hence B
Hello,
Thanks for the explanation. I was clear till here:
I was thinking that shouldn't it be Angle CBA = Angle CDB (Given by statement 2)An CBA = An CAB
Can you please assist?
Thanks,
Sri
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hey, i wrote this problem!
--
gmattesttaker2, i'm with you here -- i actually have pretty severe dyslexia, so i don't stand a chance of slogging through all those three-letter angles.
instead, you should do what srcc25anu did above, and draw a picture with the matching angles between the two triangles. (kudos to srcc25anu, by the way, for drawing an awesome picture -- and in full color, too!)
in that picture, we can see that ...
* y corresponds to 9 (since both are across from the brown angle marked "x")
* 16 corresponds to y (since both are across from the blue angle)
... and that's enough to set up the proportion.
--
gmattesttaker2, i'm with you here -- i actually have pretty severe dyslexia, so i don't stand a chance of slogging through all those three-letter angles.
instead, you should do what srcc25anu did above, and draw a picture with the matching angles between the two triangles. (kudos to srcc25anu, by the way, for drawing an awesome picture -- and in full color, too!)
in that picture, we can see that ...
* y corresponds to 9 (since both are across from the brown angle marked "x")
* 16 corresponds to y (since both are across from the blue angle)
... and that's enough to set up the proportion.
Ron has been teaching various standardized tests for 20 years.
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Hello Ron,lunarpower wrote:hey, i wrote this problem!
--
gmattesttaker2, i'm with you here -- i actually have pretty severe dyslexia, so i don't stand a chance of slogging through all those three-letter angles.
instead, you should do what srcc25anu did above, and draw a picture with the matching angles between the two triangles. (kudos to srcc25anu, by the way, for drawing an awesome picture -- and in full color, too!)
in that picture, we can see that ...
* y corresponds to 9 (since both are across from the brown angle marked "x")
* 16 corresponds to y (since both are across from the blue angle)
... and that's enough to set up the proportion.
Hope all is well. Thank you very much for the explanation. It is clear now.
Best Regards,
Sri
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Hey,macattack wrote:how do we establish the ration of the sides in similar triangles: ie why did we do AC/BC= etc.... thxxx
Well the rule is that if all angles are equal in 2 triangles, then they are similar trinagles and then the corresponding sides are in a ratio.
Regards
Anup
Regards
Anup
The only lines that matter - are the ones you make!
https://www.youtube.com/watch?v=kk4sZcG ... ata_player
Anup
The only lines that matter - are the ones you make!
https://www.youtube.com/watch?v=kk4sZcG ... ata_player
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Thx Anup but I know this information however what i do not know is how we establish the corresponding sides! plz helpanuprajan5 wrote:Hey,macattack wrote:how do we establish the ration of the sides in similar triangles: ie why did we do AC/BC= etc.... thxxx
Well the rule is that if all angles are equal in 2 triangles, then they are similar trinagles and then the corresponding sides are in a ratio.
Regards
Anup
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I have attached the solution in the attachment.
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Regards
Anup
The only lines that matter - are the ones you make!
https://www.youtube.com/watch?v=kk4sZcG ... ata_player
Anup
The only lines that matter - are the ones you make!
https://www.youtube.com/watch?v=kk4sZcG ... ata_player
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sri, you're welcome.
macattack, that's the basic proportion when you have similarity. you can do one of two things here:
1/
you can think about what similarity means, and then figure out which fractions would be the same.
for instance, let's say you had a model car, and then a real car that's 18 times the size of the model car in every dimension. then, for instance,
(length of door on real car)/(length of door on model car) = 18/1
(length of real car)/(length of model car) = 18/1
(height of real car)/(height of model car) = 18/1
etc.
all these ratios are the same.
in this case i've given an actual ratio (18/1) just to make the whole idea more concrete, but the point is that you can set the ratios equal to each other even if you don't know what they are.
2/
since this is a general fact, you should just google stuff like "similar triangles proportion", and read a bunch of the pages you get as hits.
this forum is a great place to ask about specific gmat problems, but, when it comes to general principles/fundamentals, you'll get much faster (and more thorough) feedback by just searching the web.
macattack, that's the basic proportion when you have similarity. you can do one of two things here:
1/
you can think about what similarity means, and then figure out which fractions would be the same.
for instance, let's say you had a model car, and then a real car that's 18 times the size of the model car in every dimension. then, for instance,
(length of door on real car)/(length of door on model car) = 18/1
(length of real car)/(length of model car) = 18/1
(height of real car)/(height of model car) = 18/1
etc.
all these ratios are the same.
in this case i've given an actual ratio (18/1) just to make the whole idea more concrete, but the point is that you can set the ratios equal to each other even if you don't know what they are.
2/
since this is a general fact, you should just google stuff like "similar triangles proportion", and read a bunch of the pages you get as hits.
this forum is a great place to ask about specific gmat problems, but, when it comes to general principles/fundamentals, you'll get much faster (and more thorough) feedback by just searching the web.
Ron has been teaching various standardized tests for 20 years.
--
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hey,macattack wrote:Thx Anup but I know this information however what i do not know is how we establish the corresponding sides! plz help
* again, the internet is your friend here. seek and you shall find.
* basically, the idea is that you should visually match up the two similar triangles, so that they are "facing the same way".
i used the example of a model car and a real car. that's a particularly easy example to grasp:
-- the doors on the model correspond to the doors on the real car
-- the pedals in the model correspond to the pedals in the real car
-- the roof of the model corresponds to the roof of the real car
etc.
the point is that, in your head, you imagine the model car and the real car in the same position, facing the same way, and then you just think about parts that are in the same place on both of them.
when you see a pair of similar triangles, just think of one of them as a "model" of the other one. then it's the same kind of reasoning.
the trick here is that you'll sometimes have to rotate or "flip" one of the triangles to get them to line up, as you do here. that can stretch your capacity for visualization a bit, but give it a shot -- you should be able to figure it out if you play around with it.
if you still absolutely don't get it, after messing around with it on paper for a while, then make the figure (to scale!) on paper, literally cut it out, and get the triangles to line up.
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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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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Learn more about ron
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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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