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A squirrel climbs a straight wire from point A to point C. If B is the
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30 Mar 2015, 04:19
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Re: A squirrel climbs a straight wire from point A to point C. If B is the
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30 Mar 2015, 04:29
A squirrel climbs a straight wire from point A to point C. If B is the midpoint of AC, how far above the ground is point B?
(1) Point C is 40 feet above the ground.
(2) x = 60
1) Insufficient, no information about the line 2) Insufficient, no information about the highest point 1+2) Information about the line (it's linear, so it's rising equally?) and the highest point is 40. Since you know B is the midpoint, 40/2 = 20. B is 20 feet above the ground.
C.



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Re: A squirrel climbs a straight wire from point A to point C. If B is the
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30 Mar 2015, 04:52
Option A. Statement 1 alone is sufficient.
1. point C is 40 feet above ground.
Let \(C_1\) by point on ground directly below \(C\)and \(B_1\) be point directly below \(B\).
\(\triangle ABB_1\) is similar to \(\triangle ACC_1\)
\(Therefore \frac{AB}{AC} = \frac{BB_1}{CC_1}\)
Therefore \(BB_1\) = 20
2. x = 60, alone is not sufficient.



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Re: A squirrel climbs a straight wire from point A to point C. If B is the
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06 Apr 2015, 04:27



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Re: A squirrel climbs a straight wire from point A to point C. If B is the
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12 Apr 2015, 09:49
Hi, Is this a rule? As in; if the squirrel had travelled 1/4 of the distance would he stand 10m above the ground?



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Re: A squirrel climbs a straight wire from point A to point C. If B is the
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19 Apr 2015, 10:08
tgubbay1 wrote: Hi, Is this a rule? As in; if the squirrel had travelled 1/4 of the distance would he stand 10m above the ground? hi tgubbay, It is called properties of similar triangles i.e if two triangles are similar than their angles are equal and their sides are in proportion suppose c makes a point c1 on ground and b makes a point b1 on ground then ab/bb1= ac/cc1 and the answer to your question is yes b will be 10 just plug in the values in the ratios. hope it helps



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Re: A squirrel climbs a straight wire from point A to point C. If B is the
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20 Apr 2015, 10:04
anshul2014 wrote: tgubbay1 wrote: Hi, Is this a rule? As in; if the squirrel had travelled 1/4 of the distance would he stand 10m above the ground? hi tgubbay, It is called properties of similar triangles i.e if two triangles are similar than their angles are equal and their sides are in proportion suppose c makes a point c1 on ground and b makes a point b1 on ground then ab/bb1= ac/cc1 and the answer to your question is yes b will be 10 just plug in the values in the ratios. hope it helps Hi tgubbay1  let me explain to you how the ratio of the sides was arrived at. If you look at the figure, you can observe that in triangle ACE which is a right angled triangle, we can write \(Sin x = \frac{CE}{AC}\) Similarly, in triangle ABD which is also a right angled triangle, we can write \(Sin x = \frac{BD}{AB}\) From the above two equations, we can write \(\frac{CE}{AC} = \frac{BD}{AB}\). Since we know that \(AB= \frac{AC}{2}\), we can simplify the expression to \(BD = \frac{CE}{2}\). Thus, to find the value of BD, we need to know the value of CE, which is given to us by stI. Hence, had AB been equal to \(\frac{AC}{4}\), BD would also have been equal to \(\frac{CE}{4}\). Hope it helps! Regards Harsh
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Re: A squirrel climbs a straight wire from point A to point C. If B is the
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16 Dec 2017, 22:40
niks18 Bunuel regarding approach of shreyastQuote: 1. point C is 40 feet above ground.
Let \(C_1\) by point on ground directly below \(C\)and \(B_1\) be point directly below \(B\).
\(\triangle ABB_1\) is similar to \(\triangle ACC_1\)
\(Therefore \frac{AB}{AC} = \frac{BB_1}{CC_1}\)
Therefore \(BB_1\) = 20 We can only take ratio of two sides in same triangle, did the user missed this step (although end result would be the same) \(\frac{AC}{CC_1} = \frac{AB}{BB_1}\) Regarding approach by EgmatQuantExpert trigonometry is not included in GMAT but concept of similar triangles by using side and angles property is.
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Re: A squirrel climbs a straight wire from point A to point C. If B is the
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16 Dec 2017, 23:51
adkikani wrote: niks18 Bunuel regarding approach of shreyastQuote: 1. point C is 40 feet above ground.
Let \(C_1\) by point on ground directly below \(C\)and \(B_1\) be point directly below \(B\).
\(\triangle ABB_1\) is similar to \(\triangle ACC_1\)
\(Therefore \frac{AB}{AC} = \frac{BB_1}{CC_1}\)
Therefore \(BB_1\) = 20 We can only take ratio of two sides in same triangle, did the user missed this step (although end result would be the same) \(\frac{AC}{CC_1} = \frac{AB}{BB_1}\) Regarding approach by EgmatQuantExpert trigonometry is not included in GMAT but concept of similar triangles by using side and angles property is. If \(ABC\) is similar to \(A'B'C'\), then \(\frac{AB}{A'B'}=\frac{AC}{A'C'}=\frac{BC}{B'C'}\) from here it follows that \(\frac{AB}{AC}=\frac{A'B'}{A'C'}\), \(\frac{AB}{BC}=\frac{A'B'}{B'C'}\), \(\frac{AC}{BC}=\frac{AC}{B'C'}\).
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Re: A squirrel climbs a straight wire from point A to point C. If B is the
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29 Jan 2019, 07:33
Bunuel wrote: A squirrel climbs a straight wire from point A to point C. If B is the midpoint of AC, how far above the ground is point B? (1) Point C is 40 feet above the ground. (2) x = 60 Kudos for a correct solution.Attachment: squirrelonwire_figure.PNG Use the midpoint theorem we studied in school/college. In the triangle provided consider the points from B&C hitting the ground as D&E. Hence we have two triangles > ACE & ABD. Since B is the midpoint of AC and CEBD from the midpoint theorem we know that D is the midpoint of AE and also CE=2BD. Actually, the midpoint theorem stems from the concept of similar triangles and all the sides in the 2 triangles will be in the ratio 1:2 hence CE = 2BD, AC = 2AB = 2BC, AE = 2DE = 2AD So we can conclude if we know any one of the sides we can know the corresponding side of the other triangle. Here we know CE and hence can calculate BD. Hence A is the correct answer.
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Re: A squirrel climbs a straight wire from point A to point C. If B is the
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