A squirrel climbs a straight wire from point A to point C. If B is the

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.

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.

MAGOOSH OFFICIAL SOLUTION:

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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 st-I.

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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niks18 Bunuel

regarding approach of shreyast




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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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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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 CE||BD 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.

I think the answer is D(each answer choice is enough to get the height)
1.Option 1 is already proven in the above discussions
2.Option 2 :
Let us assume that the perpendicular dropped from B meets AD at E
We can get the height as 20 using 30-60-90 triangle rule
AD:40:2a = 1:root(3):2 ------ therefore a = 40/root(3)
Now,
AE:BE:40/root(3) = 1:root(3):2 -----therefore BE = height = 20

Please confirm if my answer is correct

It is not correct. Answer choice 2 doesn’t give you the height of point C. How could you infer that the value = 40? You literally have infinite possibilities

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