In the figure above, line AC represents a seesaw that is

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In the figure above, line AC represents a seesaw that is [#permalink]

03 Oct 2006, 05:34

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ARISE AWAKE AND REST NOT UNTIL THE GOAL IS ACHIEVED

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03 Oct 2006, 06:41

20. In the figure above, line AC represents a seesaw that is touching level ground at point A. If B is the midpoint of AC, how far above the ground is point C?
(1) x = 30
(2) Point B is 5 feet above the ground.

A & B are unsufficient.

Combining

Sin 30(degree) = 5/AB
so AB can be determine and hence AC, since AC=2AB.

If we draw a perpendicula from line A which touches point C and call that point (on line A) D.

So, Sin 30 = AD/2AB
Answer is C.

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04 Oct 2006, 15:59

Answer is B.
Statement 2 alone is suff.
lets say: BA= x
then CA = 2x (since B is the mid point of CA)
Triangle ACD(D is the point where the perpendicular dropped from C touches the ground) is similar to triangle ABE(E is the point where the perpendicular dropped from C touches the ground).
Therefore x/5=2x/z
or z=10

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04 Oct 2006, 16:21

This one had me stumped. But, B it is, as others have explained before.

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05 Oct 2006, 01:03

It is B.
Apply property of similar triangles.........
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Last edited by cicerone on 25 Sep 2008, 01:30, edited 1 time in total.

Manager

Joined: 11 Jan 2006

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Location: Arkansas, US

WE 1: 2.5 yrs in manufacturing

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05 Oct 2006, 21:13

Gud explanation 800_gal...

Thanks for that..hahaa..completely forgot about the similar triangles properties...gotto revise it now...
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[#permalink] 05 Oct 2006, 21:13

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In the figure above, line AC represents a seesaw that is

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In the figure above, line AC represents a seesaw that is

In the figure above, line AC represents a seesaw that is

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