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500 ds

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by jayhawk2001 » Wed Apr 04, 2007 9:12 pm
We have to use properties of similar triangles here. Draw lines
from B and C to the ground. Lets call these points D and E respectively
on the ground. BD and CE are parallel

We are given that AB = BC.

(2) says that BD = 5. We can use properties of similar triangles now.
BD / CE = AB / AC. We can hence compute CE

So B is the correct answer.
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by dunkin77 » Thu Apr 05, 2007 7:50 am
Thanks for explanation Jay.

I am still a bit confused cause we only know AB=AC and BD=5,
so, BD / CE = AB / AC (let's say AB=AC=3)

5 / CE = 3 / 3

5/CE=1

CE=5..........?? :cry:

I would appreciate it if you could explain a bit more... thanks again for your help!!
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by gabriel » Thu Apr 05, 2007 10:01 am
dunkin77 wrote:Thanks for explanation Jay.

I am still a bit confused cause we only know AB=AC and BD=5,
so, BD / CE = AB / AC (let's say AB=AC=3)

5 / CE = 3 / 3

5/CE=1

CE=5..........?? :cry:

I would appreciate it if you could explain a bit more... thanks again for your help!!

dunkin u have got evrything right except for the part that AB= AC ... AB is actually = BC ... therefore AC=2AB ... so we have 5/CE = AB/AC .... 5/CE = 1/2 .. therefore CE = 10 .. hence the answer is B
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by vk.neni » Thu Apr 05, 2007 10:52 am
Hi,
Couldn't we solve this using information in (i) x=30? With this info the small triangle (ABD) becomes a 30-60-90 and we know the length of one of the sides, so we can figure the other two sides. Once, we've this info, we can use the similar triangle method to solve it. So, the answer would be D.
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Re: 500 ds

by myprepgmat » Sat Apr 07, 2007 5:06 am
Ans is B

I'll explain...

Let the height of point B is h and the point is H. Let the length of seesaw ie AC is 2d..

Now draw perpendicular lines to ground from point B and C. Let they intersect at ground at D ( from Point B) & E(point C).

Now we will get two similar triangles.. ie ABD and ACE..

we have BD/AB = CE/AC

ie h/d= H/2d given AB=BC we assumed AC=2d

i.e. h=H/2

given h=5ft

therefore H=10ft

case 1 doesnot give any unique value as the length of AC or height of the point B is not given.
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by gabriel » Sat Apr 07, 2007 9:51 am
vk.neni wrote:Hi,
Couldn't we solve this using information in (i) x=30? With this info the small triangle (ABD) becomes a 30-60-90 and we know the length of one of the sides, so we can figure the other two sides. Once, we've this info, we can use the similar triangle method to solve it. So, the answer would be D.
dude one of the most basic mistake for a DS... carrying information from one statement to the other .. the length of the side is mentioned only in the second statement.. so cant use it for the first ..
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