Geometry 8 : GMAT Problem Solving (PS)
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# Geometry 8

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02 Oct 2009, 10:16
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Q.
If 2 equal circles of radius 5cm have 2 common tangents AB and CD which touch the circle on A,C and B,D respectively and if CD= 24 cm, find the length of AB.
1. 27cm
2. 25cm
3. 26cm
4. 30cm
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04 Oct 2009, 13:13
Two line interception - K. Center of first circle O, of second P.

OC/KC=(KC+CD)/PD --> OC=PD=R=5 --> 5/KC=(24+KC)/5 --> Solving for KC --> KC=1.

KC=AK=1, KD=KB=KC+CD=1+24=25 --> AB=AK+KB=1+25=26
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16 Sep 2010, 07:44
virtualanimosity
What is the source of these problems ?
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16 Sep 2010, 21:12
Bunuel,

OC/KC=(KC+CD)/PD

Where does this come from? Thanks
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16 Sep 2010, 21:18
Bunuel wrote:
Two line interception - K. Center of first circle O, of second P.

OC/KC=(KC+CD)/PD --> OC=PD=R=5 --> 5/KC=(24+KC)/5 --> Solving for KC --> KC=1.

KC=AK=1, KD=KB=KC+CD=1+24=25 --> AB=AK+KB=1+25=26

Bunuel, I did not understand your explanation. Can you please be a bit elaborate?

Specifically, not sure of how you arrived at -- OC/KC = (KC+CD)/PD and KC = AK = 1.
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16 Sep 2010, 22:15
Take the intersection of line AB and CD as point E.

EB = ED as tangents from a single point to a circle are equal

EB = EC + CD take EC = x
=> EB = x+ 24

Also AE=EC = x using same explanation above.

AB = AE+EB = x+x+24 = 2x+24 = even number- if you are short of time you can easily guess.

Now triangle ECO is similar to triangle triangle EDO' where O and O' are center of 1st and 2nd respectively.

=>$$\frac{EC}{OD} = \frac{O'D}{ED}$$

=> $$\frac{x}{5}= \frac{5}{(x+24)}$$

=> $$x^2 +24x -25 = 0$$
=> $$(x +25)*(x-1)= 0$$ => $$x =1$$

=> AB$$= 2x*+24 = 2+24 = 26$$
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25 Sep 2010, 22:40
Dear Gurpreet,

I could not understand below mentioned remark of yours and the formula you arrived with this can you pls clarify

Now triangle ECO is similar to triangle triangle EDO' where O and O' are center of 1st and 2nd respectively.
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Re: Geometry 8   [#permalink] 25 Sep 2010, 22:40
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