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# In triangle ABC to the right, if BC = 3 and AC = 4, then

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Senior Manager
Joined: 11 Nov 2005
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In triangle ABC to the right, if BC = 3 and AC = 4, then [#permalink]  29 Jan 2006, 15:52
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In triangle ABC to the right, if BC = 3 and AC = 4, then what is the length of segment CD?

3
15/4
5
16/3
20/3
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triangle ABC.doc [35 KiB]

Manager
Joined: 28 Dec 2005
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I pick A, 3. If this is correct I can explain my reasoning.
VP
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This is a tricky one. Don't go by the picture!

Infact the traingle ACD and ABC are similar traingles but the common angle is BAC = ADC.

So the ratios are

BC/AC = AC/CD

3/4 = 4/CD

or CD = 16/3

(Try to draw the triangles separately and orient them in the same direction, you will get it!)
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Senior Manager
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YES YOU ARE RIGHT, this was tricky......

Thanks for the explanation Giddi77

giddi77 wrote:
This is a tricky one. Don't go by the picture!

Infact the traingle ACD and ABC are similar traingles but the common angle is BAC = ADC.

So the ratios are

BC/AC = AC/CD

3/4 = 4/CD

or CD = 16/3

(Try to draw the triangles separately and orient them in the same direction, you will get it!)
Senior Manager
Joined: 13 Jun 2005
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16/3

this is how i solved it . let CD = x

AB = 5.

also (3+x)^2 = 25 + 16 + x^2

=> x = 16/3
CEO
Joined: 20 Nov 2005
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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D

AB = 5 can be found easily.

Now triangles ABC and ACD are similar triangles.
So

BC/AC = AD/AB = AC/CD we know values of BC and AC.

BC/AC = AC/CD so 3/4 = 4/CD i.e. CD = 16/3
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

SVP
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16/3.

Just one more way:

Let CD = x & AD = y

We can get AB = 5 from right angle triangle ABC.
Now, ABD too is a right angle triangle, so we can write,

=> (3+x)^2 = 25 + y^2
=> 9 + x^2 + 6x = 25 + y^2 --------- (1)

Also, ADC is right angle triangle,
y^2 = x^2 + 16 -------------------------(2)

Put (2) in (1) for y^2 & we'll get (3)

9 + x^2 + 6x = 25 + x^2 + 16 ---------------(3)

Solve this & we get x = 16/3
Manager
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Thanx vivek..........
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