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# The area of the right triangle ABC is 4 times greater than

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Director
Joined: 23 May 2008
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The area of the right triangle ABC is 4 times greater than [#permalink]

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18 Aug 2008, 17:18
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The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?

(1) Angles ABC and KLM are each equal to 55 degrees.

(2) LM is 6 inches.
Senior Manager
Joined: 23 May 2006
Posts: 324

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18 Aug 2008, 18:11
The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?

(1) Angles ABC and KLM are each equal to 55 degrees.

(2) LM is 6 inches.

S1. Tells you that the two triangles ABC and KLM are similar traingles
A1/A2 = AB^2/KL^2
4/1 = AB^2/10^2
AB = SQ.RT 400 = 20
SUFFICIENT

S2. Tells you the lenght of LM. This does not really help us since it does not prove that the triangles are similar
SVP
Joined: 17 Jun 2008
Posts: 1547

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19 Aug 2008, 04:24
1
KUDOS
S1. Tells you that the two triangles ABC and KLM are similar traingles
A1/A2 = AB^2/KL^2
4/1 = AB^2/10^2
AB = SQ.RT 400 = 20
SUFFICIENT

What is the concept behind this?
Manager
Joined: 23 Jul 2008
Posts: 194

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19 Aug 2008, 05:31
shud be A

1) tells us the two triangles are similar hence hyp of abc can be found sufficient
2) we can only find the area of ABC doesn t tell anything else ---insufficient
VP
Joined: 18 May 2008
Posts: 1261

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07 Oct 2008, 02:13
It should be A.
(1) Since all angles are same these are similar triangles. We know that ratio of area of similar triangles is square the ratio of corresponding length. Thus Ab can be found. Sufficient
(2) we can knw the product of BC and AC but othing can be known abt AB. Insufficient
Director
Joined: 23 May 2008
Posts: 806

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07 Oct 2008, 10:11
scthakur wrote:
S1. Tells you that the two triangles ABC and KLM are similar traingles
A1/A2 = AB^2/KL^2
4/1 = AB^2/10^2
AB = SQ.RT 400 = 20
SUFFICIENT

What is the concept behind this?

why do we need to use the squares of the hyp's.? Are we some how using the pythagorean theorem?
Manager
Joined: 15 Apr 2008
Posts: 164

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07 Oct 2008, 10:32

statement one proves that the triangles are similar. hence is sufficient to find the hypotenuse of ABC.
Director
Joined: 27 Jun 2008
Posts: 542
WE 1: Investment Banking - 6yrs

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07 Oct 2008, 11:31
A

(1) you can get to know the sides and the area. - Suff
(2) Q Stem says, 4 times GREATER not 4 times - Insuff
Senior Manager
Joined: 29 Mar 2008
Posts: 348

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07 Oct 2008, 12:22
bigtreezl wrote:
scthakur wrote:
S1. Tells you that the two triangles ABC and KLM are similar traingles
A1/A2 = AB^2/KL^2
4/1 = AB^2/10^2
AB = SQ.RT 400 = 20
SUFFICIENT

What is the concept behind this?

why do we need to use the squares of the hyp's.? Are we some how using the pythagorean theorem?

Let there be two similar triangles ABC and PQR. Let us call the ratio of corresponding sides as scale factor (SF).
PPTY(1) The perimeters of two similar triangles are in the ratio of their scale factor.
PPTY (2) The areas of two similar triangles are in the same ratio as the square of their scale factors.

According to PPTY (2), in the given problem, the scale factor (SF) happens to be the ratio of the hypotenuse. It doen't have to be the hypotenuse always (could be any side).
Hence, A1/A2 = AB^2/KL^2

http://www.pinkmonkey.com/studyguides/s ... 505701.asp
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Re: DS: Triangles   [#permalink] 07 Oct 2008, 12:22
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