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

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bigtreezl
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The area of the right triangle ABC is 4 times greater than [#permalink] New post   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.



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Re: The area of the right triangle ABC is 4 times greater than [#permalink] New post   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
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Re: The area of the right triangle ABC is 4 times greater than [#permalink] New post   19 Aug 2008, 04:24
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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?
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Re: The area of the right triangle ABC is 4 times greater than [#permalink] New post   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
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Re: The area of the right triangle ABC is 4 times greater than [#permalink] New post   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
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Re: The area of the right triangle ABC is 4 times greater than [#permalink] New post   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?
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ALD
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Re: The area of the right triangle ABC is 4 times greater than [#permalink] New post   07 Oct 2008, 10:32
answer should be A

statement one proves that the triangles are similar. hence is sufficient to find the hypotenuse of ABC.
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Re: The area of the right triangle ABC is 4 times greater than [#permalink] New post   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
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Re: The area of the right triangle ABC is 4 times greater than [#permalink] New post   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

You can find more information on similar triangles from here:
https://www.pinkmonkey.com/studyguides/ ... 505701.asp



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Re: The area of the right triangle ABC is 4 times greater than [#permalink]
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