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

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The area of the right triangle ABC is 4 times greater than [#permalink] New post 10 May 2010, 19:48
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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.

[Reveal] Spoiler:
I thought the answer was D because Statement 1 tells us that the triangles are similar and statement 2 tells us that KLM = a right 6-8-10 triangle. So the area of KLM is 24 which would make ABC’s area = 96 and which makes ABC = 12-16-20 triangle.


OPEN DISCUSSION OF THIS QUESTION IS HERE: the-area-of-the-right-triangle-abc-is-4-times-greater-than-127070.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Jul 2013, 01:23, edited 1 time in total.
Edited the question and added the OA
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Re: MGMAT DS Help [#permalink] New post 11 May 2010, 03:34
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TXTDryFly wrote:
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?

“INSERT A DRAWING OF TWO RIGHT TRIANGLES”

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

(2) LM is 6 inches.

“INSERT A DRAWING OF TWO RIGHT TRIANGLES”

[Reveal] Spoiler:
A

I thought the answer was D because Statement 1 tells us that the triangles are similar and statement 2 tells us that KLM = a right 6-8-10 triangle. So the area of KLM is 24 which would make ABC’s area = 96 and which makes ABC = 12-16-20 triangle.


The above solution is not right.

Properties of Similar Triangles:

• Corresponding angles are the same.
• Corresponding sides are all in the same proportion.
• It is only necessary to determine that two sets of angles are identical in order to conclude that two triangles are similar; the third set will be identical because all of the angles of a triangle always sum to 180º.
• In similar triangles, the sides of the triangles are in some proportion to one another. For example, a triangle with lengths 3, 4, and 5 has the same angle measures as a triangle with lengths 6, 8, and 10. The two triangles are similar, and all of the sides of the larger triangle are twice the size of the corresponding legs on the smaller triangle.
• If two similar triangles have sides in the ratio \frac{x}{y}, then their areas are in the ratio \frac{x^2}{y^2}.
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: \frac{AREA}{area}=\frac{SIDE^2}{side^2}.

For more on triangles please check Triangles chapter of the Math Book (link in my signature).

Back to original question:
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 --> ABC and KLM are similar triangles --> \frac{AREA_{ABC}}{area_{KLM}}=\frac{4}{1}, so the sides are in ratio 2/1 --> hypotenuse KL=10 --> hypotenuse AB=2*10=20. Sufficient.

(2) LM is 6 inches --> KM=8 --> area_{KLM}=24 --> AREA_{ABC}=96. But just knowing the are of ABC is not enough to determine hypotenuse AB. For instance: legs of ABC can be 96 and 2 OR 48 and 4 and you'll get different values for hypotenuse. Not sufficient.

Answer: A.

Hope it helps.
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Re: MGMAT DS Help [#permalink] New post 10 May 2010, 21:47
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TXTDryFly wrote:
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?

“INSERT A DRAWING OF TWO RIGHT TRIANGLES”

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

(2) LM is 6 inches.

“INSERT A DRAWING OF TWO RIGHT TRIANGLES”

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

[Reveal] Spoiler:
A





I thought the answer was D because Statement 1 tells us that the triangles are similar and statement 2 tells us that KLM = a right 6-8-10 triangle. So the area of KLM is 24 which would make ABC’s area = 96 and which makes ABC = 12-16-20 triangle.


First of all Triangles ABC & KLM are right angled at C & M resp as KL is hypot..
Area of ABC = 1/2 * AC * CB = 4 * Area of KLM = 4 * 1/2 * KM * ML

Therefore AC * CB = 4 KM * ML

1. Angles ABC and KLM = 55 deg
Therefore AC = KM as similar triangles and sides opp equal angles..

Above equation becomes CB = 4 ML or CB / ML = 4 / 1 which is further = AB / KL
as triangles are similar.. As KL is 10 , therefore AB = 40.. sufficient..

2. LM = 6 inches . Therefore KM = 8 by pythogorus theorem.. ( 10^2 - 6^2 )
But no option to find out hypot AB.. Therefore , insufficient..

Ans : A..
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Re: MGMAT DS Help [#permalink] New post 11 May 2010, 04:31
Thanks Bunuel for Correct n Nice explanation... +1...
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Re: MGMAT DS Help [#permalink] New post 15 May 2010, 11:29
nice explanation bunuel....... kudos !!
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Re: MGMAT DS Help [#permalink] New post 04 May 2011, 01:01
a.(AB/ KL)^2 = Area (abc)/ Area (klm)

AB/KL = 2 Hence AB = 20 Thus sufficient.

b. km = 8 area ABC = 4 * area (klm) = 4*24 = 96.

thus ac *bc = 196 will give different values for AB. Hence not sufficient.

A.
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Re: MGMAT DS Help [#permalink] New post 04 May 2011, 02:50
(1) ABC and KLM are similar

Area of ABC AB^2
----------- = ------
Area of KLM KL^2

Sufficient

(2)

KM can be found.

No information about ABC, so not sufficient.

Answer - A
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Re: MGMAT DS Help   [#permalink] 04 May 2011, 02:50
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