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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 05 Feb 2012, 04:07
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Attachment:
KLM Triangle.GIF
KLM Triangle.GIF [ 2.03 KiB | Viewed 5621 times ]
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:
Ok - my reasoning for picking answer A is based on Bunuel's explanation on similar triangles. Thanks Bunuel. But can someone please let me know how to actually calculate side AB?

Considering Statement 1

Angles ABC and KLM are each equal to 55 degrees and both triangles have angle ACB and angle KML as 90 degrees. Therefore, the third angle i.e Angle CAB and angle MKL should be equal. This is because the sum of the angles of a triangle is 180. This concludes that the two triangles are similar triangles. Therefore, the corresponding sides are in proportion. This will give us

\(\frac{AC}{KM}\) = \(\frac{BC}{LM}\)=\(\frac{AB}{KL}\) ---> So is the side AB will be 10??? ----------------(Here I am stuck)

Considering Statement 2

This is clearly insufficient as Just knowing LM will not give us the lenght of AB.
[Reveal] Spoiler: OA

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Re: Hypotenuse AB [#permalink] New post 05 Feb 2012, 04:12
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enigma123 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?

(1) Angles ABC and KLM are each equal to 55 degrees.
(2) LM is 6 inches.


Below are the properties of similar triangles you might find useful for the GMAT and the solution which calculates the actual value of AB.

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.
• 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: \(\frac{AREA}{area}=\frac{SIDE^2}{side^2}\).

Back to original question:
Attachment:
KLM Triangle.GIF
KLM Triangle.GIF [ 2.03 KiB | Viewed 4933 times ]
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: The area of the right triangle ABC is 4 times greater than [#permalink] New post 24 May 2013, 03:21
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Re: The area of the right triangle ABC is 4 times greater than [#permalink] New post 27 Sep 2013, 23:02
Bannual, I am having trouble understanding this step in your explanation -
"• 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}."

When doing this problem, I chose C,

I was able to figure out that the triangles are similar but I was unfamiliar with the relationship between both hypotenuses. I thought we would need to know another side length in order to establish a ratio between one triangle and the other.

Is the relationship you described a property that we simply must memorize, or was there a way to figure it out? Can you quickly derive it? I do not understand why the lengths squared ratio is equal to the proportion of the area of both triangles.
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Re: The area of the right triangle ABC is 4 times greater than [#permalink] New post 28 Sep 2013, 03:48
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laserglare wrote:
Bannual, I am having trouble understanding this step in your explanation -
"• 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}."

When doing this problem, I chose C,

I was able to figure out that the triangles are similar but I was unfamiliar with the relationship between both hypotenuses. I thought we would need to know another side length in order to establish a ratio between one triangle and the other.

Is the relationship you described a property that we simply must memorize, or was there a way to figure it out? Can you quickly derive it? I do not understand why the lengths squared ratio is equal to the proportion of the area of both triangles.


Of course there is a way to derive this property but I recommend just to memorize it.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis ; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) ; 12. Tricky questions from previous years.

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Re: The area of the right triangle ABC is 4 times greater than [#permalink] New post 21 Oct 2014, 06:56
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Re: The area of the right triangle ABC is 4 times greater than [#permalink] New post 06 Nov 2014, 14:20
Bunuel wrote:
enigma123 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?

(1) Angles ABC and KLM are each equal to 55 degrees.
(2) LM is 6 inches.


Below are the properties of similar triangles you might find useful for the GMAT and the solution which calculates the actual value of AB.

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.
• 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: \(\frac{AREA}{area}=\frac{SIDE^2}{side^2}\).

Back to original question:
Attachment:
KLM Triangle.GIF
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.


Hi Bunuel,

Question regarding the above property.

For triangles -- if the if the sides have the ratio of x then the area will be squared. Correct? Does this imply that ALL the sides in Triangle ABC are "x times" All the sides in triangle KLM. Meaning, what happens if the sides have different ratios? base could be 2x while the height would be 3x -- what happens in a situation like that?

Thanks.
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Re: The area of the right triangle ABC is 4 times greater than [#permalink] New post 07 Nov 2014, 03:34
Expert's post
russ9 wrote:
Bunuel wrote:
enigma123 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?

(1) Angles ABC and KLM are each equal to 55 degrees.
(2) LM is 6 inches.


Below are the properties of similar triangles you might find useful for the GMAT and the solution which calculates the actual value of AB.

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.
• 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: \(\frac{AREA}{area}=\frac{SIDE^2}{side^2}\).

Back to original question:
Attachment:
KLM Triangle.GIF
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.


Hi Bunuel,

Question regarding the above property.

For triangles -- if the if the sides have the ratio of x then the area will be squared. Correct? Does this imply that ALL the sides in Triangle ABC are "x times" All the sides in triangle KLM. Meaning, what happens if the sides have different ratios? base could be 2x while the height would be 3x -- what happens in a situation like that?

Thanks.


Again your question is not very clear... In similar triangles corresponding sides are in the same ratio.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis ; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) ; 12. Tricky questions from previous years.

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: The area of the right triangle ABC is 4 times greater than [#permalink] New post 27 Mar 2015, 14:23
Bunuel wrote:
enigma123 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?

(1) Angles ABC and KLM are each equal to 55 degrees.
(2) LM is 6 inches.


Below are the properties of similar triangles you might find useful for the GMAT and the solution which calculates the actual value of AB.

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.
• 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: \(\frac{AREA}{area}=\frac{SIDE^2}{side^2}\).

Back to original question:
Attachment:
KLM Triangle.GIF
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.



Hi Bunuel,

the area of triangle should be 5 times the area of other triangle..,isn't it?
because,it mention as "4 times greater than other triangle".....
and,not as--"4 times the other triangle"

Please have a look into it..
Waiting for your expertise...

Thanks!!
Re: The area of the right triangle ABC is 4 times greater than   [#permalink] 27 Mar 2015, 14:23
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