Author 
Message 
TAGS:

Hide Tags

Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 538
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

The area of the right triangle ABC is 4 times greater than [#permalink]
Show Tags
05 Feb 2012, 05:07
3
This post received KUDOS
8
This post was BOOKMARKED
Question Stats:
61% (01:57) correct
39% (01:15) wrong based on 502 sessions
HideShow timer Statistics
Attachment:
KLM Triangle.GIF [ 2.03 KiB  Viewed 11804 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. 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.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Math Expert
Joined: 02 Sep 2009
Posts: 39695

Re: Hypotenuse AB [#permalink]
Show Tags
05 Feb 2012, 05:12
5
This post received KUDOS
Expert's post
7
This post was BOOKMARKED
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 [ 2.03 KiB  Viewed 9816 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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Math Expert
Joined: 02 Sep 2009
Posts: 39695

Re: The area of the right triangle ABC is 4 times greater than [#permalink]
Show Tags
24 May 2013, 04:21



Manager
Joined: 07 Jul 2013
Posts: 96

Re: The area of the right triangle ABC is 4 times greater than [#permalink]
Show Tags
28 Sep 2013, 00:02
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.



Math Expert
Joined: 02 Sep 2009
Posts: 39695

Re: The area of the right triangle ABC is 4 times greater than [#permalink]
Show Tags
28 Sep 2013, 04:48
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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15978

Re: The area of the right triangle ABC is 4 times greater than [#permalink]
Show Tags
21 Oct 2014, 07:56
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Senior Manager
Joined: 15 Aug 2013
Posts: 311

Re: The area of the right triangle ABC is 4 times greater than [#permalink]
Show Tags
06 Nov 2014, 15:20
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.



Math Expert
Joined: 02 Sep 2009
Posts: 39695

Re: The area of the right triangle ABC is 4 times greater than [#permalink]
Show Tags
07 Nov 2014, 04:34
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 the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 17 Jan 2015
Posts: 13

Re: The area of the right triangle ABC is 4 times greater than [#permalink]
Show Tags
27 Mar 2015, 15:23
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!!



Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 3460
GPA: 3.82

The area of the right triangle ABC is 4 times greater than [#permalink]
Show Tags
29 Dec 2015, 23:06
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution In the figure, angle C and angle M are right angles, and KL = 10. If the area of triangle ABC is four times the area of triangle KLM, what is the length AB? (1) Angles ABC and KLM have the same measure. (2) LM is 6 inches. We need to modify the original condition and the question: First, as we can see above, KL=10. Also, taking a square root of the ratio of areas yields the ratio of lengths. Then, if the area of triangle ABC is four times the area of triangle KLM, since root of 4 is 2, we know that the length of AB is twice longer than the length of KL. This can only be true under the assumption that two triangles have the same measure. Essentially, we can infer that the question is asking if two triangles have the same measure. Hence, the condition 1) is sufficient and the correct answer is A. Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
Attachments
d.jpg [ 2.89 KiB  Viewed 4173 times ]
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Find a 10% off coupon code for GMAT Club members. “Receive 5 Math Questions & Solutions Daily” Unlimited Access to over 120 free video lessons  try it yourself See our Youtube demo



Intern
Joined: 10 Aug 2015
Posts: 29
Location: India
GPA: 3.5
WE: Consulting (Computer Software)

Re: The area of the right triangle ABC is 4 times greater than [#permalink]
Show Tags
01 May 2016, 00:18
enigma123 wrote: 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. (2) LM is 6 inches. So here are my 2 cents. Let's say Bigger triangle = A and smaller one = B. Now area of A = 4 * Area of B. and we know KL=10. Now statement 1 says they have one angle equal and we know they both have right angles. From this we can derive that 3rd angle will also be equal. Now from this we know they are similar triangles and their sides have equal ratios. Let say AC/KM=BC/ML=a. a is our scale factor. Now AC=a*Km and BC=a*ML Area of KML= 1/2*KM*ML Area of ABC = 1/2AB*BC=1/2*a*KM*a*ML= 1/2 a^2 KM&ML. You can see from here that if scale factor of sides = a then area increases by a^2. Now lets apply this to our question. Scale factor of area = 4 then scale factor of sides will be sqrt of 4=2. Now B has KL=h=10 then AB = 2*10 = 20. Statement2 To prove this insufficient is very easy.(Find out why)



Senior Manager
Joined: 26 Oct 2016
Posts: 463
Location: United States
Concentration: Marketing, International Business
GPA: 4
WE: Education (Education)

Re: The area of the right triangle ABC is 4 times greater than [#permalink]
Show Tags
30 Dec 2016, 01:54
First, recall that in a right triangle, the two shorter sides intersect at the right angle. Therefore, one of these sides can be viewed as the base, and the other as the height. Consequently, the area of a right triangle can be expressed as one half of the product of the two shorter sides (i.e., the same as one half of the product of the height times the base). Also, since AB is the hypotenuse of triangle ABC, we know that the two shorter sides are BC and AC and the area of triangle ABC = (BC × AC)/2. Following the same logic, the area of triangle KLM = (LM × KM)/2. Also, the area of ABC is 4 times greater than the area of KLM: (BC × AC)/2 = 4(LM × KM)/2 BC × AC = 4(LM × KM) (1) SUFFICIENT: Since angle ABC is equal to angle KLM, and since both triangles have a right angle, we can conclude that the angles of triangle ABC are equal to the angles of triangle KLM, respectively (note that the third angle in each triangle will be equal to 35 degrees, i.e., 180 – 90 – 55 = 35). Therefore, we can conclude that triangles ABC and KLM are similar. Consequently, the respective sides of these triangles will be proportional, i.e. AB/KL = BC/LM = AC/KM = x, where x is the coefficient of proportionality (e.g., if AB is twice as long as KL, then AB/KL = 2 and for every side in triangle KLM, you could multiply that side by 2 to get the corresponding side in triangle ABC). We also know from the problem stem that the area of ABC is 4 times greater than the area of KLM, yielding BC × AC = 4(LM × KM), as discussed above. Knowing that BC/LM = AC/KM = x, we can solve the above expression for the coefficient of proportionality, x, by plugging in BC= x(LM) and AC = x(KM): BC × AC = 4(LM × KM) x(LM) × x(KM) = 4(LM × KM) x2 = 4 x = 2 (since the coefficient of proportionality cannot be negative) Thus, we know that AB/KL = BC/LM = AC/KM = 2. Therefore, AB = 2KL = 2(10) = 20 (2) INSUFFICIENT: This statement tells us the length of one of the shorter sides of the triangle KLM. We can compute all the sides of this triangle (note that this is a 6810 triangle) and find its area (i.e., (0.5)(6)(8) = 24); finally, we can also calculate that the area of the triangle ABC is equal to 96 (four times the area of KLM). We determined in the first paragraph of the explanation, above, that the area of ABC = (BC × AC)/2. Therefore: 96 = (BC × AC)/2 and 192 = BC × AC. We also know the Pythagorean theorem: (BC)2 + (AC)2= (AB)2. But there is no way to convert BC × AC into (BC)2 + (AC)2 so we cannot determine the hypotenuse of triangle ABC. The correct answer is A.
_________________
Thanks & Regards, Anaira Mitch




Re: The area of the right triangle ABC is 4 times greater than
[#permalink]
30 Dec 2016, 01:54







