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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]

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05 Feb 2012, 05:07
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KLM Triangle.GIF [ 2.03 KiB | Viewed 17990 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.

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05 Feb 2012, 05:12
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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 [ 2.03 KiB | Viewed 15153 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.

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

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

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

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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.

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..

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

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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 9486 times ]

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

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

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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 6-8-10 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.
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Re: The area of the right triangle ABC is 4 times greater than [#permalink]

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02 Jul 2017, 13:50
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.

Hope it helps.

Hi Bunuel
i have a doubt in this question as it says ...The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. so it implies that area ABC = 5* area of KLM

or the sentence should be 4 times of area of KLM

I want to clear this "greater than " dilemma

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

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06 Jul 2017, 12:45
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Bunuel

I took greater than as >

Area of one triangle > 4(Area of another triangle)
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Re: The area of the right triangle ABC is 4 times greater than [#permalink]

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08 Sep 2017, 18:05
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.

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.

Statement 1

Clearly establishes similarity- if we know the proportion of the areas and that the triangles are similar we can then find the length of the hypotenuse simply by multiplying 2. For example, consider two similar Pythagorean triplets: 3-4-5 and 6-8-10. If take any side other than the hypotenuse and square it such as

3^2/ 6^2 OR 4^2/8^2 we would get the ratio of the smaller triangle to the larger triangle... 1: 4. We then simply multiply by 2 or divide by two in order to find side lengths of the other triangle- so if we are given 8 then we divide by 2 or if we are given 4 we multiply by 2. Also consider the two triangles 3-4-5 and 12-16-20. If we all we knew was that the ratio of the larger to the smaller triangle was 1/16 and we knew the triangles were similar and were given the hypotenuse length 20 we could still find the hypotenuse of the other triangle by dividing by 4, or the square root of 16. Here's the math

4^2/16^2 = 16/256 = 1/4^2 = 1/16

Statement 2

If we knew that the triangles were similar then this would be sufficient but we do not know that the triangles are actually similar.
Insuff

A
Re: The area of the right triangle ABC is 4 times greater than   [#permalink] 08 Sep 2017, 18:05
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