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MGMAT geometry

by papgust » Sat Mar 06, 2010 8:01 am
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.

OA: A

[See image for diagram]
Image

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I didn't like MGMAT's explanation for this problem. It is needlessly too long. I solved this by a different approach during the test. Just want to check whether my approach is right.

My Solution:
ABC and KLM are right triangles. We are not sure whether they are similar triangles.

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

The other angle is 90. So, we could find the third angle.
Now, all angles are equal in both triangles. This means that they are similar triangles.
The ratio of areas of triangles is a^2:b^2 and The ratio of sides of triangles is a:b
The ratio of areas of triangles here is 4:1 (Since one triangle is 4 times greater than the other).
Therefore, ratio of sides is 2:1.

Now, KL = 10 inches. Since Side of ABC:Side of KLM :: 2:1, AB = 20 inches.

Sufficient

(2) We don't know whether triangles are similar. Insufficient.


Is this approach correct for this problem?
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by thephoenix » Sat Mar 06, 2010 8:13 am
papgust 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.

OA: A

[See image for diagram]
Image

---

I didn't like MGMAT's explanation for this problem. It is needlessly too long. I solved this by a different approach during the test. Just want to check whether my approach is right.

My Solution:
ABC and KLM are right triangles. We are not sure whether they are similar triangles.

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

The other angle is 90. So, we could find the third angle.
Now, all angles are equal in both triangles. This means that they are similar triangles.
The ratio of areas of triangles is a^2:b^2 and The ratio of sides of triangles is a:b
The ratio of areas of triangles here is 4:1 (Since one triangle is 4 times greater than the other).
Therefore, ratio of sides is 2:1.

Now, KL = 10 inches. Since Side of ABC:Side of KLM :: 2:1, AB = 20 inches.

Sufficient

(2) We don't know whether triangles are similar. Insufficient.


Is this approach correct for this problem?
i am not sure which method the MGMAT used but my appraoch was similar to yours , and its flawless
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by papgust » Sat Mar 06, 2010 8:39 am
Hi,

Thanks for confirming this approach. And here is the MGMAT explanation,
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.

The correct answer is A.
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by bpgen » Sat Mar 06, 2010 9:32 am
Ok, my take also A but having different approach:

As we established both triangles are similar triangles and not isosceles, therefore, side ratio would be 3:4:5,
so for KLM, it would be 3*2: 4*2: 5*2 i.e 6:8:10, (as KL=10, multiplying 5 by 2 to make it 10.)

Now for ABC triangle, side ratio would be, let's say, 3*a: 4*a: 5*a
As per area concerns(ABC is 4 times KLM) , equation would be:
1/2*(3*a * 4*a )=4*(1/2*(6*8))
=>12a^2=4*6*8
=>a=4
so side ratio would be 12:16:20, so AB=20 inches
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by hai1 » Sat Mar 06, 2010 1:58 pm
Why can't one use formular of cosine and sin to determine length of KL if the KLM=55 degrees?

Should we assume here that we cannot use these cosine and sin values?

Bottom line, 2 is sufficient if we can use sine and cosine.
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by bpgen » Sat Mar 06, 2010 7:04 pm
Much Appreciated hai1 for thinking about another approach!

Please note, still we need ARG-1, to determine Triangle ABC and KLM are similar triangles, without this we can't compare TanX value of ABC with KLM.

If you don't think so, Kindly explain how to reach using ARG-2 alone?

Here is detail, how we could reach solution using TanX...
Sin(x) = Opposite / Hypotenuse
Cos(x) = Adjacent / Hypotenuse
Tan(x) = Opposite / Adjacent

for angle x=55, SinX, CosX, TanX are 0.81915, 0.57358, 1.42815 Not easy to calculate, we need to depend on trigonometry functions table value, which I think for GMAT, impossible to check.)

Therefore sides of triangle KLM would be 8.1915~8, 5.7358~6...coool!

Now for Triangle ABC,
TanX=a/b(a,b are two side of right angle of ABC)

Therefore, 1.42815=a/b (Since ABC and KLM are similar triangle)

as per area concerns:
1/2*ab=4*(1/2*(8*6))
=> 1.42815b^2=4*8*6
=>b=11.5948~12
=>a=~16
Then (using famous Pythagoras theorem ) Hypotenuse of ABC would be 20!, Ooooops :-)
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by kstv » Mon Mar 08, 2010 6:16 am
Diagramatically if we can fit traingle KLM in triangle ABC cos <KLM = <ABC = 55° and <C=<M = 90°. Then K will be a pt. on AB and M is a pt. on CB. Notice KL || to AC.
Let KM = x and ML = 2y area of KLM = xy
Area of ABC = 4xy
The triangles are similar so the ratio of the sides are so KM/AC= ML/CB = KL/AB
possible if x/2x = 2y/4y = 1/2
KL/AB=1/2 , AB = 10*2=20
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by bpgen » Mon Mar 08, 2010 7:05 am
Appreciate putting old wine in a new bottle :-)

Ultimately you using side ratio, right? if you using arg-1:Angles ABC and KLM are equal to 55 degrees. Do we still need "cos <KLM = <ABC = 55°" ?
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by [email protected] » Fri Mar 16, 2012 11:53 pm
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.

OA: A

[See image for diagram]


First and foremost let me mention that this question is very well inside the domain of the actual gmat.

Statement 1: It proves that the two trianlges ABC and KLM are similar.

Now there is point that all of you missed out on...

The Ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

i.e A(ABC)/A(KLM) = (AB)^2/(KL)^2 = (BC)^2/(LM)^2 = (AC)^2/(KM)^2

hence when you apply this equation it is given that KL the hypotenuse is 10 inches...

therefore 1/4 = AC/KL hence you get the answer...


Statement 2: it gives the area of the triangle ABC but does not give the answer for the hypotenuse AB.

Hence only statement 1 is sufficient and not both of them...


Hope this really helped you man...

First and foremost le t
IT IS TIME TO BEAT THE GMAT

LEARNING, APPLICATION AND TIMING IS THE FACT OF GMAT AND LIFE AS WELL... KEEP PLAYING!!!

Whenever you feel that my post really helped you to learn something new, please press on the 'THANK' button.
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