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In parallelogram ABCD, angle A = 60. Find the the ratio of [#permalink]
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14 Oct 2003, 02:24
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In parallelogram ABCD, angle A = 60. Find the the ratio of the smaller diagonal to the larger diagonal?



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hi..
In a parallelogram,as you know, diagonals bisect each other . if AC and BD are daigonals and they bisect at point E we get 4 triangles AED , DEC ,CED and AEB.
in triangle AED angle A = 30, D=60 and E=90. since it is a 306090 triangle, if AE (The longer leg) = x then shorter leg is
x/square root(3) and hypotenuse is 2x/square root(3).
so AE = x
=> AC = 2x
DE = x/square root(3)
=> DB = 2x/square root(3)
so BD : AC = 1 : square root(3)
Hope I am right



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Adhi wrote: hi.. In a parallelogram,as you know, diagonals bisect each other . if AC and BD are daigonals and they bisect at point E we get 4 triangles AED , DEC ,CED and AEB. in triangle AED angle A = 30, D=60 and E=90. since it is a 306090 triangle, if AE (The longer leg) = x then shorter leg is x/square root(3) and hypotenuse is 2x/square root(3). so AE = x => AC = 2x DE = x/square root(3) => DB = 2x/square root(3)
so BD : AC = 1 : square root(3)
Hope I am right
Diagonals of parallelogram do bisect each other; however, the angle E is not necessarily right. If E=90, then ABCD has to be either a rhombus, or a square, or a diamond with different sides. Moreover, parallelogram diagonals do not bisect according angles. Therefore, AED is not a 306090 triangle.
The problem is much complicated then it seems at the first glance.
Last edited by stolyar on 15 Oct 2003, 00:30, edited 1 time in total.



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stolyar is correct.... the diagonals of a parallelogram are not angle bisectors... look for a different approach.
thanks



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Yes. you guys are correct. I am sorry.This problem requires different approach.
Adhi



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Re: PS: geometry [#permalink]
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15 Oct 2003, 14:37
Vicky wrote: In parallelogram ABCD, angle A = 60. Find the the ratio of the smaller diagonal to the larger diagonal?
I'm not sure I read the question correctly but I believe the answer is dependent on the shape of the parallelogram. If so, I think the solution is:
[(3 + (2k  1)^2) / (3 + (2k + 1)^2)]^1/2 where k = the ratio of the longer side to the shorter side of the parallelogram. This reduces to 1/Sqrt(3) or Sqrt(3)/3 in the special case of a rhombus where k = 1.
Comments: Vicky, you submit some interesting problems. IMHO, however, I hope that you and others will take the time to supply 5 answer choice as does the real GMAT. This is a GMAT forum, not a math forum, hence, the student can gain valuable information from the choices offered and learning how to do so is something that should be part of the learning experience. Also, coming up with appropriate incorrect answers can be challenging in itself  especially if you are trying to anticipate wrong answers as I do )
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Diagonals in a parallelogram obey the following formula:
D^2+d^2=2(a^2+b^2), where a and b are sides
Also, Area of a parellelogram = Dd/2=(a^2)sinA
I tried to play with the abovebut in vain
I feel that the given information is not sufficient to solve.



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stolyar wrote: Diagonals in a parallelogram obey the following formula:
D^2+d^2=2(a^2+b^2), where a and b are sides Also, Area of a parellelogram = Dd/2=(a^2)sinA
I tried to play with the abovebut in vain I feel that the given information is not sufficient to solve.
We have plenty of information to solve the problem, and, in fact, if one simply draws a diagram, the solution jumps right out!
Once again, my aversion to learning formulas is reinforced. Stolyar looked up two nice formulas for parallelograms, both of which are useless for this problem.
In the GMAT, you can solve virtually any geometry problem using basic principles of triangles, pythagorean theorem, angles, and circles, simply by drawing appropriate construction lines and using basic principles. Really. Let's see how this problem can be reduced to baby basic geometry:
Observe the attached diagram. The short side of the parallelogram is assumed to have length 2X and the length of the long leg is K(2X) where K = ratio of the long leg to the short.
Suppose we let the short leg of the parallelogram = 2X = segment AD. If we draw a line from point A down to the baseline, we can construct a right triangle with legs X and Sqrt(3)X (remember that angle DAB = 60 and the proportions of a 306090 triangle).
Now we want to find the ratio of the lines DB over AC. If we forget about using fancy parallelogram formulas for a second, we may notice that AC and DB are simply the hypotenueses of two right triangles that we can easily compute in terms of X and K.
Hence,
AC = Sqrt[(line AQ)^2 + (line QC)^2]
= Sqrt[ (Sqrt(3)X)^2 + (X + 2KX)^2 ] and
DB = Sqrt[((line AQ)^2 + (line AB  line QD)^2]
= Sqrt[ (Sqrt(3)X)^2 + (2KX  X)^2 ]
Since, we are only concerned about the ratio of DB over AC, we can simply set X = 1 and we get:
DB/AC = Sqrt[(3 + (2K  1)^2 ) / (3 + (2K + 1)^2] where K = ratio of AB to AD.
You can see that for a rhombus where K = 1, the above reduces to Sqrt(3 / 9) = 1 / Sqrt(3) which is easy to verify.
Vicky, is this the answer you wanted?
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So, the initial info is not suff to answer, yes?



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stolyar wrote: So, the initial info is not suff to answer, yes?
Correct. If my analysis is correct (can you find any mistakes?), you can only answer it in terms of K. I think if you draw a few parallelograms using simple numbers, you should be able to determine that a nonrhombus will have a different ratio than a rhombus.
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thanks to all of you for an animated discussion... espec. akamaibarah, stolyar.
akamai.. u are right that i should be posting answer choices cause that's what is asked at GMAT. From now onwards i will take care of that. thanks for the input.
Now as for this question. Akamai your solution is perfect given that i had not posted the answer choices.
It makes more sense if i would have posted it following way:
In parallelogram ABCD, angle A = 60. Find the the ratio of the smaller diagonal to the larger diagonal?
A) 1/2
B) 1/[2*sqrt(3)]
C) 1/sqrt(3)
D) 2/3
E) 1/3
adhi's approach is not vorrect cause it makes wrong assumptions about parallelogram as stolyar pointed out.
IMHO, the best approach (if choices are not given) is as akamai did. When choices are given we know that if something is true for a parallelogram (in this case ratio of diagonals), it has to be true for a rhombus cause it is also a parallelogram. Thus find this ratio for rhombus and then we have diagonals meeting at right angles.
thanks



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In case of a rhombus, the initial problem could be solved practically in no time, right in a head. Correct?
D/d=sqrt 3



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Vicky wrote: thanks to all of you for an animated discussion... espec. akamaibarah, stolyar. akamai.. u are right that i should be posting answer choices cause that's what is asked at GMAT. From now onwards i will take care of that. thanks for the input. Now as for this question. Akamai your solution is perfect given that i had not posted the answer choices.
It makes more sense if i would have posted it following way: In parallelogram ABCD, angle A = 60. Find the the ratio of the smaller diagonal to the larger diagonal?
A) 1/2 B) 1/[2*sqrt(3)] C) 1/sqrt(3) D) 2/3 E) 1/3
adhi's approach is not vorrect cause it makes wrong assumptions about parallelogram as stolyar pointed out. IMHO, the best approach (if choices are not given) is as akamai did. When choices are given we know that if something is true for a parallelogram (in this case ratio of diagonals), it has to be true for a rhombus cause it is also a parallelogram. Thus find this ratio for rhombus and then we have diagonals meeting at right angles. thanks
Hi Vicky!
With all due respect, I don't think the giving specific answer choices is kosher for your problem since you don't specify up front that the figure is a rhombus. IMO, you must either specify that the parallelogram is indeed a rhombus or specify that the diagonals cross at right angles (in which case it must be a rhombus), or give answer choices in terms of the ratio of the sides.
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



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akamai.. i fully understand your point of view.
IMHO there is no need to specify that the figure is a rhombus.
Something that is true for a prallelogram must be true for a figure that belongs to its class, in our case, a rhombus. We cannot consider other parallelogram's (rect, square, etc) as a test example cause the angles data (60 degree) will be violated.



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praetorian123 wrote: Vicky wrote: akamai.. i fully understand your point of view. IMHO there is no need to specify that the figure is a rhombus. Something that is true for a prallelogram must be true for a figure that belongs to its class, in our case, a rhombus. We cannot consider other parallelogram's (rect, square, etc) as a test example cause the angles data (60 degree) will be violated. i agree.
Yes. What is true for a parallelogram is also must be true for a rhombus.
HOWEVER, what is true for a rhombus is NOT necessarily true for a parallelogram!
Your answer is only true for a rhombus. Hence, it does not hold true for a general parallelogram, and your question was regarding a parallelogram with not specified "length". BTW, you can have an infinite number of parallelograms all with 60 angles for A. If you look at my diagram, just kept points A and D anchor in place, then increase the length of segments AB and CD buy the same length. The parallelogram just gets "longer" but all of the angles stay the same.
This argument can be settled easily. Simple use my diagram and set AD to 1 and AB to anything else but 1 (use 2 as a simple example), calculate both AC and BD, then find the ratio. I guarantee that you will not get the same answer as the case where AB = AD = 1, hence unless you specify that the figure is a rhombus, you will NOT get the answer you spefied.
If you use a little imagination, you can see that the ratio of the short diagonal to the long diagonal for ANY nonrectangular parallelogram approaches 1 at the limit as the "length" of the parallelogram approaches either 0 or infinity, but is less than 1 for a rhombus with the same angle for A, another way of showing that the ratio does indeed change for different length non=rectangular parallelograms with constant angle at vertex A.
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Hi akamai
I would have to agree with you. very convincing and cogent explanation.
I whole heartedly agree with your remarks that question should have givem rhombus in the problem statement and it is incorrect to say what is true for rhombus will be true for parallelogram.
The issue that i am trying to address is what if question does't mention it? if one faces such kind of a problem and is given choices. Then, the best and fastest approach is to use rhombus approach and be done with it.
thanks



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Vicky wrote: Hi akamai I would have to agree with you. very convincing and cogent explanation. I whole heartedly agree with your remarks that question should have givem rhombus in the problem statement and it is incorrect to say what is true for rhombus will be true for parallelogram. The issue that i am trying to address is what if question does't mention it? if one faces such kind of a problem and is given choices. Then, the best and fastest approach is to use rhombus approach and be done with it. thanks
I give you points for tenacity, but I don't know if I agree with your conclusion.
Consider this argument:
I have a rectangle. ONe side is 1. What is the adjacent side?
A) .5
B) 1
C) 2
D) 3
E) 10
Answer: well, in a rectangle, the adjacent side could be anything. Since we don't know, lets just make it a square and call the best answer B.
IMO, for GMAT quant problems, the "correct" answer is the "best" answer.
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