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please explain how to solve it.

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by ketkoag » Sat May 02, 2009 4:54 am
An irregular quadrilateral, the up-left angle is a right angle, and the length of the two sides adjacent to it both is x. The under-right angle is 75 degree and the under-left angle is 60 degree. What is the perimeter of the quadrilateral?
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by deepoe » Sat May 02, 2009 5:04 am
you got a picture?
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by ketkoag » Sat May 02, 2009 5:15 am
deepoe wrote:you got a picture?
i'm afraid no..
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by DanaJ » Sat May 02, 2009 5:39 am
The key here is indeed the picture. I attached one so you can easily follow my explanations.

Now, angle BAD is a right angle, and sides AB and AD are equal in length. This means that triangle BAD is a right isosceles triangle. Its hypotenuse will be:

BD^2 = AB^2 + AD^2 = 2*x^2 ----------BD = x*sqrt(2).

Now, let's look at angle ADC. Since we're talking about a quadrilateral, the sum of ABCD's interior angles will be 360. Since BAD is 90, ABC is 60 and BCD is 75, then you get that ADC will be 360 - 90 - 60 - 75 = 135.

Notice that ADC is BDC + ADB = 135. Since ADB is one of the angles of right isosceles triangle BAD, then its measurement will be 45. This makes BDC = 135 - 45 = 90. This is why triangle BCD is also a right triangle, with BDC the right angle, BCD 75 and DBC 15 degrees.

Now we use sine and cosine to solve the problem, since we have one of the sides of triangle BCD and the measurements of its angles.

Here you apply some special formulas:

1 + cos(a) = 2[cos(a/2)]^2

1 - cos(a) = 2[sin(a/2)]^2.

You then use sine and cosine of 15 to find out DC and BC, but the calculations involved make me think that this is not your average GMAT problem or that maybe I'm not doing it right. Maybe someone has better ideas....

Edit: If the problem was smth like the under-left angle is 75 and the under-right angle was 60, then the problem would be very easy and there would be no need for complicated formulas and calculations. How about you double-check to make sure?
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by dumb.doofus » Sat May 02, 2009 6:58 am
I reallly have my doubts on this problem.. anyways here is the solution..



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by DanaJ » Sat May 02, 2009 7:59 am
Image not loading - or maybe it's just my wireless connection...
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by deepoe » Sat May 02, 2009 8:11 am
just go to the link : https://www.postimage.org/image.php?v=gxgS_9r than it will work DanaJ

Is this a GMAT problem :o?
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by DanaJ » Sat May 02, 2009 9:17 am
Picture finally loaded. Your solution is pretty much the same as mine: we both used pretty advanced trigonometry formulas. This is why I'm inclined to say that this is not a question you'd see in the real thing....
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by keizer Soze » Sat May 02, 2009 10:08 am
Hi, I try to solve it this way, but I don´t know that if it is right.
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by DanaJ » Sat May 02, 2009 10:15 am
The problem with your approach is that not all Pythagorean triples follow the rapport 3:4:5.
There are many other numbers like these: 5, 12 and 13 is only one example. For more, check out the wikipedia page of the Pythagorean triples.

This is why you can't assume that, just because triangle BCD is a right triangle, its sides follow that rapport.
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by keizer Soze » Sat May 02, 2009 10:31 am
Yeap, your are right, big mistake!
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by dumb.doofus » Sat May 02, 2009 12:33 pm
Agree with DanaJ.. my solution is on the similar lines as DanaJ's and too complex stuff to be in GMAT.. anyone got a simpler solution? Experts please pitch in..

Also, Ketkoag, please let us know the source of this question too..
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by Brent@GMATPrepNow » Sat May 02, 2009 3:07 pm
dumb.doofus wrote:Agree with DanaJ.. my solution is on the similar lines as DanaJ's and too complex stuff to be in GMAT.. anyone got a simpler solution? Experts please pitch in..

Also, Ketkoag, please let us know the source of this question too..
Yes, the source would be nice. I'm guessing the source is a highschool trig book and the forum is helping someone do his/her homework :)
Suffice it to say, when sine and cosine enter the picture, we are out of the realm of official GMAT questions. At least, I certainly can't see how this could be a real GMAT question unless the question is incorrectly worded.
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by dumb.doofus » Sat May 02, 2009 3:17 pm
Brent Hanneson wrote:
dumb.doofus wrote:Agree with DanaJ.. my solution is on the similar lines as DanaJ's and too complex stuff to be in GMAT.. anyone got a simpler solution? Experts please pitch in..

Also, Ketkoag, please let us know the source of this question too..
Yes, the source would be nice. I'm guessing the source is a highschool trig book and the forum is helping someone do his/her homework :)
Suffice it to say, when sine and cosine enter the picture, we are out of the realm of official GMAT questions. At least, I certainly can't see how this could be a real GMAT question unless the question is incorrectly worded.
LOL :D Very well said Brent
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