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zubin.desai
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Post subject: Advanced Geometry Problem Set #6  Posted: Thu Apr 08, 2010 11:52 am |
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Hi I got this problem correct using a different method than what is in the solutions. I used properties of similar triangles to get the answer, which frankly is the better method and I'm not sure why MGMAT doesn't include this solution in their book.
However I am trying to understand the book's solution and just do not understand it. What triangle has base 7 and height x? I do not see this in the diagram. I want to assume this is a typo but this method also got to the correct answer! Please help, my official test is in 2 weeks. Thank you!
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zubin.desai
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Post subject: Re: Advanced Geometry Problem Set #6 Posted: Thu Apr 08, 2010 11:53 am |
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By the way this is from page 107 in the 5th Edition Geometry strategy guide.
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zubin.desai
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Post subject: Re: Advanced Geometry Problem Set #6 Posted: Thu Apr 08, 2010 11:54 am |
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sorry 4th edition, brain fart
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Ben Ku
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Post subject: Re: Advanced Geometry Problem Set #6 Posted: Sat May 01, 2010 1:57 am |
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I'm assuming you're asking about question 6.
Your approach of using similar triangles is a good one. I assume you set up something like this:
(3 / x) = (7 / 12)
In this way, x = 36/7. I'll suggest this to the developers of the Strategy Guide.
The solution shown in the Strategy Guide finds the area of the obtuse triangle bonded by 7, 12, and the unmarked side. If we let the base be 7, then the height of the triangle will be the distance between the far vertex to the line extended by the base, which is x.
You can refer to the diagram near the top of page 34.
I hope this makes sense.
_________________
Ben Ku
Instructor
ManhattanGMAT
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akatzi
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Post subject: Re: Advanced Geometry Problem Set #6 Posted: Mon Jan 17, 2011 7:09 pm |
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Hello,
Could somebody explain to me, how Ben gets to this (3 / x) = (7 / 12)?? I can't see the similar triangles needed for this equation...
Thx!!
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jnelson0612
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Post subject: Re: Advanced Geometry Problem Set #6 Posted: Sun Jan 23, 2011 5:43 pm |
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akatzi wrote: Hello,
Could somebody explain to me, how Ben gets to this (3 / x) = (7 / 12)?? I can't see the similar triangles needed for this equation...
Thx!! Probably the easiest way for me to explain this is to recommend that you go ahead and cross multiply what Ben has written. You get 3*12 = 7*x. Since we are calculating the area of this triangle we can use 12 as the base and 3 as the height. 3*12=36 * 1/2 gives us an area of 18. I can also use 7 as my base and x as my height, since x is an altitude. We know that 7 * x will also give us 36, because it is the exact same triangle. I hope this helps you see how Ben derived this.
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Jamie Nelson
ManhattanGMAT Instructor
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econ_lobo
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Post subject: Re: Advanced Geometry Problem Set #6 Posted: Sat Aug 20, 2011 12:02 am |
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I believe some of the confusion is based on the placement of the number 7. If we look at the 12 as a previous base it is centered on the base of the triangle. Our 7 in question looks more like a hypotenuse for a smaller triangle (12-n, 3, 7). I did note the previous description from page 34, but do believe this is not a typical technique and think it adds to the confusion that 7 is the base and not a part of the base.
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econ_lobo
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Post subject: Re: Advanced Geometry Problem Set #6 Posted: Sat Aug 20, 2011 12:12 am |
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Actually the review on page 34 is what I think nails this problem. My assumption of 7 as the base of the "larger" triangle was incorrect. The three triangles at the top of page 34 break down how to attack this problem.
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jnelson0612
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Post subject: Re: Advanced Geometry Problem Set #6 Posted: Sat Sep 17, 2011 10:06 pm |
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Glad it has been cleared up.
_________________
Jamie Nelson
ManhattanGMAT Instructor
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