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gphil
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Post subject: The perimeter of a certain isosceles right triangle is 16 +  Posted: Sat Nov 10, 2007 5:55 pm |
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Could pleae somebody help with the following problem? Thanks!
The perimeter of a certain isosceles right triangle is 16 + 16 * sq. rt 2. What is the length of the hypotenuse of the triangle.
A) 8
B) 16 - correct
C) 4* sq.rt.2
D) 8* sq.rt.2
E) 16* sq.rt.2
I began to solve the problem, but didn't know how to finish.
If a is a side of isosceles right triangle, then
16 + 16 * sq. rt 2 = 2*a + a*sq. rt. 2
We need to find a*sq. rt. 2
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Nov1907
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Post subject: Posted: Mon Nov 12, 2007 12:42 pm |
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Your equation is set up perfectly, just the algebra with the sqrt signs might take longer. You can go 2 ways assume the equal sides to be a or assume the hypotenuse to be a. The second turns out much easier to solve because it is easier to compare it to the perimeter answer given.
Let the hyp. be a. Then the sides are a/sqrt(2). Sum of the 2 sides = 2*a/sqrt(2) = a*sqrt (2)
So your equation becomes a+a*sqrt(2) = 16+16*sqrt(2). Here it is obvious that a = 16.
Hope this helps. :)
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gphil
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Post subject: Posted: Mon Nov 12, 2007 1:41 pm |
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I would not figure it out for sure. It really makes sence to make a hypotenuse = a. Thanks a lot!
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RonPurewal
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Post subject: Posted: Fri Nov 16, 2007 6:53 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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This is one of those problems where ESTIMATION can really save the day: you should memorize the fact that root(2) is approximately equal to 1.4.
Then your equation - in its original form - becomes:
16 + about 22.4 = 2a + about 1.4a
--> about 38.4 = about 3.4a
--> a equals a little less than 12
--> hypotenuse = a times 1.4 = a little less than 16.8.
Looks like 16 is the best choice.
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Dylan
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Post subject: The perimeter of a certain isosceles right triangle is 16 + Posted: Sun Apr 20, 2008 12:19 pm |
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This problem appeared on the practice GMATPrep exam from GMAC. I came up with the same equation, as the original poster (see below).
16 + 16 * sq. rt. (2) = 2a + a * sq. rt. (2)
We need to find a * sq. rt. (2)
Can someone walk me through the math to calculate a or a * sq. rt. (2)?
Thanks!
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AG
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Post subject: you can also substitute the answers Posted: Sun Apr 20, 2008 3:19 pm |
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lets start from E and go up
1. H = 16sqrt2 means sides = 16 doesn't work for the perimeter total 32+16srt2
2. H = 8sqrt2 means side = 8 does not work for the perimeter total 16+8sqrt2
3. H = 4sqrt2 means side = 4 does not work for the perimeter total 8+4sqrt2
4. H = 16 means side = 16/sqrt2 which means perimeter = 16sqrt2+16 voila you got the ans
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RonPurewal
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Post subject: Re: The perimeter of a certain isosceles right triangle is 1 Posted: Thu Apr 24, 2008 3:32 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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Dylan wrote: This problem appeared on the practice GMATPrep exam from GMAC. I came up with the same equation, as the original poster (see below).
16 + 16 * sq. rt. (2) = 2a + a * sq. rt. (2)
We need to find a * sq. rt. (2)
Can someone walk me through the math to calculate a or a * sq. rt. (2)?
Thanks!
well, my first answer is 'see the above posts'. but if you're looking for a really mechanical way to solve, then you can always do this:
16 + 16√2 = 2a + a√2
16 + 16√2 = a(2 + √2)
so
a = (16 + 16√2) / (2 + √2)
you can rationalize the denominator by multiplying by its conjugate, (2 - √2), making the denominator into a difference of squares: (2 + √2)(2 - √2) = 4 - 2 = 2.
therefore
a = (16 + 16√2)(2 - √2) / 2
multiply out --> = (32 - 16√2 + 32√2 - 32) / 2
= 16√2 / 2
= 8√2
so hypotenuse = 8√2 x √2 = 8 x 2 = 16
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Dylan
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Post subject: Thank you Ron! Posted: Tue Apr 29, 2008 5:56 pm |
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Your explanation was much better than GMAC's!
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RonPurewal
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Post subject: Re: Thank you Ron! Posted: Wed Apr 30, 2008 4:23 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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Dylan wrote: Your explanation was much better than GMAC's!
well thank you. but how did you get a gmac explanation? as far as i know, the gmatprep questions don't come with answer explanations...?
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kevincan
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Post subject: Posted: Wed May 14, 2008 5:22 pm |
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Make sure you know four right triangles inside out: (3-4-5, 5-12-13, 1-sqrt(3)-2, 1,1,sqrt(2)) -If the hypotenuse is h, each leg is h x sqrt(2)/2, so the perimeter would be h + h x sqrt(2). Thus h=16
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rfernandez
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Post subject: Posted: Fri Jun 06, 2008 4:24 am |
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| ManhattanGMAT Staff |
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Posts: 386
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paulprior_ire
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Post subject: Re: Posted: Mon Apr 25, 2011 8:06 am |
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kevincan wrote: Make sure you know four right triangles inside out: (3-4-5, 5-12-13, 1-sqrt(3)-2, 1,1,sqrt(2)) -If the hypotenuse is h, each leg is h x sqrt(2)/2, so the perimeter would be h + h x sqrt(2). Thus h=16 Sorry to drag up an old post; I spotted the above but given the ratio of x:x:x.sqrt(2). Would this not lend itself more to the Hypotenuse being x.sqrt(2)? so if each leg is 8 the hypotenuse would be x.sqrt(2)? Thanks,
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RonPurewal
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Post subject: Re: Re: Posted: Wed Apr 27, 2011 5:04 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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paulprior_ire wrote: Sorry to drag up an old post; I spotted the above but given the ratio of x:x:x.sqrt(2). Would this not lend itself more to the Hypotenuse being x.sqrt(2)? so if each leg is 8 the hypotenuse would be x.sqrt(2)?
Thanks, in that post, the poster defined the variable to be h = hypotenuse. with that definition in place, each leg is h/√2, which reduces to h√2/2. if you allow x to be the length of each leg, then, yes, the hypotenuse is x√2. see here for an easier solution: post36393.html#p36393
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