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Need Help With Math Problem: Triangles

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Need Help With Math Problem: Triangles [#permalink] New post 18 Nov 2010, 07:33
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Question Stats:

50% (02:11) correct 50% (00:29) wrong based on 10 sessions
The perimeter of a certain isoceles right triangle is 16 + (16 * Square Root (2)). What is the length of the hypotenuse.

8
16 (correct answer)
4 * Square Root (2)
8 * Square Root (2)
16 * Square Root (2)

Thanks!
[Reveal] Spoiler: OA
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Re: Need Help With Math Problem: Triangles [#permalink] New post 18 Nov 2010, 08:34
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ANS : B

lets say we have an isoceles triangle ABC with B = 90 degress and AB=BC=X and CA=Y.

now applying rt angle formula sqr Y = sqr X + sqr X = 2(sqr X).......................1
also, perimeter = 2X + Y = 16 + 16(sqrt of 2).................2

solve both the equations for Y.. in the end, we would get the equation something like:

Y + Y(sqrt 2) = 16 + 16(sqrt 2).

From this, its easily seen that Y=16. :)

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Re: Need Help With Math Problem: Triangles [#permalink] New post 18 Nov 2010, 09:51
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Since we are dealing with isosceles triangle, we know that its 2 legs are equal.

The equation for its perimeter = hypotenuse + 2*leg.

and since isosceles right triangle is a 45 45 right triangle,

leg = \frac{hypotenuse}{\sqrt{2}}

which can be simplified to leg = \frac{hypotenuse}{2}\sqrt{2}

So the perimeter equation becomes

perimeter = hypotenuse + 2\frac{hypotenuse}{2}\sqrt{2}

perimeter = hypotenuse + hypotenuse\sqrt{2}

and since we are given

perimeter = 16 + 16\sqrt{2}

therefore hypotenuse = 16.
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Re: Need Help With Math Problem: Triangles [#permalink] New post 18 Nov 2010, 16:02
Remember the equation for isoscelleous right triangle sides- x:x:x*sqrt(2)

Let one leg= a Thus,
2a+sqrt(2)*a= 16+16*sqrt(2)

Readjusting
2a+sqrt(2)*a= 16*sqrt(2)+16

thus a = 8 *sqrt(2)
And hypotenus= 16

Ans:- B
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Re: Need Help With Math Problem: Triangles [#permalink] New post 22 Nov 2010, 13:45
messed up....:(
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Re: Need Help With Math Problem: Triangles [#permalink] New post 22 Nov 2010, 18:14
Hey guys,

Nice solution - just one thing I like to point out on these:

Think like the testmaker!

If I were writing the test and knew that everyone studies the 45-45-90 ratio as: 1, 1, sqrt 2

I would make a living off of making the shorter sides a multiple of sqrt 2 so that the long side is an integer. People aren't looking for that! And they often won't trust themselves enough to calculate correctly...they'll look at the answer choices and see that 3 of them are Integer*sqrt 2, and they'll think they screwed up somehow because the right answer "should" have a sqrt 2 on the end.

So...keep in mind that with the Triangle Ratios:

45-45-90
x: x: x*sqrt 2

30-60-90
x : x*sqrt3 : 2x

One of the easiest tricks up the GMAT author's sleeve is to make x equal to a multiple of the radical so that the radical appears on the side you're not expecting and the integer shows up where you think it shouldn't!

Also, as you go through questions like these, ask yourslf "how could they make that question a little harder" or "how could they test this concept in a way that I wouldn't be looking for it". Training yourself to look out for unique cases, from the testmaker's perspective, helps you to get a real mastery of the GMAT from a high level.
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Re: Need Help With Math Problem: Triangles   [#permalink] 22 Nov 2010, 18:14
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