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GMAT prep question-right triangle

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by TrizMA » Sun Jun 29, 2008 11:47 am
sides are: x, x, √2x where hyp=√2x

perimeter= 2x+ √2x= √2x(√2+1)

√2x (√2+1)= 16+16√2= 16(1+√2)

√2x= 16(1+√2)/ (1+√2) = 16

hyp=√2x=16

You can also use POE to get the answer
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by anjaligeorge1 » Sun Jun 29, 2008 11:53 am
followed the same method.the answer is 16
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by andreasonlinegr » Sun Jun 29, 2008 1:16 pm
thank you for your quick response.
I still don't get why hyp is √2x, should not be x√x ??is not right that an isosceles right triangle has the following sides: x, x and x√2 ?

also, you mentioned that you can solve this by POE. What's POE?

thanks again .
much appreciated.
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by TrizMA » Sun Jun 29, 2008 1:33 pm
take equal side to be x.

so hyp²= x² + x²= 2x²
hyp= √(2x²)=x√2

POE is process of elimination. You work backwards i.e you select answer choice and see which one satisfies.

For example in this case:

pick D i.e. 8√2 as an answer choice. I selected this becaue this is a middle number amongst answer choices

if hyp=8√2; x= hyp/ √2= 8; let's see perimeter= 16+ 8√2
So D doesn't give correct answer & to get correct answer we need to have a bigger hyp

i pick next one i.e hyp=16

hyp=16; x=16/√2= (16* √2)/ (√2*√2)= 8√2

perimeter= hyp+2x= 16+ 2(8√2)= 16+ 16√2.

So with hyp 16 the question is solved. If u still want to check other answer choices u will find that this is the only one which solves equation. This is working backwards n is esp useful for algebra.
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by TrizMA » Sun Jun 29, 2008 1:39 pm
yes, u r right. hyp= x√2. i meant the same thing when i wrote √2x. it's just sometimes diff to write with symbols on pc.

I hope now u can understand the solution. plus hope POE process also helps
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by andreasonlinegr » Sun Jun 29, 2008 11:13 pm
thank you Trizma. I can now solve this through both ways. Thank you again !!
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