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OG12 Q. Maybe easy for many, but 'think it is in 700 level 02 Nov 2009, 09:13
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OG12 Q. Maybe easy for many, but 'think it is in 700 level as the approach needs to strike in test-conditions to get a solution!

The hypotenuse of a right triangle is 10cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.



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OG12 Q. Maybe easy for many, but 'think it is in 700 level 02 Nov 2009, 09:55
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IMO D.

Statement 1) \(S1*S2 = 50.\) and \(S1^2+S2^2 = 100.\) Solving both we can get the perimeter.
Statement 2) \(S1 = S2\) and \(S1^2+S2^2 = 100\). Solving both we can get the perimeter.
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OG12 Q. Maybe easy for many, but 'think it is in 700 level 02 Nov 2009, 23:08
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hgp2k
IMO D.

Statement 1) \(S1*S2 = 50.\) and \(S1^2+S2^2 = 100.\) Solving both we can get the perimeter.
Statement 2) \(S1 = S2\) and \(S1^2+S2^2 = 100\). Solving both we can get the perimeter.
Show more

Thanks for the explanation. I see there aren't any more attempts :)
OA:
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D

The point here is to realize that we have to put the values of \(S1*S2\) and \(S1^2+S1^2\) into the equation of \((S1+S2)^2\) to get to the solution...
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OG12 Q. Maybe easy for many, but 'think it is in 700 level 02 Nov 2009, 23:12
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D.

1) B*H = 50 , B^2 + 2BH + H^2 = 100 + 100 , (B+H)^2 = 200 ==> Suff.
2) 2 Side same - 45 - 45 - 90 : 1 : 1 : \sqrt{2} ==> Suff.
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OG12 Q. Maybe easy for many, but 'think it is in 700 level Updated on: 06 Nov 2009, 05:51
Originally posted by Economist on 06 Nov 2009, 05:20.
Last edited by Economist on 06 Nov 2009, 05:51, edited 2 times in total.
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gmattokyo
hgp2k
IMO D.

Statement 1) \(S1*S2 = 50.\) and \(S1^2+S2^2 = 100.\) Solving both we can get the perimeter.
Statement 2) \(S1 = S2\) and \(S1^2+S2^2 = 100\). Solving both we can get the perimeter.

Thanks for the explanation. I see there aren't any more attempts :)
OA:
Show Spoiler
D

The point here is to realize that we have to put the values of \(S1*S2\) and \(S1^2+S1^2\) into the equation of \((S1+S2)^2\) to get to the solution...
Show more
True, \(S1*S2 = 50, (S1 + S2)^2 = 200 , S1 + S2 = \sqrt{200 }\)(corrected ) ---eq 1.
\(S1*S2 = 50\)---eq 2.
Solve, and get S1 and S2.
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OG12 Q. Maybe easy for many, but 'think it is in 700 level 06 Nov 2009, 05:25
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gmattokyo
hgp2k
IMO D.

Statement 1) \(S1*S2 = 50.\) and \(S1^2+S2^2 = 100.\) Solving both we can get the perimeter.
Statement 2) \(S1 = S2\) and \(S1^2+S2^2 = 100\). Solving both we can get the perimeter.

Thanks for the explanation. I see there aren't any more attempts :)
OA:
Show Spoiler
D

The point here is to realize that we have to put the values of \(S1*S2\) and \(S1^2+S1^2\) into the equation of \((S1+S2)^2\) to get to the solution...
True, \(S1*S2 = 50, (S1 + S2)^2 = 200 => S1 + S2 = \sqrt{200}\) ---eq 1.
\(S1*S2 = 50\)---eq 2.
Solve, and get S1 and S2.
Show more

Hey buddy
Just one small correction :)
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OG12 Q. Maybe easy for many, but 'think it is in 700 level 06 Nov 2009, 05:53
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Thanks hpg2k :), a bit rusty, came back to forums after a week of some personal $hIT
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OG12 Q. Maybe easy for many, but 'think it is in 700 level 06 Nov 2009, 06:24
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Thanks hpg2k :), a bit rusty, came back to forums after a week of some personal $hIT
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Hope everything is fine. How is your preparation going on?
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OG12 Q. Maybe easy for many, but 'think it is in 700 level 06 Nov 2009, 20:59
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Hope this doesn't sound redundant - coming back after a long hiatus.

1 - Essentially b * h = 50 so b = 50/h . Using Pythagorus theorem , one unknown one equation - Sufficient
2 - Two legs are of the same length - Again one unknown, one equation - Sufficient.

Hence D
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OG12 Q. Maybe easy for many, but 'think it is in 700 level 06 Nov 2009, 21:14
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gmattokyo
hgp2k
IMO D.

Statement 1) \(S1*S2 = 50.\) and \(S1^2+S2^2 = 100.\) Solving both we can get the perimeter.
Statement 2) \(S1 = S2\) and \(S1^2+S2^2 = 100\). Solving both we can get the perimeter.

Thanks for the explanation. I see there aren't any more attempts :)
OA:
Show Spoiler
D

The point here is to realize that we have to put the values of \(S1*S2\) and \(S1^2+S1^2\) into the equation of \((S1+S2)^2\) to get to the solution...
True, \(S1*S2 = 50, (S1 + S2)^2 = 200 , S1 + S2 = \sqrt{200 }\)(corrected ) ---eq 1.
\(S1*S2 = 50\)---eq 2.
Solve, and get S1 and S2.
Show more


You cannot say that: \((S1 + S2)^2 = 200, .......... S1 + S2 = \sqrt{200 }\) ............(corrected) ---eq 1.


That relationship cannot be established from the statement given in the question.

The question states only h = 10 and it doesnot tell (b+h) = sqrt (200) = 10 sqrt (2). That is only possible if b = h, which is only provided in statement 2.

The original solution by "hgp2k" is flawless and complete.
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OG12 Q. Maybe easy for many, but 'think it is in 700 level 06 Nov 2009, 22:01
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GMAT TIGER


You cannot say that: \((S1 + S2)^2 = 200, .......... S1 + S2 = \sqrt{200 }\) ............(corrected) ---eq 1.

That relationship cannot be established from the statement given in the question.

The question states only h = 10 and it doesnot tell (b+h) = sqrt (200) = 10 sqrt (2). That is only possible if b = h, which is only provided in statement 2.

The original solution by "hgp2k" is flawless and complete.
Show more
I don't get this. h=hyp, a and b are the legs.
given: \(1/2 * a * b = 25\) and\(h=10.\)

\(h^2 = a^2 + b^2 = 100\) (pythagoras)
Now, we have \((a+b)^2 = a^2 + b^2 + 2*a*b\)
Here,\(a*b = 50, a^2 + b^2 = 100.\)

So \((a+b)^2 = 100 + 2*50 = 200\)
Would like to know the error :)
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OG12 Q. Maybe easy for many, but 'think it is in 700 level 06 Nov 2009, 22:22
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Economist
GMAT TIGER


You cannot say that: \((S1 + S2)^2 = 200, .......... S1 + S2 = \sqrt{200 }\) ............(corrected) ---eq 1.

That relationship cannot be established from the statement given in the question.

The question states only h = 10 and it doesnot tell (b+h) = sqrt (200) = 10 sqrt (2). That is only possible if b = h, which is only provided in statement 2.

The original solution by "hgp2k" is flawless and complete.

I don't get this. h=hyp, a and b are the legs.
given: \(1/2 * a * b = 25\) and\(h=10.\)

\(h^2 = a^2 + b^2 = 100\) (pythagoras)
Now, we have \((a+b)^2 = a^2 + b^2 + 2*a*b\)
Here,\(a*b = 50, a^2 + b^2 = 100.\)

So \((a+b)^2 = 100 + 2*50 = 200\)
Would like to know the error :)
Show more


Oh thats ok now....................
I thought you arrived to that solution considering statement 2.

However the original solution by "hgp2k" is flawless and complete.



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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