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Hayabusa
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We have a right triangle. The hypotenuse is 16. The sum of 24 Jul 2007, 13:01
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We have a right triangle. The hypotenuse is 16. The sum of the other two sides is 18. What is the area?



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Hayabusa
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We have a right triangle. The hypotenuse is 16. The sum of 24 Jul 2007, 13:11
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cruiser
34
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You are half way there! Show some work. Not sure how you got 34.
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ashkrs
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We have a right triangle. The hypotenuse is 16. The sum of 24 Jul 2007, 13:36
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Hayabusa
We have a right triangle. The hypotenuse is 16. The sum of the other two sides is 18. What is the area?
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Well
Let x be one of the side .So another side is 18-x
So
x^2 + (18-x)^2 = 16^2
boils down to
a^2 - 18a + 34 = 0
or a*(18-a) = 34

Now we know area of traingle 1/2 * a * (18-a )
so dividing the above equation by 2 gives area as 17 .
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Hayabusa
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We have a right triangle. The hypotenuse is 16. The sum of 24 Jul 2007, 13:41
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ashkrs
Hayabusa
We have a right triangle. The hypotenuse is 16. The sum of the other two sides is 18. What is the area?

Well
Let x be one of the side .So another side is 18-x
So
x^2 + (18-x)^2 = 16^2
boils down to
a^2 - 18a + 34 = 0
or a*(18-a) = 34

Now we know area of traingle 1/2 * a * (18-a )
so dividing the above equation by 2 gives area as 17 .
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Awesome work! Very efficient.
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AugiTh
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We have a right triangle. The hypotenuse is 16. The sum of 24 Jul 2007, 13:46
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Hayabusa
We have a right triangle. The hypotenuse is 16. The sum of the other two sides is 18. What is the area?
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given
x^2 + y^2 = 16^2
and x+y = 18

Area = 1/2 * base* height
i.e. 1/2*x*y

now x+y=18
(x+y)^2 = 18^2
x^2+y^2+2xy=18^2
16^2+2xy=18^2
2xy=18^2-16^2
2xy=68
xy=34

Area = 1/2xy=17!!!
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Hayabusa
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We have a right triangle. The hypotenuse is 16. The sum of 24 Jul 2007, 13:56
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Great job guys. I realize that the problem itself is not terribly hard.
I followed the same logic as ashkrs and got to the quadratic equation.
The mistake that I made was that I tried to solve it. I still got the answer, but solving quadratic equations is not a good gmat strategy. I failed to notice that if I factor out b, I get a.b=34. All I need to do is divide it by 2.

Keep up the good work.
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Robin in NC
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We have a right triangle. The hypotenuse is 16. The sum of 24 Jul 2007, 16:51
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Even after reading and understanding the explanations for this problem, it took me almost 5 minutes to solve it from start to finish, because of the arithmetic (finding the values of 18^2 and 16^2). I would like to see what all the answer choices were, to know if there was a way to ballpark this one.
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We have a right triangle. The hypotenuse is 16. The sum of 24 Jul 2007, 18:41
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Robin in NC
Even after reading and understanding the explanations for this problem, it took me almost 5 minutes to solve it from start to finish, because of the arithmetic (finding the values of 18^2 and 16^2). I would like to see what all the answer choices were, to know if there was a way to ballpark this one.
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Actually you will not need to calculate 18^2 or 16^2 . Because here we are dealing with the difference . ie 18^2 - 16^2 which can be calculated easily as (18-16)(18+16).
But yeah I agree with your point if answer choices are there we can probably ballpark .



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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 Quantitative Questions 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!