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Re: M16, #15 [#permalink]
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13 Mar 2013, 05:49
Igor010 wrote: ExecMBA2010 wrote: Someone pl explain why A cannot be the answer? If we map the triangle on a coordinate plane we have the base on Xaxis and apex on Yaxis. Going by formula: Area = 1/2 (base x height) we have the base as 5 units and in order to have the area bigger than 15 we need the height to be >6 units. S1 says A<3, so height of triangle is <3 too hence area is less than 15. S1 SUFFICIENT  consider \(A=10\), what is the area then?
From S2 we know the triangle is right hence if the side on Xaxes is 5 units the other 2 sides will be 3 & 4 units. Again, area <15. S2 SUFFICIENT If above are correct then answer is D. Comments welcome. Hope that helps! This certainly is very helpful, thank you & Bunuel!!



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Re: M16, #15 [#permalink]
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13 Mar 2013, 20:52
[quote="ritula"]Vertices of a triangle have coordinates x(1,0), y(4,0) and z(0,A). Is the area of the triangle bigger than 15 ?
(1) A<3 (2) The triangle is right
/quote]
Area of triangle \(=\) (1/2) \(*5*A\)
I Since, we have given \(A\) \(<\) \(3\)
We can not conclude form this condition that the area of the triangle is bigger than 15 or lesser than 15...
II We have given, the triangle is right angle triangle,
So angle xzy will be right angle and using formula, \(xz^2 + yz^2 = xy^2\) we can find A.
\((A^2+1) + (A^2 +16)\) \(=\)\(25\)
=> \(A= 2\)
So, Area will be lesser than 15 and answer is B.



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Re: M16, #15 [#permalink]
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26 Feb 2014, 03:21
Hi GMATians
I suppose here D would be logical option My apporach Statement 1 A<3 now according to distance formula 1,0 and 4,0 distance is 5 therefore area 5A/2>15 , means A<6 hence SUFFICENT sTATEMENT 2RIGHT ANGLE if we take 3,4,5 triplet the area will be 3*4/2=6 and any subsequent relation gives area <15 Hence Sufficient Therefore D (Correct me if I am wrong )



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Re: M16, #15 [#permalink]
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26 Feb 2014, 03:27



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Re: M16, #15 [#permalink]
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27 Apr 2014, 02:06
From the coordinates of 3 vertices, the triangle's area = [4(1)]a/2 = 5a/2 (a: length of OA) 1) A<3: if A = 10 > area = 25>15 but if A = 2 > area = 5<15. Hence insufficient 2) The triangle is right: distances from (0,A) to (1,0) and to (4,0) are sqrt(a^2+1) and sqrt (a^2+16) respectively In a right triangle, we have: area = [sqrt(a^2+1) x sqrt (a^2+16)]/2 = 5a/2 => (a^2+1)(a^2+16) = 25a^2 => a^2 = 4 => a = 2 and area = 5<15 => sufficient
Hence choose B







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