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M16, #15

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Re: M16, #15 [#permalink] New post 13 Mar 2013, 04:49
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 X-axis and apex on Y-axis. 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 X-axes 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! :-D


This certainly is very helpful, thank you & Bunuel!!
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Re: M16, #15 [#permalink] New post 13 Mar 2013, 19:52
[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] New post 26 Feb 2014, 02:21
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] New post 26 Feb 2014, 02:27
Expert's post
Vibhor89 wrote:
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 )


Nope. The correct answer is B, not D. Check here: if-vertices-of-a-triangle-have-coordinates-87344.html#p656628

Hope it helps.
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Re: M16, #15 [#permalink] New post 27 Apr 2014, 01:06
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
Re: M16, #15   [#permalink] 27 Apr 2014, 01:06
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