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Re: M16-15  [#permalink]

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New post 09 Feb 2017, 00:03
nelliegu wrote:
Hi Bunuel,

Regarding Statement 2, if I consider that (0,A) is on the point (0,2) or (0,10) I could form a right triangle and I would get different answers for the question. Why wouldn't these points be an option?

Triangle with vertices (-1, 0), (4, 0), and (0, 10) is NOT right, it's acute.

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Re: M16-15 Alternate Solution.  [#permalink]

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New post 06 Mar 2017, 12:25
I think another way to approach this problem is by considering the following.

If you draw out the triangle you notice that (−1,0), (4,0) Lie on the X axis with a distance of 5. The third point lies at some number "a" the Y axis.

(1) a>3

Not Sufficient because if a = 4 then the area is 4*5*1/2 = 10 which is less than 15. Now if a = 10 then the area is 5*10*1/2 = 25 which is greater than 15.

(2) The Triangle is Right.

Sufficient. If you quickly plot the points, you will realize that in order for the triangle to be right the side that measures 5 on the X axis has to be the hypotenuse.

Using Pythagorean's Theorem substituting 5 for C you can see that a^2 + b^2 = 5^2 This tells us that the Base and Height must be less than 5. Therefore no combination of a*b*1/2 will give you a number greater than 15.

So the Answer Choice Must be B.
Joined: 15 Jan 2017
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Re: M16-15  [#permalink]

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New post 11 Mar 2017, 03:47
martinjudge91 wrote:
Hello Bunuel,

correct me if I'am wrong about my explanation with satement II.
As we know that there has to be a right angle at point D or C of your chart we can determine the length of the legs by applying the Pythagoras' theorem, which defines the lenght of the hypotheneuse that is 5 in this example. Thus the only combination of the both legs is a 3-4-5 trinagle which yields an area smaller than 15. Thus, statement 2 is sufficient on its own.

this is great solution! thank you!
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M16-15  [#permalink]

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New post 20 Sep 2017, 19:06
I interpreted question stem as: is 1/2*5*|A|>15?
If A>0 => 5A/2>15 or is A>6?
If A<0 => -5A/2>15 or is A<-6

(1) A<3. Doesn't rule out A=2,1,-1,-2.. etc, so insufficient

(2) Right triangle. Neither of the vertices on AB are at a perpendicular to vertex C (AC and BC aren't vertical), therefore the right angle must be at ACB. Thus, per Pythagoras, 5^2 = AC^2 + BC^2, meaning that both AC & BC < 5 and by extension, Area = 1/2*AC*BC < 15. Sufficient.
(p.s. the various methods on here for assessing statement (2) are all great)
M16-15 &nbs [#permalink] 20 Sep 2017, 19:06

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