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M16-15

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If vertices of a triangle have coordinates \((-1, 0)\), \((4, 0)\), and \((0, A)\), is the area of the triangle greater than 15 ?


(1) \(A \lt 3\)

(2) The triangle is right
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Statement (1) by itself is insufficient. If \(A\) is close to 0, the area of the triangle is small, but if \(A\) is a large negative (for example, \(A = -1000\)), the area of the triangle is large.

Statement (2) by itself is sufficient. Only the angle at vertex \((0, A)\) can be right. As \(|A|\) increases, the value of the angle gets smaller. At some point, the angle becomes equal to 90 degrees and we can snapshot the triangle at this moment. After that, we can find the area of the triangle and answer the question. Please see the image below.

Image


Answer: B
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Re: M16-15 [#permalink]

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New post 12 Feb 2015, 08:31
Query regarding Stmnt 1

How can the Area be Negative? Max value of A can be 2.99 or 0.01. Either case, isnt the statement sufficient?

I am still unable to understand explanation for Stmnt 2

Pls clarify.
Thanks

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New post 12 Feb 2015, 08:34
buddyisraelgmat wrote:
Query regarding Stmnt 1

How can the Area be Negative? Max value of A can be 2.99 or 0.01. Either case, isnt the statement sufficient?

I am still unable to understand explanation for Stmnt 2

Pls clarify.
Thanks


A in the firs statement is the y-coordinate of the third vertex (0, A), so it can be negative.

As for (2) please elaborate what is unclear there.

Thank you.
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Re: M16-15 [#permalink]

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New post 12 Feb 2015, 09:00
Bunuel wrote:
buddyisraelgmat wrote:
Query regarding Stmnt 1

How can the Area be Negative? Max value of A can be 2.99 or 0.01. Either case, isnt the statement sufficient?

I am still unable to understand explanation for Stmnt 2

Pls clarify.
Thanks


A in the firs statement is the y-coordinate of the third vertex (0, A), so it can be negative.

As for (2) please elaborate what is unclear there.

Thank you.



Ok I understood my error in Stmnt 1

Took me quite sometime to re-read and understand Stmnt 2 - So what we're saying is that at some point, vertex (0,A) will be 90 degrees and we'll be able to answer the question in Yes / No - However we do not have a definite answer - Is my understanding correct?

Just pushing it further, if the 3rd vertex was say (-1, A) , then Stmnt 2 would be Insufficient - Is this understanding correct?

Thanks again

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Re: M16-15 [#permalink]

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New post 12 Feb 2015, 09:03
buddyisraelgmat wrote:
Bunuel wrote:
buddyisraelgmat wrote:
Query regarding Stmnt 1

How can the Area be Negative? Max value of A can be 2.99 or 0.01. Either case, isnt the statement sufficient?

I am still unable to understand explanation for Stmnt 2

Pls clarify.
Thanks


A in the firs statement is the y-coordinate of the third vertex (0, A), so it can be negative.

As for (2) please elaborate what is unclear there.

Thank you.



Ok I understood my error in Stmnt 1

Took me quite sometime to re-read and understand Stmnt 2 - So what we're saying is that at some point, vertex (0,A) will be 90 degrees and we'll be able to answer the question in Yes / No - However we do not have a definite answer - Is my understanding correct?

Just pushing it further, if the 3rd vertex was say (-1, A) , then Stmnt 2 would be Insufficient - Is this understanding correct?

Thanks again


I think the following post might help: if-vertices-of-a-triangle-have-coordinates-87344.html#p656628
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Re: M16-15 [#permalink]

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Bunuel - the way I resolved statement 2 is if angle ADB is the right angles then the product of the slopes of AD and DB will be -1.
slope of AD X slope of DB = -1
\(\frac{A}{1}\) X \(\frac{A}{-4}\) = -1

or \(A^2 = 4\)
or \(|A| = 2\)
or A= +2 or A= -2

from there we can calculate the are of the triangle to be \(\frac{1}{2}\)X5X2 = 5 - sufficient.

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Re: M16-15 [#permalink]

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New post 10 Aug 2016, 18:19
I'm confused how we know that Only the angle at vertex (0,A)(0,A) can be right.

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New post 19 Aug 2016, 07:31
@Bunnel,
For the second part if we find out the minimum height of the triangle required to make the area 15.
Then knowing the second part we can find out that the base if 5 and it also hypotenuse of the triangle ..
Minimum height required is 6. if the hypotenuse is 5 that means the altitude cannot be greater than 5 . And we can not have a triangle with area greater than 15.

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rishabh16 wrote:
@Bunnel,
For the second part if we find out the minimum height of the triangle required to make the area 15.
Then knowing the second part we can find out that the base if 5 and it also hypotenuse of the triangle ..
Minimum height required is 6. if the hypotenuse is 5 that means the altitude cannot be greater than 5 . And we can not have a triangle with area greater than 15.


The question asks: is the area of the triangle greater than 15? A NO answer is still a definite answer, so it's still sufficient.
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Re: M16-15 [#permalink]

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New post 24 Aug 2016, 01:43
my logic was
for right triangle coordinates of (0,A) can be (0,0) or any point on Y Axis
so can't determine whether area is greater or lesser?
correct option should E

please correct me, if I'm wrong?

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Re: M16-15 [#permalink]

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New post 24 Aug 2016, 04:52
r19 wrote:
my logic was
for right triangle coordinates of (0,A) can be (0,0) or any point on Y Axis
so can't determine whether area is greater or lesser?
correct option should E

please correct me, if I'm wrong?


The OA is B, so you must be wrong somewhere... If you re-read the solution you will find out that there is only ONE value of A for which the triangle is right. It cannot be (0, 0) be cause in this case the triangle is not right, in fact in this case you won't even have a triangle, you'l have three points on a line.
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Re: M16-15 [#permalink]

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New post 06 Sep 2016, 22:05
i think the answer should be D,

STATEMENT (2) The triangle is right
GIVEN vertices of a triangle have coordinates (−1,0)(−1,0), (4,0)(4,0), and (0,A)(0,A)

AB = 5 , POINT D will be on the y axis all the time (either in the positive y-direction or in the negative y-direction so point C as per the figure given in the explainantion)

point A and B on the both side of point D or C so right angle will be on D or C

AD^2= AO^2 +OD^2 = 1^2 ?(SQUARE) +Y^2

SIMILARLY

BD^2= BO^2 +OD^2 ( O IS THE origin (0,0) ) = 16+ y^2

for right angle triangle ABD

25= Y^2 +1 +Y^2+16
by solving it Y is either 2 or -2 SO POINT D (0,2 ) OR POINT C (0,-2)
either case area OF TRIANGLE ABD= BASE *HEIGHT/2= 5*2/2 = 5 is less then 15


so please correct the answer

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Re: M16-15 [#permalink]

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New post 18 Oct 2016, 23:56
Bunuel - Why is statement 1 insufficient?

By determinant method, the area of triangle comes out to be 5A. Given A<3, Clearly area <15.

Am I missing something?

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New post 19 Oct 2016, 23:48
If A=-100, implies that area is 500, right? Because A=|D|.

Am I missing something?

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New post 21 Oct 2016, 07:53
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.

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Re: M16-15 [#permalink]

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New post 08 Feb 2017, 14:07
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?
Thanks!!

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Re: M16-15   [#permalink] 08 Feb 2017, 14:07

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