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M14#10

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M14#10 [#permalink] New post 18 Mar 2009, 12:41
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Vertices of a triangle have coordinates (-2, 2), (3, 2), (x, y). What is the area of the triangle?

1. |y - 2| = 1
2. angle at the vertex (x, y) equals 90 degrees

[Reveal] Spoiler: OA
A

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Re: DS (M14,Q10): gmatclub test question:vertices of a triangle. [#permalink] New post 18 Mar 2009, 14:10
sanjay_gmat wrote:
Vertices of a triangle have coordinates (-2, 2), (3, 2), (x, y). What is the area of the triangle?

1. |y - 2| = 1
2. angle at the vertex (x, y) equals 90 degrees


This is from test M14; question Q10.

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Re: DS : gmatclub test question : vertices of a triangle. [#permalink] New post 24 Mar 2009, 04:44
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My ans is A.

Stat 1:
if we make a graph we can see that (x,y) will be either on the line Y=1 or Y=3.
This is bacause |y-2| = 1 gives Y as 3 or 1.
in both cases the area will be 1/2(distance between -2,2 and 3,2 * 1)
height = 1 comes from the fact that both line Y=1 and Y=3 are at a distance of 1 unit from Y=2 which is the line
forming the base of the triangle.

Stat 2:
Since there is no information which will give us the co-ordinates of (x,y) so not suffecient.

please post the OA !
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Re: DS : gmatclub test question : vertices of a triangle. [#permalink] New post 24 Mar 2009, 05:31
My ans is also A.

from the question we get the base length, i.e 3+2=5.

from statement 1, we get the height:
y-2=1 if y>2
or, y=3

and y-2=-1 if y<2
or, y=1

for both the cases, height=1 ... as the base line passes from y=2.
so, area will remain same.

Area = 1/2*base*height.

from statement 2, we can't find out any specific value of x or y, because there may be many possible (x,y) value, which make a right angle at the vertex.
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Re: M14#10 [#permalink] New post 13 Oct 2009, 07:30
From stmt 2, can we not deduce that it is a 3:4:5 right triangle?
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Re: M14#10 [#permalink] New post 13 Oct 2009, 10:37
There is an infinite number of points that can make that angle right. S2 only tells us that the angle is 90 degrees.
crackgmat007 wrote:
From stmt 2, can we not deduce that it is a 3:4:5 right triangle?

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Re: M14#10 [#permalink] New post 18 Dec 2009, 06:56
If it is a right angle triangle and one of the side is 5, will it always be a 3-4-5 triplet?

dzyubam wrote:
There is an infinite number of points that can make that angle right. S2 only tells us that the angle is 90 degrees.
crackgmat007 wrote:
From stmt 2, can we not deduce that it is a 3:4:5 right triangle?
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Re: M14#10 [#permalink] New post 18 Dec 2009, 07:01
I got the answer. The side 5 need not be a hypotenuse. It may also be 5-12-13 triplet.

gouribhalerao wrote:
If it is a right angle triangle and one of the side is 5, will it always be a 3-4-5 triplet?

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Re: M14#10 [#permalink] New post 18 Dec 2009, 10:26
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dzyubam wrote:
There is an infinite number of points that can make that angle right. S2 only tells us that the angle is 90 degrees.
crackgmat007 wrote:
From stmt 2, can we not deduce that it is a 3:4:5 right triangle?


Yep, that's right. There are infinite number of triangles that will satisfy this condition. Imagine a circle drawn with the line segment joining the points (-2,2) and (3,2) as the diameter. The point (x,y) can lie anywhere on the circumference of this circle (except these two points, of course) and the angle will be a right angle.
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Re: M14#10 [#permalink] New post 22 Dec 2009, 22:59
I thought ans D was right as from stmt2, we can still find the area of the triangle.
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Re: M14#10 [#permalink] New post 23 Dec 2009, 04:41
Please elaborate how you can find the area. You shouldn't be able to as the height of this triangle can be very different (too small versus long enough). So, you can't be sure of the area of the triangle. Using S2 we only know the length of the hypotenuse (triangle's base), the height is still unknown.

Do you agree?
tania wrote:
I thought ans D was right as from stmt2, we can still find the area of the triangle.

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Re: M14#10 [#permalink] New post 28 Oct 2010, 06:26
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SudiptoGmat wrote:
This is a very good question. My answer is also A.



Such a nice trick.
I have got it !

2. threre is a right triangle, in which a^2+b^2=25, a, b could be not only 3 and 4 but also non integers and square roots of figures, which when squared give in sum 25.
for 2) just consider a^2= 13 and b^2=12 and a^2=1 and b^2=24, in each of cases there wuould be different areas.
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Re: M14#10 [#permalink] New post 04 Nov 2010, 11:21
/I would like to add one more observation.

The gratest area of a right triangle is obtained if two legs of triangle are equal so in our case the maximum area of triangle is when legs are equal each being \sqrt{12.5}.
the lowest area is when legs are equal to \sqrt{0.1} and \sqrt{24.9}
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Re: M14#10 [#permalink] New post 27 Dec 2010, 07:52
Pkit wrote:
/I would like to add one more observation.

The gratest area of a right triangle is obtained if two legs of triangle are equal so in our case the maximum area of triangle is when legs are equal each being \sqrt{12.5}.
the lowest area is when legs are equal to \sqrt{0.1} and \sqrt{24.9}



your observation is good but here it just asks about the Area of Triangle not the Max Area.

statement 1 is sufficient to find the area of triangle.
statement 2 might be sufficient but i don't find a solution to find the area.
so, for me. statement 2 is insufficient.
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Re: M14#10 [#permalink] New post 27 Dec 2010, 15:19
im kind of confused about statement 1, so we're saying for (x,1) or (x,3) no matter what x is, the height will always be length 1 for this triangle?
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Re: M14#10 [#permalink] New post 27 Dec 2010, 15:33
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sanjay_gmat wrote:
Vertices of a triangle have coordinates (-2, 2), (3, 2), (x, y). What is the area of the triangle?

1. |y - 2| = 1
2. angle at the vertex (x, y) equals 90 degrees

[Reveal] Spoiler: OA
A

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gtr022001 wrote:
im kind of confused about statement 1, so we're saying for (x,1) or (x,3) no matter what x is, the height will always be length 1 for this triangle?


Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).
Attachment:
render.php (1).png
render.php (1).png [ 17.42 KiB | Viewed 4887 times ]

(1) |y-2|=1 --> y=3 or y=1 --> vertex C could be anywhere on the blue line y=3 or anywhere on the red line y=1. But in ANY case the are of ABC will be the same --> area=\frac{1}{2}*base*height so base=AB=5 and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> area=\frac{1}{2}*base*height=\frac{5}{2}. Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it heps.
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Re: M14#10 [#permalink] New post 27 Dec 2010, 15:42
the graph really helps, +1!
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Re: M14#10 [#permalink] New post 28 Dec 2010, 05:16
thanks for the explanation. i went for B. :(
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Re: M14#10 [#permalink] New post 03 Jan 2012, 22:22
Thanks for the explanation Bunuel. Looks much easier after that explanation.
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Re: M14#10 [#permalink] New post 25 Dec 2013, 21:50
Ans is A (ist is sufficient since it provides the height of triangle as 1

if y-2= 1 then y=3 since y=2 for the base the height of the triangle is 1
and if y-2= -1 then y=1 and again, since y=2 for the base the height of the triangle is 1

now one can easily apply the formula for area= 1/2 *base *height
Re: M14#10   [#permalink] 25 Dec 2013, 21:50
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