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This topic is locked. If you want to discuss this question please repost it in the respective forum. 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 Source: GMAT Club Tests  hardest GMAT questions



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Re: DS (M14,Q10): gmatclub test question:vertices of a triangle. [#permalink]
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18 Mar 2009, 15:10
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. thanks



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Re: DS : gmatclub test question : vertices of a triangle. [#permalink]
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24 Mar 2009, 05:44
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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 y2 = 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 coordinates of (x,y) so not suffecient.
please post the OA !



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Re: DS : gmatclub test question : vertices of a triangle. [#permalink]
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24 Mar 2009, 06:31
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: y2=1 if y>2 or, y=3
and y2=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]
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13 Oct 2009, 08:30
From stmt 2, can we not deduce that it is a 3:4:5 right triangle?



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Re: M14#10 [#permalink]
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13 Oct 2009, 11:37
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]
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18 Dec 2009, 07:56
If it is a right angle triangle and one of the side is 5, will it always be a 345 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]
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18 Dec 2009, 08:01
I got the answer. The side 5 need not be a hypotenuse. It may also be 51213 triplet. gouribhalerao wrote: If it is a right angle triangle and one of the side is 5, will it always be a 345 triplet?



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Re: M14#10 [#permalink]
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18 Dec 2009, 11:26
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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]
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22 Dec 2009, 23:59
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]
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23 Dec 2009, 05:41
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]
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28 Oct 2010, 07:26
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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]
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04 Nov 2010, 12:21
/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]
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27 Dec 2010, 08:52
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]
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27 Dec 2010, 16:19
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]
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27 Dec 2010, 16:33
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 Source: GMAT Club Tests  hardest GMAT questions 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 [ 17.42 KiB  Viewed 5926 times ]
(1) y2=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]
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27 Dec 2010, 16:42
the graph really helps, +1!



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Re: M14#10 [#permalink]
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28 Dec 2010, 06:16
thanks for the explanation. i went for B.



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Re: M14#10 [#permalink]
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03 Jan 2012, 23:22
Thanks for the explanation Bunuel. Looks much easier after that explanation.



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Re: M14#10 [#permalink]
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25 Dec 2013, 22:50
Ans is A (ist is sufficient since it provides the height of triangle as 1
if y2= 1 then y=3 since y=2 for the base the height of the triangle is 1 and if y2= 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










