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m08 q#6

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m08 q#6 [#permalink] New post 25 Nov 2008, 10:19
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The vertices of a triangle have coordinates (x, 1) , (5, 1) , and (5, y) where x < 5 and y > 1 . What is the area of the triangle?

1. x = y
2. Angle at the vertex (x, 1) = angle at the vertex (5, y)

[Reveal] Spoiler: OA
C

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I disagree with OA..

from stmnt 1 and 2, we can tell the triangle is issoc. but the area can 8 or 2...
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Re: m08 q#6 [#permalink] New post 25 Nov 2008, 18:52
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From the stem, we know for sure that it's a right triangle, which doesn't mean it is not isosceles, but we aren't for sure that it is isosceles.

From Statement 1, it could be isoceles, but not necessarily. I'll draw a picture to explain...headed to Photoshop.

In statement 1, the same color X's go together. Because x = y. As you can see, the area of the triangle if the BLUE X is used is 1.5. if the Purple/Pink X is used, the area is 2, and if the Orange X is used, the area is back to 1.5 S1 cannot be sufficient because we've found different values of area when we also satisfy the rule presented in S1

In Statement 2, we do (as freshina12 says) have an isosceles triangle. This can only be true when we use the pink X's as this will create two 45 degree angles. The area here is 2, but there are infinite number of other options. For the X along y=1, x can be anything upto the value of 5. As for the line x=5, the y value can be anything as long as it is > 1. So as long as we have 2-45 degree angles, S2 will be satisfied, but we will get different values for the area of the triangle formed.

Only when we consider the statements together do we get a single value, which is seen at the PINK X in the picture below. I'm not sure what you mean the area could be 8. I don't see that. I do see how it could be 2 though.

I say answer is C.

Attachment:
TriangleGraph.jpg
TriangleGraph.jpg [ 40.19 KiB | Viewed 6559 times ]


fresinha12 wrote:
The vertices of a triangle have coordinates (x, 1) , (5, 1) , and (5, y) where x < 5 and y > 1 . What is the area of the triangle?

1. x = y
2. Angle at the vertex (x, 1) = angle at the vertex (5, y)

I disagree with OA..

from stmnt 1 and 2, we can tell the triangle is issoc. but the area can 8 or 2...

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Re: m08 q#6 [#permalink] New post 25 Nov 2008, 19:30
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fresinha12 wrote:
The vertices of a triangle have coordinates (x, 1) , (5, 1) , and (5, y) where x < 5 and y > 1 . What is the area of the triangle?

1. x = y
2. Angle at the vertex (x, 1) = angle at the vertex (5, y)

I disagree with OA..

from stmnt 1 and 2, we can tell the triangle is issoc. but the area can 8 or 2...


The line from points (x,1) and (5,1) is parallel to the x axis and the line from points (5,1) and (5,y) is parallel to y axis making it perpendicular to the former line. so these points make a right angle triangle.

1: If x = y, x and could have any value >1 but <5. so nsf.
2: If angles at (x,1) and (5,y) are equal, then 5-x = y-1.
so x+y = 6
x and y could be any value given the constraints are met. nsf.

1&2: 5-x = y-1.
x+y = 6
2x = 6
x = 3
so area = (1/2) (5-x) (y-1) = 1/2 (2x2) = 2

seems C if I have not missed anything as I was trying to get done by each statement but not.........
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Re: m08 q#6 [#permalink] New post 26 Nov 2008, 12:22
JM..thanks I realized my mistake..and its a stupid one..i mis-read stmnt 1 to say (5-x)=(y-1)..

which is true but we have stmnt 1 saying point x is equal to point y..

I can draw an isoceles triangle when x-5=4 and y-1=4

jallenmorris wrote:
From the stem, we know for sure that it's a right triangle, which doesn't mean it is not isosceles, but we aren't for sure that it is isosceles.

From Statement 1, it could be isoceles, but not necessarily. I'll draw a picture to explain...headed to Photoshop.

In statement 1, the same color X's go together. Because x = y. As you can see, the area of the triangle if the BLUE X is used is 1.5. if the Purple/Pink X is used, the area is 2, and if the Orange X is used, the area is back to 1.5 S1 cannot be sufficient because we've found different values of area when we also satisfy the rule presented in S1

In Statement 2, we do (as freshina12 says) have an isosceles triangle. This can only be true when we use the pink X's as this will create two 45 degree angles. The area here is 2, but there are infinite number of other options. For the X along y=1, x can be anything upto the value of 5. As for the line x=5, the y value can be anything as long as it is > 1. So as long as we have 2-45 degree angles, S2 will be satisfied, but we will get different values for the area of the triangle formed.

Only when we consider the statements together do we get a single value, which is seen at the PINK X in the picture below. I'm not sure what you mean the area could be 8. I don't see that. I do see how it could be 2 though.

I say answer is C.

Attachment:
TriangleGraph.jpg


fresinha12 wrote:
The vertices of a triangle have coordinates (x, 1) , (5, 1) , and (5, y) where x < 5 and y > 1 . What is the area of the triangle?

1. x = y
2. Angle at the vertex (x, 1) = angle at the vertex (5, y)

I disagree with OA..

from stmnt 1 and 2, we can tell the triangle is issoc. but the area can 8 or 2...
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Re: m08 q#6 [#permalink] New post 19 Feb 2010, 06:32
GMAT TIGER wrote:
2: If angles at (x,1) and (5,y) are equal, then 5-x = y-1.
so x+y = 6
x and y could be any value given the constraints are met. nsf.


GT.

Great explaination. I almost have my head wrapped around it. Could I get you to explain how the equation in the second stem was used. Whats the rule there? Can you elaborate?

Thanks.

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Re: m08 q#6 [#permalink] New post 19 Feb 2010, 06:52
Since the angle at vertex (x,1)=(5,y)...this infers that the sides opposite to those angles are equal...so it is a Right Isosceles Triangle and hence we get 5-x=y-1 -> x+y=6

Since x=y, the value of x & y is 3 and the length of each side is 2.

So Area of triangle = 1/2 * 2 * 2 = 2 sq. units
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Re: m08 q#6 [#permalink] New post 19 Feb 2010, 11:56
A(3;1), C(5;1), B(5;3)
C=90°

S = ( 2∙2)/2
S = 2
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Re: m08 q#6 [#permalink] New post 19 Feb 2010, 21:32
stmt1 : x=y

stmt2 : angle (x,1) = angle (y,1)
this means sides opposite to that angle are equal

to identifty the distance b/w the sides (y2-y1)^2+(x2-x1)^2

so (5-x)^2 +(1-1)^2=(5-5)^2+(y-1)^2

(5-x)^2=(y-1)^2 taking square root ob both sides

5-x=y-1 so x+y=6

not sufficient

by combining both 1 & 2 we get

x=3 and y=3

so ans is C
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Re: m08 q#6 [#permalink] New post 24 Feb 2011, 20:15
[quote="jallenmorris"]From the stem, we know for sure that it's a right triangle, which doesn't mean it is not isosceles, but we aren't for sure that it is isosceles.

From Statement 1, it could be isoceles, but not necessarily. I'll draw a picture to explain...headed to Photoshop.

In statement 1, the same color X's go together. Because x = y. As you can see, the area of the triangle if the BLUE X is used is 1.5. if the Purple/Pink X is used, the

Good explanation. +1.
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Re: m08 q#6 [#permalink] New post 22 Apr 2012, 01:07
Expert's post
If the vertices of a triangle have coordinates (x,1), (5,1), and (5,y) where x<5 and y>1, what is the area of the triangle?

Look at the diagram below:
Image
Notice that vertex (x,1) will be somewhere on the green line segment and the vertex (5,y) will be somewhere on the blue line segment. So, in any case our triangle will be right angled, with a right angle at vertex (5, 1). Next, the length of the leg on the green line segment will be 5-x and the length of the leg on the blue line segment will by y-1. So, the area of the triangle will be: area=\frac{1}{2}*(5-x)*(y-1)

(1) x=y --> since x<5 and y>1 then both x and y are in the range (1,5): 1<(x=y)<5. If we substitute y with x we'll get: area=\frac{1}{2}*(5-x)*(y-1)=\frac{1}{2}*(5-x)*(x-1), different values of x give different values for the area (even knowing that 1<x<5). Not sufficient.

(2) Angle at the vertex (x,1) is equal to angle at the vertex (5,y) --> we have an isosceles right triangle: 5-x=y-1. Again if we substitute y-1 with 5-x we'll get: area=\frac{1}{2}*(5-x)*(y-1)=\frac{1}{2}*(5-x)*(5-x), different values of x give different values for the area. Not sufficient.

(1)+(2) x=y and 5-x=y-1 --> solve for x: x=y=3 --> area=\frac{1}{2}*(5-3)*(3-1)=2. Sufficient.

Answer: C.
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Re: m08 q#6 [#permalink] New post 26 Apr 2012, 09:00
Bunuel wrote:
If the vertices of a triangle have coordinates (x,1), (5,1), and (5,y) where x<5 and y>1, what is the area of the triangle?

Look at the diagram below:
Image
Notice that vertex (x,1) will be somewhere on the green line segment and the vertex (5,y) will be somewhere on the blue line segment. So, in any case our triangle will be right angled, with a right angle at vertex (5, 1). Next, the length of the leg on the green line segment will be 5-x and the length of the leg on the blue line segment will by y-1. So, the area of the triangle will be: area=\frac{1}{2}*(5-x)*(y-1)

(1) x=y --> since x<5 and y>1 then both x and y are in the range (1,5): 1<(x=y)<5. If we substitute y with x we'll get: area=\frac{1}{2}*(5-x)*(y-1)=\frac{1}{2}*(5-x)*(x-1), different values of x give different values for the area (even knowing that 1<x<5). Not sufficient.

(2) Angle at the vertex (x,1) is equal to angle at the vertex (5,y) --> we have an isosceles right triangle: 5-x=y-1. Again if we substitute y-1 with 5-x we'll get: area=\frac{1}{2}*(5-x)*(y-1)=\frac{1}{2}*(5-x)*(5-x), different values of x give different values for the area. Not sufficient.

(1)+(2) x=y and 5-x=y-1 --> solve for x: x=y=3 --> area=\frac{1}{2}*(5-3)*(3-1)=2. Sufficient.

Answer: C.


i used substitution method, but Bunuel...ur methods are really awesome.. it has been a great week in learning so many stuffs.. but i want to work more on Probability, Standard deviation, fractions and surds kinds of problem, i want to see more of them... thanks SIR
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Re: m08 q#6 [#permalink] New post 21 Apr 2014, 08:22
This is a square triangle and the area is (5-x)(y-1)/2.
1) x=y: insufficient
2) The two equal angles make it a isosceles triangle, but still insufficient
Combine 2 stats -> 5-x=y-1 and x=y -> x=y=3 -> sufficient

Choose C
Re: m08 q#6   [#permalink] 21 Apr 2014, 08:22
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