What is the area of the region enclosed by x>=2, y>=3, : PS Archive
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What is the area of the region enclosed by x>=2, y>=3,

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What is the area of the region enclosed by x>=2, y>=3, [#permalink]

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19 May 2008, 05:45
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What is the area of the region enclosed by x>=2, y>=3, and y=-x+9?

Can you please explain me, what is the method to solve such kind of problems?
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Re: PS- Area of the Region [#permalink]

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19 May 2008, 07:11
On an X,Y grid, those values x>=2, y>=3 and Y=-x+9 are going to form a triangle. You would need to figure out where they intersect, the length of the base and height, multiple and divide by 2. Or maybe it will form a quadrilateral. You need to figure out the length of the sides, what shape it is and do the math to find the area. Plot the data on a grid to find the shape. I don't have time to go through to find an answer, but I believe this should help you out at least a little bit.

After going through the problem, it appears that it will indeed form a triangle. One line will be vertical at 3 on the x axis. The other will be horizontal at 2 on the y-axis. then the y = -x+9 is a diagonal line that crosses the Y axis at point +9 and then the slope is -1/1. So it goes down 1 for each 1 it goes right. This creates a triangle with coordinates of {3,2}, {3,6} and {7,2}. Because it's >= think of a shaded area above 2 and to the right of 3 and the hypotnuse is created by y=-x+9. So you have a shaded triangle. If the extension of the hypotnuse goes through +9 on the y-axis, that means that to cross y>=3 it will have to move 3 spaces to the right, and 3 down. So it if starts at +9, then goes down 3 for each 3 it goes right, the coordinates where it crosses x>=2 should be {3,6} (Y axis @ 9 - 3 =6). Then it continues down 1 and right 1 until it crosses x>=2. So if the coordinates of the point on the y-axis is {3,6}, then you take 6 - 2 to get 4 as the length of the left, vertical side of the triangle. We already know that the slope of the hypotnuse is -1/1. that creates two 45-degree angles where the line crosses x>=2 and y>=3 and these 2 lines create a 90 degree angle. We know this is a 1-1-sqrt{2} triangle. If one side is 4 then the area is 4*4*1/2 = 8.

Check me on this to see if I'm right. I'm 90% confident on this answer.

Value wrote:
What is the area of the region enclosed by x>=2, y>=3, and y=-x+9?

Can you please explain me, what is the method to solve such kind of problems?

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings CEO Joined: 17 Nov 2007 Posts: 3589 Concentration: Entrepreneurship, Other Schools: Chicago (Booth) - Class of 2011 GMAT 1: 750 Q50 V40 Followers: 530 Kudos [?]: 3455 [0], given: 360 Re: PS- Area of the Region [#permalink] Show Tags 19 May 2008, 07:35 1. x>=2, y>=3, Y=-x+9 means that triangle is right with two 45 angles. 2. Y=-x+9 intersects y=3 at x=6. the length of a leg is 6-2=4, therefore area=12*4^2=8 _________________ HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame SVP Joined: 30 Apr 2008 Posts: 1887 Location: Oklahoma City Schools: Hard Knocks Followers: 40 Kudos [?]: 564 [0], given: 32 Re: PS- Area of the Region [#permalink] Show Tags 19 May 2008, 07:43 Thanks. Now I'd like to edit my other post to read "I'm 100% sure of this answer!" lol walker wrote: 1. x>=2, y>=3, Y=-x+9 means that triangle is right with two 45 angles. 2. Y=-x+9 intersects y=3 at x=6. the length of a leg is 6-2=4, therefore area=12*4^2=8 _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: PS- Area of the Region [#permalink]

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19 May 2008, 07:57
walker wrote:
1. x>=2, y>=3, Y=-x+9 means that triangle is right with two 45 angles.
2. Y=-x+9 intersects y=3 at x=6. the length of a leg is 6-2=4, therefore area=12*4^2=8

Walker could you please explain this? I am lost.
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Re: PS- Area of the Region [#permalink]

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19 May 2008, 08:55
I guess it is better to show a drawing
Attachments

t64179.png [ 1.34 KiB | Viewed 729 times ]

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Re: PS- Area of the Region [#permalink]

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19 May 2008, 08:59
hey walker, did you recognize the approach to the problem right away or did you substitute numbers and realized it half way through?
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Re: PS- Area of the Region [#permalink]

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19 May 2008, 09:06
When I did this problem, I realized it was a triangle from the beginning. They give you 3 lines, one is vert, one is horizontal, and one is a slope. That has to mean 3-sided (i.e., triangle) or they would have given you a 4th line. It's just a matter of finding one side and doing the math.
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**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Director Joined: 23 Sep 2007 Posts: 789 Followers: 5 Kudos [?]: 184 [0], given: 0 Re: PS- Area of the Region [#permalink] Show Tags 19 May 2008, 09:09 jallenmorris wrote: When I did this problem, I realized it was a triangle from the beginning. They give you 3 lines, one is vert, one is horizontal, and one is a slope. That has to mean 3-sided (i.e., triangle) or they would have given you a 4th line. It's just a matter of finding one side and doing the math. Awesome advice. Recognizing the approach is 90% of the work, the rest is just simple math. SVP Joined: 30 Apr 2008 Posts: 1887 Location: Oklahoma City Schools: Hard Knocks Followers: 40 Kudos [?]: 564 [0], given: 32 Re: PS- Area of the Region [#permalink] Show Tags 19 May 2008, 09:16 It's actually a longer process to the 3 sided triangle than a quadrilateral. A quad would look like this: 4 >= x >= 2, 6>= y >= 3. That's a small rectangle. 2 x 3 = 6. Much easier than figuring out where the base of the triangle is, the slope, how long one side is. Some would then figure out the end point of the other side rather than seeing the slope makes two 45-degree angles so each base of the triangle is the same. If the test makers can make you spend even 1 minute longer on a problem than necessary, that's 1 minute less you have to work on later questions and they've won a small battle in the war that is the test (lame example, but I've watched a lot of war movies lately. ). Jarod _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: PS- Area of the Region [#permalink]

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19 May 2008, 09:30
gmatnub wrote:
hey walker, did you recognize the approach to the problem right away or did you substitute numbers and realized it half way through?

My thoughts I have posted above as they are in my mind. I recognized a right-triangle-problem right away.
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Re: PS- Area of the Region [#permalink]

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20 May 2008, 18:05
this is going to sound silly, but how do you know where y=-x+9 intersects the other lines ?
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Re: PS- Area of the Region [#permalink]

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20 May 2008, 18:55
It's not silly if you need to know. A line always has the same format with respect the x,y. It's y = (change in y) / (change in x) x + 9

Example: $$y = \frac{2}{5}x + 3$$

Since you are asking about where y = -x + 9 intersects the other lines you seem to already understand the basic structure of the format y = -x + 9 and what each part means.

the -x is the same as saying $$\frac{-1}{1}$$ so the change in y is -1 and the change in x is +1. So, from +9 on the Y-axes, we move -1 on the y-axis (or down) and +1 on the X-axis, or right. It doesn't matter if you do the X first, as long as you remember to look to the denominator for the change in X.

We know that if we move 1 down and 1 right, that's the same as a daigonal line. {Look at the picture in prior posts here}. In a $$1, 1, \sqrt{2}$$ triangle, we know that both sides are equal. Same thing with a $$\frac{-1}{1} or \frac{1}{1}$$ slope. If we start out at +9 on the Y-axis, when we have gone down 1, right 1 nine times, we will cross the X-axis, and that will cross at +9 on the X-axis. So that would create a big triangle with bases of 9 and hypotnuse of $$9sqrt{2}$$.

When x>=2 and y>=3, X>=2 creates a vertical line at +2 on the X-axis. So starting from the point of {0,9}, we subtract 1 from the y and add 1 to the x. We do this until we get to x = 2. Starting from the beginning {0,9}...{1,8}...{2,7}. We're at the line x>=2. Y>=3 is a horizontal line at +3 on the y-axis. The point where the 2 lines (x>=2 & y>=3) intersect is {2,3}. The difference in y coordinates gives the legth of one side. 7 - 3 = 4. Because we know anytime (change in y) = (Change in X) creates a 45 degree angle, we know that if one side is 4, the other side must be 4 aso. h * w * 1/2 = area of trianlge. 4 * 4 * 1/2 = 8.

pmenon wrote:
this is going to sound silly, but how do you know where y=-x+9 intersects the other lines ?

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Re: PS- Area of the Region   [#permalink] 20 May 2008, 18:55
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