Find all School-related info fast with the new School-Specific MBA Forum

It is currently 30 Aug 2014, 20:34

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Equation |x| + |y| = 5 encloses a certain region on the

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Director
Director
avatar
Joined: 12 Jun 2006
Posts: 539
Followers: 1

Kudos [?]: 13 [0], given: 1

GMAT Tests User
Equation |x| + |y| = 5 encloses a certain region on the [#permalink] New post 17 Feb 2007, 14:04
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

50% (01:03) correct 50% (01:09) wrong based on 3 sessions
Equation |x| + |y| = 5 encloses a certain region on the coordinate plane. What is the area of this region?

5
10
25
50
100
SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1816
Followers: 8

Kudos [?]: 88 [0], given: 0

GMAT Tests User
 [#permalink] New post 17 Feb 2007, 15:07
(D) for me :)

It's a square with vertices at (5,0), (0,-5), (-5,0) and (0,5).

So the area = 4 * triangles (one of them : (0,0), (5,0), (0,5)) = 5*5/2 * 4 = 50 :)
Director
Director
avatar
Joined: 12 Jun 2006
Posts: 539
Followers: 1

Kudos [?]: 13 [0], given: 1

GMAT Tests User
 [#permalink] New post 17 Feb 2007, 15:13
Fig wrote:
(D) for me :)

It's a square with vertices at (5,0), (0,-5), (-5,0) and (0,5).

So the area = 4 * triangles (one of them : (0,0), (5,0), (0,5)) = 5*5/2 * 4 = 50 :)
Fig, I don't follow. Please explain the bolded portion.
SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1816
Followers: 8

Kudos [?]: 88 [0], given: 0

GMAT Tests User
 [#permalink] New post 17 Feb 2007, 15:20
ggarr wrote:
Fig wrote:
(D) for me :)

It's a square with vertices at (5,0), (0,-5), (-5,0) and (0,5).

So the area = 4 * triangles (one of them : (0,0), (5,0), (0,5)) = 5*5/2 * 4 = 50 :)
Fig, I don't follow. Please explain the bolded portion.


To know the area, I build up it by using the area of 4 similar triangles :).... Then, I calculate the area of one of them. This area is also the area of the 3 other triangles. :)

Have a look on the Fig 1, the triangle in flashy green is the one described in my calculation of the area. :)
Attachments

Fig1_1 Triangle_The square.gif
Fig1_1 Triangle_The square.gif [ 4.85 KiB | Viewed 1114 times ]

Manager
Manager
avatar
Joined: 12 Feb 2006
Posts: 117
Followers: 1

Kudos [?]: 8 [0], given: 0

GMAT Tests User
 [#permalink] New post 18 Feb 2007, 00:36
I another way you can go if you want to stick to squares is 5sqrt(2) x 5sqrt(2).
Director
Director
avatar
Joined: 12 Jun 2006
Posts: 539
Followers: 1

Kudos [?]: 13 [0], given: 1

GMAT Tests User
 [#permalink] New post 18 Feb 2007, 00:58
ok Fig,

I got ya'. but could we do the same with 1,4; -1,-4; -1,4; 1,-4 or 3,2;
-3,-2; -3,2; 3,-2. I tried them cursorily. they didn't seem to work.
SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1816
Followers: 8

Kudos [?]: 88 [0], given: 0

GMAT Tests User
 [#permalink] New post 18 Feb 2007, 01:25
ggarr wrote:
ok Fig,

I got ya'. but could we do the same with 1,4; -1,-4; -1,4; 1,-4 or 3,2;
-3,-2; -3,2; 3,-2. I tried them cursorily. they didn't seem to work.


The 2 shapes u design like this are rectangles. :)... Notice the x and y coordonates of the point u have built up : their absolute value is not equal (|1| != |4|) :).... If they were equal, then we have a square :)

Notice also that sides of such rectangle are parallel to the X and Y axes :)

If gives a and b the absolute values of the x and y coordonaites of each vertex from a same rectangle, then we have:

The aera of such a rectangle = (2*a) * (2*b)

In your cases,
o Aera 1 = 2*1 * 2*4 = 16
o Aera 2 = 3*1 * 2*2 = 12
Director
Director
avatar
Joined: 12 Jun 2006
Posts: 539
Followers: 1

Kudos [?]: 13 [0], given: 1

GMAT Tests User
 [#permalink] New post 18 Feb 2007, 02:52
|x| + |y| = 5
Quote:
The 2 shapes u design like this are rectangles. ... Notice the x and y coordonates of the point u have built up : their absolute value is not equal (|1| != |4|
In the stem, we need the abs. val. of x and y to equal 5. Why would we need the abs. val. of x and y (|1| and |4| in this case) to be equal. Don't they just need to sum to 5?
SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1816
Followers: 8

Kudos [?]: 88 [0], given: 0

GMAT Tests User
 [#permalink] New post 18 Feb 2007, 03:06
ggarr wrote:
|x| + |y| = 5
Quote:
The 2 shapes u design like this are rectangles. ... Notice the x and y coordonates of the point u have built up : their absolute value is not equal (|1| != |4|
In the stem, we need the abs. val. of x and y to equal 5. Why would we need the abs. val. of x and y (|1| and |4| in this case) to be equal. Don't they just need to sum to 5?


No :).... It is 2 different things.... Your example are not equivalent to the problem :)

|x| + |y| = b (b>0) will always create a square such that it is shown on my Fig 1.

Your examples create rectangles with sides that are parellel to the X and Y axes :).... Draw them, u will see what I mean :)

An equivalent system of equations for your examples is :
o |x| = c
o |y| = d
Manager
Manager
User avatar
Joined: 22 Jul 2009
Posts: 192
Followers: 4

Kudos [?]: 188 [0], given: 18

GMAT Tests User
Re: Coordinate Geometry problem 2 [#permalink] New post 21 Aug 2009, 17:26
Four different approaches to reach to a quick answer:

1) Diagonal of the square. d=10 -> area = 10^2 / 2 = 50

2) Phytagoras, 45:45:90. Hyp=side=5*sqrt(2) -> area = (5*sqrt(2))^2 = 50

3) Phytagoras, calculation. Hyp=side=sqrt(5^2+5^2)=sqrt(50) -> area = (sqrt(50))^2 = 50

4) 4 triangles. areaT = (b*h)/2 = (5*5)/2 = 25/2 -> area = areaT*4 = 25/2 * 4 = 50
Re: Coordinate Geometry problem 2   [#permalink] 21 Aug 2009, 17:26
    Similar topics Author Replies Last post
Similar
Topics:
If equation |x|+|y|= 5 encloses a certain region on the dzelkas 4 05 Jun 2008, 16:45
1 If equation |x| + |y| = 5 encloses a certain region on the suntaurian 8 08 Mar 2008, 12:18
1 Experts publish their posts in the topic If equation |x| + |y| = 5 encloses a certain region on the dominion 7 17 Jan 2008, 23:38
Experts publish their posts in the topic If the equation |x| + |y| = 5 encloses a certain region on GK_Gmat 4 20 Nov 2007, 21:09
if the equation |x|+ |y| = 5 encloses a certain region on bmwhype2 2 09 Nov 2007, 09:50
Display posts from previous: Sort by

Equation |x| + |y| = 5 encloses a certain region on the

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.