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Area of a region

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Area of a region

by hk » Thu May 21, 2009 6:32 pm
If the equation |x/2| + |y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region?

A. 20
B. 50
C. 100
D. 200
E. 400

OA after some discussions.
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by dumb.doofus » Thu May 21, 2009 8:48 pm
I think 200 is the answer..

You just need to find the x and y intercepts.. those come out to be (10,0), (-10,0), (0,10) and (0,-10)..

so its basically a square with side as 10root(2).

So area is 200
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by scoobydooby » Thu May 21, 2009 9:00 pm
am getting 50

|x/2| + |y/2| = 5 is the equation of a line, with the x and y intercepts being (10, 10)
the x and the y intercepts enclose a triangle
so the area=1/2*10*10=50

hence, B
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by aj5105 » Thu May 21, 2009 9:35 pm
Can you please explain how to find the x and y intercepts?
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by agoyal2 » Thu May 21, 2009 10:14 pm
Agree with Doofus.

|X| + |y| = 10

So, when x = 0
|Y|=10 Which implies Y = +10, -10

Similarly for X. Gives you the fours coordinates of the square.
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by PAB2706 » Fri May 22, 2009 1:46 am
yeh 200 is wht i am getting..


x co-ordinate +10 and -10

y co-ordinate +10 and -10

forms a square.... ie four triangular regions each with area 50

Thus total area =200sq units.
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by hk » Fri May 22, 2009 8:13 am
OA is D

|x/2| + |y/2| = 5 is same as |x|+|y|=10.

Substitute values for x, y and you'll find that the Area is a square with side 10sqrt(2) = sqrt(200)

So area = sqrt(200) * sqrt(200) = 200
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by mike22629 » Fri May 22, 2009 10:55 am
The best way to do these problems are to find out what each variable is equal to when the other equals 0.

So

When y = 0

x = 10 or -10

When x = 0

y = 10 or -10

So the four points are
(10,0) (-10,0) (0,10) (0,-10)

Answer two hundred.
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by maihuna » Sat May 23, 2009 5:45 am
Plotting on xy axis with +y value one may see an triangle with x ranging from -10 to 10 with height of 10, similarly with -y values another triangle with -y axis with similar shape as above...

the resulting shape will not be an square, in my opininon, but for a quadrilateral with base 20 height 10 or two trinagles with base 10 height 10 area will come s 20*10 =200
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by DarkKnight » Sat May 23, 2009 8:28 am
Folks,

Why isn't anyone considering following pairs:

(5,5), (-5,5), (5,-5) and (-5,-5) in which case we will get a square with side length 10 and area 100.

thanks in advance.
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by dumb.doofus » Sat May 23, 2009 12:45 pm
DarkKnight wrote:Folks,

Why isn't anyone considering following pairs:

(5,5), (-5,5), (5,-5) and (-5,-5) in which case we will get a square with side length 10 and area 100.

thanks in advance.
not just these values .. there are so many other values too that would satisfy the equation.. for instance, (3,7) or (4,6).. take whatever value of x and y that adds up to 10... and it will satisfy the equation..

important point to see here is that when you join (-5,5) and (5,5), the y-intercept would be at point (0,5) and this doesn't satisfy the equation... so you can't create a square like that.. only way you can create a square is by finding y and x intercept..

what you have created is an area that is part of the bigger square.. just draw the figure and you'll see..
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by DarkKnight » Mon May 25, 2009 1:09 pm
doofus, very good explanation. Thanks for clarifying it. Much appreciated.
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