|x|+|y| = 10, represents a square with side length equal to 10*sqrt(2),

So area = 100*2 = 200

Try to plot them using four points [(10,0),(0,10),(-10,0),(0,-10)] on a co-ordinate plane.

anshumishra wrote:|x|+|y| = 10, represents a square with side length equal to 10*sqrt(2),

is this some property ?

So area = 100*2 = 200

Try to plot them using four points [(10,0),(0,10),(-10,0),(0,-10)] on a co-ordinate plane.

if i plot these points the side of square comes 20

bblast wrote:

anshumishra wrote:|x|+|y| = 10, represents a square with side length equal to 10*sqrt(2),

is this some property ?

So area = 100*2 = 200

Try to plot them using four points [(10,0),(0,10),(-10,0),(0,-10)] on a co-ordinate plane.

if i plot these points the side of square comes 20

No, the area is bounded by these four lines :
x+y = 10
x-y = 10
-x+y = 10
-x-y=10

I think, you have calculated the diagonal. Can you redraw the diagram, and check? [Else, I may be making mistake....sneaking the time out of office a bit...]

Hi there!

Anshu´s solution is correct, but let me help you (bblast) exploring the problem a bit more...

Please note that |x| + |y| = 10 must (at first sight and not counting symmetries) be considered in 4 scenarios:

(1) x <=0 and y >=0 :: in this case we get -x + y = 10 and this is a line that passes through points (-10,0) and (0,10)

This line is indicated in red in the figure I created.

(2) x <= 0 and y <= 0

(3) x >= 0 and y >=0

(4) x >=0 and y <= 0

Please work cases (2) to (4) to realize that the region mentioned in the question stem is (Anshu´s comment) "a square with side length equal to 10*sqrt(2)" and, more than that, with an area that is four times the area of the red right isosceles triangle in my figure!

We get 4* [(10*10)/2] = 200 and we are done.

Regards,
Fabio.

wow this was a tricky problem, did not realize the solution at first, thanks guys.

bblast wrote:wow this was a tricky problem, did not realize the solution at first, thanks guys.

Thanks for the "thanks", bblast. Beautiful problem, by the way.