Write a system of inequalities to describe the shaded figure in the picture here: http://i91.photobucket.com/albums/k305/Suger_Pie/dafds.jpgI...

At the site below,

href="http://i91.photobucket.com/albums/k305/Suger_Pie/dafds.jpg">http://i91.photobucket.com/albums/k305/Suger_Pie/dafds.jpg,

we
notice a trapezium whose base is streteched  equally about the origin 3 units on either
sides of the origin along the  x axis.

If we go up from
along the y axis by 3 units we see the other parallel side of the trapezium which
streches by 2 units on either side of the y axis (perpenducular to y axis and || to x
axis.

The shade part is the area of the trapezium  with
vertices at A(-3 , 0) , B(3,0), C(2 , 3) and D(-2,3).

The
area of the trapezium = (1/2)(sum of parallel sides)(distance between the paralle sides)
= (1/2){(3+3)+(2+2)}(3) =15.

The inequality here we can try
to make out  is :

Area of  the trapezium ABCD  <
area of the outer rectangle with vertices at (-3,0) , (3,0), ( 3,3) and (-3,3) = 6*3 =
18 sq unit.

The area of the trapezium ABCD > Area of
the inner rectangle  with vertices at (-2,0) ,(2,0), (2,3) and (-2,3) =  4*3 = 12
sqrt.

Of course the average of are  inner and outer
rectangles = (18+12)/2 = 15 sq units = area of the trapezium
ABCD.