Work done by gas

How do we find the work done by a gas (physics review chapter 4)?  I don't see how the area under the curve can be broken down into a triangle on top of a rectangle...
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The work done in a PV graph is the area under the curve.  If we're trying to find the area done by A, we should focus only on the line labeled A (and thus, ignore B and C).  The area under A extends from the line, A, all the way to the x-axis.  You'll see that this creates a trapezoid bound by the following points:  (1, 5), (6, 2), (6, 0), and (1, 0).  Thus, the area under this curve can be done using the equation for the area of a trapezoid, or by adding the areas of the triangle bound by (5, 1), (6, 2), and (1, 2) and the rectangle (2, 1), (6, 2), (6,0), (1, 0).  Since I like simpler math, I'll go with the trapezoid.

The area of a trapezoid is (b1 + b2)/2 * h -- the average of the bases times the height.  Well, one base here is 5 Pa long (from the x-axis to the top point of line A) and another is 2 Pa long (from the x-axis to the lowest point on line A).  The height is therefore 5 m^3 -- the length on the x-axis.  So the area is (5 Pa + 2 Pa)/2 * 5 m^3 = 3.5 Pa * 5 m^3 = 17.5 J.