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The solution is to use the fact that the irregular hexagon is made up of several smaller regular polygons, that have easy to find areas.
In the very center of the table is a rectangle with the dimensions of w=56cm, and l=132cm. The Area of a Rectangle is l x w. 56 x 132= 7,392.
Along the top edge of the table, it is broken into a trapezoid and a right triangle. Principles of vertical lines tell us that if we draw a parallel line at the other end of the 56 cm leg, it will also be perpendicular to the dashed horizontal that makes the base of the triangle. This breaks the trapezoid into two right triangles and another rectangle, dimensions 56cm x 38cm. The area of this rectangle would be 2,128 cm sq.
The two right triangles formed from the trapezoid could be placed edge to edge to form a triangle with `b=132-56` and `h=38` The Area of a triangle is 1/2bh, that would come out to 1,444 sq cm.
Finally the drawing's lack of labels along the bottom edge indicates that the table is symmetrical, which means the trapezoid at the bottom of the table breaks into a rectangle and two triangles congruent to the ones on the top, and consequently with the same areas of 2,128 sq cm, and 1,444 sq cm. respectively.
Now that we have the areas of all the polygons that made up the hexagon, we add them all together to find the area of the hexagon.
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