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Find the area of pentagon ABCDE. |
Solution: You can tell by looking, that this pentagon is not a "regular" pentagon. But, that is not a problem as the "box" method works for all forms of polygons.
• Draw the box.
• The outer triangles have been numbered with integers this time, and ABCDE will be called X.

The answer is 34 square units. |
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This particular problem lends itself to also being easily solved by "decomposing",
so let's take a look at that solution also.
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Find the area of this pentagon by decomposing it into recognizable shapes. |
Solution: The pentagon can be divided into ΔABC and trapezoid ACDE. The trapezoid can then be broken down further into two triangles and a rectangle.
• The area of ΔABC = ½(6)(3) = 9 sq. units.
• The area of the rectangular part of the trapezoid = 4 x 5 = 20 sq. units.
• Each of the small Δs in the trapezoid has an area of ½(1)(5) =2.5 sq. units.
The total area is
9 + 20 + 2(2.5) = 34 square units.
(Same result as was seen in Example 1.)
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Find the area of this composite figure made up of an irregular hexagon and a square. |
Solution: Finding the area of the square is easy.
• Area of square = 3 x 3 = 9 sq. units
Finding the area of the hexagon is a bit more tricky.
Let's compose further and form a rectangle around the outside of the hexagon.
• The 4 right triangles surrounding the hexagon are all of the same area: ½(3)(2) = 3 sq. units each.
• The area of the rectangle we drew is 6 x 7 = 42.
• Subtract the area of the four right triangle from the area of the rectangle to get the area of the hexagon. 42 - 4(3) = 42 - 12 = 30 sq. units.
• The area of the hexagon plus the area of the square = 30 + 9 = 39 sq. units.
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