Find the area of trapezoid ABCD with bases AB and CD; m<A=60 degrees, AD=18, and CD=16

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Assuming that the trapezoid is symmetrical, the trapezoid is composed of two triangles and a rectangle made by drawing perpedindicular lines from points c and d to the line segment AB.  The base of the triangles would be 9 cm because AD is the hypotenuse of a right triangle, and it's base is adjacent to the 60 degree angle.  `cos 60=1/2=b/18` b=9.  The height of the triangle can be estimated by taking the sin of 60 degrees because h is opposite the 60 degree angle.  `sin60=h/18~~.866` Multiplying by 18 we get `h~~15.59` The area of the triangles combined is`2xx(1/2xx9xx15.59)=9xx15.59=140.3` And the Area of the rectangle is `16*15.59=249.42` We add the areas together to get the area of the trapezoid, which is approximately 265 `cm^2`

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