In trapezoid ABCD, AB = CD = 2 cm, BC = 6 cm, segment AD is parallel to segment BC, and m BAD = 60˚. What is the area of the trapezoid?

Question

In trapezoid ABCD, AB = CD = 2 cm, BC = 6 cm, segment AD is parallel to segment BC, and m BAD = 60˚. What is the area of the trapezoid? Express your answer in simplest radical form.

Answer 1

First we have to picture the trapezoid.  Since AD is parallel to BC they must be the top and base.  Since sides AB and CD are equal, it is an isosceles trapezoid, meaning the base angles are the same and are <90°, while the top angles are the same and are >90°.  Since BAD is 60°, the base is AD.  Drop a line down from B that forms a 90° angle with AD and we form a 30-60-90 right triangle with hypotenuse 2 cm.  That line would then be √3cm and the side within AD would be 1 cm.  This means AD is 1+1=2 cm longer than BC, so now we can find the area of the trapezoid:

 

A = (BC + AD)/2 x height

A = (6 + 8)/2 x √3

A = 7√3