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**dudeosu** question: an equilateral triangle abc sits atop an isosceles trapezoid abde which has a height of 6 cm. if the area of the trapezoid abde is twice the area of the triangle abc, what is the area of the trapezoid?

this for the most part seems simple to me. the area of the trapezoid is twice the area of the equilateral triangle, thus the area of the trapezoid= ((3^2)/2)(L^2) with L being a length of one of the sides of the equilateral triangle. Now for the trapezoid, we can also construct another equation for its area. we know that in the area lies a box 6cm x L. the rest is the two triangles on the side. Where I am stuck is what the area of these two triangles are??? if we set ((3^2)/2)(L^2) = 6cm x L + (area of two triangles) then couldn't we find the area of the trapezoid? I just don't know how to find the base of these triangles.