SOLUTION: The area of an isosceles trapezoid is three times than that of an equilateral triangle. If the heights of the trapezoid and the triangles are both equal to h = 12 sqrt(3), what is

Question 1176016: The area of an isosceles trapezoid is three times than that of an equilateral triangle. If the heights of the trapezoid and the triangles are both equal to h = 12 sqrt(3), what is the length of the median/ midline of the trapezoid in meters?
Answer by greenestamps(13209) (Show Source): You can put this solution on YOUR website!

The height of the equilateral triangle is 12*sqrt(3). Use your knowledge about equilateral triangles and 30-60-90 right triangles to determine that the side length of the equilateral triangle is 24. Then the area of the triangle, one-half base times height, is 144*sqrt(3).

The area of the trapezoid is the length of the midline of the trapezoid times the height. The area of the trapezoid is 3 times the area of the equilateral triangle, which is 432*sqrt(3). And the height of the trapezoid is 12*sqrt(3), so the length of the midline of the trapezoid is 432/12 = 36.

ANSWER: The length of the midline of the trapezoid is 36.

A picture can help show that this answer makes sense.

If the trapezoid has 60 degree and 120 degree angles, then it can be viewed as being composed of three of the equilateral triangles. The length of the midline of the trapezoid is then three times the "midline" of each equilateral triangle; and since the triangles have side length 24, the length of each midline is 12.

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