Another way to solve this problem is to consider the area to be made of one rectangle and two triangles. If the shorter width is w, the longer width is 2w. If the base of one of the triangles is x, then the base of the other is 2w-x. Area of rectangle=8w because the sides of the rectangle are 8 and w metres. The areas of the triangles are ½(8x)=4x and ½(8*(2w-x))=4(2w-x). Add these areas together: 8w+4x+8w-4x=16w, so 16w=60 sq m and w=15/4=3.75m. If we double the area and keep w the same and replace 8 by h, we get 2hw=120. So h=120/2w=60/w=60/3.75=16m, double the original depth of 8m.
The disadvantage of this method is that it's longer than the other method. The advantage is that it is not formulaic, doesn't use the formula for finding the area of a trapezoid. This is because it breaks down the figure into simpler figures for which area can be calculated from first principles. So it is an intuitive method, compared with a formulaic method. It also shows why the placement of the triangles is irrelevant, because of the terms 4x which cancel out, x being the length of the base of one triangle.