If we have a trapezoid with parallel sides of a & b then the length of the parallel line that bisects the trapezoid is = rms(a,b)= [(a^2+b^2)/2]^(1/2)
In this case the parallel lines are of lengths 3*m & 9*m and thus the length of the bisector// to the y-axis=m*(45)^(1/2)
Thus, mx=m*(45)^(1/2) or the vertical line x = (45)^(1/2) bisects the given trapezoid.
The surprising non-contributory factor is, m = the slope of the line y=mx.
In the modified case, area of the trapezoid = (3m+9m)*6/2 =9*(m+3)
Thus, k*6 =9*(m+3)/2 or k=(3/4)*(m+3)
Consider the region in Quadrant 1 totally bounded by the 4 lines: x = 3, x = 9, y = 0, and y = mx (where m is positive). Determine the value of c such that the vertical line x = c bisects the area of that totally bounded region.
Needless to say, your first task should be to draw a representative diagram for the problem that has been described. Also, after determining the value of c, be sure to comment on the (probably) surprising non-contributing factor in this problem.
Bonus: What if the problem were modified such that a horizontal line, y = k was to be the area bisector of the totally bounded region. Determine what the value of k would be in that case.
2 answers
tkia