Asked by borat
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.
Answers
Answered by
Topquark
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)
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)
Answered by
hggj
tkia
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.