This works if the whole of the parabolic region lies within the rectangle, but aren't there situations when this is not the case ?
Show that if we take a parabolic region and inscribe a triangle whose base is the line segment that bounds the region and whose apex is located at the point where the tanget line to the parabola is parallel to the base, then the area of the triangle is more than half of the area of the parabolic region.
Is it sufficient to say the following:
If we draw a perpendicular line at each endpoint of of the line segment which is the base of the triangle up to the same height as the apex of the parabola, then we have a rectangle. We know that the area of the parabola is smaller than this rectangle, and we also know that the area of the triangle is exactly half of the rectangle. Therefore, the area of the triangle must be more than half the area of the parabola.
This question is in the section on "Infinite summations" under the subtopic "Archimedes' Argument." Is there another, more relevant way to prove this?
Assume the parabola has the equation of , and let the two edge points of the base line be .
Then we have
So the slope of the base is
Let be the third vetex, the tagent line at it is
, with a slope .
Since we have , we get .
That is, if we draw the median on the base, it is parallel to the x-axis.
Now draw two lines parallel to the x-axis, starting from the two base points
, we get a parallelogram, which covers the parabola region obviously and we're done.
I realized that we don't need to say that the median is parallel to the symmetry axis.
We need only the last sentence:
Draw two lines parallel to symmetry axis of the parabola, starting from the two base points, we get a parallelogram, which covers the parabola region obviously and we're done.