Let be the midpoint of the arc . Pick at random and pick such that (where denotes perpendicular). Then
Solutions in category Geometry
An isosceles triangle has sides that are #sqrt125#, #sqrt125#, and 10 units. What is its area?
the area of a triangle is found through #A=1/2bh#. We have the base, but not the length of the height.
Imagine that an altitude is drawn down the center of the triangle, perpendicularly bisecting the base and bisecting the vertex angle. This creates two congruent right triangles inside the original isosceles triangle.
You know each of these triangles has a hypotenuse of #sqrt125# and a base of #5#, since they're half the original base.
We can use the Pythagorean theorem to figure out the length of the missing side, which is the height of the triangle.
So, we now know that the base is #10# units and the height is #10# units.
Thus, #A=1/2(10)(10)=50 " units"^2#
Given two ordered pairs (1,-2) and (3,-8), what is the equation of the line in slope-intercept form?
The general equation for a line in slope-intercept form is:
The point on a line segment dividing it into two segments of equal length. The midpoint of a line segment is easy to locate by first constructing a lens using circular arcs, then connecting the cusps of the lens. The point where the cusp-connecting line intersects the segment is then the midpoint (Pedoe 1995, p. xii). It is more challenging to locate the midpoint using only a compass (i.e., a Mascheroni construction).For the line segment in the plane determined by and , the midpoint can be calculated as(1)Similarly, for the line segment in space determined by and , the midpoint can be calculated as(2)In a right triangle, the midpoint of the hypotenuse is equidistant from the three polygon vertices (Dunham 1990).In the figure above, the trilinear coordinates of the midpoints of the triangle sides are , , and .The midpoint of a line segment with endpoints and given in trilinear coordinates is , where(3)(4)(5)(left as an exercise in Kimberling..