Find the radius of a circle with the given circumference. 6 square root of 2 pi in.
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We have the following function that is a quadratic function:
So the graph of this function is shown in the figure below. This is a <em>parabola</em> as you can see. The roots of this functions, that is, the x-intercepts are:
As you can see in the figure. This function decreases from
and increases from 
Finally, another thing we can see from the graph is that the vertex is the point:
So the graph of this function is shown in the figure below. This is a <em>parabola</em> as you can see. The roots of this functions, that is, the x-intercepts are:
As you can see in the figure. This function decreases from
Finally, another thing we can see from the graph is that the vertex is the point:
Focus on triangles DFC and DFE:
- They are both right triangles because DF is perpendicular to EC
- DC=DE
- DF is a common side
So, the two triangles are congruent. It follows that CF=FE, and in particular
And since x=3, the length of EF is