The circumference of a regular polygon is 46.4 cm and the length of the radius of the described circle is 7 cm, what is the surface of the polygon?

Question

I need help, if You could please show the whole proces…? Thank You!

Answer 1

If the circle is inscribed it is easier (no trig needed). You need to derive a general case equation for the area of a regular polygon based on the given radius and perimeter relationship. General case means an n sided polygon.

To get started you might draw a 5 sided polygon. Inscribe a circle. Cut the polygon into triangles all meeting at the center of the polygon. For the pentagon you will have 5 triangles.

In general this is always true: the height of the triangles is the radius of the circle; the base of the triangle is related to the perimeter, that is P/n.

The total area of the polygon is n * Area of each triangle. Quite conveniently, you will see that you have an n in the numerator and an n in the denominator. n's cancel and you should now be able to relate the Area of the polygon to the radius of the circle and the Perimeter.

For the circumscribed case, it is similar, but to get the height of each triangle you will need a little trig.

I've been dabbling in manim to make illustrations. Let me know if you'd like to see an illustration of this problem.

Answer 2

Hi Dar M.,

You should be able to address the other responder's comment easily, without even going back to ypur problem set! A circle of radius 7 has circumference ~ 43.98. Does that mean the circle is circumscribed, or inscribed, with respect to the polygon? Think about that geometrically. What is the shortest distance between two points (such as, vertices of the polygon) "defined" as?

That intuition firmly in hand, start trying to solve the perimeter of the polygon, using right triangles drawn to the centers of the faces (i.e., the intersections with the circle). You will need to possibly use a little trig, if you get other than n = 3, 4, or 6. Unless you really have a head for half-angle sines and the like!

-- Cheers, --Mr. d.