Our mission is to write the perimeters of the two polygons in terms of r.
Perimeter of the Outer Polygon
As we can see, the outer polygon is a square. To find the length of its sides, we will draw a diameter of the circle.
From the above we conclude that the length of the sides of the square is l = 2r. We are ready to find its perimeter.
P_1 = 4l ⇒ P_1 &= 4(2r)
⇒ P_1 &= 8r
Perimeter of the Inner Polygon
The inner polygon is a regular hexagon, so its perimeter is 6 times the length of each side. Now, let's draw the segments from the vertices of the hexagon to the center of the circle.
Since all the drawn radii has the same length, we have that all these triangles are isosceles and the base angles are all congruent.
Additionally, we have that the vertex angle of each triangle has a measure of 360^(∘)6, or 60^(∘). We can now use the Triangle Angle-Sum Theorem to find the measure of the base angles.
60^(∘) + 2(Base Angles) = 180^(∘)
⇓
Base Angles = 60^(∘)
Since the base angles have a measure of 60^(∘) each, we conclude that each triangle is equilateral. Therefore, the sides of the hexagons have a length of s = r. Knowing this, we are able to find the perimeter of the hexagon.
P_2 = 6s ⇒ P_2 = 6r
b
The circumference of the circle is equal to 2Ï€ times the radius.
C = 2Ï€ rFrom Part A, the perimeter of the circumscribed polygon is 8r and the perimeter of the inscribed polygon is 6r. By using a calculator, let's find 2Ï€ and compare C to these two perimeters.
2π ≈ 6.28
⇓
C ≈ 6.28r
↙ ↘
C < 8r C > 6r
Finally, we write a compound inequality that represents the two inequalities above.
6r < C < 8r
c
In Part B we wrote the following inequality.
6r < C < 8rSince the diameter of a circle is twice the radius, we can rewrite the perimeters of the polygons as follows.
d=2r ⇒
6r = 3(2r) = 3d
8r = 4(2r) = 4d
Let's substitute the new expressions for the perimeters into the compound inequality.
3d < C < 4d
The last inequality implies that the circumference of a circle is between 3 and 4 times its diameter.
d
In Part C, we wrote the inequality 3d< C < 4d. Let's increase the number of sides in each polygon.
As we can see, as the number of sides increases the lower limit increases and the upper limit decreases. In fact, we can see that both of them approach to π d, which implies that C=π d.
Thus, we have seen that the circumference of a circle is C=Ï€ d or C = 2Ï€ r.
