c Notice that each radius bisects the angles of the polygon. Use the properties of a 30^(∘)-60^(∘)-90^(∘) triangle.
A
a l=2a
B
b l = 2sqrt(3)/3a
C
c l = 2sqrt(3)a
Practice makes perfect
a Let's begin by drawing a square of side length l, and let's draw its center, apothem, and a radius.
Since the diagonals of a square bisect the angles, we get that the triangle above is a 45^(∘)-45^(∘)-90^(∘) triangle.
In consequence, the triangle marked above is isosceles, and therefore, the apothem is equal to half the length of the sides of the square.
a = l/2 or
l = 2a
b In this part, we will draw a regular hexagon of side length l, and let's draw its center, apothem, and a radius.
Notice that the radii of a regular hexagon divide it into 6 equilateral triangles. This means that one acute angle of the right triangle drawn above has a measure of 60^(∘). In consequence, the other acute angle has a measure of 30^(∘).
By the properties of 30^(∘)-60^(∘)-90^(∘) triangles, we have that the longer leg, in our case the apothem, is equal to sqrt(3) times the shorter leg.
a = sqrt(3)*l/2
or
l = 2sqrt(3)/3a
c Lastly, we consider an equilateral triangle of side length l, and let's draw its center, apothem, and a radius.
Notice that the radii of an equilateral triangle bisect each angle. Consequently, one of the acute angles of the right triangle drawn above has a measure of 30^(∘) implying that the other has a measure of 60^(∘).
By using the properties of 30^(∘)-60^(∘)-90^(∘) triangles, we get that the longer leg is equal to sqrt(3) times the shorter leg.
l/2 = sqrt(3)a ⇒
l = 2sqrt(3)a
