By considering dodecagons inscribed and circumscribed about a unit disk, establish the inequalities

First, we draw some pictures of the situation for reference.

(* Note: I don’t know a way to do this without using trig functions, which haven’t been introduced in the text yet. If you have an alternative approach without them, please leave a comment and let us know about it. *)

Since these are dodecagons, the angle at the origin of the circle of each triangular sector is , and the angle of the right triangles formed by splitting each of these sectors in half (shown in the diagrams) is then . Then we use the fact that

Now, for the circumscribed dodecagon we have the area of the right triangle with base 1 in the diagram on the left given by

Since there are 24 such triangles in the dodecahedron, we then have the area of the circumscribed dodecahedron given by

For the inscribed dodecagon we consider the right triangle with hypotenuse 1 in the diagram. The length of one of the legs is then given by and the other is given by . So the area of the triangle is

Since there are 24 such triangles in the inscribed dodecahedron, we then have,

Since the area of the unit circle is, by definition, , and it lies in between these two dodecahedrons, we have,

Here’s an approach to Question 18 in Section 2.4 the uses only basic geometry:

A dodecagon is a 12-sides polygon. Consider one of the 12 uniform triangles that make up a dodecagon inscribed in a unit circular disk. Let’s call it triangle ABC, where segments AB and BC are unit length (i.e., length of one) and angle ABC is 30 degrees (30 degrees = 360 degrees / 12). Draw a line through point A and perpendicular to segment BC. Call the intersection point on segment BC point D, where point D lies between points B and C. Consider triangle ABD. If angle ABC is 30 degrees, so is angle ABD. Angle BDA is 90 degrees, so angle BAD must be 60 degrees. For a 30-60-90 triangle, if the hypotenuse (segment AB) is length 1, the short leg (segment AD) must be length 1/2. Consider again triangle ABC, where segment AD is the height and segment BC is the base. The area of that triangle is (1/2)*1*(1/2) = 1/4. The area of 12 of those triangles (the inscribed dodecagon) is 12*(1/4) = 3. That proves pi > 3.

Now consider the dodecagon circumscribed about the unit circular disk. Consider one of the 12 uniform triangles that make up that larger dodecagon. Let’s call it triangle EFG, where segments EF and FG are equal length (and extend past the edge of the unit circular disk) and angle EFG is 30 degrees (30 degrees = 360/12). Draw a point H that bisects segment EG. Segment FH is unit length (i.e., length of one). Consider triangle EFH. Angle EFH is 15 degrees (30 degrees / 2 = 15). Angle FHE is 90 degrees, so angle FEH is 75 degrees. For a 15-75-90 triangle, if the long leg (segment FH) is length 1, the short leg (segment EH) must be length 2-SQRT(3). Consider again triangle EFG. If segment EH is length 2-SQRT(3), segment EG must be twice that length, or 4-2*SQRT(3). If segment EG is the base and segment FH is the height, then the area of that triangle is 1/2*(4-2*SQRT(3))*1 = 2-SQRT(3). The area of 12 of those triangles is 12*(2-SQRT(3)). This proves pi < 12*(2-SQRT(3)). Combing this with the earlier proof, we get 3 < pi < 12*(2-SQRT(3)), as desired.