1. State the area formulas for the following polygons:
   1. An equilateral triangle with side s
   2. A square with side s
2. A regular polygon has two features: the sides are all the same length, and the interior angles all have the same measure.

   

   Use the following to find a formula for the area of a regular pentagon with side length s.

   1. First, show that you can cut the pentagon into five congruent triangles. This may be easy to see, with an accurate drawing, but it’s not so easy to prove! The only things you know about the pentagon are the equal angle measures and the equal side lengths.
      You can’t assume anything about symmetry, for example.
      
   2. For one of the congruent triangles, let the side that is the side of the pentagon be the base of the triangle. Find the height to that base.
   3. Now find the area of the pentagon.
3. Use the same idea to find the area of a regular hexagon.
4. Find a general formula for the area of a regular n-gon. Verify the formula for triangles and rectangles.

Hint to problem 2a. One way to do this is to describe how to construct the five triangles, and show they are congruent as you go. Here’s a partial outline for how to do that:

1. Show that the angle bisector at a vertex is the perpendicular bisector of the side opposite that vertex.
   1. First draw diagonals from the vertex to the endpoints of the opposite side. Show that two of the resulting triangles are congruent.
   2. Draw the angle bisector. Show that you have another pair of congruent triangles.
2. Draw another angle bisector, for a vertex adjacent to the one already bisected. Draw segments from the intersection of the two bisectors to the three opposite vertices. Show the two triangles are congruent.
3. All five congruent triangles are already showing on your figure. Explain how you know they are congruent.

Hint to problem 2b. For an angle of measure x, the ratio is tan x.

Hint to problem 3. Here’s a partial outline of a proof that you can divide a hexagon into six congruent triangles:

1. Show that the angle bisector at a vertex is the angle bisector of the opposite vertex.
   1. First draw diagonals from the vertex to the vertices two away, forming two triangles and a quadrilateral between them. Show that the triangles are congruent.
   2. Draw the remaining diagonal, to the vertex directly opposite. This cuts the quadrilateral into two triangles. Show those triangles are congruent.
   3. Explain why this means that last diagonal bisects both interior angles.
2. Draw another angle bisector from a vertex adjacent to the one already bisected to the vertex opposite. Show the two triangles formed are congruent.
3. Now draw segments from the intersection of the two bisectors to the remaining vertices. Show that the triangles formed are congruent to the other two (and to each other).

1. 1. Area = s²
   2. Area = s²
2. 1. See the solutions.
   2. The height is .
   3. Area = s²
3. See the solution for the proof that the hexagon can be cut into six congruent triangles. Area = s², or an equivalent form such as s² or s².
4. Area_n = s²
   For triangles, n = 3, giving
   
   For rectangles, n = 4, giving
   

Solution for problem 2a.

1. Show that the angle bisector at a vertex is the perpendicular bisector of the side opposite that vertex.
   1. First draw diagonals from the vertex to the endpoints of the opposite side. Triangles ABC and AED are congruent by the SAS congruence theorem. So ACAD and EADBAC.
      
      
   2. Draw the bisector of the angle at the same vertex (in this case, A). Since EADBAC and AF bisects BAE, F ACF AD. Triangles F AC and F AD are congruent, again by SAS.
      
      
      This means AF CAF D, so AF CD. Since F C = F D, the angle bisector of A is also the perpendicular bisector of CD.
2. Draw another angle bisector, for a vertex adjacent to the one already bisected. Draw segments from the intersection of the two bisectors to the three opposite vertices.
   
   
   Since G is on the perpendicular bisector of CD, it’s equidistant from C and D. Since G is on the perpendicular bisector of BC, it’s equidistant from C and B, also. So GD = GC = GB. By the SSS congruence theorem, GDCGBC.
3. Note that GDC and GBC are isosceles as well as congruent to each other. This means that the following angles are congruent: GDF , GCF , GCB, and GBC. Since each of these is half the interior angle of the pentagon, and segments bisecting CAE and AED have been constructed, all the angles at the vertices are congruent.
   
   
   So triangles BGA, AGE, and EGD are isosceles. That means AG = BG = DG = EG, and the five triangles (AGB, BGC, CGD, DGE, and EGA) are congruent to each other by SSS.

Solution for problem 3.

1. Show that the angle bisector at a vertex is the angle bisector of the opposite vertex.
   1. First draw diagonals from the vertex to the vertices two away, forming two triangles and a quadrilateral between them. The triangles are congruent by the SAS congruence theorem.
      
      
   2. Draw the remaining diagonal, to the vertex directly opposite. This cuts the quadrilateral into two triangles. These are congruent by the SSS congruence theorem.
      
      
   3. From the triangle congruence above you see that AD bisects CDE. Since BACF AE and CADEAD, it must be true that BADF AD.
2. Draw another angle bisector. The triangles formed are congruent to each other by the ASA congruence theorem.
   
   
3. Now draw segments from the intersection of the angle bisectors to the remaining vertices.
   
   
   The triangles in the previous step were isosceles (because the base angles are congruent). All six triangles are congruent to one another by SAS.