2.  Find areas of regular polygons.  Find areas of circles.  Bet ya didn’t see THAT coming!

3.  Regular polygons, have all sides CONGRUENT.  Properties of central, and inscribed angles.  The degree measure of one circle = 360°  30° – 60° – 90°

4. When thinking of regular polygons, imagine them being inscribed in a circle. This will give you the basis to find initial angles for different problems. When a regular polygon is inscribed into a circle, from the center of the circle to a vertex of the polygon is a radius of the circle.

5. APOTHEM: A segment, from the center of a circle, that is perpendicular to a side of the inscribed polygon. N B AC D Segment ND is an APOTHEM

6. C N B AC D In ∆ ABC, Segments NB and NC are congruent, because they’re radii, making ∆CNB an isosceles triangle. When all of the other radii are drawn they break the big triangle into three congruent mini triangles. Now to find the areas. A= ½ bh, OR, A= ½ sa

7. C N B AC D To calculate the area of the triangles, you use the normal polygon area formula, using the apothem for the height, and the side for the base. Now to find the areas. A= ½ bh, OR, A= ½ sa

8. C N B A D Notice, the area of one small triangle is ½ sa square units. so the area of the BIG triangle is 3( ½ sa) Now notice that the perimeter of the BIG triangle is 3s units. we can substitute P (perimeter) for 3s in the area formula. C So A= 3( ½ sa) becomes A= ½ Pa This formula can be used for the area of any regular polygon.

9. EXAMPLE #1( Remember the properties of a circle) A B C D E Find area of a regular pentagon with a perimeter of 50 centimeters The central angles of a regular pentagon are all congruent. So, the measure of each angle is 360/5 or 72 P Q

10. EXAMPLE #1( Remember the properties of a circle) A B C D E Find area of a regular pentagon with a perimeter of 50 centimeters P Segment PQ is an apothem of pentagon ABCDE. It bisects angle CPB and is a perpendicular bisector of segment BC. Therefore, measure of angle BPQ = ½(72) or 36. Since the perimeter is 50 cm each side is 10cm, which makes segment BQ = 5 cm. Q

11. A C D E P Q B Write a trig ratio to find the length of segment PQ. Tan of angle BPQ = opposite/adjacent Tan 36° = 5/PQ (PQ) tan 36° = 5 PQ = 5/tan 36° PQ ≈ 6.9. Area = ½Pa A = ½(50)(6.9) A = 172.5

12. Area of a Circle A = πr² (r = radius)

13. 4 Find the area of the shaded region. Assume that the triangle is equilateral. First find the are of the circle. A = πr² = π(4)² ≈ 50.3

14. 4 To find the area of the triangle, use 30° - 60° - 90° rules. First find the length of the base. The hypotenuse is 4, so MP is 2. MB is 2√3. AB is 4√3. Next, find the height, MC. MC = 2√3(√3) or 6 Use the formula to find the area of a triangle. A = ½bh = ½(4√3)(6) ≈ 20.8 The area of the shaded region is 50.3 – 20.8 or 29.5 A B C M P 60°

15. Assignment: Pg. 613 #8-22 evens, 26, 27, 30-33, 39 - 44