2.  Find areas of regular polygons  Find areas of circles

3. Find Areas of Regular Polygons

4. Apothem – a segment that is drawn from the center of a regular polygon perpendicular to the side of the polygon. If all radii were drawn, there would be 6 congruent isosceles triangles. An area of a hexagon is determined by adding areas of the triangles inside. A B C D E F G H

5. Area of a triangle= ½bh = ½sa If the area of a triangle = ½sa, then the area of a hexagon = 6(½sa)

6. Find the area of a regular pentagon with a perimeter of 40 cm. P K L MQN J P

7. We first need to find the apothem. Since it is regular, all of the central angles are congruent, or equal to 360/5 or 72°. That makes angle NPQ equal to 36°. Side NM = 8 since the perimeter is 40, so NQ = 4. Now, we can solve the problem: tan NPQ = QN / PQ tan 36 = 4 / PQ PQ = 4 / TAN 36 PQ = 5.5 Area = ½ Pa or ½ (40)(5.5) = 110 So the area of the pentagon is 110 cm. P K L MQN J P

8. Find Areas of Circles

9. Area of a circle = r²

10. 48 in 34in A caterer has a 48-inch diameter table that is 34 inches tall. She wants a tablecloth that will touch the floor. Find the area of the tablecloth. 34 + 48 + 34 = ll6 divided by 2 = 58 so the radius = 58. A = pi(58)² = 10,568 inches 2

11. Find the area of the shaded region. Assume that the triangle is equilateral and the radius of the circle is 4. 4m

12. A = pi * r² A = pi(4)² = 50.3  Area of Circle Construct ΔABC, a right Δ. Use 30-60-90 rules to find the lengths of the sides and then to find the height of the equilateral triangle. AB = 2 BC = 2√3 DB = 2√3(√3) = 6  Height of ΔCDE One side of ΔCDE is equal to 2(2√3) = 4√3 So, A = ½ (4√3)(6) = 20.8  Area of ΔCDE Now, area of shaded region is 50.3 – 20.8 which is equal to 29.5 m². 4m A BC E D

13.  Pre-AP Pg. 613 #8 – 22, 24, 30, 32, 39 - 44  Geometry Pg. 613 #8 – 22, 24, 30, 32, 39 - 41