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11.3 Areas of Polygons and Circles .

11.3 Areas of Polygons and Circles. Objectives:. Find areas of regular polygons. Find areas of circles. Bet ya didn’t see THAT coming!. THINGS TO REMEMBER. Regular polygons, have all sides CONGRUENT. Properties of central, and inscribed angles.

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11.3 Areas of Polygons and Circles .

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  1. 11.3Areas of Polygons and Circles.

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

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

  4. Areas of POLYGONS. 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. Now for the new stuff. APOTHEM: A segment, from the center of a circle, that is perpendicular to a side of the inscribed polygon. B Segment ND is an APOTHEM D N A C

  6. Now to find the areas. 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. A= ½ bh, OR, A= ½ sa B D N C A C

  7. Now to find the areas. 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. A= ½ bh, OR, A= ½ sa B D N C A C

  8. 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. B So A= 3( ½ sa) becomes A= ½ Pa This formula can be used for the area of any regular polygon. D N C A C

  9. EXAMPLE #1( Remember the properties of a circle) 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 B Q A C P E D

  10. EXAMPLE #1( Remember the properties of a circle) 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. Find area of a regular pentagon with a perimeter of 50 centimeters B Q A C P E D

  11. 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 B Q A C P E D

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

  13. 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 4

  14. 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 M B A 60° 4 60° P C

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

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