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Area of Regular Polygons and Circles continued …

Area of Regular Polygons and Circles continued …. Do NOW 04-15-2014. Take out your notes from yesterday and work on the two area of a polygon problems using the A = (1/2)Pa formula. Homework: Irregular Polygons Worksheet TEST FRIDAY!!!!. Area of a Regular Polygon:.

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Area of Regular Polygons and Circles continued …

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  1. Area of Regular Polygons and Circles continued …

  2. Do NOW 04-15-2014 Take out your notes from yesterday and work on the two area of a polygon problems using the A = (1/2)Pa formula. Homework: Irregular Polygons Worksheet TEST FRIDAY!!!!

  3. Area of a Regular Polygon: Let’s begin with an example using a regular pentagon: What is the Perimeter of MNOPQ when QP = 12 inches? SOHCAHTOA What is the Area? = 247.80 in2 P = # of sides  side length = 5 (12 in.) = 60 inches How do we find the apothem(a)? 360/ 5 = 72o 72/ 2 = 36o Tan (36o) = (6/a) a = (6/(tan(36o))) = 8.26 cm

  4. How can we use all of this information to find the area of a circle? Can we get to the area of a circle from the equation for Perimeter?

  5. Area of a Circle Can we use the area of a regular polygon? If a circle has an area of A square units and a radius of r units, then: A = πr2 • Yes! • A = (1/2) Pa • PC = 2πr • Plug in! • A = (1/2)(2πr)a • What’s our “a”? It’s our radius!! • A = (1/2)(2πr)(r) = πr2

  6. Let’s Try One! P = 2πr = 2π(9) = 18π cm ≈ 56.55 cm A = πr2 = π(9)2 = (99)π cm2 = 81π cm2 ≈ 254.47 cm2 Your Turn!! Let the circle shown below have a radius of 9 centimeters, what is the perimeter and the area of the circle?

  7. What if we want to know the area of the shaded region around an inscribed polygon? Area of a Circle – Area of a Square: (490.87 in2) – (312.5 in2) 178.37 in2 Area of the Square: = s2 s2 + s2= (2r)2, where r = 12.5 2s2 = (25)2 = 625 s2 = 312.5 in2 Square root both sides! s = 17.68 inches Area of a Circle: = πr2 = π(12.5)2 = 156. 25π in2 ≈ 490.87 in2 Let r = 12.5 inches What do we need to calculate??

  8. Areas of Irregular Figures Objective: TLW find areas of irregular figures on and off the coordinate plane. SOL G.14b

  9. So we’ve already discussed regular polygons … what if we have an irregular figure? What does irregular mean?

  10. An Irregular Figure is a figure that: But how do we solve for the area? • Is not comprised of equal side and angle measurements • Can it be classified like the other polygons we’ve studied? • NO • Do irregular figures exist in our everyday lives? What are some examples?

  11. The area of an Irregular figure is the: • Sum of all it’s distinct parts • What do I mean by distinct parts?? • Non-overlapping figures that can be combined to create the irregular figure • These parts can be made up of rectangles, squares, triangles, circles, and other polygons! • Let’s think about our original figure, how could we break it up? How would breaking our image up help us to find the area?

  12. Let’s evaluate that same figure! • Do we have all of the measurements we need? • What is missing? • The base of our bottom Δ • How could we find this? • The Pythagorean Theorem!! • AE = 6 cm • Now what can we do?

  13. Let’s evaluate that same figure! Area of ΔCFB = (1/2) bh = (1/2) (8 cm) (4 cm) = 16 cm2 Area of ☐BFDE = (sides)2 = (8 cm)2 = 64 cm2 Area of ΔBEA = (1/2) bh = (1/2) (6 cm) (8 cm) = 24 cm2 Now add it all together to get your Area! A = 104 cm2

  14. What if we have a figure that is missing a portion of it’s area?

  15. Consider the image from the previous slide … • Can this figure be separated into other figures? • Yes • What are they?

  16. Consider the image from the previous slide … I’m going to set up the problem for the area of a rectangle + a semi-circle, minus the triangle. Also, assume the triangle is an equilateral. Area of the rectangle = lw = 19 in 6 in = 114 in2 YOUR TURN!!!! Area of the semi-circle = (1/2)πr2 = (1/2)π (3 in)2 = 4.5π in2 Area of the figure = (114 in2 – 9√3 in2) + 4.5π in2 = 112.5 in2 Area of the Δ = (1/2) bh = (1/2) (6) (3√3) = 9√3 in2

  17. What if we were on the coordinate plane? What do we need to know about a figure to evaluate it? • Consider the image to your left! • How can we break this image into two portions that we can evaluate with our prior knowledge? Area of Trapezoid: = (1/2) h(b1 + b2) = (1/2)(7) (8+6) = 49 u2 Area of Triangle: = (1/2) bh = (1/2)(6)(3) = 9 u2 Area of Irregular Figure: = 49 u2 + 9 u2 = 58 u2

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