2. Do NOW 04-15-2014Do NOW 04-15-2014

3. Area of a Regular Polygon: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? P = # of sides  side length = 5  (12 in.) = 60 inches What is the Area? = 247.80 in 2 SOHCAHTOA How do we find the apothem(a)? 360/ 5 = 72 o 72/ 2 = 36 o Tan (36 o ) = (6/a) a = (6/(tan(36 o ))) = 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. Can we use the area of a regular polygon?  Yes!  A = (1/2) Pa  P C = 2πr  Plug in!  A = (1/2)(2πr)a  What’s our “a”? It’s our radius !!  A = (1/2)(2πr)(r) = πr 2 If a circle has an area of A square units and a radius of r units, then: A = πr 2

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

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

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:An Irregular Figure is a figure that:  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? But how do we solve for the area?

11.  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!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!Let’s evaluate that same figure! Area of Δ CFB = (1/2) b  h = (1/2) (8 cm) (4 cm) = 16 cm 2 Area of ☐ BFDE = (sides) 2 = (8 cm) 2 = 64 cm 2 Area of Δ BEA = (1/2) b  h = (1/2) (6 cm) (8 cm) = 24 cm 2 Now add it all together to get your Area ! A = 1 0 4 c m 2

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

15. Consider the image from the previous slide …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 …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 in 2 Area of the semi-circle = (1/2) π r 2 = (1/2)π (3 in) 2 = 4.5π in 2 Area of the Δ = (1/2) b  h = (1/2) (6) (3√3) = 9√3 in 2 Area of the figure = (114 in 2 – 9√3 in 2 ) + 4.5π in 2 = 112.5 in 2

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(b 1 + b 2 ) = (1/2)(7) (8+6) = 49 u 2 Area of Triangle: = (1/2) b  h = (1/2)(6)(3) = 9 u 2 Area of Irregular Figure: = 49 u 2 + 9 u 2 = 58 u 2