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Do- Now 04-14-2014Do- Now 04-14-2014  Warm UP  Go over Test  Review beginning slides from before break  TLW find areas of regular polygons and circles.

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Presentation on theme: "Do- Now 04-14-2014Do- Now 04-14-2014  Warm UP  Go over Test  Review beginning slides from before break  TLW find areas of regular polygons and circles."— Presentation transcript:

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2 Do- Now 04-14-2014Do- Now 04-14-2014  Warm UP  Go over Test  Review beginning slides from before break  TLW find areas of regular polygons and circles.  TEST FRIDAY!!  No Homework

3 Find the area of each Figure Below:Find the area of each Figure Below: 6.79 cm 2 214.16 yd 2 54 u 2

4  What is the definition of a regular polygon?  A Regular Polygon is a convex polygon in which all angles are congruent and all the sides are congruent  Context of the definition; What does convex refer to?  Every internal angle is less than or equal to 180 o  Every line segment between two vertices is in or on the polygon  How could this information assist us in finding the area of a regular polygon?

5 Look at the Hexagon given below: Consider the following questions:  What could we do to hexagon ABCDEF in order to find the height of h ?  Inscribe the polygon in a circle  How does this help us?  We can now evaluate the area using Radius  What would be our radii? GE & GF  How does this help us?  We can now use SOHCAHTOA to find lengths and angle measures h

6 Using the Hexagon given below:  Looking at the image, line segment GH is drawn:  from the center of the regular polygon perpendicular to a side of the polygon.  This perpendicular segment, or height, is called the apothem. (Labeled by the “ h ”)  How does labeling this segment help us find the area of the regular hexagon?  What do we know about the central angles of a regular polygon?  Does this help us? If so, how? h h

7 Using the Hexagon given below: By labeling and defining h as our perpendicular bisector, we know that Δ EGF is an Isosceles Triangle. We know that both of the radii are congruent in our triangle, that h bisects angle EGF making those angles congruent, and they share congruent right angles. h What would happen if we drew line segments from the center t each vertex on the hexagon? we end up with 6 congruent Δ ’s

8 How can we use this information to determine the area of our hexagon in square units?  If we have 6 repeating triangles, how can we used this information to find the area of a hexagon ?  Remember the area of a rhombus h Rhombus: (1/2) b  h = [(1/2)((1/2) d 1 )((1/2) d 2 )]  4 = (1/2) d 1  d 2 Hexagon: (let a = h ) = [(1/2) a  b ]  # of sides = [(1/2) a  b ]  6 = (1/2) 6 b  awhat is 6b?

9 Area of a Regular Polygon:Area of a Regular Polygon:  If a regular polygon has an area of A square units, a perimeter of P units, and an apothem of a units, then: 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 A = (1/2)P  a

10 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?

11 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

12 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!!

13 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


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