1. Tuesday 3/19 or Wednesday 3/20
9-2
Tuesday 3/19 or Wednesday 3/20

2. Bell Work
Find the unknown side lengths in each special right triangle.
1. a 30°-60°-90° triangle with hypotenuse 2 ft
2. a 45°-45°-90° triangle with leg length 4 in.
3. a 30°-60°-90° triangle with longer leg length 3m

3. A circle is the locus of points in a plane that are a fixed distance from a point called the center of the circle. A circle is named by the symbol  and its center. A has radius r = AB and diameter d = CD.
The irrational number 
is defined as the ratio of
the circumference C to
the diameter d, or
Solving for C gives the formula
C = d. Also d = 2r, so C = 2r.

4. You can use the circumference of a circle to find its area
You can use the circumference of a circle to find its area. Divide the circle and rearrange the pieces to make a shape that resembles a parallelogram.
The base of the parallelogram is about half the circumference, or r, and the height is close to the radius r. So A   r · r =  r2.
The more pieces you divide the circle into, the more accurate the estimate will be.

6. Example 1A: Finding Measurements of Circles
Find the area of K in terms of .
A = r2
Area of a circle.
Divide the diameter by 2 to find the radius, 3.
A = (3)2
A = 9 in2
Simplify.

7. Example 1B: Finding Measurements of Circles
Find the radius of J if the circumference is (65x + 14) m.
Circumference of a circle
Substitute (65x + 14) for C.
Divide both sides by 2.
C = 2r
(65x + 14) = 2r
r = (32.5x + 7) m

8. Example 1C: Finding Measurements of Circles
Find the circumference of M if the area is
25 x2 ft2
Step 1 Use the given area to solve for r.
Area of a circle
Substitute 25x2 for A.
Divide both sides by .
Take the square root of both sides.
A = r2
25x2 = r2
25x2 = r2
5x = r

9. Example 1C Continued
Step 2 Use the value of r to find the circumference.
Substitute 5x for r.
Simplify.
C = 2(5x)
C = 10x ft
C = 2r

10. On your own! # 1
Find the area of A in terms of  in which
C = (4x – 6) m.
A = r2
Area of a circle.
A = (2x – 3)2 m
Divide the diameter by 2 to find the radius, 2x – 3.
A = (4x2 – 12x + 9) m2
Simplify.

11. Always wait until the last step to round.
The  key gives the best possible approximation for  on your calculator.
Always wait until the last step to round.
Helpful Hint

12. Example 2: Cooking Application
A pizza-making kit contains three circular baking stones with diameters 24 cm, 36 cm, and 48 cm. Find the area of each stone. Round to the nearest tenth.
24 cm diameter
36 cm diameter
48 cm diameter
A = (12)2
≈ cm2
A = (18)2
≈ cm2
A = (24)2
≈ cm2

13. On your own! # 2
A drum kit contains three drums with diameters of 10 in., 12 in., and 14 in. Find the circumference of each drum.
10 in. diameter in. diameter in. diameter
C = d
C = (10)
C = 31.4 in.
C = d
C = (12)
C = 37.7 in.
C = d
C = (14)
C = 44.0 in.

14. The center of a regular polygon is equidistant from the vertices
The center of a regular polygon is equidistant from the vertices. The apothem is the distance from the center to a side. A central angle of a regular polygon has its vertex at the center, and its sides pass through consecutive vertices. Each central angle measure of a regular n-gon is

15. Regular pentagon DEFGH has a center C, apothem BC, and central angle DCE.

16. To find the area of a regular n-gon with side length s and apothem a, divide it into n congruent isosceles triangles.
The perimeter is P = ns.
area of each triangle:
total area of the polygon:

18. Example 3A: Finding the Area of a Regular Polygon
Find the area of regular heptagon with side length 2 ft to the nearest tenth.
Step 1 Draw the heptagon. Draw an isosceles triangle with its vertex at the center of the heptagon. The central angle is .
Draw a segment that bisects the central angle and the side of the polygon to form a right triangle.

19. Example 3A Continued
Step 2 Use the tangent ratio to find the apothem.
The tangent of an angle is
opp. leg
adj. leg
Solve for a.

20. Example 3A Continued
Step 3 Use the apothem and the given side length to find the area.
Area of a regular polygon
The perimeter is 2(7) = 14ft.
Simplify. Round to the nearest tenth.
A  14.5 ft2

21. The tangent of an angle in a right triangle is the ratio of the opposite leg length to the adjacent leg length. See page 525.
Remember!

22. Example 3B: Finding the Area of a Regular Polygon
Find the area of a regular dodecagon with side length 5 cm to the nearest tenth.
Draw a segment that bisects the central angle and the side of the polygon to form a right triangle.
Step 1 Draw the dodecagon. Draw an isosceles triangle with its vertex at the center of the
dodecagon. The central angle is

23. Example 3B Continued
Solve for a.
The tangent of an angle is
opp. leg
adj. leg
Step 2 Use the tangent ratio to find the apothem.

24. Example 3B Continued
Step 3 Use the apothem and the given side length to find the area.
Area of a regular polygon
The perimeter is 5(12) = 60 ft.
Simplify. Round to the nearest tenth.
A  cm2

25. On your own! # 3
Find the area of a regular octagon with a side length of 4 cm.
Step 1 Draw the octagon. Draw an isosceles triangle with its vertex at the center of the octagon. The
central angle is
Draw a segment that bisects the central angle and the side of the polygon to form a right triangle.

26. On your own! #3 Continued
Step 2 Use the tangent ratio to find the apothem
The tangent of an angle is
opp. leg
adj. leg
Solve for a.

27. On your own! #3 Continued
Step 3 Use the apothem and the given side length to find the area.
Area of a regular polygon
The perimeter is 4(8) = 32cm.
Simplify. Round to the nearest tenth.
A ≈ 77.3 cm2

28. Lesson Quiz: Part I
Find each measurement.
1. the area of D in terms of 
2. the circumference of T in which A = 16 mm2

29. Lesson Quiz: Part II
Find each measurement.
3. Speakers come in diameters of 4 in., 9 in., and 16 in. Find the area of each speaker to the nearest tenth.
Find the area of each regular polygon to the nearest tenth.
4. a regular nonagon with side length 8 cm
5. a regular octagon with side length 9 ft

30. Tuesday 3/19 or Wednesday 3/20
9-3
Tuesday 3/19 or Wednesday 3/20

31. Bell Work
Find the area of each figure.
1. a rectangle in which b = 14 cm and
h = 5 cm
2. a triangle in which b = 6 in. and h = 18 in.
3. a trapezoid in which b1 = 7 ft, b2 = 11 ft, and
h = 3 ft

32. A composite figure is made up of simple
shapes, such as triangles, rectangles,
trapezoids, and circles. To find the area of a composite figure, find the areas of the simple shapes and then use the Area Addition Postulate.

33. Example 1A: Finding the Areas of Composite Figures by Adding
Find the shaded area. Round to the nearest tenth, if necessary.
Divide the figure into parts.
area of half circle:

34. Example 1A Continued
area of triangle:
area of the rectangle:
A = bh = 20(14) = 280 mm2
shaded area:
50 ≈ mm2

35. Example 1B: Finding the Areas of Composite Figures by Adding
Find the shaded area. Round to the nearest tenth, if necessary.
Divide the figure into parts.
area of parallelogram:
A = bh = 8(5)= 40ft2
area of triangle:
shaded area:
= 65 ft2

36. On your own! #1
Find the shaded area. Round to the nearest tenth, if necessary.
Area of rectangle:
A = bh = 37.5(22.5)
= m2
Area of triangle:
Total shaded area is about m2.
= m2

37. Example 2: Finding the Areas of Composite Figures by Subtracting
Find the shaded area. Round to the nearest tenth, if necessary.
area of a triangle:
area of the half circle:
Subtract the area of the half circle from the area of the triangle.
area of figure:
234 –  ≈ ft2

38. Example 2: Finding the Areas of Composite Figures by Subtracting
Find the shaded area. Round to the nearest tenth, if necessary.
area of circle:
A = r2 = (10)2 = 100 cm2
area of trapezoid:
area of figure:
100 –128  cm2

39. On your own! # 2
Find the shaded area. Round to the nearest tenth, if necessary.
area of circle:
A = r2 = (3)2  28.3 in2
area of square:
A = bh  (4.24)(4.24)  18 in2
area of figure: 28.3 – 18 = 10.3 in2

40. Example 3: Fabric Application
A company receives an order for 65 pieces of fabric in the given shape. Each piece is to be dyed red. To dye 6 in2 of fabric, 2 oz of dye is needed. How much dye is needed for the entire order?
To find the area of the shape in square inches, divide the shape into parts.
The two half circles have the same area as one circle.

41. Example 3 Continued
The area of the circle is
(1.5)2 = 2.25 in2.
The area of the square is
(3)2 = 9 in2.
The total area of the shape is 2.25 + 9 ≈ 16.1 in2.
The total area of the 65 pieces is 65(16.1) ≈ in2.
The company will need ≈ 348 oz of dye for the entire order.

42. On your own! # 3
The lawn that Katie is replacing requires 79 gallons of water per square foot per year. How much water will Katie save by planting the xeriscape garden?
Area times gallons of water
375.75(79) = 29,684.25
Subtract water used
29, – 6, =
23,296.5 gallons saved.

43. To estimate the area of an irregular shape, you can sometimes use a composite figure.
First, draw a composite figure that resembles the irregular shape.
Then divide the composite figure into simple shapes.

44. Example 4: Estimating Areas of Irregular Shapes
Use a composite figure to estimate the shaded area. The grid has squares with a side length of 1 ft.
Draw a composite figure that approximates the irregular shape. Find the area of each part of the composite figure.

45. Example 4 Continued
area of triangle a:
area of triangle b:
area of rectangle c:
area of trapezoid d:
A = bh = (2)(1) = 2 ft2
Area of composite figure:
= 5 ft2
The shaded area is about 5 ft2.

46. On your own! #4
Use a composite figure to estimate the shaded area. The grid has squares with side lengths of 1 ft.
Draw a composite figure that approximates the irregular shape. Find the area of each part of the composite figure.

47. On your own! #4 Continued
area of triangle:
area of half circle:
area of rectangle:
A = lw = (3)(2) = 6 ft2
The shaded area is about 12 ft2.

48. Lesson Quiz: Part I
Find the shaded area. Round to the nearest tenth, if necessary. 1. 2.

49. Lesson Quiz: Part II 3.
Mike is remodeling his kitchen. The countertop he wants costs $2.70 per square foot. How much will Mike have to spend on his remodeling project?

50. 9-4
Thursday 3/21

51. Bell Work
Use the slope formula to determine the slope of each line. 1. 2.
3. Simplify

52. Example 1A: Estimating Areas of Irregular Shapes in the Coordinate Plane
Estimate the area of the irregular shape.

53. Example 1A Continued
Method 1: Draw a composite figure that approximates the irregular shape and find the area of the composite figure.
The area is approximately
= 30 units2.

54. Example 1A Continued
Method 2: Count the number of squares inside the figure, estimating half squares. Use a  for a whole square and a for a half square.
There are approximately 24 whole squares and 14 half squares, so the area is about

55. On your own! #1
Estimate the area of the irregular shape.
There are approximately 33 whole squares and 9 half squares, so the area is about 38 units2.

56. Example 2: Finding Perimeter and Area in the Coordinate Plane
Draw and classify the polygon with vertices E(–1, –1), F(2, –2), G(–1, –4), and H(–4, –3). Find the perimeter and area of the polygon.
Step 1 Draw the polygon.

57. Example 2 Continued
Step 2 EFGH appears to be a parallelogram. To verify this, use slopes to show that opposite sides are parallel.

58. Example 2 Continued
slope of EF =
slope of FG =
slope of GH =
slope of HE =
The opposite sides are parallel, so EFGH is a parallelogram.

59. Example 2 Continued
Step 3 Since EFGH is a parallelogram, EF = GH, and FG = HE.
Use the Distance Formula to find each side length.
perimeter of EFGH:

60. Example 2 Continued
To find the area of EFGH, draw a line to divide EFGH into two triangles. The base and height of each triangle is 3. The area of each triangle is
The area of EFGH is 2(4.5) = 9 units2.

61. On your own! #2
Draw and classify the polygon with vertices H(–3, 4), J(2, 6), K(2, 1), and L(–3, –1). Find the perimeter and area of the polygon.
Step 1 Draw the polygon.

62. On your own! #2 Continued
Step 2 HJKL appears to be a parallelogram. To verify this, use slopes to show that opposite sides are parallel.

63. On your own! #2 Continued
are vertical lines.
The opposite sides are parallel, so HJKL is a parallelogram.

64. On your own! #2 Continued
Step 3 Since HJKL is a parallelogram, HJ = KL, and JK = LH.
Use the Distance Formula to find each side length.
perimeter of EFGH:

65. On your own! #2 Continued
To find the area of HJKL, draw a line to divide HJKL into two triangles. The base and height of each triangle is 3. The area of each triangle is
The area of HJKL is 2(12.5) = 25 units2.

66. Example 3: Finding Areas in the Coordinate Plane by Subtracting
Find the area of the polygon with vertices A(–4, 1), B(2, 4), C(4, 1), and D(–2, –2).
Draw the polygon and close it in a rectangle.
Area of rectangle:
A = bh = 8(6)= 48 units2.

67. Example 3 Continued
Area of triangles:
The area of the polygon is 48 – 9 – 3 – 9 – 3 = 24 units2.

68. On your own! #3
Find the area of the polygon with vertices
K(–2, 4), L(6, –2), M(4, –4), and N(–6, –2).
Draw the polygon and close it in a rectangle.
Area of rectangle:
A = bh = 12(8)= 96 units2.

69. The area of the polygon is 96 – 12 – 24 – 2 – 10 = 48 units2.
On your own! #3 Continued
Area of triangles: a b d c
The area of the polygon is 96 – 12 – 24 – 2 – 10 = 48 units2.

70. Example 4: Problem Solving Application
Show that the area does not change when the pieces are
rearranged.

71. Find the area of each shape.
Left figure
Right figure
top triangle:
top triangle:
top rectangle:
top rectangle:
A = bh = 2(5) = 10 units2
A = bh = 2(5) = 10 units2

72. Lesson Quiz: Part I
1. Estimate the area of the irregular shape.
2. Draw and classify the polygon with vertices L(–2, 1), M(–2, 3), N(0, 3), and P(1, 0). Find the perimeter and area of the polygon.

73. Lesson Quiz: Part II
3. Find the area of the polygon with vertices
S(–1, –1), T(–2, 1), V(3, 2), and W(2, –2).
4. Show that the two composite figures cover the same area.