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9-2. Tuesday 3/19 or Wednesday 3/20. 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. The irrational number 

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9-2

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  1. 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. The irrational number  is defined as the ratio of the circumference C to the diameter d, or A circleis 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. Ahas radius r= ABand diameter d= CD. 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. 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.

  5. 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.

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

  7. 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. A = r2 Area of a circle 25x2 = r2 Substitute 25x2 for A. 25x2 = r2 Divide both sides by . Take the square root of both sides. 5x = r

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

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

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

  11. 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 A = (18)2 A = (24)2 ≈ 452.4 cm2 ≈ 1017.9 cm2 ≈ 1809.6 cm2

  12. 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 12 in. diameter 14 in. diameter C = d C = d C = d C = (10) C = (12) C = (14) C = 31.4 in. C = 37.7 in. C = 44.0 in.

  13. The center of a regular polygonis equidistant from the vertices. The apothemis the distance from the center to a side. A central angle of a regular polygonhas its vertex at the center, and its sides pass through consecutive vertices. Each central angle measure of a regular n-gon is

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

  15. area of each triangle: total area of the polygon: 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.

  16. 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.

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

  18. 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

  19. Remember! 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.

  20. Step 1 Draw the dodecagon. Draw an isosceles triangle with its vertex at the center of the dodecagon. The central angle is . 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.

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

  22. 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 279.9 cm2

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

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

  25. 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

  26. 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

  27. 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

  28. 9-3 Tuesday 3/19 or Wednesday 3/20

  29. 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

  30. A composite figureis 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.

  31. 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:

  32. Example 1A Continued area of triangle: area of the rectangle: A = bh = 20(14) = 280 mm2 shaded area: 50 + 280 + 84 ≈ 521.1 mm2

  33. 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: 40 + 25 = 65 ft2

  34. On your own! #1 Find the shaded area. Round to the nearest tenth, if necessary. Area of rectangle: A = bh = 37.5(22.5) = 843.75 m2 Area of triangle: Total shaded area is about 1781.3 m2. = 937.5 m2

  35. 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 – 10.125 ≈ 202.2 ft2

  36. 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  186.2 cm2

  37. 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

  38. 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.

  39. 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) ≈ 1044.5 in2. The company will need 1044.5 ≈ 348 oz of dye for the entire order.

  40. 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 29,684.25 – 6,387.75 = 23,296.5 gallons saved.

  41. 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.

  42. 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.

  43. Example 4 Continued area of triangle a: area of triangle b: area of rectangle c: A = bh = (2)(1) = 2 ft2 area of trapezoid d: Area of composite figure: 1 + 0.5 + 2 + 1.5 = 5 ft2 The shaded area is about 5 ft2.

  44. 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.

  45. 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.

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

  47. 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?

  48. 9-4 Thursday 3/21

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