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Using Circumference, Arc Lengths, Perimeter and Area in the Real World

Using Circumference, Arc Lengths, Perimeter and Area in the Real World. Monday, October 13, 2014. How do we solve real world problems?. Lesson 6.7 and 6.8. M2 Unit 3: Day 11. Find the area of the shaded region. 1. 2. 15. 24 cm. 24 cm. A(shaded) = A(square) – A(16 circles)

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Using Circumference, Arc Lengths, Perimeter and Area in the Real World

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  1. Using Circumference, Arc Lengths, Perimeter and Area in the Real World Monday, October 13, 2014 How do we solve real world problems? Lesson 6.7 and 6.8 M2 Unit 3: Day 11

  2. Find the area of the shaded region. 1. 2. 15 24 cm 24 cm A(shaded) = A(square) – A(16 circles) = 576 – 16 (9 ) = 123.6 cm² A(shaded) = A(circle) – A(square) = (15² ) – (15√2*15√2) = (225 ) – (450) = 114.16 m²

  3. Find the perimeter of the region. 3. 138.56 in

  4. 4. Thread A spool of thread contains 150 revolutions of thread. The diameter of the spool is 3 centimeters. Find the length of the thread to the nearest centimeter. 6 cm 2827 cm

  5. 5. A pizza is cut into 10 equal slices. The arc length of one piece of pizza is 4 in. Find the circumference of the pizza.

  6. The dimensions of a car tire are shown at the right. To the nearest foot, how far does the tire travel when it makes15revolutions? = 15 + 2 (5.5) d C = πd Find distance traveled Tire Revolutions 6. SOLUTION Find the diameter of the tire STEP 1 = 26 in. Find the circumference of the tire STEP 2 =π(26) ≈81.68in.

  7. 1 ft 1225.2 in. = 102.1 ft 12 in. 15 81.68 in EXAMPLE 2 Use circumference to find distance traveled STEP 3 Find the distance the tire travels in 15 revolutions. In one revolution, the tire travels a distance equal to its circumference. In 15 revolutions, the tire travels a distance equal to 15 times its circumference. = 1225.2 in STEP 4 Use unit analysis. Change 1225.2 inches tofeet. ANSWER The tire travels approximately 102 feet.

  8. C = πd 12 in. 500 ft. = 6000 in. 1 ft GUIDED PRACTICE 7. A car tire has a diameter of 28 inches. How many revolutions does the tire make while traveling 500 feet? SOLUTION Find the circumference of the tire STEP 1 =π(28) ≈87.92in. STEP 2 Use unit analysis. Change 500 feet to inches.

  9. ANSWER The tire makes 68.2 revolutions. N 87.92 6000 N = 68.24 GUIDED PRACTICE Find the number of revolutions STEP 3

  10. 8. 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.

  11. The company will need 1044.5 ≈ 348 oz of dye for the entire order. 8. 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? 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.

  12. Homework: Page 229 # 24,25 Page 233 # 28,29 Page 235 # 23,26,27

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