1. MM2G3 Students will understand properties of circles. MM2G3 c Use the properties of circles to solve problems involving the length of an arc and the area of a sector. Using Circumference, Arc Lengths, Perimeter and Area in the Real World How do we solve real world problems? M2 Unit 3: Day 11 Lesson 6.7 and 6.8 Monday, September 14, 2015

2. MM2G3 Students will understand properties of circles. MM2G3 c Use the properties of circles to solve problems involving the length of an arc and the area of a sector. Find the area of the shaded region. A(shaded) = A(circle) – A(square) = (15² ) – (15√2*15√2) = (225 ) – (450) = 114.16 m² A(shaded) = A(square) – A(16 circles) = 576 – 16 (9 ) = 123.6 cm² 1. 2. 24 cm 15

3. MM2G3 Students will understand properties of circles. MM2G3 c Use the properties of circles to solve problems involving the length of an arc and the area of a sector. Find the perimeter of the region. 138.56 in 3.

4. MM2G3 Students will understand properties of circles. MM2G3 c Use the properties of circles to solve problems involving the length of an arc and the area of a sector. 2827 cm 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. 4. 6 cm

5. MM2G3 Students will understand properties of circles. MM2G3 c Use the properties of circles to solve problems involving the length of an arc and the area of a sector. 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. 5.

6. MM2G3 Students will understand properties of circles. MM2G3 c Use the properties of circles to solve problems involving the length of an arc and the area of a sector. The dimensions of a car tire are shown at the right. To the nearest foot, how far does the tire travel when it makes 15 revolutions? STEP 1 Find the diameter of the tire STEP 2 Find the circumference of the tire Tire Revolutions SOLUTION d = 15 + 2 (5.5)= 26 in. C= πd= π(26)≈ 81.68 in. Find distance traveled 6.

7. MM2G3 Students will understand properties of circles. MM2G3 c Use the properties of circles to solve problems involving the length of an arc and the area of a sector. EXAMPLE 2 Use circumference to find distance traveled STEP 3Find 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. 1581.68 in = 1225.2 in STEP 4 Use unit analysis. Change 1225.2 inches to feet. 1225.2 in. 1 ft 12 in. = 102.1 ft The tire travels approximately 102 feet.ANSWER

8. MM2G3 Students will understand properties of circles. MM2G3 c Use the properties of circles to solve problems involving the length of an arc and the area of a sector. GUIDED PRACTICE 7. A car tire has a diameter of 28 inches. How many revolutions does the tire make while traveling 500 feet? SOLUTION STEP 1 Find the circumference of the tire C= πd= π(28)≈ 87.92 in. STEP 2 Use unit analysis. Change 500 feet to inches. 500 ft. 12 in. 1 ft = 6000 in.

9. MM2G3 Students will understand properties of circles. MM2G3 c Use the properties of circles to solve problems involving the length of an arc and the area of a sector. GUIDED PRACTICE The tire makes 68.2 revolutions.ANSWER STEP 3 Find the number of revolutions N87.92 6000 = 68.24N

10. MM2G3 Students will understand properties of circles. MM2G3 c Use the properties of circles to solve problems involving the length of an arc and the area of a sector. A company receives an order for 65 pieces of fabric in the given shape. Each piece is to be dyed red. To dye 6 in 2 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. 8.

11. MM2G3 Students will understand properties of circles. MM2G3 c Use the properties of circles to solve problems involving the length of an arc and the area of a sector. The area of the circle is (1.5) 2 = 2.25 in 2. The area of the square is (3) 2 = 9 in 2. The total area of the shape is 2.25 + 9 ≈ 16.1 in 2. The total area of the 65 pieces is 65(16.1) ≈ 1044.5 in 2. The company will need 1044.5 ≈ 348 oz of dye for the entire order. A company receives an order for 65 pieces of fabric in the given shape. Each piece is to be dyed red. To dye 6 in 2 of fabric, 2 oz of dye is needed. How much dye is needed for the entire order? 8.

12. MM2G3 Students will understand properties of circles. MM2G3 c Use the properties of circles to solve problems involving the length of an arc and the area of a sector. Homework: Page 229 # 24,25 Page 233 # 28,29 Page 235 # 23,26,27