Coach Copy January 7, 2015 Warm Up 1. 62 + 82 + 102 + 32 + 92 = 36 + 64 + 100 + 9 + 81 = 290 2. If x = 5, y = 9, and z = 2, find the following: a. x3 = 5 * 5 * 5 = 125 b. y2 = 9 * 9 = 81 c. Z6 = 2 * 2 * 2 * 2 * 2 * 2 = 64 3. If x = 7 and y = 3, find the answer to the following expressions: a. x2 + 5x + 4 = 72 + (5 * 7) + 4 = 88 b. y3 + 4y2 + 7y + 5 = 33 + (4 * 3 * 3) + (7 * 3) + 5 = = 27 + 36 + 21 + 5 = 89 4. The area of a triangle = ½ * base * height Find the area of this triangle. H = 10 Area = ½ * 8 * 10 = 40 B=8 Example Exercises Perimeter 1. The perimeter of a rectangular plot of land that is going to be used as a playground is 42 meters. What are the dimensions of the playground given that the length is three meters more than the width? Make a table. Width Length Perimeter 5 8 26 7 10 34 The width is 9 m and the length is 12 m. 9 12 42 10 13 46 Circumference Note: Circumference = 2 x π x r, where r = radius. π is approximately 3.14159. You can use 3.14 or 22/7. r 2. You just biked 5.5 km. You know that each wheel of your bike has a radius of 35 cm. a. Approximately how far does each wheel travel in just one turn? b. Approximately how many turns does each wheel make in the whole 6-km trip? a. Each wheel travels 2 * π * r in each turn, which is 2 * 3.14 * 35 = 220 cm approximately. b. The wheel makes about 5500/2.2 = 2500 turns. Area of Rectangles and Squares Note: Area of a square or rectangle = length x width 3. Your mom let you build a flower garden in your backyard and gave you 36 meters of fence. How do you build the garden so that it has the largest area possible? You can make a table or guess and check. The largest area is achieved with a square , with the side = 9m, and the area of 81. Area of Circles Note: Area of a circle = π x r2 4. A tinsmith wants to make a baking pan by cutting the largest possible circular disk from a square sheet of tin that measures 30cm on each side. What will be the approximate area of the leftover scraps of tin? The largest possible pan would have a diameter of 30, so the area of the pan is roughly 15 * 15 * π = 707. The area of the square is 900. The area of the scraps is therefore 900-707 = 193 cm2. Exercises 1. Assume that the length of all three sides of a triangle are whole numbers? How many different triangles are there that have a perimeter of 10? Make a table of all the postitive integer combinations that add to 10. A B C 1 1 8 1 2 7 2 2 6 1 3 6 2 3 5 1 4 5 3 3 4 2 4 4 So the possible sides of a triangle with a perimeter of 10 are: 2, 4, 4 and 3, 3, 4. 2. A Norman window is shaped like a rectangle connected with a semicircle. What is the area and the perimeter of the window shown in the drawing below, if the length of the rectangle is 3 meters and the width is 2 meters? The area of the semicircle is 1 * 1 * π / 2 = 1.57, the area of the x 3 = 6. The area of the window is then 7.57 m2. rectangle is 2 The perimeter of the window is equal to three sides of the plus the perimeter of the semicircle. rectangle Rectangle perimeter = 3 + 2 + 3 = 8 Semicircle perimeter = (2 * π * 1)/2 = 3.14 Perimeter of the window = 8 + 3.14 = 11.14 m 3. The hour hand of the clock is 4 inches long and the minute hand is 6 in. long. How far does the tip of each hand travel in 24 hours? The hour hand goes 2 rounds in 24 hours, the tip goes (4 * 2 * π) * 2 = 50 inches. The minute hand goes 24 rounds in 24 hours, the tip goes (6 * 2 * π ) * 24 = 904 inches. 4. A rectangular swimming pool measures 12 meters by 20 meters. A concrete walk will be built around the pool and is 2 meters wide. What is the area of the walk? (2 * 2 * 24) + (2 * 2 * 12) , or (16 * 24) – (12 * 20). The answer is 144. Exercises, cont. Do the following problems if you have time. 5. The dimensions of each segment of the overlapping rectangle and circle are given in cm. Assume that the center of the circle is coincident with the vertex of the rectangle. Find the sum of the areas of the shaded regions, in sq cm. We know the radius of the circle is 2. Sum of the shaded areas = (Area of the rectangle – quarter circle) + (Area of circle – quarter circle) = (8 * 10) – ((π * 22)/4) + ((π * 22) – ((π * 22)/4)) = (80 – 3.14) + (12.57 – 3.14) = 86.3 sq cm 6. A rectangle is divided up into four smaller rectangles. The length of each segment is a whole number of cm. The areas of three of the rectangles are given in the diagram. Find the number of sq cm in the total area of rectangle ACEG. We know the common side of the left rectangles is 3. Therefore the upper left rectangle is 2 x 3, and the lower right rectangle is 3 x 3. Therefore, the upper right rectangle is 5 x 2, and the lower right rectangle is 5 x 3. Therefore the sides of rectangle ACEG are 5 x 8, and the area is 40 sq cm. Homework Answers January 7, 2015 1. ABCD is a rectangle whose area is 12 square units. How many square units are contained in the area of trapezoid EFBA? Trapezoid EFBA = ABCD – Triangle AED – Triangle BCF = 12 – (½ * 3 * 1) – (½ * 3 * 1) = 9 square units 2. A single story house is to be built on a rectangular lot 70 feet wide by 100 feet deep. The shorter side of the lot is along the street. The house must be set back 30 feet from the street. It also must be 20 feet from the back lot line and 10 feet from each side lot line. What is the greatest area that the house can have, in sq ft? Drawing a picture may help you visualize the problem. The lot is 70 ft x 100 ft. The maximum house area is (70 – 10 – 10) * (100 – 30 – 20) = 50 * 50 = 2500 ft2 3. A rectangle 4 cm by 12 cm is divided into four triangles, as shown. The areas of three of the four triangles are stated in the diagram. Find the number of sq cm in the area of the shaded triangle. Area of the rectangle = 4 * 12 = 48 sq cm Area of the shaded triangle = 48 – 16 – 8 – 18 = 6 sq cm 4. An empty carton is opened and flattened to form the figure shown. The carton has both a top and a bottom. Find the total area of the figure shown, in sq cm. The area of the figure is the sum of 6 rectangles. The two small sections on the sides are the sides of the box. The sides of the box are 7 cm long. The flap at the bottom of the figure is 10 – 7 = 3 cm tall. Therefore, the sides of the box are both 7 * 3 = 21 sq cm. The top section of the figure is 9 * 7 = 63 sq cm. The second section from the top is 9 * 3 = 27 sq cm. The third section from the top is 7 * 9 = 63 sq cm. The bottom section is 9 * 3 = 27 sq cm. The total area of the figure is (2 * 21) + (2 * 27) + (2 * 63) = 42 + 54 + 126 = 222 sq cm
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