Grade 11 Essential Mathematics Unit 6: Measurement and Geometry

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1 Grade 11 Essential Mathematics Unit 6:

2 INTRODUCTION When people first began to take measurements, they would use parts of the hands and arms. For example, a digit was the width of a thumb. This kind of measurement system lacks consistency because people s bodies are all different sizes. In this unit, you will learn about measurement. Most countries around the world, including Canada, use the metric system. This system of measurement includes units such as centimetres, grams, and litres. The United States uses the Imperial system of measurement, which includes units such as miles, feet, pounds, and gallons. This unit will examine both of these systems of measurement. Assessment: o Lesson 1 Assignment: Metric and Imperial Systems o Lesson 2 Assignment: Using Formulas o Lesson 3: Assignment 1: Manipulating Area and Perimeter Formulas o Lesson 3 Assignment 2: Using Area and Perimeter Formulas o Lesson 4 Assignment: Surface Area Prisms and Cylinders o Lesson 5 Assignment: Surface Area Spheres, Pyramids, Cones o Lesson 6 Assignment: Volume o Lesson 7 Assignment: Manipulating Formulas for Surface Area and Volume o Lesson 8 Assignment: Converting Between Systems Unit Test:

3 Measurement Introduction: History of Measurement When people first began to take measurements, they would use parts of the hands and arms. For example, a digit was the width of a thumb. This kind of measurement system lacks consistency because people s bodies are all different sizes. So, using an arm length or hand span to measure a distance wasn t the best of systems. Today, there are two main systems of measurement. 1. Systeme internationale (SI) also known as the Metric System, it is used in Canada and most of the modern world. 2. Imperial System is the system that is used in the United States LESSON 1: THE METRIC SYSTEM: The metric system is an effective system because it is based on decimal numbering and all measurements are multiples of 10. This means that you can easily convert from one measurement to another just by moving the decimal. (Because moving the decimal is the same as multiplying or dividing by a multiple of 10) The primary unit (Base Unit) for measuring length is the meter. Table 1: Base Units (Metric System) Base Quantity Base Unit Symbol Length metre m Area square metre m 2 Volume cubic metre litre m 3 L Mass gram g Time second s Temperature degree Celsius C

4 Convert 150 metres to kilometres. We want to convert from Meters to Kilometers. We start at the meters and then count how many jumps are needed to get to Km. To get from meters to kilometers it is three jumps to the LEFT. This means that we move the decimal place three times to the LEFT. We have 150 m which is a whole number this means that the decimal is right at the end of the number. Convert 3500 centimetres to metres. So 150 m = km We want to convert from Centimeters to meters. We start at the centimeters and then count how many jumps are needed to get to meters. To get from meters to centimeters it is two jumps to the LEFT. This means that we move the decimal place two times to the LEFT. We have 3500 cm which is a whole number this means that the decimal is right at the end of the number. So 3500 cm = 35 m

5 Convert 14.3 km to metres. We want to convert from Kilometers to Meters. We start at the kilometers and then count how many jumps are needed to get to meters. To get from kilometers to meters it is three jumps to the RIGHT. This means that we move the decimal place three times to the RIGHT. There will be blank spaces that we fill in with zeros Convert 15.3m to mm. So 14.3 km = m We want to convert from Meters to Millimeters. We start at the meters and then count how many jumps are needed to get to millimeters. To get from meters to millimeters it is three jumps to the RIGHT. This means that we move the decimal place three times to the RIGHT. There will be blank spaces that we fill in with zeros So 15.3 km = m

6 THE IMPERIAL SYSTEM: The disadvantage of the Imperial system of measurement is that you cannot easily convert from one measurement to another as easily as we did in the Metric System. There is no common conversion factor for the imperial system; therefore, we CANNOT just move the decimal place to convert between measurements. The primary unit (base unit) for measuring length is the foot. Imperial Units Conversion Chart 1 mile = 1760 yards = 5280 feet = inches Length 1 yard = 3 feet = 36 inches 1 foot = 12 inches 1 gallon = 4 quarts Capacity 1 quart = 2 pints 1 pints = 2 cups 1 cup = 8 fluid ounces Mass 1 ton = 2000 pounds 1 pound = 16 ounces Convert 3.25 feet into inches. Convert 6 miles into yards. x inches 12 inches = 3.25 feet 1 foot 12 in x in = ( ) (3.25 ft) = 39 inches 1 ft x yards 6 miles = 1760 yards 1 mile 1760 yd x yd = ( ) (6 mi) = yd 1 mi

7 Convert 27 inches into feet. x feet 27 inches = 1 foot 12 inches x ft = ( 1 ft ) (27 in) = 2.25 ft 12 in Convert feet into miles. x miles ft = 1 mi 5280 ft x mi = ( 1 mi ) (45628 ft) = 8.6 miles 5280 ft

8 Lesson 1 Assignment: Metric and Imperial Systems Grade 11 Essentials Math See your teacher for Lesson 1 Assignment

9 Formulas: Grade 11 Essentials Math

10

11 LESSON 2: USING FORMULAS Area and Perimeter The circular target for a parachute jump has a diameter of 22.4m. To make the area more visible, the circle is filled in with chalk. One bag of chalk can cover an area of 15.5m 2. How many bags of chalk are required? 22.4m r = d 2 = = 11.2 m A = πr 2 = (3.14)( ) = (3.14)(125.44) = m 2 # of Bags = total area area one bag covers = = 25.4 bags 15.5 Shawna has a farm that is shaped like a parallelogram. The base of her land is 525m and the height is 750m. Suppose that one truck load of seed is required to plant an area of m 2, how many bags will she need to plant the entire field? 750 cm 525 cm A = bh = (525)(750) = m 2 # of trucks = # of trucks = total area area one truck covers = 31.5 truck loads

12 Find the area of the shaded portion in the diagram below. We are going to find the area of the entire rectangle first. Then we will find the area of the circle. Once we have those, we can determine the area of the shaded portion by subtracting the area of the circles from the area of the rectangles. A rectangle = ab = (10)(10) = 100 in 2 A circle = πr 2 = (3.14)(2 2 ) = (3.14)(4) = in 2 There are 4 circles so the total area for all four is: (4)(12.56) = in 2 A shaded = = in 2

13 Find the perimeter and area of the shape in the diagram below. Let us find the Perimetre first. In order to do this we need to know the length of the hypotenuse. a = 18 4 = 14 m b = = 13 m a 2 + b 2 = c = x = x = x 2 x = 365 = 19.1 m P = = m We know can find the area of each of the sections. We have to divide the original shape into sections that we have formulas for, the original shape is an irregular shape for which we don t have a formula. A 1 = LW = (18)(26) = 468 m 2 A 2 = bh 2 = (13)(14) 2 A total = = 611 m 2 = 91 m 2 A 3 = LW = (4)(13) = 52 m 2

14 Curriculum Outcomes: 11E3.G Develop an understanding of spatial relationships Lesson 2 Assignment: Using Formulas See your teacher for Lesson 2 Assignment

15 LESSON 3: MANIPULATING AREA AND PERIMETER FORMULAS We will be using the formulas for area and perimeter to solve for one of the missing lengths given an area or perimeter. If the area of a rectangular backyard is 90m 2 and it has a length of 11.25m what is the width of the yard? a = b =? A = 90 m 2 90 = (11.25)(b) 90 = (11.25)(b) m = b If the area of a triangle is 125cm 2 and the base is 15cm long how tall is this triangle? A = 125 cm 2 h =? 15 cm A = bh = (15)(h) = (7.5)(h) h = = 16.7 cm

16 The perimeter of a rectangle is 68.8 feet and it has a length of 17.5 feet, what is the width? b =? P = 68.8 ft P = 2a + 2b a = 17.5 ft 68.8 = (2)(17.5) + (2)(b) 68.8 = 35 + (2)(b) = (2)(b) 33.8 = (2)(b) = b b = 16.9 ft A circle has an area of cm 2 what is the radius? r =? A = cm 2 A = r = (3.14)(r 2 ) = r2 36 = r 2 36 = r r = 6 cm

17 A circle has a circumference of mm, what is the diameter? r =? C = cm C = 2 r = (2)(3.14)(r) = (6.28)(r) = r r = 5.5 mm d = 2r = (2)(5.5) = 11 mm You are painting one all in your room. It measures 12 feet long and has an area of 84 ft 2. What is the height of the wall? If one can of paint covers 25 ft 2, how many cans will you need to paint your wall? H =? A = LW 84 = (12)(H) L = 12ft = H H = 7 ft # of cans of paint = total area to paint area one can will paint = 84 = 3.36 cans of paint 25

18 A rectangular shaped window has a semi-circle stained glass decorative piece attached to the top. If the area of the entire window is 215.2m 2 and the stained glass portion has a diameter of 5m what are the dimensions of the rectangular window? A total = m 2 d = 5m r = d 2 = 5 2 = 2.5m A half circle = r2 2 = (3.14)(2.52 ) = (3.14)(6.25) = = 9.8 m A rectangle = = 205.4m 2 A = LW = (5)(W) = (5)(W) m = W

19 Curriculum Outcomes: 11E3.G Develop an understanding of spatial relationships Lesson 3: Assignment: Manipulating Area and Perimeter Formulas See your teacher for Lesson 3 Assignment

20 Curriculum Outcomes: 11E3.G Develop an understanding of spatial relationships Lesson 3 Assignment 2: Using Area and Perimeter Formulas See your teacher for Lesson 3 Assignment 1

21 LESSON 4: FINDING SURFACE AREA Prisms and Cylinders Find how much paint is required to cover a wooden cabinet that has dimensions of 24 inches wide, 48 inches tall and 16 inches deep. You will need two coats of paint and one can of paint will cover 8 ft 2. SA = 2LW + 2LH + 2WH 48 in = = 4 ft SA = (2)(2)(1.33) + (2)(2)(4) + (2)(1.33)(4) SA = SA = ft 2 24 in = = 2 ft 16 in = 16 = 1.33 ft 12 Need two Coats of Paint: SA Total = (2)(31.96) = ft 2 # of cans = total area to paint area one covers # of cans = 64 8 = 8 cans of paint

22 A water tower is cylindrical in shape with a height of 8.6m and a diameter if 33m. How much surface area would be painted if the top and sides of three water towers need to be re-painted? d = 33m diameter = 33m radius = r = d 2 = 33 2 = 16.5 m H = 8.6m Reading the question it says that the top and the sides are painted which means that the bottom of the tower is not painted. So we need to make sure that we change the formula accordingly. SA = 2πrh + 2πr 2 SA = 2πrh + πr 2 SA = (2)(3.14)(16.5)(8.6) + (3.14)( ) SA = = = 1746 m 2 There are three towers to be painted: SA = (3)(1746) = 5238 m 2

23 The wood that Terrance wants to use to make a shelving unit costs $6.49/ft 2. How much will it cost him (assuming no wastage) to make a shelving unit that is 4 ft wide by 12 inches deep by 5 ft tall if there are 4 shelves (plus the top and bottom)? We don t have a formula that is in this shape so we need to look at the shapes individually. There are two sides on this bookcase. Back of the bookcase: 5 ft 5 ft 12 = 12inches = 1 foot 4 ft A = LW = (4)(5) = 20 ft 2 A sides = 2LW = (2)(5)(1) = 10 ft 2 Shelves and the top and bottom: Six in total 1 ft 4 ft A = 6(L)(W) = (6)(4)(1) = 24 ft 2 Total Surface Area = = 54 ft 2 Cost = ($6.49)(54) = $350.46

24 You are looking for a new tent (triangle prism) to go camping this summer. You have found one that has a height of 5 feet and 6 inches, is 5 feet wide and 7 feet long. You want to know how much material was used to make this tent. Again we don t have a formula for this shape. We need to split this into shapes for which we have formulas. 5.5 ft 5 ft 7 ft Front and Back: 5.5 ft SA front/back = 2 ( bh 2 ) = 2 ((5)(5.5)) = 2(13.75) = 27.5 ft ft Botttom: SA = LW = (5)(7) = 35 ft 2 7 ft 5 ft

25 Sides: We don t know that length of the slanted side of the tent. We will need to find this before we can find the area of the sides. s 7 ft In order to find the length of the slanted side we need to use Pythagorean Theorem. To do this, we need to have a right angle triangle = s ft s = s = s 2 5 ft 2.5 ft s = 36.5 = 6.04 ft Now that we have the length of the slanted side we can use it to determine the area of the sides of the tent. A sides = 2LW = (2)(7)(6.04) = 84.6 ft 2 SA total = = ft 2

26 Curriculum Outcomes: 11.E3.G.1 Solve problems that involve SI and imperial units in surface area measurements Grade 11 Essentials Math Lesson 4 Assignment: Surface Area Prisms and Cylinders See your teacher for Lesson 4 Assignment

27 LESSON 5: FINDING SURFACE AREA Pyramids, Spheres, and Cones Determine the surface area of a cone if it has a radius of 3.4m and a slant height of 12.2m. 3.4m SA = πr 2 SA = (3.14)(3.4 2 ) 12.3m SA = (3.14)(11.56) SA = 36.3 m 2 What is the surface area of a sphere with diameter of 2.5 feet? The formula for SA requires the radius, given the diamter so we need to divide by 2 to find the radius. r = d 2 = = 1.25ft SA = 4πr 2 SA = (4)(3.14)( ) SA = (4)(3.14)(1.5625) SA = 19.6 ft 2

28 A child s tent is the shape of a triangular prism. If it has a slant height of 4 feet, a width of 6 feet and a length of 6 feet, what is the surface area of the canvas required to build the tent? What would the surface area of the tent be if it were a square based pyramid in shape? We don t have a formula for this shape so we will need to find the area of each of the pieces that make up this tent. SA sides = LW = (6)(4) = 24ft 2 4ft There are two sides so: SA sides = (2)(24) = 48 ft 2 SA bottom = LW = (6)(6) = 36ft 2 6ft 6ft We now need to find the SA of the triangular sides To do this we need to know the height of the tent. We will use Pythagorean Theroem to determine the height. We will need a Right Angle Triangle to use Pythagorean Theroem a 2 + b 2 = c b 2 = b 2 = 16 b 2 = 16 9 b a = 3 ft 6 ft c = 4 ft b = 7 = 2.6 SA front back = bh 2 = (6)(2.6) 2 = 7.8 ft 2 There are two triangles so: SA front back = (2)(7.8) = 15.6 ft 2 SA total = = 99.6 ft 2

29 We have a formula for the Surface Area of this shape so we just need to use it. SA = 2sb + b 2 4ft SA = (2)(4)(6) SA = = 84 ft 2 6ft 6ft

30 Curriculum Outcomes: 11.E3.G.1 Solve problems that involve SI and imperial units in surface area measurements Grade 11 Essentials Math Lesson 5 Assignment: Surface Area Spheres, Pyramids, Cones See your teacher for Lesson 5 Assignment

31 LESSON 6: VOLUME Find the volume for the following shapes: V = LWH = (2)(2)(12) = 48cm 3 V = BHL 2 = (20)(15)(30) = 9000 = 4500 mm 2 2 2

32 We need the radius for the Volume formula. r = d 2 = 6 2 = 3 in V = πr 2 h V = (3.14)(3 2 )(4) V = (3.14)(9)(4) V = in 2 Calculate the volume of the space left after a square based pyramid has been removed from the rectangular solid. Assume that the pyramid and the rectangle are the same height. To find the amount of space left we need to know the volume of each shape. V rectangle = LWH = (6)(6)(8.5) = 306 cm 3 V pyramid = b2 h 3 = (62 )(8.5) = (36)(8.5) = 306 = 102 cm V space = = 204 cm 3

33 Paul wants to cover an area of yard that is 10.8m by 9.5 m with 10cm of topsoil. How much topsoil does he need to order? How much will it cost if the soil is $18.75/m 3? H 10cm 10.8m L 9.5m W We need to make sure that all the units are the same. Since the questions is asking for the cost and it is in m 3 it is easiest to have all the units in meters. We need to have the Height in meters, H = 10 cm = 0.10 m V = LWH = (10.8)(9.5)(0.10) = m 3 A grain hopper is in the shape of an inverted cone on the bottom of a cylinder. The hopper has a diameter of 48 inches, a height of 8 feet. The cylinder has a height of 45 inches. How much will the hopper hold once built? 48in First we need to have all the units in the same measure. Let us choose Feet but Inches would also be correct. 45 in H cylinder = 45 inches = = 3.75ft diameter = 48 in = = 4ft 8ft radius = 4 2 = 2ft H cone = = 4.25ft V cylinder = πr 2 h = (3.14)(2 2 )(3.75) = 47.1 ft 3 We need to know the height of the cone. H cone = = 4.25 ft V cone = πr2 h 3 = (3.14)(22 )(4.25) = (3.14)(4)(4.25) = = 17.8ft V total = = 64.9 ft 3

34 Tennis balls are usually sold in cylindrical containers that often hold 3 tennis balls. If each tennis ball has a radius of 3.4cm, what are the dimensions of the cylindrical container that holds three tennis balls? What is the volume of one tennis ball? What is the volume of the container? r r = 3.4 cm The radius of the tennis ball is also the radius of the cylinder Three tennis balls stacked will give us the height of the cylinder Each tennis ball has a diameter of 6.8 cm so three stacked is H cylinder H cylinder = (3)(d) = (3)(6.8) = 20.4 cm Dimensions of the Cylinder: r = 3.4cm H = 20.4 cm V tennis ball = (4)(3.14)(39.3) 3 V tennis ball = 4πr3 3 = (4)(3.14)(3.43 ) 3 = = 164.6cm 3 V cylinder = πr 2 h = (3.14)(3.4 2 )(20.4) = (3.14)(11.56)(20.4) = cm 3

35 Curriculum Outcomes: 11.E3.G.2 Solve problems that involve SI and imperial units in volume and capacity measurements Lesson 6 Assignment: Volume Grade 11 Essentials Math See your teacher for Lesson 6 Assignment

36 LESSON 7: MANIPULATION OF SURFACE AREA AND VOLUME FORMULAS: We won t always be calculating the surface area or volume of objects. Sometimes we are given this information and we need to solve for a side length or a height or the radius of the object. When we are given the Surface Area or the Volume and asked to find the missing side this requires us to manipulate the formula to solve for what we have been asked to determine. The surface area of paper required to cover a soup can is cm 2. If the radius of the soup can is 4 cm, how tall is it? r = 4cm SA = cm 2 The label on a soup can only goes around the can so we don t need to include the top and the bottom in the formula for Surface Area. /// SA = 2πr 2 + 2πrh = 2πrh Now that we have the correct formula we will substitute the values in where they belong in the formula and solve for the height SA = 2πrh = (2)(3.14)(4)(h) = (25.12)(h) = (25.12)(h) cm = h

37 What is the radius of a soup can if the height is 7 cm and its volume is cm 3? Grade 11 Essentials Math V = cm 3 h = 7in Here we are given the Volume of the can. V = πr 2 h Now substitute in the values that are given in the question = (3.14)(r 2 )(7) We want to solve for the radius so we will simplify the right hand side as much as possible = (21.98)(r 2 ) = (21.98)(r2 ) = r 2 9 = r r = 3cm

38 What is the radius of a cone if its height is 14 inches and the volume is in 3? Grade 11 Essentials Math h = 14in V = in 3 We are given the Volume of a cone: V = πr2 h Substitute in the equation the values given in the question = (3.14)(r2 )(14) 3 Simplify the Right Hand side as much as possible first = (43.96)(r2 ) = (14.65)(r 2 ) = (14.65)(r2 ) = r 2 9 = r r = 3 in What is the radius of a spherical balloon if it has a volume of in 3? We have the volume of a sphere: V = 4πr = (4)(3.14)(r3 ) = (12.56)(r3 ) = (4.19)(r 3 ) = (4.19)(r3 ) = r = r r = 5.5cm

39 What is the slant height of a square based pyramid if the surface area 4095m 2 is and the length of its base is 23m? SA = 4095m 2 SA = 2sb + b 2 s 4095 = (2)(s)(23) + (23 2 ) 23m 23m Simplify the right side as much as possible 4095 = (46)(s) We need to subtract the before we can solve for s = (46)(s) = (46)(s) = (46)(s) m = s

40 Curriculum Outcomes: 11.E3.G.3 Solve problems that require the manipulation and application of formulas related to surface area and volume Lesson 7 Assignment: Manipulating Formulas for Surface Area and Volume See your teacher for Lesson 7 Assignment

41 LESSON 8: CONVERTING BETWEEN SYSTEMS Length Conversion Metric (SI) unit Imperial unit 1 millimetre (mm) inches (in) 1 centimetre (cm) 0.39 inches (in) feet (ft) 1 metre (m) inches (in) 3.28 feet (ft) 1.09 yards (yd) 1 kilometre (km) 0.62 miles Examples (round to 1 decimal): a. Convert 2 cm to inches x inches 2 cm = 0.39 inches 1 cm x cm = (2 cm)(0.39 in) 1 cm = 0.78 cm b. Convert 1.7 m to yards x yards = c. Convert 4.5 miles to kilometres x yards 1.7 m = 1.09 yards 1 m (2 cm)(0.39 in) 1 cm x km 4.5 miles = 1 km 0.62 miles = yds x km = (1 km)(4.5 miles) 0.62 miles = 7.26 km d. Convert 6.3 feet to metres x m 6.3 feet = 1 m 3.28 ft x cm = (1 m)(6.3 ft) 3.28 ft = 1.92 m

42 Rebecca is planning to install sod in her backyard, which is 18.2 m by 9.8 m. If sod costs $0.28/ft 2, how much will it cost to sod the backyard? 18.2m 9.8m The first step is to make sure that we have matching units. The cost is given in Feet but the measurements for the yard are in Meters. We need to change the Meters to Feet. Length: x feet 3.28 ft = 18.2 m 1 m x ft = (18.2 m)(3.28 ft) 1 m = ft Width: x feet 9.8 m = 3.28 ft 1 m x ft = (9.8 m)(3.28 ft) 1 m = ft Now we can find the area of the yard: A = LW = (59.07)(32.14) = ft 2 Cost = ($0.28)( ) = $537.25

43 You would like to replace the carpet in your living room and have measured it to be 12ft by 15ft. The price of the carpet that you want is $24.99/m 2. How much carpet do you need to buy? What will this cost you (before taxes)? 12 ft 15 ft The first step is to make sure that we have matching units. The cost is given in Meters but the measurements for the yard are in Feet. We need to change the Feet to Meters. x m = (12 ft)(1 m) 3.28 ft Length: = 3.66 m x m 12 ft = 1 m 3.28 ft Width: x m 15 ft = 1 m 3.28 ft x m = (15 ft)(1 m) 3.28 ft = 4.57 m Now we can find the area of the yard: A = LW = (3.66)(4.57) = m 2 Cost = ($24.99)(16.73) = $ A landscaper needs to estimate the amount of material required for a project building a circular brick patio for a client. The diameter of the patio is 13m and one bundle of brick will cover 116ft 2. How many bundles does he need? 13m Need the units in Feet: x ft 13 m = 3.28 ft 1 m (13m)(3.28 ft) x ft = = ft 1 m We will need the Radius: r = d 2 = = ft 2 A = πr 2 = (3.14)( ) = (3.14)(454.54) = ft 2 Number of Bundles = Total area to cover area one bundle covers = = 12.3 bundles 116

44 Curriculum Outcomes: 11.E3.G.3 Solve problems that require the manipulation and application of formulas related to surface area and volume Lesson 8 Assignment: Converting Between Systems See your teacher for Lesson 8 Assignment

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