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Measurement Chapter 12

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DAY 1

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Two principal systems of Measurement The U. S. Customary (English) System – used in the United States but in almost no other country. The International System (metric) – used by all countries worldwide including the United States.

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In early times, units of measurement were defined more for convenience than accuracy. Inch – breath of thumb Inch - 3 grains of barley, taken from the middle of the ear and laid end to end. Foot – length of one’s foot Yard – Circumference of King Henry I’s waist Yard – Distance from one’s nose to thumb when arm is extended to one’s side.

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Pace – distance of two steps of a marching army Mile – 1000 paces Acre – the amount of land a yolk of oxen could plow in one day.

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Scruple = 20 grains 3 scruples = 1 dram Rod = 16 ½ feet Rod – the total length of the left feet of the first 16 men coming out of church on Sunday morning.

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How many ounces in a pound? How many ounces in a pint?

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16 ounces = 1 pound 16 ounces = 1 pint Absolutely no connection between the two!

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Drugs, gold, common things? Troy pound Apothecary pound Avoirdupois pound

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Troy pound = 12 ounces (Gold, Silver, precious metals) Apothecary pound = 12 ounces (Drugs) Avoirdupois pound = 16 ounces (Common things)

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U. S. Customary System (English) 12 inches = 1 foot 3 feet = 1 yard 5280 feet = 1 mile 8 fluid ounces = 1 cup 2 cups = 1 pint 2 pints = 1 quart 4 quarts = 1 gallon

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U. S. Customary System (English) 2000 pounds = 1 ton Water freezes at 32°F Water boils at 212°F

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An important practical purpose of measurement is communication. As commerce developed and goods were traded over increasingly large distance, the need for a standard system of units became more and more apparent.

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French mathematicians were enlisted to come up with a new system of measurement around They calculated the distance from the North Pole to the equator and divided that measurement by 10,000,000. One ten-millionth of that distance, they called “meter” which means “measure.”

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(They actually learned years later that they had calculated the distance incorrectly. The measurement of a meter did not change.)

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1840 France went totally metric. It was against the law to use anything else. At that time England and France were bitter enemies. England refused to use the new system of measurement.

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Because of the close ties that the United States had with England, we also chose to stay with the English system of measurement. Thomas Jefferson and John Quincy Adams tried to convince Congress to make they change but they were voted down.

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By the time the United States had broken such close ties to England, they had also become a world power. While the other nations were converting to the metric system, the United States felt like if they wanted to trade with us, they would use our system.

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The Metric Conversion Act was passed in the 1970’s. The Metric Conversion Act called for a gradual, voluntary change to the metric system. The United States was the only country of any size that had not changed to the metric system. Canada was waiting on us put finally changed – overnight!

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Meter (m) The basic metric unit used to measure length. A little longer than a yard About the distance from the floor to a doorknob

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1000 m = 1 kilometer (km) 100 m = 1 hectometer (hm) 10 m = 1 dekameter (dkm or dam) Meter (m) 1 m = 10 decimeters (dm) 1 m = 100 centimeters (cm) 1 m = 1000 millimeters (mm)

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1000 m = 1 km* 100 m = 1 hm 10 m = 1 dkm Meter (m)* 1 m = 10 dm 1 m = 100 cm* 1 m = 1000 mm*

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1 km is a little more than ½ mile. Compare 1 cm to ½ inch. The thickness of a dime is close to 1 mm.

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When using the metric system internationally, we use spaces instead of commas.

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1495 mm = ________ m

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1495 mm = m 29.4 cm = _____ mm

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1495 mm = m 29.4 cm = 294 mm 38,741 m = _____ km

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1495 mm = m 29.4 cm = 294 mm 38,741 m = km

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Liter (L) The liter is the basic metric unit used to measure volume or capacity. A liter is the volume of 1 tenth of a meter cubed. (1 dm 3 ) A liter is a little more than a quart.

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1000 L = 1 kL 100 L = 1 hL 10 L = 1 dkL Liter (L)* 1 L = 10 dL 1 L = 100 cL 1 L = 1000 mL*

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1 mL is about 1 drip from an eye dropper. 1 mL = 1cm 3 1 mL = one “cc”

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Gram (g) The gram is the basic metric unit for measuring weight or mass. A gram is the weight of 1 cm 3 of water. A gram is about the weight of a paper clip. A gram is about the weight of a dollar bill.

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1cm³ holds 1 mL and weighs 1 g

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1000 g = 1 kg* 100 g = 1 hg 10 g = 1 dkg Gram (g)* 1 g = 10 dg 1 g = 100 cg 1 g = 1000 mg*

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1 kg is a little over 2 pounds. 1 grain of salt weighs about 1 mg.

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5g 200kg 2kg 2mg 100g 9kg Nickel Compact Automobile Two-liter bottle of coke Recommended daily allowance of B-6 Size D Battery Large Watermelon

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Celsius Water freezes at 0°C Water boils at 100°C Normal body Temperature is 37°C Normal Room Temperature is 23°C

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Unit Analysis Unit Analysis is a procedure that will help you arrange your calculation to make it easier to know when to multiply and when to divide when converting from one unit to another.

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3.75 miles = ______ yards 5280 feet = 1 mile 3 feet = 1 yard 3.75 miles = 6600 yards

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60 miles/hour = _____ feet/sec 1 mile = 5280 feet 1 hour = 60 minutes 1 minute = 60 seconds 60 miles/hour = 88 feet/sec

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1495 mm = _____ m

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29.4 cm = _____ mm

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38,741 m = _____ km

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A fish tank at the aquarium has the shape of a rectangular prism 2m deep by 3m wide by 3m high. What is the capacity in liters?

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1 m = 100 cm 1 cm 3 = 1 mL 1000 mL = 1 L

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18,000 L

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Day 2

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Homework Questions Page 752

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Unit Analysis

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Perimeter The perimeter of any 2 dimensional object is the distance around that object. To measure distance, we need only one dimension, length. 1 centimeter:

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Use your centimeter ruler to find the perimeter of each polygon. Measure to the nearest tenth of a centimeter.

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Area The area of any two dimensional object is the amount of space in the interior of the object. In order to fill the inside of a two dimensional object, we need two dimensions, length and width. You can fill a two dimensional object with squares. 1 square centimeter:

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COUNT the number of squares inside each of the polygons to find the area. You may have to approximate. State your answer to the nearest whole square centimeter.

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Rectangular Array Model for Multiplication Finding the number of squares in a rectangle can be done more quickly by counting the squares in each row and multiplying by the number of columns of squares. If the squares are not marked, measuring the base of a rectangle will tell how many squares we could put in each row. Measuring the height of the rectangle will tell how many rows of squares we will have.

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The AREA of a rectangle (number of square units) is the measure of the BASE multiplied by the measure of the HEIGHT. Rectangle: AREA = BASE x HEIGHT

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Square is a special kind of rectangle. (All squares are rectangles) Square: AREA = BASE x HEIGHT

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Use your centimeter ruler to measure the dimensions of the rectangle and square to the nearest tenth of a centimeter. Use the formula to find the area of each.

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Parallelogram

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A PARALLELOGRAM can be dissected and rearranged to make a rectangle. Parallelogram: AREA = BASE x HEIGHT

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Trapezoid

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A TRAPEZOID can be dissected and rearranged to make a rectangle. Trapezoid: AREA = Average of the BASES x HEIGHT A = (b 1 + b 2 ) x h 2

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Use your centimeter ruler to measure the dimensions of the parallelogram and trapezoid. Use the formula to find the area. MAKE SURE THE MEASUREMENT FOR HEIGHT IS THE LENGTH OF A LINE SEGMENT PERPENDICULAR TO THE BASE AND REACHING THE HEIGHEST POINT OF THE POLYGON.

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Compare the area you got by counting to the area you computed by use of the formulas.

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Use the remaining square, rectangle, and 2 parallelograms to make triangles. Cut each quadrilateral in half diagonally to make a right-isosceles triangle, right- scalene triangle, obtuse triangle, and acute triangle. Glue ONE of each type triangle inside the quadrilateral it came from.

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Perimeter Use your centimeter ruler to measure the lengths of the sides of each TRIANGLE to the nearest tenth of a centimeter. Find the perimeter of each TRIANGLE.

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AREA Approximate the area of each TRIANGLE by COUNTING the number of squares in each. How do the measurements for area you counted compare to the measurements for area of the quadrilaterals you counted on page 1?

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Notice that the area formula for each of the quadrilaterals we used to make triangles is A = b x h. We made each triangle by cutting the quadrilaterals in half. Triangle: AREA = ½ x BASE x HEIGHT

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Use your centimeter ruler to measure the base and height of each triangle. Use the formula to find the area. MAKE SURE THE MEASUREMENT FOR HEIGHT IS THE LENGTH OF A LINE SEGMENT PERPENDICULAR TO THE BASE AND REACHING THE HEIGHEST POINT OF THE TRIANGLE.

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How do your answers for area compare when your counted the squares to when you used the formula?

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Circumference Just as the perimeter of a polygon is the distance around it, the CIRCUMFERENCE of a circle is the distance around the circle. The circumference is only one dimension, length.

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Cut a string the length of the distance around the circle. Measure the string to the nearest tenth of a centimeter. Cut a piece of string the length of the diameter. Measure to the nearest tenth of a centimeter.

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Pi (Π) Pi is defined as the ratio of the circumference of any circle to the diameter of the circle. Use your measurements from your circle and divide. Round to the nearest hundredth. Circumference ÷ diameter

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Pi Your answer should be very close to Irrational number: non-repeating, non- terminating decimal. Most common approximations are 3.14 and 22/7. If an exact answer is required, Π will be part of the answer.

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Use this formula, your measurement for diameter, and 3.14 for pi to find the circumference. How does it compare to the string you measured?

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Area Count the square centimeters in the circle to get the approximate area. Cut your circle to make a parallelogram. What part of the circle is the height of the parallelogram? What part of the circle is the base of the parallelogram?

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Parallelogram Area = base x height Area = ½ Circumference x radius Area = ½ (Πd) x r Area = Π r x r Area = Π r 2

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Use this formula, your measurement for radius to the nearest tenth of a centimeter (half the diameter you measured), and 3.14 for pi to find the area to the nearest hundredth. How does this compare to the approximation you counted.

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Example 12.9a Page 761

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Example 12.9b

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Example 12.10a

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Example 12.10b

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Find the Perimeter

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Find the Area

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Find the Area of the Yellow Region

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DAY 3

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Homework Questions Page 769

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#4a

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#4b

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#13c

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#15

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#17

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#28

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#42

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There’s Pi in My Circle!

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Pythagorean Theorem If a right triangle has legs of length a and b and its hypotenuse has length c, then a 2 + b 2 = c 2 The square on the hypotenuse is equal to the sum of the squares of the other two sides.

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Find x

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Find y

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Pythagorean Theorem Let a triangle have sides of length a, b, and c. If a 2 + b 2 = c 2, then the triangle is a right triangle and the angle opposite the side of length c is its right angle.

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Determine if the three lengths are the sides of a right triangle

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Determine if the three lengths are the sides of a right triangle. __ 1055√3

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Determine if the three lengths are the sides of a right triangle

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Two young braves and three squaws are sitting proudly side by side. The first squaw sits on a buffalo skin with her 50 pound son. The second squaw is on a deer skin with her 70 pound son. The third squaw, who weighs 120 pounds, is on a hippopotamus skin. Therefore, the squaw on the hippopotamus is equal to the sons of the squaws on the other two hides.

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Note measurements on 2c, Page 788 Note 3 dimensional measurements Note #5

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Area on Geo-Boards

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DAY 4

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Homework Questions Page 783

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#9

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#41

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Area and Perimeter Lab

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DAY 5

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Lab Questions

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Surface Area The surface area of any polyhedron is the sum of the areas of its faces.

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Find the Surface Area:

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SA= front + back + right + left + top + bottom

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bh b = 4 h = 6 4·6 24

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SA= front + back + right + left + top + bottom bh 24 b = 4 h = 6 4·6 24

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SA= front + back + right + left + top + bottom bh 24 bh b = 4 b = 5 h = 6 h = 6 4·6 5·

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SA= front + back + right + left + top + bottom bh 24 bh 30 b = 4 b = 5 h = 6 h = 6 4·6 5·

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SA= front + back + right + left + top + bottom bh 24 bh 30 bh b = 4 b = 5 b = 4 h = 6 h = 6 h = 5 4·6 5·6 4·

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SA= front + back + right + left + top + bottom bh 24 bh 30 bh20 b = 4 b = 5 b = 4 h = 6 h = 6 h = 5 4·6 5·6 4·

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SA= front + back + right + left + top + bottom bh 24 bh 30 bh 20 b = 4 b = 5 b = 4 h = 6 h = 6 h = 5 4·6 5·6 4· SA = = 148 cm²

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Find the Surface Area:

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SA = base + 4 Triangles

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bh b = 5 h = 5 5·5 25

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SA = base + 4 Triangles bh 4(½bh) b = 5 h = 5 h = 6 5·5 4(½·5·6) 25 4(3·5) 4(15) 60

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SA = base + 4 Triangles bh 4(½bh) b = 5 h = 5 h = 6 5·5 4(½·5·6) 25 4(3·5) 4(15) 60 SA = = 85 cm²

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Find the Surface Area

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SA = top + bottom + 8 little + 4 big rectangles rectangles

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TOP

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SA = top + bottom + 8 little + 4 big rectangles rectangles 800 cm²

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SA = top + bottom + 8 little + 4 big rectangles rectangles 800 cm² +

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Little Rectangle

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SA = top + bottom + 8 little + 4 big rectangles rectangles 8(50) 800 cm² cm² cm² +

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Big Rectangle

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SA = top + bottom + 8 little + 4 big rectangles rectangles 4(200) 800 cm² cm² cm² cm²

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SA = top + bottom + 8 little + 4 big rectangles rectangles 4(200) 800 cm² cm² cm² cm² SA = 2800 cm²

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Find the Surface Area

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SA = base + 4 triangles

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bh b = 10 h = 10 (10)(10) 100cm²

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SA = base + 4 triangles bh4(½bh) b = 10b = 10 h = 10h = ?? (10)(10) 100cm²

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SA = base + 4 triangles bh4(½bh) b = 10b = 10 h = 10h = ?? (10)(10) 100cm²

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SA = base + 4 triangles bh4(½bh) b = 10b = 10 h = 10h = 12 (10)(10)4(½)(10)(12) 100 cm² cm²

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SA = base + 4 triangles bh4(½bh) b = 10b = 10 h = 10h = 12 (10)(10)4(½)(10)(12) 100 cm² cm² SA = 340 cm²

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Find the Surface Area

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SA = Top + Bottom + Middle

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Πr² r = 5 Π(5)² 25Π

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SA = Top + Bottom + Middle Πr² 25Π r = 5 Π(5)² 25Π

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SA = Top + Bottom + Middle Πr² 25Πbh r = 5b = C of circle Π(5)²b = Πd 25Πd = 10 b = 10Π h = 12 (10Π)(12) 120Π

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SA = Top + Bottom + Middle Πr² 25Πbh r = 5b = C of circle Π(5)²b = Πd 25Πd = 10 b = 10Π SA = 25Π + 25Π + 120Π h = 12 SA = 170Π cm² (10Π)(12) 120Π

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Surface Area of a cone: SA = Πr² + Πrs (s = slant height) Surface Area of a sphere: SA = 4Πr²

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Grocery Lab

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DAY 6

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Find the Volume

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Area of the base:Rectangle bh (4)(5) 20 cm²

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Area of the base:20 cm² 20 cm³ will fit in the bottom of the Prism.

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Area of the base:20 cm² 20 cm³ will fit in the bottom of the Prism. The height of the Prism is 6cm so we can make 6 layers of 20 cm³.

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Area of the base:20 cm² The height of the Prism: 6cm Volume of a prism: Area of the base x height of the prism V = (20 cm²)(6cm) = 120 cm³

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Find the Volume

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How many cubes can fit in the bottom of the prism?

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Area of Base = 800 cm² How many layers will fit in the prism? Height of the Prism = 10 cm Volume of prism = Area of base x height of prism Volume = (800 cm²)(10 cm) Volume = 8000 cm³

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What is the volume in liters? Volume = 8000 cm³ Volume = 8000 cm³ = 8000 mL = 8 L

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Find the Volume

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How many cubes fit in the bottom of the cylinder?

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Area of Base = 25Π cm² How many layers fit inside the cylinder?

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How many cubes fit in the bottom of the cylinder? Area of Base = 25Π cm² How many layers fit inside the cylinder? height of cylinder = 12 cm

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Area of Base = 25Π cm² height of cylinder = 12 cm Volume = Area of base x height of cylinder V = (25Π cm²)(12 cm) V = 300 Πcm³

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Find the Volume

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How many cubes will fit on the base of the cone if the sides went 90° up?

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Area of Base = Πr² r = 5 cm A = Π(5)² A = 25Π cm²

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How many cubes will fit on the base of the cone if the sides went 90° up? Area of Base = 25Π cm² How many layers would there be?

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How many cubes will fit on the base of the cone if the sides went 90° up? Area of Base = 25Π cm² How many layers would there be? Height of the cone = 12 cm

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Area of Base = 25Π cm² Height of the cone = 12 cm Volume of a CYLINDER with the same dimensions: V = (25Π cm²)(12cm)

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The volume of a cone is ⅓ the volume of a cylinder with the same dimensions.

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Area of Base = 25Π cm² Height of the pyramid = 12 cm Volume of a CYLINDER = (25Π cm²)(12cm) Volume of the CONE = (25Π cm²)(12 cm) 3

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Area of Base = 25Π cm² Height of the pyramid = 12 cm Volume of a CYLINDER = (25Π cm²)(12cm) Volume of the CONE = (25Π cm²)(12 cm) 3 V = 100П cm³

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Find the Volume

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How many cubes will fit on the base of the pyramid if the sides went 90° up?

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Area of the Base = bh b = 9 cm h = 9 cm A = (9)(9) A = 81 cm ²

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How many cubes will fit on the base of the pyramid if the sides went 90° up? Area of the Base = 81 cm ² How many layers would there be?

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How many cubes will fit on the base of the pyramid if the sides went 90° up? Area of the Base = 81 cm ² How many layers would there be? Height of the pyramid = 10 cm

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Area of the Base = 81 cm ² Height of the pyramid = 10 cm Volume of a PRISM with the same dimensions: V = (81 cm²)(10 cm)

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The volume of a pyramid is ⅓ the volume of a prism with the same dimensions.

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Area of the Base = 81 cm ² Height of the pyramid = 10 cm Volume of PRISM = (81 cm²)(10 cm) Volume of the PYRAMID = (81 cm²)(10 cm) 3

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Area of the Base = 81 cm ² Height of the pyramid = 10 cm Volume of PRISM = (81 cm²)(10 cm) Volume of the PYRAMID = (81 cm²)(10 cm) 3 V = 270 cm³

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Volume of a sphere:

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Grocery Lab II

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DAY 7

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Homework Questions Page 803

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Can of Tennis Balls Is it taller, or wider around? How tall is the can as it relates to the tennis ball? How wide around is the can as it relates to the tennis ball?

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Loop of yarn What happens to the area inside the loop as I move my hands closer together and farther apart? What happens to the perimeter?

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LABS Cut 7 sheets 16 cm by 16 cm each Cut 2 sheets 16 cm by 12 cm Cut 1 sheet 12 cm by 12 cm

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DAY 8

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Homework Questions Measurement

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From Jurasic Park by Michael Crichton “Do mathematicians believe in intuition?” “Absolutely. Very important, intuition. Actually, I was thinking of fractals,” Malcolm said. “You know about fractals?” Grant shook his head. “Not really, no.” Fractals are a kind of geometry, associated with a man named Mandelbrot. Unlike ordinary Euclidean geometry that everyone learns in school – squares and cubes and spheres – fractal geometry appears to describe real objects in the natural world. Mountains and clouds are fractal shapes. So fractals are probably related to reality. Somehow.

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“Well, Mandelbrot found a remarkable thing with his geometry tools. He found that things looked almost identical at different scales. “At different scales?” Grant said. “For example,” Malcolm said, “a big mountain, seen from far away, has a certain rugged mountain shape. If you get closer, and examine a small peak of the big mountain, it will have the same mountain shape. In fact, you can go all the way down the scale to a tiny speck of rock, seen under a microscope – it will have the same basic fractal shape as the big mountain.”

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“It’s a way of looking at things” Malcolm said. “Mandelbrot found a sameness from the smallest to the largest. And this sameness of scale also occurs for events.” “Consider cotton prices. There are good records of cotton prices going back more than a hundred years. When you study fluctuations in cotton prices, you find that the graph of prices fluctuations in the course of a day looks basically like the graph for a week, which looks basically like the graph for a year, or for ten years.”

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“And that’s how things are. A day is like a whole life. You start out doing one thing, but end up doing something else, plan to run an errand, but never get there... And at the end of you life, your whole existence has that same haphazard quality, too. Your whole life has the same shape as a single day.” “You see, the fractal idea of sameness carries within it an aspect of recursion, a kind of doubling back on itself, which means that events are unpredictable. That they can change suddenly, and without warning.”

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Did You Know? Page 837

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Koch’s Snowflake Koch’s Curve Finite Area Infinite Perimeter Fractional dimension

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Koch’s Snowflake Stage 1 – Divide each side of the equilateral triangle into 3 equal lengths. Make an equilateral triangle on the middle of the three line segments on each of the three sides. Stage 2 – Divide each resulting line segment into 3 equal lengths. Make an equilateral triangle on the middle line segment of each of the resulting line segments. Stage 3 – repeat on each resulting line segment.

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Sierpinski’s Triangle Stage 1 – Connect the midpoint of each side of the equilateral triangle to make four smaller equilateral triangles. Stage 2 – Leaving the center triangle unaltered, repeat the process with each of the other three equilateral triangles. Stage 3 – Repeat State 4 – Repeat

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Chaos Game Roll a die to choose a vertex at random using the following guidelines. Roll 1 or 2, choose A Roll 3 or 4, choose B Roll 5 or 6, choose C Roll the die again to choose another vertex. Use your ruler to find the midpoint between the two vertices and make a point. Roll the die again to choose a vertex. Use your ruler to find the midpoint between the dot you just made and the vertex you just chose at random. Repeat for a total of 20 points.

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3 Dimensional Sierpenski’s Draw 2 intersecting diagonals on the back of a regular size envelope. From the top corner of the envelope, cut on the line to the point of intersection on each side. Remove the triangle.

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Make firm folds on the lines you drew. Fold the right side of the envelope into the left side Secure with a piece of tape.

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Use four tetrahedrons together to make the first generation of the 3 dimensional Sierpenski’s triangle. Use four first generation triangles together to make a 2 nd generation.

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Fractal Pop-Ups

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Day 9

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Measurement Test Conversion ≈ 1/3 Area and Perimeter ≈ 1/3 Surface Area and Volume ≈ 1/3

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Homework Questions Lab Questions

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Transformation of the Plane Imagine that each point P of the plane was “moved” to a new position P’ in the same plane. P’ is called the image of P. P is called the preimage of P’.

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Rigid Transformation Does not allow stretching or shrinking

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Translation (slide) All points are moved in the same direction and the same distance. Example 13.1, Page 822

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Rotation (turn) One point of the plane is held fixed and the remaining points are turned about that point the same number of degrees. Example 13.2, Page 824

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Reflection (flip or mirror reflection) A reflection is determined by a line in the plane called the line of reflection. Each point P of the plane is transformed to the point P’ on the opposite side of the line of reflection and the same distance from the line of reflection. Example 13.3, Page 826

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Glide Reflection Combines both the translation (slide) and reflection. It is required that the line of reflection be parallel to the direction of the slide. Figure 13.4, Page 827

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Name It!

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Day 10

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Make a Square! Tangrams – Ancient Chinese Puzzle Tangrams, 330 Puzzles, by Ronald C. Reed

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Tessellations TILE – a simple closed curve, together with it’s interior A set of tiles forms a TILING of a figure if the figure is completely covered by the tiles without overlapping any interior points of the tiles. Tilings are also known as TESSELLATIONS.

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Regular Tessellations All of the tiles are regular polygons of one shape. There are only three regular tessellations. Why?

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Semiregular Tessellation A tessellation made up of more than one type of regular polygon and identical vertex figures. There are 8 semiregular tessellations.

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Tessellation Lab Will any triangle tessellate the plane? Will any quadrilateral tessellate the plane?

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