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Notes for the 4 th Grading period

Notes for the 4 th Grading period. Mrs. Neal 6 th Advance and 7 th Average. Section 11.5 Area of Triangles and Trapezoids. Objective To find the areas of triangles and trapezoids Vocabulary 1. Trapezoid - quadrilateral with one pair of parallel sides.

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Notes for the 4 th Grading period

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  1. Notes for the 4th Grading period Mrs. Neal 6th Advance and 7th Average

  2. Section 11.5 Area of Triangles and Trapezoids • Objective • To find the areas of triangles and trapezoids • Vocabulary • 1. Trapezoid - quadrilateral with one pair of parallel sides. • Area of a triangle = A = 1 x b x h or b x h area = .5 x base x height 2 2 A = .5x3x7 A = 10.5 cm sq height 3cm 7 cm base

  3. Section 11.5 Area of Triangles and Trapezoids Base 1 • Area of a trapezoid expressed in square units • A = .5h(b 1 + b 2) or h (b1 + b2) 2 A = .5 x 5 (8 + 12 ) A = .5 x 5 ( 20 ) A = .5 x 100 A = 50 in squared h=height Base 2 8in 5 in 12in

  4. Section 11.7 Area of Complex Figures • Objective – To find the areas of complex figures • Vocabulary • 1. Complex Figure – a figure made p of circles, rectangles, squares and other 2– dimensional figures. To find the area of complex figures, separate it into figures you already know how to find the areas of and then add the numbers together.

  5. Section 11.7 Area of Complex Figures take the area of the rectangle and the triangle separately then add the areas together rectangle length = 6ft width = 3ft Area = 6 x 3 =18 ft sq triangle base = 2ft height = 1ft Area = .5 x 2 x 1 = 1 ft sq Rectangle area + triangle area = 18 + 1 = 19 ft sq (total) 6ft 3ft 4ft 2ft

  6. Section 11.7 Area of Complex Figures Find the area of circle and divide in half Area = 3.14 x r 2 Area = 3.14 x 11 2 = 3.14 x 121 = 379.9 m sq. then divide in half 379.9 = 189.95 m sq. 2

  7. Section 11.7 Area of Complex Figures • The pitching mound on a baseball field has a circular area of 254.47 square feet. Find its diameter. Round to the nearest foot. • Work backward through the problem • Area = 3.14 x r 2 so 254.47 = 3.14 x r 2 • 254.47 = r 2 3.14 • 81.04 = r 2 • √81.04 = r • 9 = r • d = 2r = 2x9 = 18 So the diameter of the pitcher’s mound is 18 ft

  8. Section 12.2 Volume of Rectangular Prisms • Objective • To find the volume of rectangular prisms • Vocabulary • 1. Volume- the measure of space inside of a three dimensional figure. It is measured in cubic units (meaning the units are raised to the third power unlike area which is to the second power) • Formula- V = L x W x H Volume= Length x Width x Height

  9. Section 12.2 Volume of Rectangular Prisms • 2. Rectangular Prism – a solid figure that has two parallel and congruent sides, or bases, that are rectangles. Volume = 5 x 8 x 3 = 120 cm 3 3cm 5cm 8cm

  10. 12.3 Volume of Cylinders • Objective • To find the volume of cylinders • Vocabulary • 1. Cylinder – a solid figure that has two congruent, parallel circles are its bases. • Formula – pi x r 2 x h = 3.14 x r 2 x h V = 3.14 x 3 2 x 7 V = 3.14 x 9 x 7 V = 3.14 x 63 = 197.82 =197.8 in 3 3in 7in

  11. 12.4 Surface Area of Rectangular Prisms • Objective • To find the surface areas of rectangular prisms • Vocabulary • Surface area – the sum of all of the areas, or faces, of a three-dimensional figure • Formula – SA = 2lw + 2lh + 2wh

  12. 12.4 Surface Area of Rectangular Prisms • Ex. 3cm 4cm 5cm L=5 w=4 h=3 SA = 2x5x4 + 2x5x3 + 2x4x3 SA = 40 + 30 + 24 SA = 94 cm 2

  13. 12.4 Surface Area of Rectangular Prisms • L=7in W =2in H = 9in • SA = 2x7x2 + 2x7x9 + 2x2x9 SA = 28 + 126 + 36 SA = 190in2

  14. 12.5 Surface Area of Cylinders • Objective • To find the surface areas of cylinders • Vocabulary • Circumference – the distance around a circle • Formula SA = 2x(3.14xr2) + (2x3.14xr)xh The area of two circles +circumference ofbase times the height

  15. 12.5 Surface Area of Cylinders A = 3.14xr 2 = 3.14 x 4 2 = 50.2 in sq R= 4in H =2in A = 2x3.14xrxh 2x3.14x4x2 157.8 in sq A = 3.14xr 2 = 3.14 x 4 2 = 50.2 in sq SA =2(50.2) + 157.8 = 100.5 + 157.8 = 258.3in sq

  16. 7-1 Ratios • Ratios – a comparison of two numbers by division • Equivalent Ratios – 2 ratios that have the same value • Can be written in several different ways • 2:8 or 2/8 or 2 to 8 • Can also be simplified • Best way- treat as a fraction and reduce to lowest terms • Can not compare numbers in a ratio unless they have the same unit, if not you must convert one unit to the other. • Examples • 1. 9 to 12 = 9/12 = ¾ • 2. 27:15 = 27/15 = 9/5 • 3. 4 feet:4 yards = 4 feet: 12 feet =4/12 = 1/3

  17. 7-1 Ratios • Comparisons of two ratios • 3/8 compared to 6/12 • Turn into decimals • .375 and .5 • So 6/12 is greater

  18. 7-2 Rates • Rate – a ratio that compares 2 quantities with different kinds of units • Unit rate – a rate simplified to a denominator of 1 unit • Examples • 1. rate=300 tickets in 6 days Unit rate=50 tickets in 1 day • 2. rate=220miles on 8 gallons Unit rate= 27.5 miles on 1 gallon

  19. 7-2 Rates • Comparison of rates – which is the better buy? • 8 oz jar of pickles for $1.14 or a 12 oz jar for $1.75? • $1.12/8 = $0.14 per oz • $1.75/12 =$0.15 per oz • 8 oz jar is the better buy

  20. 7-3 Solving Proportions • Proportion – an equation stating two ratios are equivalent • Cross product- the product of the numerator of one ratio and the denominator of the other ratio. If equal-then it is a proportion • Cross multiply and divide is the method of finding a missing number in a proportion.

  21. 7-3 Solving Proportions • Do these ratios form a proportion? • 1. 4/9 and 2/3 • cross products 4x3=12 and 9x2=18 • Cross products not equal therefore not a proportion • 2. 15/9 and 10/6 • Cross products 15x6=90 and 9x10=90 • Cross products are equal therefore it is a proportion • Solve the proportion 3. 1/5 = x/60 1x60÷5=x=12 4. 10/k=2.5/4 10x4÷2.5 =k=16

  22. 10-6 Similar Figures • Similar Figures – figures that have the same shape but not necessarily the same size. • Indirect Measurement-when you use similar figures to find the length, width, or height of objects too difficult to measure directly. • Solving method – use a proportion – take the given measurements of one figure and set it equal to the given and missing measurements of the other figure. 5/15 =4/x 5x4÷15=x X=12m 6m 18m 5m 15m 4m x

  23. 10-6 Similar Figures • 7/8=x/1.5 • 7x1.5÷8=x • X=1.3 cm 7cm x 1.5cm 8cm

  24. 7-4 Scale Drawings • 7-4 Scale Drawings • Scale Drawings - a drawing that represents something too small or too large to be drawn at actual size. • Scale - gives the relationship between the distance between the distance on the map and the actual distance. • Scale Factor – a scale written as a ratio in simplest form with the same units • Ex. 1cm = 2.5 m - scale factor is 1cm = 250 cm ( 100 cm in 1 meter)

  25. 7-4 Scale Drawings • Scale model – a model used to represent something that is too small or too large for an actual sized model (cell in the body, layout of a city) • Method to solve – set up as a proportion and solve for missing number. Use the given scale as the first fraction. • Ex. The distance from a city in Indiana to another city in Illinois is 8cm on a map. If the scale factor of the map is 1cm = 25km, find the actual distance between the cities. • 1/25 = 8/a so 25 x 8 ÷ a = 200 km as the actual distance.

  26. 7-7,7-8,8-2 Percent Proportion • 7-8 Percent Proportion • 7-7 Percent of a number • 8-2 Percent Equation • All of these sections measure the same thing • Uses proportion set-up to find a missing number • Percent Proportion – compares a part of a quantity to the whole quantity (base) • %/100 = is /of or %/100 =part/base

  27. 7-7,7-8,8-2 Percent Proportion • Usually the number that is the part has the word (is) in front of it and the number that is a base has the word (of) in front of it. • The number 100 is always in the proportion and the percent number is always on top of the 100. • Solve by cross multiplying then dividing to find the missing number. • Ex. 52% of 160 = 52/100 = ?/160 so 52 x 160 ÷ 100 = 83.2 • Ex. What percent of 60 is 15 = ?/100 = 15/60 so 15 x 100 ÷ 60 = 25%

  28. 8-4 Percent of Change • Percent of change – a ratio that compares the change in quantity to the original amount • Equation percent of change = amount of change original amount • Percent of increase – if the original quantity is increased • Percent of decrease – if the original quantity is decreased • Method of solving – find the difference between the new and original measures or amounts, this number is the numerator and the original number is the denominator. Divide the difference by the original then multiply by 100 to make it a percent. Then identify it as a decrease or a increase

  29. 8-4 Percent of Change • Examples • Original : $75 New: $15 75-15 = 60 then 60/75= .8 then .8 x 100 = 80% decrease since the price went down • Last year the chess club had 24 members. This year the club has 30 members. Find the percent of decrease in the number of members. 30-24 = 6 then 6/30 = .2 then .2 x 100 = 20% increase since the membership went up.

  30. 8-5 Sales Tax and Discount • Sales tax – an additional amount of money charged on items that people buy. The total cost of an item is the regular price plus the sales tax • Discount – the amount by which the regular price of an item is reduced. The sale price is the regular price minus the discount. • Examples • Sales tax – graphing calculator costs $90, sales tax= 4.25% find what 4.25% of 90 is = .0425 x 90 = 3.83 then add it to 90 = 90 + 3.83 = $93.83

  31. 8-5 Sales Tax and Discount • Sales tax – socks $2.99 sales tax 5.5% .055 x 2.99 = .16 2.99 + .16 = $3.15 • Discount – snowboard = $169 on sale 35% discount .35 x 169 = 59.15 169 – 59.15 = $109.85 Percent of discount - guitar original price= $299.95 new price = $179.99 299.95-179.99=119.96 then 119.96÷299.95 = .399 = .40 .40 x 100 = 40%

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