Geometry Area of Figures

Transcription

1 Slide 1 / 152

2 Slide 2 / 152 Geometry Area of Figures

3 Slide 3 / 152 Table of Contents Area of Rectangles Click on the topic to go to that section Area of Triangles Law of Sines Area of Parallelograms Area of Regular Polygons Area of Circles & Sectors Area of Other Quadrilaterals Area & Perimeter of Figures in the Coordinate Plane PARCC Sample Questions

4 Slide 4 / 152 Throughout this unit, the Standards for Mathematical Practice are used. MP1: Making sense of problems & persevere in solving them. MP2: Reason abstractly & quantitatively. MP3: Construct viable arguments and critique the reasoning of others. MP4: Model with mathematics. MP5: Use appropriate tools strategically. MP6: Attend to precision. MP7: Look for & make use of structure. MP8: Look for & express regularity in repeated reasoning. Additional questions are included on the slides using the "Math Practice" Pull-tabs (e.g. a blank one is shown to the right on this slide) with a reference to the standards used. If questions already exist on a slide, then the specific MPs that the questions address are listed in the Pull-tab.

5 Slide 5 / 152 Area of Rectangles Return to Table of Contents

6 Slide 6 / 152 Area of a Rectangle The area of a rectangle is defined to be the number of squares of area "1" that can fit within it. In the below drawing, 6 unit squares fit within the below rectangle of height 2 and based

7 Slide 7 / 152 Area of a Rectangle In general, the area of a rectangle is equal to its base times its height. This can also be referred to as its length times its width. A rectangle = length x width (lw) = base x height (bh) h b

8 Slide 8 / 152 Area of a Rectangle Sometimes, the dimensions will not be given, so you will need to calculate them before calculating the area. Since a rectangle is a quadrilateral with 4 right angles, 2 right triangles can be formed when drawing one of its diagonals. Therefore, Pythagorean Theorem can be a helpful formula. Another helpful formula is the perimeter formula for a rectangle: P = 2l + 2w There might also be questions asking you about the population density of a town, state, or country. It is the ratio that represents the number of people living per square mile and can be found by dividing the total population by the total area.

9 Slide 9 / 152 Example The diagonal of a rectangle is 34 feet and its length is 14 feet more than its width. Find the length, width, and area of the rectangle. We know that the width is unknown, so let's call it "x". Therefore, the length will be "x + 14". Using the Pythagorean Theorem & our Algebra skills, we can solve for x. x 2 + (x + 14) 2 = 34 2 x 2 + x x = 1,156 2x x = 0 2(x x - 480) = 0 2(x + 30)(x - 16) = 0 x + 30 = 0 or x - 16 = 0 x = -30 or x = 16 x ft Since x is the length of a side, it has to be positive. Therefore, our final answer is x = 16 ft = width length = 30 ft Area = 480 ft 2 x

10 Slide 10 / What is the area of a rectangle that has a length of 8.4 cm and a width of 3.7 cm?

11 Slide 11 / Televisions, are advertised using the length of the diagonal. For example, a 26" TV could have a length of 24" and a width of 10", as shown below. 24" 10" 26" What is the area of an 80" TV if the length is 69.3"?

12 Slide 12 / The population density is the amount of people living per square mile. If the town of Geometryville is a rectangular town that has a length of 24 miles and a width of 13 miles, and its population is 2,500 people, what is the population density of the town?

13 Slide 13 / The diagonal of a rectangle is 10 feet and its width is 2 feet less than its length. What is the length of the rectangle? A 4 feet B 6 feet C 8 feet D 9 feet

14 Slide 14 / The diagonal of a rectangle is 10 feet and its width is 2 feet less than its length. What is the width of the rectangle? A 4 feet B 6 feet C 8 feet D 9 feet

15 Slide 15 / The diagonal of a rectangle is 10 feet and its width is 2 feet less than its length. What is the area of the rectangle? A 80 ft 2 B 60 ft 2 C 48 ft 2 D 24 ft 2

16 Slide 16 / 152 Area of Triangles Return to Table of Contents

17 Slide 17 / 152 Area of a Right Triangle Clearly, the area of each of the below right triangles is equal to half the area of the rectangle they comprise. Since the area of the rectangle is bh, the area of each right triangle is 1/2 bh. A Δ = 1/2 bh h b

18 Slide 18 / 152 Area of Any Triangle Now, let's find the area formula for an arbitrary scalene triangle ABC with base "b" and height "h." Keep in mind that the height is always measured perpendicular to the base that is opposite the vertex. A h C b B

19 Slide 19 / 152 Area of Any Triangle Let's draw right triangle ACD such that the new triangle plus the original triangle form a larger right triangle ABD. ΔADC is a right Δ with base "b' " and height "h". ΔADB is a right Δ with base "b' + b" and height "h". A h D b' C b B

20 Area ΔADC = 1/2 b'h Slide 20 / 152 Area of Any Triangle Area ΔADB = 1/2 (b' + b) h Area ΔABC = Area ΔADB - Area ΔADC Area ΔABC = 1/2 (b' + b) h - 1/2 b'h Area ΔABC = 1/2 b'h + 1/2bh - 1/2b'h Area ΔABC = 1/2 bh A h D b' C b B

21 Slide 21 / 152 Area of Any Triangle So, the area of any triangle, not just right triangles, is given by : Area Δ = 1/2 bh A h C b B

22 Slide 22 / 152 Area of Any Triangle While the formula is the same for any triangle: Area Δ = 1/2 bh For triangles that are not right triangles, the height is usually not directly given. Instead, you may be given the lengths of one or more sides and the measures of one or more angles. A For instance, in this case 8 how would you find the 5 height...and then the area? 40 C 4 B

23 Slide 23 / 152 Area of Any Triangle Draw an altitude from the vertex to the base, creating a right triangle with the same height as our original triangle. ΔABD has the same height as our original triangle ΔABC. And both triangles share a side, AB, and an angle, B. Which trig function would allow us to find the value of "h"? A h 5 8 D C 4 40 B

24 Slide 24 / 152 Area of Any Triangle sin θ = opposite / hypotenuse = opp / hyp opp = (hyp)(sin θ) h = (8)(sin 40 ) h = (8)(0.64) h = 5.1 units Now that you have the base and the height of ΔABC, how would you find the area? h A D 5 8 C 4 40 B

25 Slide 25 / 152 Area of Any Triangle A = 1/2 bh A = 1/2 (4)(8)(sin 40 ) A = 1/2 (4)(8)(0.64) A = 10.2 units 2 A h D C 4 B

26 Slide 26 / 152 Area of Any Triangle Examining our solution, we can see that the area of the triangle is given by half the product of two sides multiplied by the sine of the angle between them. In this case, the sides were of length 4 and 8 and the angle is 40. A = 1/2 bh A = 1/2 (4)(8)(sin 40 ) A = 1/2 (4)(8)(0.64) A A = 10.2 units 2 h D C 4 B

27 A Δ = 1/2 ac sinb Slide 27 / 152 Area of Any Triangle Replacing the numbers by labeling the angles with upper case letters and the sides opposite them with matching lower case letters yields this formula. But, there's nothing special about those sides and that angle, so it will also be true that: A Δ = 1/2 ab sinc Or A Δ = 1/2 bc sina A The area of a triangle is equal to the product of any two sides and the sine of the included angle. b C c a B

28 Slide 28 / 152 Example Find the area of ΔABC. In this case, the altitude you draw will be within the triangle. A C 9 B

29 Slide 29 / Find the area of ΔDEF. Round your answer to the nearest hundredth. E 8 in. D 12 in. 34 F

30 Slide 30 / Find the area of ΔGHI. Round your answer to the nearest hundredth. H 16 in. G 9 in. 28 I

31 Slide 31 / Find the area of ΔJKL. Round your answer to the nearest hundredth. K 15 in. J 32 6 in. L

32 Slide 32 / Find the area of ΔPQR. Round your answer to the nearest hundredth. Q 9 in. P in. R

33 Slide 33 / 152 Law of Sines Return to Table of Contents

34 Slide 34 / 152 We just learned that we can find the area of any triangle with any of these three formulas: A Δ = 1/2 ac sinb A Δ = 1/2 ab sinc A Δ = 1/2 bc sina Law of Sines Since the area of a triangle will be the same regardless of which formula we use, these three formulas must be equal. Setting them equal to one another will give us a general relationship between the sides and the angles of a triangle. b A c C a B

35 Slide 35 / 152 Law of Sines A Δ = 1/2 ac sinb = 1/2 ab sinc = 1/2 bc sina Let's look at one pair at a time and simplify. 1/2 ac sinb = 1/2 ab sinc AND 1/2 ab sinc = 1/2 bc sina ac sinb = ab sinc c sinb = b sinc c / sinc = b / sinb ab sinc = bc sina a sinc = c sina a / sina = c / sinc a b c = = sina sinb sinc

36 Slide 36 / 152 Law of Sines This relationship between the sides and angles of a triangle will be true for all triangles. a b c = = sina sinb sinc OR sina sinb sinc = = a b c A b c C a B

37 Slide 37 / 152 Example Find the missing segment lengths and angle measures in ΔABC if m B = 28, m C = 103, and b = 26 cm. c 28 a B Since we have a triangle, we know that the sum of the interior angles is 180. Therefore, we can find the m A using the Triangle Sum Theorem. m A = = 49 A 26 cm 103 C

38 Slide 38 / 152 Example Find the missing segment lengths and angle measures in ΔABC if m B = 28, m C = 103, and b = 26 cm. c 28 a B Now that we have the measurements of all of the angles and one side length given, we can find the remaining sides using the Law of Sines. 26 sin28 = a sin49 26 sin28 = c sin103 A 26 cm 103 C 26sin49 = asin28 a = 26sin49 sin28 26sin103 = csin28 c = 26sin103 sin28 a = cm to reveal c = cm to reveal

39 Slide 39 / 152 Example Find the missing segment lengths and angle measures in ΔABC if m C = 122, a = 12 cm and c = 18 cm. B In this problem, we need to determine one of the missing angles first. However, we can't use the Triangle Sum Theorem yet, since we only know the measurement of 1 angle. Therefore, we need to use the Law 18 cm of Sines. 12 cm 18 = 12 sin122 sina A b 122 C 18sinA = 12sin122 sina = m A = sin -1 12sin ( 12sin122 ) 18 m A to = reveal 34.43

40 A Slide 40 / 152 Example Find the missing segment lengths and angle measures in ΔABC if m C = 122, a = 12 cm and c = 18 cm. B Now, that we know 2 angle measures, we can calculate the measurement of our final missing angle. m B = = cm b 122 C 12 cm Finally, we can calculate the length of our final side. 18 = b sin122 sin sin23.57 = bsin122 b = 18sin23.57 sin122 b = 8.49 cm to reveal

41 Slide 41 / Find the missing segment lengths and angle measures in ΔABC if m A = 70, m B = 64, and c = 5 yd. What is m C? B A 36 B 46 5 yd 64 a C 56 D 136 A 70 b C

42 Slide 42 / Find the missing segment lengths and angle measures in ΔABC if m A = 70, m B = 64, and c = 5 yd. What is the value of a? Round your answer to the nearest hundredth. B 5 yd 64 a A 70 b C

43 Slide 43 / Find the missing segment lengths and angle measures in ΔABC if m A = 70, m B = 64, and c = 5 yd. What is the value of b? Round your answer to the nearest hundredth. B 5 yd 64 a A 70 b C

44 Slide 44 / Find the missing segment lengths and angle measures in ΔABC if m C = 111, b = 3 in., and c = 5 in. What is m B? Round your answer to the nearest hundredth. B 5 in. a A 3 in. 111 C

45 Slide 45 / Find the missing segment lengths and angle measures in ΔABC if m C = 111, b = 3 in., and c = 5 in. What is m A? Round your answer to the nearest hundredth. B A B C in. a D A 3 in. 111 C

46 Slide 46 / Find the missing segment lengths and angle measures in ΔABC if m C = 111, b = 3 in., and c = 5 in. What is the value of a? Round your answer to the nearest hundredth. B 5 in. a A 3 in. 111 C

47 Slide 47 / 152 Area of Parallelograms Return to Table of Contents

48 Slide 48 / 152 Area of a Parallelogram From these results, we can find the area formula for any parallelogram by dividing the parallelogram into two triangles. h b

49 Slide 49 / 152 Area of a Parallelogram We can see below that a parallelogram is comprised of two triangles. Since the area of each triangle is 1/2 bh. The area of a parallelogram is bh. A parallelogram = bh h b

50 Slide 50 / 152 Area of a Parallelogram As with triangles, the height of a parallelogram is often not given. The height must be found by drawing an altitude and using trigonometry to find the height of the parallelogram. A parallelogram = bh h b

51 Slide 51 / 152 Find the area of ABCD. Example This time, use the cosine rather than the sine function, just for practice. Either can be used depending on the angle you choose. 20 To use cosine, our 1st step will be to drop down the altitude, or height Since we know that the entire obtuse angle is 110 and the right angle measures 90, the remaining acute angle is 20

52 Slide 52 / 152 Find the area of ABCD. Example This time, use the cosine rather than the sine function, just for practice. Either can be used depending on the angle you choose Now, let's find the height of our parallelogram. cos20 = h 5 h = 5cos20 h = 4.70 units And its area A = 8(4.70) A = units 2

53 Slide 53 / A diagonal parking space creates a parallelogram. Its length is 20.5 feet, its distance along the curb is 9 feet 4 inches, and the acute angle that is made with the curb is 75. Find the height of the parking space. A 2.4 ft 9 ft 4 in. 9 ft 4 in B 9 ft 20.5 ft C 9.7 ft D 34.8 ft

54 Slide 54 / A diagonal parking space creates a parallelogram. Its length is 20.5 feet, its distance along the curb is 9 feet 4 inches, and the acute angle that is made with the curb is 75. Find the area of the parking space. A ft 2 B ft ft 9 ft 4 in. 9 ft 4 in C ft 2 D ft 2

55 Slide 55 / A window frame is in the shape of a parallelogram. Its base is 3 feet long, its other side length is 2.5 feet long, and the obtuse angle created between two of the sides is 114. How much glass would be required to fill the window? Round your answer to the nearest hundredth.

56 Slide 56 / Mrs. Polygon is creating a quilt for her grandson. The quilt is created by stitching together parallelograms that are different colors. Each parallelogram is 4 inches long its other side is 3 inches long, and the acute angle created between the two sides is 32. What is the area of each parallelogram used to make the quilt? Round your answer to the nearest hundredth.

57 Slide 57 / If the quilt requires 5 parallelograms horizontally and 10 parallelograms vertically, how much material will Mrs. Polygon need to make the quilt?

58 Slide 58 / 152 Area of Regular Polygons Return to Table of Contents

59 Slide 59 / 152 Area of Regular Polygons A regular polygon is a polygon that has all its sides and angles congruent: it is both equilateral & equiangular.

60 Slide 60 / 152 Circumscribing a Regular Polygon A polygon is circumscribed by drawing around it the smallest circle on which lie all the vertices of the polygon. That circle is called the circumcircle of the polygon. C r

61 Slide 61 / 152 Circumcenter and Circumradius The center of a regular polygon is the center of its circumscribed circle, shown here as "C." This is called the circumcenter of the polygon. The radius of a regular polygon is the radius of the circumscribed circle, shown here as "r." This is called the circumradius. C r

62 Slide 62 / 152 Central Angle of a Regular Polygon A central angle of a regular polygon is an angle with one vertex at the circumcenter and two vertices on the circumcircle. The sides of the central angle are radii of the circle. The degrees of the central angle can be found using the formula Central angle = 360 n where n is the number of sides in the regular polygon. r C r

63 Slide 63 / 152 The Apothem a Regular Polygon The apothem of a Regular Polygon is the shortest distance from the center of the polygon to one of its sides. An apothem is perpendicular to a side of the polygon. An apothem is also the altitude (height) of the isosceles triangle formed by the sides of a central angle and the side of the polygon that is opposite the central angle. r C a r

64 Slide 64 / 152 Area of Regular Polygons A polygon of n sides is comprised of n triangles. Each triangle has a height equal to the apothem, "a." The base of each triangle is equal to the perimeter, P, divided by the number of sides, n: b = P/n. What is the area of a triangle whose base is P/n and whose height is a? a r

65 Slide 65 / 152 Area of Regular Polygons A Δ = 1/2 bh A Δ = 1/2(P/n)a There are n triangles in a regular polygon, so the area of the polygon is given by: A = na Δ A = (n)(1/2)(p/n)a A = 1/2Pa r r a a a r a r

66 Slide 66 / 152 Area of Regular Polygons The formula for the area of a polygon is simple: A = 1/2Pa But, often we are not given both P and a. We have to find one or both of them using trigonometry. r a a r a r a r

67 Slide 67 / 152 Example Let's find the area of a regular pentagon whose sides have a length of 7.

68 Slide 68 / 152 Example Let's find the area of a regular pentagon whose sides have a length of 7. A = 1/2 Pa 7 For a pentagon, n = 5, so it's perimeter will be 5 times the length of a side. In this case, P = (5)(7) = 35. But, how do we find the apothem? Let's look at one of the five triangles that comprise the pentagon.

69 Slide 69 / 152 Example Let's find the area of a regular pentagon whose sides have a length of 7. A = 1/2 Pa 7 For a pentagon, n = 5, so it's perimeter will be 5 times the length of a side. In this case, P = (5)(7) = 35. But, how do we find the apothem? Let's look at one of the five triangles that comprise the pentagon.

70 Slide 70 / 152 Example Let's find the area of a regular pentagon whose sides have a length of 7. The central angle is given by: 7 m C = 360/n = 360/5 = 72 r C 72 r Since the s must add to 180 and the measures of the base s of an isosceles triangle are equal, those base s must = The apothem, a, is the altitude of this triangle.

71 Slide 71 / 152 Example Let's find the area of a regular pentagon whose sides have a length of 7. 7 The two legs of the right triangle shown are a and 3.5. Tangent θ = opposite / adjacent tan θ = opp / adj r 36 r a opp = adj (tan θ) a = 3.5 (tan 54 ) a = 3.5 (1.38) a = 4.8 units

72 Slide 72 / 152 Example Let's find the area of a regular pentagon whose sides have a length of 7. 7 We now know that P = 35 units and a = 4.8 units. A = 1/2 Pa A = 1/2 (35)(4.8) r 72 r A = 84 units

73 Slide 73 / 152 Example Find the area of a regular octagon whose sides have a length of 8. 8 What is the question asking? To find the area of the octagon How can you represent the problem with symbols and numbers? A = 1/2 Pa How could you start this problem? Is there anything that you can find right away? P = 8(8) = 64 units

74 Slide 74 / 152 Example Find the area of a regular octagon whose sides have a length of 8. 8 How do we find the apothem? How could you use a drawing to show your way of thinking? Create an equation to find the apothem. 4 tan 67.5 = a 4 a = 4 tan 67.5 a = 9.66 units a

75 Slide 75 / 152 Example Find the area of a regular octagon whose sides have a length of 8. 8 Now that we know all of the needed measurements, find the area of the regular octagon. a = 9.66 units P = 8(8) = 64 units A = 1/2 (9.66)(64) A = units 2

76 Slide 76 / Calculate the apothem of the regular polygon shown in the figure below. 16

77 Slide 77 / Calculate the side length of the regular polygon shown in the figure below. 16

78 Slide 78 / Calculate the perimeter & area of the regular polygon shown in the figure below. 16

79 Slide 79 / Calculate the side length of the regular polygon shown in the figure below. 15

80 Slide 80 / Calculate the perimeter & area of the regular polygon shown in the figure below. 15

81 Slide 81 / Calculate the apothem of the regular polygon shown in the figure below. 7

82 Slide 82 / Calculate the perimeter of the regular polygon shown in the figure below. 7

83 Slide 83 / Calculate the area of the regular polygon shown below. 7

84 Slide 84 / 152 Area of Circles and Sectors Return to Table of Contents

85 Slide 85 / 152 Area of a Circle Interestingly, the formula, A = 1/2Pa, also leads to the formula for the area of a circle. If you let n go to infinity, the regular polygon approaches the shape of a circle. The apothem, a, approaches the radius of the circle, r. And, the perimeter of the polygon approaches the circumference of the circle: 2πr. Then, A = 1/2Pa approaches A = 1/2(2πr)(r) A = πr 2 r a

86 Slide 86 / Find the area of a circle that has a radius of 8 in. A 4π in 2 B 8π in 2 C 16π in 2 D 64π in 2

87 Slide 87 / Find the area of a circle that has a diameter of 17 in. A 8.5π in 2 B 17π in 2 C 72.25π in 2 D 289π in 2

88 Slide 88 / Find the area of a circle that has a circumference of 18π in. A 324π in 2 B 81π in 2 C 18π in 2 D 9π in 2

89 Slide 89 / 152 Sectors of Circles A sector of a circle is the portion of the circle enclosed by two radii and the arc that connects them. B Minor Sector Major Sector A C

90 Slide 90 / Which arc borders the minor sector? A Arc AB B Arc AC A C Arc ADB C B D

91 Slide 91 / Which arc borders the major sector? A Arc AB B Arc AC A C Arc ADB C B D

92 Slide 92 / 152 The Area of a Sector A A sector is part of a circle. The area of a complete circle is given by A circle = πr 2 C θ r B Similar to the arc length, we have to find the fraction of the circle in the sector and multiply this fraction by the area of the entire circle to find the area of a sector D If the central angle of the sector is given in degrees, that's just the measure of that angle divided by 360 degrees, yielding: A sector = (θ/360 )(πr 2 ) when θ is the central angle of the sector measured in degrees

93 Slide 93 / 152 The Area of a Sector In this case, we know that the measure of our central angle, or m C = 110. But, if we are also told that the radius of the circle is 20 cm, we can determine the area of the sector enclosed by AB, AC & CB. Using our formula, we know that θ = 110, and r = 20 cm. A D C cm If we substitute these numbers into our formula, it will give us the area of our sector. A sector = (110 /360 )π(20) 2 A sector = cm 2 B

94 Slide 94 / 152 The Area of a Sector A Alternatively, we could just set this up as a proportion and solve it in one step. Area of Sector Area of Circle = Central angle 360 D C cm B A sector = 110 πr A sector = 110 (πr 2 ) 360 A sector = 110 π(20) A sector = 1100 π cm 2 = cm 2 9

95 Slide 95 / Find the area of the sector. Leave your answer in terms of π. A B 3 45 C A sector = (θ/360 )(πr 2 )

96 Slide 96 / Find the area of the major sector. Leave your answer in terms of π. C 8 cm A 60 T A sector = (θ/360 )(πr 2 )

97 Slide 97 / Find the area of the minor sector of the circle. Round your answer to the nearest hundredth. C 5.5 cm 30 T A

98 Slide 98 / Find the Area of the major sector for the circle. Round your answer to the nearest thousandth. C 12 cm A 85 T

99 Slide 99 / What is the central angle for the major sector of the circle? C 15 cm A 120 G

100 Slide 100 / Find the area of the major sector. Round to the nearest thousandth. C 15 cm A 120 G

101 Slide 101 / If a circle is divided into 2 sectors, one major and one minor, then the sum of the areas of the 2 sectors is equal to the total area of the circle. True False

102 Slide 102 / A group of 10 students order pizza. They order 5 12" pizzas, that contain 8 slices each. If they split the pizzas equally, how many square inches of pizza does each student get (to the nearest hundredth)?

103 Slide 103 / You have a circular sprinkler in your yard. The sprinkler has a radius of 25 ft. How many square feet does the sprinkler water if it only rotates 120? Round your answer to the nearest hundredth.

104 Slide 104 / 152 Area of Other Quadrilaterals Lab: Area of Other Quadrilaterals Return to Table of Contents

105 Slide 105 / 152 Area of Other Quadrilaterals So far, we have discussed calculating the area of rectangles, triangles, parallelograms, circles, sectors and regular polygons, and the relationships between each formula. In this section, we are going to determine how 3 additional area formulas are related to those that we already know and to each other.

106 Slide 106 / 152 Area of a Trapezoid Let's start off by deriving the area formula of a trapezoid. If you will recall from our unit on Quadrilaterals, a trapezoid is a quadrilateral with 1 pair of opposite sides that are parallel, called bases and 1 pair of opposite sides that are not parallel, called legs. If we label our bases as b 1 and b 2 and draw an altitude, or the height (h), we have the figure given to the right. h b 1 b 2

107 Slide 107 / 152 Area of a Trapezoid We can split the trapezoid into 2 triangles by drawing the diagonal from the upper left corner to the lower right corner. Click on the bases to reveal each new triangle (given in red). h b 1 Now, we can separate these two triangles. Using the triangles, how could you calculate the area of a trapezoid? Add up the triangle areas b 2

108 Slide 108 / 152 Area of a Trapezoid b 1 b 1 h h + = h b 2 b 2 We know that the area of a triangle is 1/2 bh and that the sum of these two triangles will be the area of the trapezoid. Create an equation using this fact and the variables provided. A Trapezoid = 1/2 b 1 h + 1/2 b 2 h What algebra facts can be used to simplify the equation? How does it simplify? Distributive Property or Factoring A Trapezoid = 1/2 h(b 1 + b 2 )

109 Slide 109 / 152 Area of a Rhombus Next, we are going to derive the area formula of a rhombus. If you will recall from our unit on Quadrilaterals, a rhombus is a parallelogram with congruent sides and diagonals that are perpendicular and bisect each other. Since it is a parallelogram, the formula A = bh will work for this shape. But what if we are given the diagonal lengths instead? What connections do you see? Can we figure out a formula for this case? If we label our diagonals as d 1 and d 2, we have the figure given to the right. d 2 d 1

110 Slide 110 / 152 The diagonals split the rhombus into 4 congruent triangles. If we move two the bottom triangles to the top, but on opposite sides (i.e. move the bottom right triangle to the upper left corner), we will create another shape and determine the area formula. Area of a Rhombus d 1 d 2 Use the following steps to show the animation in the diagram: Click on the hash marks in the 2 bottom triangles to reveal each new triangle (given in red). Move the new red triangles to the opposite corners of the shape. Click the bottom of the original rhombus (shown in black) to hide it.

111 Slide 111 / 152 Area of a Rhombus d 2 d 1 What shape has been made? Rectangle What is the area formula for this shape? A = bh or A = lw Using these connections and the variables given above, create an equation to represent the area of a rhombus. A Rhombus = 1/2 d 1 d 2

112 Slide 112 / 152 Area of a Kite Since the shape of a kite is very similar to a rhombus, you are going to explain how the Area of a Kite formula is the same as the Area of a Rhombus formula for homework. A Kite = 1/2 d 1 d 2 d 2 d 1

113 Slide 113 / 152 Example You are constructing a desk to fit into your bedroom using wood for the flat top surface and metal bars for the legs. In order to save space, you determine that an isosceles trapezoid would be the best shape. The bases of the trapezoid will be 3 feet and 7 feet and the angle formed by the short base and each leg of the trapezoid is 135. How much wood is required to make the flat top surface of your desk? 3 ft ft

114 Slide 114 / In order to make a kite, you need to cut enough wrapping paper, based on the lengths of the spars, or the sticks used as the frame of the kite (see diagram to the right). If the longest spar is 36 inches & the shortest spar is 24 inches, how much wrapping paper do you need to make your kite? Spars

115 Slide 115 / One diagonal of a rhombus is 1/3 times as long as the other. The area of the rhombus is 0.24m 2. What are the lengths of the diagonals? A 1.2 m and 1.2 m B 1.2 m and 0.4 m C 1.8 m and 0.6 m D 0.6 m and 0.4 m

116 Slide 116 / A glass window in the shape of an isosceles trapezoid has bases that measure 8 inches and 12 inches. If the angle between the longest base and the legs is 60, what is the area of the window? Round your answer to the nearest hundredth. 12 in in.

117 Slide 117 / 152 Area of Complex Figures In most real-world problems, you will need to calculate the area of complex figures, or shapes that are a combination of 2 or more shapes. In order to calculate the area of these types of shapes, we need to either add or subtract the areas of the primary shapes involved. Add the area of a rectangle & 1/2 the area of a circle. Subtract the area of the rectangles from the area of the trapezoid.

118 Slide 118 / 152

119 Example Slide 119 / 152 In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color. Find the amount of wall space that will be painted. If one quart-size can of paint covers 87.5 ft 2, then how many cans are required to paint 2 coats of the accent color? If one quart-size can of paint costs $15, how much money do you need to spend to paint your wall?

120 Example Slide 120 / 152 In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color. Find the amount of wall space that will be painted. Wall Area: A = 13.5(8.5) A = ft 2 Doorway Area: A = 7(3) = 21 ft 2-2 doorways, so total area not included is 42 ft 2 Painted Area: A = A = ft 2

121 Example Slide 121 / 152 In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color. If one quart-size can of paint covers 87.5 ft 2, then how many cans are required to paint 2 coats of the accent color? Total painted Area = 72.75(2) = ft 2 Quart-sized Cans of Paint = 145.5/87.5 = cans

122 Example Slide 122 / 152 In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color. If one quart-size can of paint costs $15, how much money do you need to spend to paint your wall? $15(2) = $30

123 Slide 123 / What is the area of the shaded region? A 20 cm 2 B 30 cm 2 C 100 cm 2 D 200 cm 2

124 Slide 124 / What is the area of the entire figure? Round your answer to the nearest tenth. A 25.1 yd 2 B 28.9 yd 2 C 54 yd 2 D 79.1 yd 2

125 Slide 125 / A plant vase is formed by combining a square base with 4 isosceles trapezoids (see figure below). The square base has an area of 25 in 2, the long bases of the trapezoids are 7 in., and the heights of the trapezoids are 4 in. How much glass was needed to make the entire vase? 7 in. 4 in.

126 Slide 126 / The Bisect Building Company has created a building plan for the new garage at the Heptagon home, shown in the figure. The roof of the garage is made from 2 trapezoids and 2 isosceles triangles. What is the area of the entire roof? A 700 ft 2 B 600 ft 2 20 ft 10 ft C 250 ft 2 10 ft D 100 ft 2 20 ft 30 ft

127 Slide 127 / The Bisect Building Company has created a building plan for the new garage at the Heptagon home, shown in the figure. The roof of the garage is made from 2 trapezoids and 2 isosceles triangles. Each bundle of shingles can cover approximately 40 ft 2, and shingles must be purchased in full bundles. How many bundles of shingles are required to cover the roof? 20 ft A 15 B 17 C 18 D ft 10 ft 10 ft 30 ft

128 Slide 128 / The Bisect Building Company has created a building plan for the new garage at the Heptagon home, shown in the figure. The roof of the garage is made from 2 trapezoids and 2 isosceles triangles. Each bundle of shingles costs $ How much should the Bisect Building Company budget for the shingles? 20 ft 10 ft 10 ft 20 ft 30 ft

129 Slide 129 / 152 Area & Perimeter of Figures in the Coordinate Plane Return to Table of Contents

130 Slide 130 / 152 Area & Perimeter of Figures in the Coordinate Plane Some problems will show your shapes in a coordinate plane. In these cases, you will need to calculate the side lengths, or diagonal lengths, of your shapes using the distance formula. Remember that the distance formula is d = After calculating the desired distances, you will either need to add them to calculate the perimeter or multiply them to calculate the area, using the appropriate formula.

131 Slide 131 / 152 Example Calculate the perimeter and area of rhombus JKLM y J K x L M Since JKLM is a rhombus, all of the sides are congruent. How would we calculate the perimeter? Calculate one side length using the distance formula & then multiply that value by 4 to calculate our perimeter. JK = (7-3) 2 + (3-1) 2 = = = 20 = Perimeter = 4(2 5) = 8 5 units units

132 Slide 132 / 152 Example Calculate the perimeter and area of rhombus JKLM. y Since JKLM is a rhombus, A = 1/2 d 1 d K 2 J x L M -6 Therefore, we need to calculate the lengths of the diagonals using the distance formula. JL = (5-3) 2 + (-1-1) 2 = (-2) 2 = = 8 = 2 2 KM = (7-1) 2 + (3 - (-3)) 2 = = = 72 = 6 2 A = 1/2 (2 2)(6 2) A = 12 units 2

133 Slide 133 / 152 Example The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles. Based on the information given, how many miles is the perimeter of North Dakota? At the end of 2010, the population of North Dakota was 672,591 people. Based on the information given, what was the population density at the end of 2010? Miles (hundreds) y 4 3 E F 2 1 H 0 G x Miles (hundreds)

134 Slide 134 / 152 The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles. Based on the information given, how many miles is the perimeter of North Dakota? How do we calculate the perimeter? Example add up all of the side lengths What strategies are you going to use? counting & distance formula Miles (hundreds) 4 y 3 E 2 F 1 H 0 G x Miles (hundreds)

135 Slide 135 / 152 The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles. Based on the information given, how many miles is the perimeter of North Dakota? EF = 3.3 = 330 miles GH = 3.4 = 340 miles EH = 2.11 = 211 miles FG = ( ) 2 + (0-2.11) 2 = (0.1) 2 + (-2.11) 2 = Example = = 211 miles Miles (hundreds) 4 y 3 E 2 F 1 H 0 G x Miles (hundreds) Perimeter: = 1092 miles

136 Slide 136 / 152 Example The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles. At the end of 2010, the population of North Dakota was 672,591 people. Based on the information given, what was the population density at the end of 2010? What does population density mean? ratio that represents the number of people living per square mile. How do we find it? dividing the total population by the total area Miles (hundreds) 4 y 3 E 2 F 1 H 0 G x Miles (hundreds)

137 Slide 137 / 152 Example The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles. At the end of 2010, the population of North Dakota was 672,591 people. Based on the information given, what was the population density at the end of 2010? Area = 1/2(211)( ) = 1/2(211)(670) = 70,685 square miles Population Density: 672,591 = 9.51 people per 70,685 square mile Miles (hundreds) 4 y 3 E 2 F 1 H 0 G x Miles (hundreds)

138 Slide 138 / Calculate the perimeter of square PQRS. y A 10 units B 20 units C 25 units D units P x 6 8 Q -2 S -4-6 R

139 Slide 139 / Calculate the area of square PQRS. y A units 2 B 20 units 2 C 25 units D 50 units P x 6 8 Q -2 S -4-6 R

140 Slide 140 / 152 Series of SMART Response Questions The picture below shows the yard of the Fractal family. What is the perimeter of (0, 150) (50, 150) their yard? Every spring, Mr. Fractal fertilizes the lawn. What is the area of it? This year, the Fractals are going to have a fence installed, represented by the red (50, 25) (0, 20) dotted line. Determine the amount of fencing (140, 10) required. (0, 100) garden (50, 75) (180, 75) House (140, 25) Driveway (180, 5) (200, 150) (180, 25) (200, 0) In the future, the Fractals want to install a circular pool centered at (115, 105). If the pool must be at least 10 feet away from the house, which of the listed measurements could be the area of the pool? Select all that apply. A 2, ft 2 C 1, ft 2 E ft 2 B 1, ft 2 D ft 2

141 Slide 141 / The picture below shows the yard of the Fractal family. What is the perimeter of their yard? (0, 150) (0, 100) garden (50, 150) (50, 75) (180, 75) (200, 150) (0, 20) (50, 25) House (140, 25) Driveway (180, 25) (140, 10) (180, 5) (200, 0)

142 Slide 142 / The picture below shows the yard of the Fractal family. Every spring, Mr. Fractal fertilizes the lawn. What is the area of it? (0, 150) (0, 100) garden (50, 150) (50, 75) (180, 75) (200, 150) (0, 20) (50, 25) House (140, 25) Driveway (180, 25) (140, 10) (180, 5) (200, 0)

143 Slide 143 / The picture below shows the yard of the Fractal family. This year, the Fractals are going to have a fence installed, represented by the red dotted line. Determine the amount of fencing required. (0, 150) (0, 100) garden (50, 150) (50, 75) (180, 75) (200, 150) (0, 20) (50, 25) House (140, 25) Driveway (180, 25) (140, 10) (180, 5) (200, 0)

144 Slide 144 / The picture below shows the yard of the Fractal family. In the future, the Fractals want to install a circular pool centered at (115, 105). If the pool must be at least 10 feet away from the house, which of the listed measurements could be the area of the pool? Select all that apply. A 2, ft 2 B 1, ft 2 C 1, ft 2 D ft 2 E ft 2 (0, 150) (0, 100) (0, 20) garden (50, 150) (50, 75) (180, 75) (50, 25) House (140, 25) (140, 10) Driveway (180, 5) (200, 150) (180, 25) (200, 0)

145 Slide 145 / 152 PARCC Sample Questions The remaining slides in this presentation contain questions from the PARCC Sample Test. After finishing this unit, you should be able to answer these questions. Good Luck! Return to Table of Contents

146 Question 13/25: Part B Slide 146 / 152 Topic: Area of a Rectangle 59 A steel pipe in the shape of a right circular cylinder is used for drainage under a road. The length of the pipe is 12 feet and its diameter is 36 inches. The pipe is open at both ends. A wire screen in the shape of a square is attached at one end of the pipe to allow water to flow through but to keep people from wandering into the pipe. The length of the diagonals of the screen are equal to the diameter of the pipe. The figure represents the placement of the screen at the end of the pipe. Select from each set of choices to correctly complete the sentence. The perimeter of the screen is approximately inches, A 72 B 201 C 125 and the area of the screen is square inches. D 324 E 648 F 1,018 PARCC Sample Question - EOY

147 Question 15/25 Slide 147 / 152 Topic: Area & Perimeter of a Figures in the Coordinate Plane 60 The figure shows rectangle ABCD in the coordinate plane with point A at (0, 2.76), B at (3.87, 2.76), C at (3.87, 0) and D at the origin. Rectangle ABCD can be used to approximate the size of the state of Colorado with x and y scales representing hundreds of miles. Part A Based on the information given, how many miles is the perimeter of Colorado? PARCC Sample Question - EOY

148 Slide 148 / 152 Question 15/25 Topic: Area & Perimeter of a Figures in the Coordinate Plane 61 The figure shows rectangle ABCD in the coordinate plane with point A at (0, 2.76), B at (3.87, 2.76), C at (3.87, 0) and D at the origin. Rectangle ABCD can be used to approximate the size of the state of Colorado with x and y scales representing hundreds of miles. Part B At the end of 2010, the population of Colorado was 5,029,196 people. Based on the information given, what was the population density at the end of 2010? A 25 people per square mile B 47 people per square mile C 2,269 people per square mile D 7,586 people per square mile PARCC Sample Question - EOY

149 Question 24/25 Slide 149 / 152 Topic: Area & Perimeter of a Figures in the Coordinate Plane 62 Luke purchased a warehouse on a plot of land for his business. The figure represents a plan of the land showing the location of the warehouse and parking area. The coordinates represent points on a rectangular grid with units in feet. Part A What is the perimeter of the plot of land. Express your answer to the nearest tenth of a foot. PARCC Sample Question - EOY

150 Question 24/25 Slide 150 / 152 Topic: Area & Perimeter of a Figures in the Coordinate Plane 63 Luke purchased a warehouse on a plot of land for his business. The figure represents a plan of the land showing the location of the warehouse and parking area. The coordinates represent points on a rectangular grid with units in feet. Part B What is the area of the plot of land that does not include the warehouse and the parking area? PARCC Sample Question - EOY

151 Slide 151 / 152 Question 24/25 Topic: Area & Perimeter of a Figures in the Coordinate Plane 64 Luke purchased a warehouse on a plot of land for his business. The figure represents a plan of the land showing the location of the warehouse and parking area. The coordinates represent points on a rectangular grid with units in feet. Part C Luke is planning to put a fence along two interior sides of the parking area. The sides are represented in the plan by the legs of the trapezoid. What is the total length of fence needed? Express your answer to the nearest tenth of a foot. PARCC Sample Question - EOY

152 Slide 152 / 152 Topic: Area & Perimeter of a Question 24/25 Figures in the Coordinate Plane 65 Luke purchased a warehouse on a plot of land for his business. The figure represents a plan of the land showing the location of the warehouse and parking area. The coordinates represent points on a rectangular grid with units in feet. Part D In the future, Luke has plans to construct a circular storage bin centered at coordinates (50, 40) on the plan. Which of the listed measurements could be the diameter of the bin that will fit on the plot and be at least 2 feet away from the warehouse? Select all that apply. A 10 feet D 22 feet B 15 feet E 25 feet C 18 feet PARCC Sample Question - EOY