Geometry. Slide 1 / 152. Slide 2 / 152. Slide 3 / 152. Area of Figures Table of Contents

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1 Slide 1 / 152 Slide 2 / 152 Geometry rea of Figures Table of ontents Slide 3 / 152 rea of Rectangles rea of Triangles Law of Sines rea of Parallelograms rea of Regular Polygons rea of ircles & Sectors rea of Other Quadrilaterals lick on the topic to go to that section rea & Perimeter of Figures in the oordinate Plane PR Sample Questions

2 Throughout this unit, the Standards for Mathematical Practice are used. MP1: Making sense of problems & persevere in solving them. MP2: Reason abstractly & quantitatively. MP3: onstruct viable arguments and critique the reasoning of others. MP4: Model with mathematics. MP5: Use appropriate tools strategically. MP6: ttend to precision. MP7: Look for & make use of structure. MP8: Look for & express regularity in repeated reasoning. dditional 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. 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: onstruct viable arguments and critique the reasoning of others. MP4: Model with mathematics. MP5: Use appropriate tools strategically. MP6: ttend to precision. MP7: Look for & make use of structure. MP8: Look for & express regularity in repeated reasoning. Math Practice dditional 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. Slide 4 () / 152 Slide 5 / 152 rea of Rectangles Return to Table of ontents

3 rea of a Rectangle Slide 6 / 152 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 rea of a Rectangle Slide 7 / 152 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. rectangle = length x width (lw) = base x height (bh) h b rea of a Rectangle Slide 8 / 152 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. nother 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.

4 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". x ft x Slide 9 / 152 Using the Pythagorean Theorem & our lgebra 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 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 rea = 480 ft 2 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. x + 14 We know that the width Questioning is unknown, to help address MP standards: What information do you have? (MP1) so let's call it "x". What is this problem asking? (MP1) What strategies are you going to use? x Therefore, the length (MP1) will be "x + 14". 34 ft How can you represent the problem with Using the Pythagorean numbers Theorem and symbols? & (MP2) reate an equation to represent the our lgebra skills, we problem. can solve (MP2) for x. x 2 + (x + 14) 2 = 34 What 2 labels (or units of measurement) should you use? (MP6) x 2 + x x Does = anyone 1,156 have the same answer, but a 2x 2 different way to explain it? (MP7) + 28x = 0 What concepts that Since we have x learned is the length before of a side, it 2(x x - 480) were = 0 useful in solving [This object has this? is a pull to tab] be (MP8) positive. Therefore, our final answer is 2(x + 30)(x - 16) = 0 x = 16 ft = width x + 30 = 0 or x - 16 = 0 length = 30 ft x = -30 or x = 16 rea = 480 ft 2 Math Practice Slide 9 () / What is the area of a rectangle that has a length of 8.4 cm and a width of 3.7 cm? Slide 10 / 152

5 1 What is the area of a rectangle that has a length of 8.4 cm and a width of 3.7 cm? Slide 10 () / 152 = cm 2 2 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. Slide 11 / " 10" 26" What is the area of an 80" TV if the length is 69.3"? 2 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. Slide 11 () / " 10" w 26" = in = 2, in 2 What is the area of an 80" [This TV object if is a the pull tab] length is 69.3"?

6 3 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? 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? 8 people per square mile Slide 12 () / 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? 4 feet Slide 13 / feet 8 feet D 9 feet

7 4 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? 4 feet Slide 13 () / feet 8 feet D 9 feet 5 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? 4 feet Slide 14 / feet 8 feet D 9 feet 5 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? 4 feet Slide 14 () / feet 8 feet D 9 feet

8 6 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? 80 ft 2 Slide 15 / ft 2 48 ft 2 D 24 ft 2 6 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? 80 ft 2 Slide 15 () / ft 2 48 ft 2 D 24 ft 2 Slide 16 / 152 rea of Triangles Return to Table of ontents

9 rea of a Right Triangle Slide 17 / 152 learly, 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. Δ = 1/2 bh h b rea of ny Triangle Slide 18 / 152 Now, let's find the area formula for an arbitrary scalene triangle with base "b" and height "h." Keep in mind that the height is always measured perpendicular to the base that is opposite the vertex. h b rea of ny Triangle Slide 19 / 152 Let's draw right triangle D such that the new triangle plus the original triangle form a larger right triangle D. ΔD is a right Δ with base "b' " and height "h". ΔD is a right Δ with base "b' + b" and height "h". h D b' b

10 rea ΔD = 1/2 b'h rea of ny Triangle Slide 20 / 152 rea ΔD = 1/2 (b' + b) h rea Δ = rea ΔD - rea ΔD rea Δ = 1/2 (b' + b) h - 1/2 b'h rea Δ = 1/2 b'h + 1/2bh - 1/2b'h rea Δ = 1/2 bh h D b' b rea of ny Triangle rea ΔD = 1/2 b'h Questioning to help address MP standards: rea ΔD = 1/2 (b' + How b) hcan we prove that the area of a triangle is equal to 1/2 bh for all triangles? (MP3) rea Δ = rea ΔD What - rea strategies ΔD are you going to use? (MP1) rea Δ = 1/2 (b' + How b) h can - 1/2 you b'h represent the problem with numbers and symbols? (MP2) rea Δ = 1/2 b'h + 1/2bh - 1/2b'h reate an equation to represent the problem. (MP2) rea Δ = 1/2 bh What concepts that you have learned about before were useful in writing this algebraic proof? (MP8) Math Practice Slide 20 () / 152 h D b' b rea of ny Triangle Slide 21 / 152 So, the area of any triangle, not just right triangles, is given by : rea Δ = 1/2 bh h b

11 rea of ny Triangle Slide 22 / 152 While the formula is the same for any triangle: rea 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. For instance, in this case 8 how would you find the 5 height...and then the area? 40 4 rea of ny Triangle Slide 22 () / 152 While the formula is the same for any triangle: rea 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. Math Practice Question to the left of the triangle addresses MP1, MP3, and MP7. For instance, in this case how would you find the 5 8 height...and then the area? 40 4 rea of ny Triangle Draw an altitude from the vertex to the base, creating a right triangle with the same height as our original triangle. Slide 23 / 152 ΔD has the same height as our original triangle Δ. nd both triangles share a side,, and an angle,. Which trig function would allow us to find the value of "h"? h 5 8 D 4 40

12 rea of ny Triangle Draw an altitude from the vertex to the base, creating a right triangle with the same height as our original triangle. Slide 23 () / 152 ΔD has the same height as our original triangle Δ. Math Practice nd both triangles share a side,, and an angle,. Question above the triangle addresses Which trig function MP7. would allow us to find the value of "h"? h 5 8 D 4 40 rea of ny Triangle Slide 24 / 152 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 Δ, how would you find the area? h D rea of ny Triangle Slide 24 () / 152 sin θ = opposite / hypotenuse = opp / hyp Questioning to help address MP standards: opp = (hyp)(sin θ) What strategies are you going to use? (MP1) h = (8)(sin 40 ) How can you represent the problem with numbers and symbols? (MP2) h = (8)(0.64) reate an equation to represent the problem. (MP2) What concepts that you have learned about h = 5.1 units before were useful? (MP8) Now that you Question to the left of the triangle addresses MP1. have the base and the height of Δ, how would h 5 8 you find the area? 40 Math Practice D 4

13 rea of ny Triangle Slide 25 / 152 = 1/2 bh = 1/2 (4)(8)(sin 40 ) = 1/2 (4)(8)(0.64) = 10.2 units 2 h D 4 rea of ny Triangle Slide 26 / 152 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. = 1/2 bh = 1/2 (4)(8)(sin 40 ) = 1/2 (4)(8)(0.64) = 10.2 units 2 h D 4 rea of ny 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. Slide 27 / 152 Δ = 1/2 ac sin ut, there's nothing special about those sides and that angle, so it will also be true that: Δ = 1/2 ab sin Or Δ = 1/2 bc sin The area of a triangle is equal to the product of any two sides and the sine of the included angle. b c a

14 Example Slide 28 / 152 Find the area of Δ. In this case, the altitude you draw will be within the triangle Example Slide 28 () / 152 Find the area of Δ. sin 50 = h 4 In this case, the 4 altitude you 8 draw will be h = within 4sin50 the triangle. h = 1/2(9)(4sin50 ) = units Questioning to help address MP standards: How could you start this problem? (MP1) 4 8 How can you represent the problem with numbers and symbols? (MP2) reate an equation to represent the problem. (MP2) 50 What do you know about trigonometry that could apply to [This this object situation? is a pull tab] (MP7) 9 7 Find the area of ΔDEF. Round your answer to the nearest hundredth. E Slide 29 / in. D 12 in. 34 F

15 7 Find the area of ΔDEF. Round your answer to the nearest hundredth. E Slide 29 () / 152 D 12 in. 8 in. 34 F = in 2 8 Find the area of ΔGHI. Round your answer to the nearest hundredth. Slide 30 / 152 H 16 in. G 28 9 in. I 8 Find the area of ΔGHI. Round your answer to the nearest hundredth. Slide 30 () / 152 H 16 in. G 9 in. 28 I = in 2

16 9 Find the area of ΔJKL. Round your answer to the nearest hundredth. K Slide 31 / in. J 32 6 in. L 9 Find the area of ΔJKL. Round your answer to the nearest hundredth. K Slide 31 () / in. = in 2 J 32 6 in. L 10 Find the area of ΔPQR. Round your answer to the nearest hundredth. Slide 32 / 152 Q 9 in. P in. R

17 10 Find the area of ΔPQR. Round your answer to the nearest hundredth. Slide 32 () / 152 Q P 9 in in. R = in 2 Slide 33 / 152 Law of Sines Return to Table of ontents Law of Sines Slide 34 / 152 We just learned that we can find the area of any triangle with any of these three formulas: Δ = 1/2 ac sin Δ = 1/2 ab sin Δ = 1/2 bc sin 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 c a

18 Law of Sines Slide 35 / 152 Δ = 1/2 ac sin = 1/2 ab sin = 1/2 bc sin Let's look at one pair at a time and simplify. 1/2 ac sin = 1/2 ab sin ND 1/2 ab sin = 1/2 bc sin ac sin = ab sin c sin = b sin c / sin = b / sin ab sin = bc sin a sin = c sin a / sin = c / sin a b c = = sin sin sin Law of Sines Slide 35 () / 152 Δ = 1/2 ac sin = 1/2 ab sin = 1/2 bc sin Questioning to help address MP Let's look at one pair standards: at a time and simplify. What concepts that we have learned 1/2 ac sin = 1/2 before ab sin are useful ND in simplify 1/2 ab each sin = 1/2 bc sin algebraic equation? (MP8) ac sin What = ab strategies sin are you ab going sin = to bc use sin to get the same letters on the same side of c sin the = b equation? sin (MP1) a sin = c sin Math Practice c / sin = b / sin [This object is a pull tab] a b c = = sin sin sin a / sin = c / sin Law of Sines Slide 36 / 152 This relationship between the sides and angles of a triangle will be true for all triangles. a b c = = sin sin sin OR sin sin sin = = a b c b c a

19 Example Slide 37 / 152 Find the missing segment lengths and angle measures in Δ if m = 28, m = 103, and b = 26 cm. c 28 a Since we have a triangle, we know that the sum of the interior angles is 180. Therefore, we can find the m using the Triangle Sum Theorem. m = = cm Example Slide 37 () / 152 Find the missing segment lengths and angle measures in Δ if m = 28, m = 103, and b = 26 cm. Filling in the blanks on this slide accomplish MP1. Since we have a triangle, we know that the 28 lternative sum questions: of the interior angles is 180. What information Therefore, do we you can have? find the (MP1) m using the c What strategies Triangle Sum are you Theorem. going to use? (MP1) a m = = 49 an you do this mentally? (MP1 & MP5) Math Practice cm Example Slide 38 / 152 Find the missing segment lengths and angle measures in Δ if m = 28, m = 103, and b = 26 cm. c 28 a 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 sin cm 26sin49 = asin28 a = 26sin49 sin28 a = cm to reveal 26sin103 = csin28 c = 26sin103 sin28 c = cm to reveal

20 Example Slide 38 () / 152 Find the missing segment lengths and angle measures in Δ if m = 28, m = 103, and b = 26 cm. Questioning to address the MP standards: How could you start this problem? (MP1) How can you Now represent that we the have problem the measurements with of all numbers and of the symbols? angles (MP2) and one side length given, we reate 28 a representation can find the remaining of the problem. sides using the Law (MP2) of Sines. c What labels could you use? (MP6) a 26 = a 26 = c sin28 sin49 sin28 sin103 Math Practice cm 26sin49 [This object is = a pull asin28 tab] a = 26sin49 sin28 a = cm to reveal 26sin103 = csin28 c = 26sin103 sin28 c = cm to reveal Example Find the missing segment lengths and angle measures in Δ if m = 122, a = 12 cm and c = 18 cm. 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 sin b sin = 12sin122 sin = m = sin -1 12sin m to = reveal sin122 ( ) 18 Slide 39 / 152 Example Find the missing segment lengths and angle measures in Δ if m = 122, a = 12 cm and c = 18 cm. In this problem, we need to determine one 18 cm of the missing angles first. However, we Questioning to address the MP can't use the Triangle Sum Theorem yet, standards: since we only know the measurement of 1 How could you start this problem? (MP1) angle. Therefore, we need to use the Law How can you represent the problem with of Sines. numbers 12 cmand symbols? (MP2) reate a representation 18 = 12 of the problem. (MP2) sin122 sin b What labels could you use? (MP6) sin = 12sin122 sin = 12sin Math Practice m = sin -1 m to = reveal sin122 ( ) 18 Slide 39 () / 152

21 Example Find the missing segment lengths and angle measures in Δ if m = 122, a = 12 cm and c = 18 cm. Now, that we know 2 angle measures, we can calculate the measurement of our final missing angle. m = = cm b 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 Slide 40 / 152 Example Find the missing segment lengths and angle measures in Δ if m = 122, a = 12 cm and c = 18 cm. Filling in the blank on this slide accomplishes Now, that MP1. we know 2 angle measures, we can calculate the measurement of our final lternative questions: What missing information angle. do you have? (MP1) What m strategies = 180 are - you 122 going to use? = (MP1) 18 cm an you Finally, do this we mentally? can calculate (MP1 & MP5) the length of our 12 Questioning cm final side. to address the MP standards (finding the value of b): How could you start 18 this problem? = b (MP1) How can you represent sin122 the sin23.57 problem with 122 numbers and symbols? (MP2) b reate a representation of the problem. (MP2) 18sin23.57 = bsin122 What labels could [This object you is use? a pull tab] (MP6) b = 18sin23.57 sin122 Math Practice b = 8.49 cm to reveal Slide 40 () / Find the missing segment lengths and angle measures in Δ if m = 70, m = 64, and c = 5 yd. What is m? Slide 41 / yd 64 a 56 D b

22 11 Find the missing segment lengths and angle measures in Δ if m = 70, m = 64, and c = 5 yd. What is m? Slide 41 () / D yd b a 12 Find the missing segment lengths and angle measures in Δ if m = 70, m = 64, and c = 5 yd. What is the value of a? Round your answer to the nearest hundredth. Slide 42 / yd 64 a 70 b 12 Find the missing segment lengths and angle measures in Δ if m = 70, m = 64, and c = 5 yd. What is the value of a? Round your answer to the nearest hundredth. Slide 42 () / yd 64 a a = 6.53 yd 70 b

23 13 Find the missing segment lengths and angle measures in Δ if m = 70, m = 64, and c = 5 yd. What is the value of b? Round your answer to the nearest hundredth. Slide 43 / yd 64 a 70 b 13 Find the missing segment lengths and angle measures in Δ if m = 70, m = 64, and c = 5 yd. What is the value of b? Round your answer to the nearest hundredth. Slide 43 () / yd 64 a b = 6.25 yd 70 b 14 Find the missing segment lengths and angle measures in Δ if m = 111, b = 3 in., and c = 5 in. What is m? Round your answer to the nearest hundredth. Slide 44 / in. a in.

24 14 Find the missing segment lengths and angle measures in Δ if m = 111, b = 3 in., and c = 5 in. What is m? Round your answer to the nearest hundredth. Slide 44 () / in. a m = in. 15 Find the missing segment lengths and angle measures in Δ if m = 111, b = 3 in., and c = 5 in. What is m? Round your answer to the nearest hundredth. Slide 45 / in. a D in. 15 Find the missing segment lengths and angle measures in Δ if m = 111, b = 3 in., and c = 5 in. What is m? Round your answer to the nearest hundredth. Slide 45 () / in. a D in.

25 16 Find the missing segment lengths and angle measures in Δ if m = 111, b = 3 in., and c = 5 in. What is the value of a? Round your answer to the nearest hundredth. Slide 46 / in. a in. 16 Find the missing segment lengths and angle measures in Δ if m = 111, b = 3 in., and c = 5 in. What is the value of a? Round your answer to the nearest hundredth. Slide 46 () / 152 a = 3.07 in. 5 in. a in. Slide 47 / 152 rea of Parallelograms Return to Table of ontents

26 rea of a Parallelogram Slide 48 / 152 From these results, we can find the area formula for any parallelogram by dividing the parallelogram into two triangles. h b rea of a Parallelogram Slide 49 / 152 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. parallelogram = bh h b rea of a Parallelogram Slide 50 / 152 s 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. parallelogram = bh h b

27 Find the area of D. Example Slide 51 / 152 This time, use the cosine rather than the sine function, just for practice. Either can be used depending on the angle you choose 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 Find the area of D. Example Slide 52 / 152 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 nd its area = 8(4.70) = units 2 Find the area of D. Example Slide 52 () / 152 This time, use the cosine rather than the sine function, just for practice. Either can be used depending on the angle you choose. 5 Math Practice Now, let's find the height of Questioning to address the MP 20 our standards: parallelogram. How could you start this problem? (MP1) How can you represent the problem cos20 with = h 5 5 numbers and symbols? (MP2) reate a representation of the problem. h = 5cos20 (MP2) h = 4.70 units 110 What labels could you use? (MP6) 8 nd its area = 8(4.70) = units 2

28 17 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. 2.4 ft 9 ft 4 in. 9 ft 4 in Slide 53 / ft 20.5 ft 9.7 ft D 34.8 ft 17 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. 2.4 ft 9 ft 9.7 ft 9 ft 4 in. 9 ft 4 in ft Slide 53 () / 152 D 34.8 ft 18 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. Slide 54 / ft ft 2 9 ft 4 in. 9 ft 4 in ft ft 2 D ft 2

29 18 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. Slide 54 () / ft ft 2 9 ft 4 in. 9 ft 4 in ft ft 2 D ft 2 19 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. Slide 55 / 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. Slide 55 () / ft 2

30 20 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. 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 in 2 Slide 56 () / If the quilt requires 5 parallelograms horizontally and 10 parallelograms vertically, how much material will Mrs. Polygon need to make the quilt? Slide 57 / 152

31 21 If the quilt requires 5 parallelograms horizontally and 10 parallelograms vertically, how much material will Mrs. Polygon need to make the quilt? Slide 57 () / in 2 Slide 58 / 152 rea of Regular Polygons Return to Table of ontents rea of Regular Polygons Slide 59 / 152 regular polygon is a polygon that has all its sides and angles congruent: it is both equilateral & equiangular.

32 ircumscribing a Regular Polygon Slide 60 / 152 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. r ircumcenter and ircumradius Slide 61 / 152 The center of a regular polygon is the center of its circumscribed circle, shown here as "." 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. r entral ngle of a Regular Polygon Slide 62 / 152 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 entral angle = 360 n where n is the number of sides in the regular polygon. r r

33 The pothem a Regular Polygon Slide 63 / 152 The apothem of a Regular Polygon is the shortest distance from the center of the polygon to one of its sides. n apothem is perpendicular to a side of the polygon. n 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 a r rea of Regular Polygons Slide 64 / 152 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 rea of Regular Polygons Slide 64 () / 152 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. Math Practice Question next to the figure addresses MP2 What is the area of a triangle and MP4. whose base is P/n and whose height is a? a r

34 rea of Regular Polygons Slide 65 / 152 Δ = 1/2 bh Δ = 1/2(P/n)a There are n triangles in a regular polygon, so the area of the polygon is given by: = n Δ = (n)(1/2)(p/n)a = 1/2Pa r r a a a r a r rea of Regular Polygons Slide 65 () / 152 Δ = 1/2 bh Questions to address the MP standards: Δ = 1/2(P/n)a What is this problem asking? (MP1) How could you start this problem? (MP1) How can you represent the problem with There are n triangles in numbers a and symbols? (MP2) regular polygon, so the reate area a representation of the problem. of the polygon is given (MP2) by: Write an algebraic representation to describe this situation. (MP4) = n Δ What concepts that we have learned r abefore were useful in solving this? (MP8) = (n)(1/2)(p/n)a a = 1/2Pa r a Math Practice r a r rea of Regular Polygons Slide 66 / 152 The formula for the area of a polygon is simple: = 1/2Pa ut, 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

35 Example Slide 67 / 152 Let's find the area of a regular pentagon whose sides have a length of 7. Example Let's find the area of a regular pentagon whose sides have a length of 7. Slide 68 / 152 = 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. ut, how do we find the apothem? Let's look at one of the five triangles that comprise the pentagon. Example Slide 69 / 152 Let's find the area of a regular pentagon whose sides have a length of 7. = 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. ut, how do we find the apothem? Let's look at one of the five triangles that comprise the pentagon.

36 Example Slide 70 / 152 Let's find the area of a regular pentagon whose sides have a length of 7. The central angle is given by: 7 m = 360/n = 360/5 = 72 r 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. Example Let's find the area of a regular pentagon whose sides have a length of 7. Slide 71 / 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 Example Slide 72 / 152 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. = 1/2 Pa = 1/2 (35)(4.8) r 72 r = 84 units

37 Example Find the area of a regular octagon whose sides have a length of 8. Slide 73 / What is the question asking? To find the area of the octagon How can you represent the problem with symbols and numbers? = 1/2 Pa How could you start this problem? Is there anything that you can find right away? P = 8(8) = 64 units Example Find the area of a regular octagon whose sides have a length of 8. Slide 73 () / Math Practice Questions 1 & 3 address MP1. What is the question asking? To find the area of the octagon How can you represent the problem Question 2 addresses MP2. with symbols and numbers? = 1/2 Pa How could you start this problem? Is there anything that you can find right away? P = 8(8) = 64 units 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? Slide 74 / reate an equation to find the apothem. tan 67.5 = a 4 a = 4 tan 67.5 a = 9.66 units a 67.5

38 Example Find the area of a regular octagon whose sides have a length of 8. 8 Math Practice How do we find the apothem? How could you use a drawing to show your The 1st set of questions address way of MP5. thinking? 8 "reate an equation..." addresses MP dditional Q's to address Math Practices: Would it help to draw a picture? a (MP4 & MP5) 45 What tools do you need? (MP5) 4 4 What labels could you use? (MP6) reate an equation to find the [This object is a pull apothem. tab] tan 67.5 = a 4 a = 4 tan 67.5 a = 9.66 units Slide 74 () / 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 = 1/2 (9.66)(64) = units 2 Slide 75 / 152 Example Find the area of a regular octagon whose sides have a length of 8. 8 Math Practice 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 dditional Q's to address Math Practices: What information do you = have? 1/2 (9.66)(64) (MP1) What do you need to find = out? (MP1) units 2 What tools do you need? (MP5) What labels could you use? (MP6) Slide 75 () / 152

39 22 alculate the apothem of the regular polygon shown in the figure below. Slide 76 / alculate the apothem of the regular polygon shown in the figure below. Slide 76 () / 152 m = 72 θ = 54 a = 16 sin(54) a = units alculate the side length of the regular polygon shown in the figure below. Slide 77 /

40 23 alculate the side length of the regular polygon shown in the figure below. Slide 77 () / 152 m = 72 θ = 54 b = 16 cos(54) b = 9.40 units s = units alculate the perimeter & area of the regular polygon shown in the figure below. Slide 78 / alculate the perimeter & area of the regular polygon shown in the figure below. Slide 78 () / 152 a = units s = units P = 5(18.80) P = 94 units = 1 / 2 (12.94)(94) = units 2 16

41 25 alculate the side length of the regular polygon shown in the figure below. Slide 79 / alculate the side length of the regular polygon shown in the figure below. Slide 79 () / m = 18 θ = 81 b = 15/tan(81) b = 2.38 units s = 4.76 units 26 alculate the perimeter & area of the regular polygon shown in the figure below. Slide 80 /

42 26 alculate the perimeter & area of the regular polygon shown in the figure below. Slide 80 () / s = 4.76 units P = 10(4.76) P = 47.6 units = 1 / 2 (15)(47.6) = 357 units 2 27 alculate the apothem of the regular polygon shown in the figure below. Slide 81 / alculate the apothem of the regular polygon shown in the figure below. Slide 81 () / 152 m = 60 θ = 60 a = 3.5tan(60) a = 6.06 units Note: , so a = =

43 28 alculate the perimeter of the regular polygon shown in the figure below. Slide 82 / alculate the perimeter of the regular polygon shown in the figure below. Slide 82 () / 152 P = 6(7) P = 42 units 7 29 alculate the area of the regular polygon shown below. Slide 83 / 152 7

44 29 alculate the area of the regular polygon shown below. Slide 83 () / 152 a = 6.06 = P = 6(7) P = 42 units = 1 / 2 (6.06)(42) = units 2 = units 2 7 Slide 84 / 152 rea of ircles and Sectors Return to Table of ontents rea of a ircle Interestingly, the formula, = 1/2Pa, also leads to the formula for the area of a circle. Slide 85 / 152 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. nd, the perimeter of the polygon approaches the circumference of the circle: 2πr. Then, = 1/2Pa approaches = 1/2(2πr)(r) = πr 2 r a

45 30 Find the area of a circle that has a radius of 8 in. Slide 86 / 152 4π in 2 8π in 2 16π in 2 D 64π in 2 30 Find the area of a circle that has a radius of 8 in. Slide 86 () / 152 4π in 2 8π in 2 16π in 2 D 64π in 2 D = 64π in 2 31 Find the area of a circle that has a diameter of 17 in. Slide 87 / π in 2 17π in π in 2 D 289π in 2

46 31 Find the area of a circle that has a diameter of 17 in. Slide 87 () / π in 2 17π in π in 2 D 289π in 2 = 72.25π in 2 32 Find the area of a circle that has a circumference of 18π in. 324π in 2 Slide 88 / π in 2 18π in 2 D 9π in 2 32 Find the area of a circle that has a circumference of 18π in. 324π in 2 Slide 88 () / π in 2 18π in 2 D 9π in 2 = 81π in 2

47 Sectors of ircles Slide 89 / 152 sector of a circle is the portion of the circle enclosed by two radii and the arc that connects them. Minor Sector Major Sector 33 Which arc borders the minor sector? Slide 90 / 152 rc rc rc D D 33 Which arc borders the minor sector? Slide 90 () / 152 rc rc rc D D

48 34 Which arc borders the major sector? Slide 91 / 152 rc rc rc D D 34 Which arc borders the major sector? Slide 91 () / 152 rc rc rc D D sector is part of a circle. The rea of a Sector Slide 92 / 152 The area of a complete circle is given by circle = πr 2 θ r 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: sector = (θ/360 )(πr 2 ) when θ is the central angle of the sector measured in degrees

49 The rea of a Sector Slide 93 / 152 In this case, we know that the measure of our central angle, or m = 110. ut, if we are also told that the radius of the circle is 20 cm, we can determine the area of the sector enclosed by, &. Using our formula, we know that θ = 110, and r = 20 cm. D cm If we substitute these numbers into our formula, it will give us the area of our sector. sector = (110 /360 )π(20) 2 sector = cm 2 The rea of a Sector Slide 93 () / 152 In this case, we know that the measure of our central angle, or m = 110. ut, if we are also told that the radius of 110 the circle is 20 cm, we can determine the 20 cm area of the sector enclosed by, &. The blank spaces help the students "reate a representation of D the Using our formula, we know problem" (MP2). that θ = 110, and r = 20 cm. If we substitute these numbers into our formula, it will give us the area of our sector. sector = (110 /360 )π(20) 2 sector = cm 2 Math Practice The rea of a Sector Slide 94 / 152 D cm lternatively, we could just set this up as a proportion and solve it in one step. rea of Sector rea of ircle sector = = 110 πr entral angle 360 sector = 110 (πr 2 ) 360 sector = 110 π(20) sector = 1100 π cm 2 = cm 2 9

50 The rea of a Sector Slide 94 () / 152 D lternatively, we could just set this up as a proportion and solve it in one step. rea of Sector entral angle Questions to address the MP standards: = rea of ircle 360 How can you represent the problem with numbers and symbols? (MP2) 110 What concepts that we have sector learned = 110before 20 cmwere useful in solving this? πr 2 (MP8) 360 How is the area of a sector related to the arc length? (MP7) sector = 110 (πr 2 ) 360 Math Practice [This object is a pull tab] sector = 110 π(20) sector = 1100 π cm 2 = cm Find the area of the sector. Leave your answer in terms of π Slide 95 / 152 sector = (θ/360 )(πr 2 ) 35 Find the area of the sector. Leave your answer in terms of π units 2 Slide 95 () / 152 sector = (θ/360 )(πr 2 )

51 36 Find the area of the major sector. Leave your answer in terms of π. Slide 96 / cm 60 T sector = (θ/360 )(πr 2 ) 36 Find the area of the major sector. Leave your answer in terms of π. Slide 96 () / cm 60 cm 2 T sector = (θ/360 )(πr 2 ) 37 Find the area of the minor sector of the circle. Round your answer to the nearest hundredth. Slide 97 / cm 30 T

52 37 Find the area of the minor sector of the circle. Round your answer to the nearest hundredth. Slide 97 () / cm 30 T cm 2 38 Find the rea of the major sector for the circle. Round your answer to the nearest thousandth. Slide 98 / cm 85 T 38 Find the rea of the major sector for the circle. Round your answer to the nearest thousandth. Slide 98 () / cm 85 cm 2 T

53 39 What is the central angle for the major sector of the circle? Slide 99 / cm 120 G 39 What is the central angle for the major sector of the circle? Slide 99 () / cm 120 G 40 Find the area of the major sector. Round to the nearest thousandth. Slide 100 / cm 120 G

54 40 Find the area of the major sector. Round to the nearest thousandth. Slide 100 () / cm 120 cm 2 G 41 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. Slide 101 / 152 True False 41 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. Slide 101 () / 152 True False True

55 42 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)? Slide 102 / 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 Each student hundredth)? gets 4 pieces, which is 1/2 a pizza. Slide 102 () / 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. Slide 103 / 152

56 43 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. Slide 103 () / 152 Slide 104 / 152 rea of Other Quadrilaterals Lab: rea of Other Quadrilaterals Return to Table of ontents rea of Other Quadrilaterals Slide 105 / 152 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.

57 rea of a Trapezoid Let's start off by deriving the area formula of a trapezoid. Slide 106 / 152 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 rea of a Trapezoid Slide 107 / 152 We can split the trapezoid into 2 triangles by drawing the diagonal from the upper left corner to the lower right corner. lick 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? dd up the triangle areas b 2 rea of a Trapezoid Slide 107 () / 152 We can split the trapezoid into 2 triangles by drawing the diagonal from the upper left corner to the lower right corner. lick on the bases to reveal each new triangle (given in red). The final question addresses MP1. Math Practice Now, we dditional can separate Questions these two that could be triangles. used: What connections do you see Using the between triangles, the how two could triangles you & the calculate trapezoid? the area of (MP4) a trapezoid? dd up the triangle areas h b 1 b 2

58 rea of a Trapezoid Slide 108 / 152 b1 b1 h h + = h b2 b2 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. reate an equation using this fact and the variables provided. Trapezoid = 1/2 b 1h + 1/2 b 2h What algebra facts can be used to simplify the equation? How does it simplify? Distributive Property or Factoring Trapezoid = 1/2 h(b 1 + b 2) rea of a Trapezoid Slide 108 () / 152 b1 b1 h h + = h b2 The sentence "reate an equation..." b2 addresses MP2. We know that the area of a triangle is 1/2 bh and that the sum of these two triangles will The be final the two area questions of the trapezoid. address reate an equation using MP7. this fact and the variables provided. Trapezoid = 1/2 b 1h + 1/2 b 2h What algebra facts can be used to simplify the equation? How does it simplify? Distributive Property or Factoring Trapezoid = 1/2 h(b 1 + b 2) Math Practice rea of a Rhombus Next, we are going to derive the area formula of a rhombus. Slide 109 / 152 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 = bh will work for this shape. ut what if we are given the diagonal lengths instead? What connections do you see? an 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 1 d 2

59 rea of a Rhombus Next, we are going to derive the area formula of a rhombus. Slide 109 () / 152 If you will recall from our unit on Quadrilaterals, a rhombus is a parallelogram with congruent The questions sides and in diagonals the 3rd paragraph that are perpendicular and bisect (Since each it other. is a...) address MP4. Math Practice Since it is a parallelogram, the formula = bh will work for this shape. ut what if we are given the diagonal lengths instead? What connections do you see? an 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 1 d 2 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. rea of a Rhombus d 1 d 2 Slide 110 / 152 Use the following steps to show the animation in the diagram: lick 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. lick the bottom of the original rhombus (shown in black) to hide it. rea of a Rhombus The diagonals split the rhombus into 4 congruent Diagrams are given below to show the end diagram for each step in the bulleted list. triangles. If we move two the bottom triangles to the top, but d2 on opposite sides (i.e. move the bottom right triangle to the d1 d upper left corner), we will 1 d2 create another shape and determine the area formula. Teacher Notes Use the following steps to show the animation in the diagram: d2 lick on the hash marks in the 2 bottom triangles to reveal each new triangle (given in red). d1 Move the new red triangles to the opposite corners of the shape. lick the bottom of the original rhombus (shown in black) to hide it. d1 d 2 Slide 110 () / 152

60 rea of a Rhombus Slide 111 / 152 d 2 d 1 What shape has been made? Rectangle What is the area formula for this shape? = bh or = lw Using these connections and the variables given above, create an equation to represent the area of a rhombus. Rhombus = 1/2 d 1d 2 rea of a Rhombus Slide 111 () / 152 dditional Questions that could be used: d 1 What shape has been an made? anyone think of another method that might have worked? (MP3) Rectangle - Sample : alculate the area What is the area formula of 1 for of the this right shape? triangles & multiply by 4 = bh or = lw Using these connections and the variables given above, create an equation to represent the area of a rhombus. Rhombus = 1/2 d 1d 2 Math Practice d 2 The questions on this slide address MP2 & MP4. rea of a Kite Slide 112 / 152 Since the shape of a kite is very similar to a rhombus, you are going to explain how the rea of a Kite formula is the same as the rea of a Rhombus formula for homework. Kite = 1/2 d 1d 2 d 2 d 1

61 Example Slide 113 / 152 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 Example Slide 113 () / 152 You are constructing a desk to fit into your bedroom using wood for the Questions to address Math Practices: flat top surface What and is metal this problem bars for asking? the legs. (MP1) In order to save space, you determine How that could an isosceles you start this trapezoid problem? would (MP1) be the best shape. The bases of the How trapezoid could you will use the be drawing 3 feet and to show 7 feet your and the angle formed by the short thinking? base (MP5) and each leg of the trapezoid is 135. How much wood is required What do you to know make about the flat right top triangles surface & trapezoids of your desk? that you can apply to this situation? (MP7) an you find a shortcut to solve this problem? How could your shortcut make 3 ft this problem easier? (MP8) ft ft ft ft 45 2 ft 7 ft 2 ft [This = 1/2(2)(3 object is a pull + 7) tab] = 10 ft 2 44 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 Slide 114 / 152

62 44 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? 432 in 2 or 3 ft 2 Spars Slide 114 () / 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? Slide 115 / m and 1.2 m 1.2 m and 0.4 m 1.8 m and 0.6 m D 0.6 m and 0.4 m 45 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? Slide 115 () / m and 1.2 m 1.2 m and 0.4 m 1.8 m and 0.6 m D 0.6 m and 0.4 m

63 46 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. 60 Slide 116 / in. 46 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 2 Slide 116 () / in. rea of omplex Figures Slide 117 / 152 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. dd the area of a rectangle & 1/2 the area of a circle. Subtract the area of the rectangles from the area of the trapezoid.

64 Slide 118 / 152 Slide 118 () / 152 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?

65 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 rea: = 13.5(8.5) = ft 2 Doorway rea: = 7(3) = 21 ft 2-2 doorways, so total area not included is 42 ft 2 Painted rea: = = ft 2 Example Slide 120 () / 152 In order to redesign your bedroom, you decide to paint one of the walls with an accent color, Questions leaving to the address remaining Math Practices: walls with the same. The wall to be painted measures What information 13' 6" do by you 8' 6" have? and (MP1) contains 2 doorways, each measuring 7' high What and do you 3' need wide. to The find out? doorways (MP1) are not going to be What strategies are you going to use? (MP1) painted with the accent Would color. it help to draw a picture? (MP4) How is this related to the different area Find the amount formulas of wall that space we have that learned will be about? painted. (MP7) How can you represent the problem w/ Wall rea: symbols = 13.5(8.5) & numbers? (MP2) = ft 2 What tools do you need to solve this problem? (MP5) Doorway What rea: labels = could 7(3) you = 21 use? ft 2 (MP6) - 2 doorways, What concepts so total that area we not have learned before were useful included is 42 ft 2 in [This solving object is this? a pull tab] (MP8) Math Practice Painted rea: = = ft 2 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 rea = 72.75(2) = ft 2 Quart-sized ans of Paint = 145.5/87.5 = cans

66 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 Questions can of paint to address covers 87.5 Math ft 2 Practices:, then how many cans are required to What paint 2 information coats of the do accent you have? color? (MP1) What do you need to find out? (MP1) Total painted What strategies rea = 72.75(2) are you going to use? (MP1) = ft 2 How can you represent the problem Quart-sized w/ symbols & numbers? (MP2) ans of What Paint tools = 145.5/87.5 do you need to solve this problem? = (MP5) cans What labels could you use? (MP6) Math Practice 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 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. Questions to address Math Practices: If one quart-size What can information of paint costs do $15, you have? how much (MP1) money do you need to spend What to paint do your wall? need to find out? (MP1) What strategies are you going to use? $15(2) (MP1) = $30 How can you represent the problem w/ symbols & numbers? (MP2) What tools do you need to solve this problem? (MP5) What labels could you use? (MP6) Math Practice

67 47 What is the area of the shaded region? Slide 123 / cm 2 30 cm cm 2 D 200 cm 2 47 What is the area of the shaded region? Slide 123 () / cm 2 30 cm cm 2 D 200 cm 2 48 What is the area of the entire figure? Round your answer to the nearest tenth. Slide 124 / yd yd 2 54 yd 2 D 79.1 yd 2

68 48 What is the area of the entire figure? Round your answer to the nearest tenth. Slide 124 () / yd yd 2 54 yd 2 D 79.1 yd 2 D 49 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? Slide 125 / in. 4 in. 49 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. 121 in 2 Slide 125 () / in.

69 50 The isect uilding ompany 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. Slide 126 / 152 What is the area of the entire roof? 700 ft 2 20 ft 600 ft 2 10 ft 250 ft 2 10 ft D 100 ft 2 20 ft 30 ft 50 The isect uilding ompany 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. Slide 126 () / 152 What is the area of the entire roof? 700 ft 2 20 ft 600 ft ft 2 D 100 ft 2 20 ft 10 ft 10 ft 30 ft 51 The isect uilding ompany 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. Slide 127 / 152 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 D ft 10 ft 10 ft 30 ft

70 51 The isect uilding ompany 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. Slide 127 () / 152 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 D ft 10 ft 10 ft 30 ft 52 The isect uilding ompany 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. Slide 128 / 152 Each bundle of shingles costs $ How much should the isect uilding ompany budget for the shingles? 20 ft 10 ft 10 ft 20 ft 30 ft 52 The isect uilding ompany 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. Slide 128 () / 152 Each bundle of shingles costs $ How much should the isect uilding ompany budget for the shingles? $ ft 10 ft 10 ft 20 ft 30 ft

71 Slide 129 / 152 rea & Perimeter of Figures in the oordinate Plane Return to Table of ontents rea & Perimeter of Figures in the oordinate 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. Slide 130 / 152 Remember that the distance formula is d = fter 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. Example alculate the perimeter and area of rhombus JKLM. y Since JKLM is a rhombus, 10 all of the sides are congruent. How would we calculate the 8 perimeter? alculate one side length using 6 the distance formula & then multiply that value by 4 to 4 K calculate our perimeter. 2 J JK = (7-3) 2 + (3-1) 2 = x L M = = 20 = Perimeter = 4(2 5) = 8 5 units units Slide 131 / 152

72 Example alculate the perimeter and area of rhombus JKLM. y Since JKLM is a rhombus, 10 all of the sides are congruent. 8 Question about How calculating would the we perimeter calculate the addresses MP1. perimeter? alculate one side length using 6 dditional Q's to the address distance Math formula Practices: & then What information multiply do you that have? value (MP1) by 4 to What do you need to find out? (MP1) 4 calculate our perimeter. How Kcan you represent the problem w/ symbols & numbers? JK = (MP2) (7-3) 2 + (3-1) 2 What tools do you need to solve this J problem? (MP5) = What labels could you use? (MP6) x = L6 8-2 = 20 = Perimeter = 4(2 5) -4 M = 8 5 units -6 Math Practice units Slide 131 () / 152 Example alculate the perimeter and area of rhombus JKLM. y Since JKLM is a rhombus, = 1/2 d 1d 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 = 1/2 (2 2)(6 2) = 12 units 2 Slide 132 / 152 Example alculate the perimeter and area of rhombus JKLM. y Since JKLM is a rhombus, = 1/2 d 1d 2 10 Therefore, we need to calculate 8 Questions about the area formula the lengths & the of the diagonals lengths of the diagonals address using MP1. the distance formula. 6 dditional Q's to address Math JL Practices: = (5-3) 2 + (-1-1) 2 4 What information do you have? (MP1) = (-2) 2 What do you need Kto find out? (MP1) How can you represent the problem = 4 w/ symbols J & numbers? (MP2) = 8 = 2 2 What tools do you need to solve this problem? (MP5) KM = (7-1) (3 - (-3)) 2 2What 4labels could you xuse? (MP6) L6 8 = 6-2 When calculating the distances, can anyone think of another method that might = have worked or been easier? (MP7 & MP8) M - ns: triangles could be = used 72 = = 1/2 (2 2)(6 2) = 12 units 2 Math Practice Slide 132 () / 152

73 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. ased on the information given, how many miles is the perimeter of North Dakota? t the end of 2010, the population of North Dakota was 672,591 people. ased on the information given, what was the population density at the end of 2010? Miles (hundreds) 4 3 E 2 1 y H 0 G x Miles (hundreds) F 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. ased on the information given, how many miles is the perimeter of North Dakota? How do we calculate the perimeter? 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) Slide 134 / 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. Questions about the perimeter & the ased on the strategies information used address MP1. y given, how many miles is the perimeter 4 dditional of North Q's to address Math Dakota? Practices: 3 How do How we calculate can you the represent the problem E F perimeter? w/ symbols & numbers? (MP2) 2 What tools do you need to add up all of the side lengths 1solve this problem? (MP5) H What strategies What labels are you could going you use? (MP6) 0 G x to use? Miles (hundreds) counting & distance formula Math Practice Miles (hundreds) Slide 134 () / 152

74 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. ased 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 = = = 211 miles Miles (hundreds) 4 y 3 E 2 F 1 H 0 G x Miles (hundreds) Perimeter: = 1092 miles Slide 135 / 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. Questions about the perimeter & the ased on the y strategies information used address MP1. given, how many miles is 4 the perimeter dditional of North Q's to address Math Dakota? 3 Practices: E How can you represent the F 2problem w/ symbols & numbers? (MP2) EF = 3.3 = 330 miles 1 What tools do you need to solve this GH = 3.4 = 340 problem? miles (MP5) H What labels could you use? EH = 2.11 = 211 miles 0(MP6) G x FG = ( ) 2 + (0-2.11) 2 Miles (hundreds) = (0.1) 2 + (-2.11) 2 Perimeter: = = 1092 miles = = 211 miles Math Practice Miles (hundreds) Slide 135 () / 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. t the end of 2010, the population of North Dakota was 672,591 people. ased 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) Slide 136 / 152

75 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. Questions about population density t the end of 2010, the address population MP6. y of North Dakota was 672,591 people. ased on the dditional information Q's to address 4 Math given, what was the Practices: population 3 density at the end of How 2010? can you represent E the problem F w/ symbols & numbers? 2 (MP2) What does population What density tools do you need to solve this mean? 1 problem? (MP5) ratio that represents What the labels number could you use? H (MP6) of people living per square mile. [This object is a 0pull tab] G x Miles (hundreds) How do we find it? dividing the total population by the total area Math Practice Miles (hundreds) 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. t the end of 2010, the population of North Dakota was 672,591 people. ased on the information given, what was the population density at the end of 2010? rea = 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) 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. Questions about population density t the end of 2010, the address population MP6. y of North Dakota was 672,591 people. ased on the dditional information Q's to address 4 Math given, what was the Practices: population 3 density at the end of How 2010? can you represent E the problem F w/ symbols & numbers? 2 (MP2) rea = 1/2(211)(340 What + 330) tools do you need to solve this 1 = 1/2(211)(670) problem? (MP5) H = 70,685 square What miles labels could you use? (MP6) [This object is a 0pull tab] G x Miles (hundreds) Population Density: 672,591 = 9.51 people per 70,685 square mile Math Practice Miles (hundreds) Slide 137 () / 152

76 53 alculate the perimeter of square PQRS. y units 8 20 units 6 25 units 4 D units P 2 Slide 138 / x 6 8 Q -2 S -4-6 R 53 alculate the perimeter of square PQRS. y 10 units 20 units 25 units D units P Slide 138 () / x 6 8 Q -2 S -4-6 R 54 alculate the area of square PQRS. y units units units 2 4 D 50 units 2 2 P Slide 139 / x 6 8 Q -2 S -4-6 R

77 54 alculate the area of square PQRS. y units 2 20 units 2 25 units D 50 units P Slide 139 () / x 6 8 Q -2 S -4-6 R Series of SMRT 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 required. (0, 100) garden (50, 75) (180, 75) House (140, 25) Driveway (200, 150) (140, 10) (180, 5) (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. 2, ft 2 1, ft 2 E ft 2 1, ft 2 D ft 2 (180, 25) Slide 140 / The picture below shows the yard of the Fractal family. What is the perimeter of their yard? Slide 141 / 152 (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)

78 55 The picture below shows the yard of the Fractal family. What is the perimeter of their yard? Slide 141 () / 152 (0, 150) (0, 100) (0, 20) garden (50, 150) (200, 150) Left side = 130 ft ack = (50, 75) (180, ) ft Right side = 150 ft House Front Diagonal = 40,400 (140, 25) ft (50, 25) Perimeter = = 681 ft Driveway (180, 25) (140, 10) (180, 5) (200, 0) 56 The picture below shows the yard of the Fractal family. Every spring, Mr. Fractal fertilizes the lawn. What is the area of it? Slide 142 / 152 (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) 56 The picture below shows the yard of the Fractal family. Every spring, Mr. Fractal fertilizes the lawn. What is the area of Plot it? of Land: Trapezoid = 28,000 ft 2 (0, 150) (0, 100) (0, 20) garden (50, 150) House: Rectangle (200, 150) = 6,500 ft 2 Driveway: Trapezoid = 700 ft 2 (50, 75) Garden: (180, 75) Triangle = 1,250 ft 2 HouseLawn: (140, = 25) 28,000 - (6, ,250) = 19,550 [This ftobject is a pull tab] (50, 25) 2 Driveway (180, 25) (140, 10) (180, 5) (200, 0) Slide 142 () / 152

79 57 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. Slide 143 / 152 (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) 57 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. Slide 143 () / 152 (0, 150) (0, 100) (0, 20) garden (50, 150) (50, 25) House (140, 25) Driveway Left side = 125 ft ack (200, = 150) 200 ft Right side = 125 ft Front = 70 ft (50, 75) (180, 75) Total = 520 ft (180, 25) (140, 10) (180, 5) (200, 0) 58 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. 2, ft 2 1, ft 2 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) Driveway (200, 150) (180, 25) (140, 10) (180, 5) (200, 0) Slide 144 / 152

80 58 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. 2, ft 2 1, ft 2 1, ft 2 D ft 2 E ft 2 (0, 150) (0, 100) (0, 20) garden (50, 150) (50, 75) (180, 75) (50, 25), D & E House (140, 25) Driveway (200, 150) (180, 25) (140, 10) (180, 5) (200, 0) Slide 144 () / 152 PR Sample Questions Slide 145 / 152 The remaining slides in this presentation contain questions from the PR Sample Test. fter finishing this unit, you should be able to answer these questions. Good Luck! Return to Table of ontents Question 13/25: Part Slide 146 / 152 Topic: rea of a Rectangle 59 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. 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, and the area of the screen is square inches. D 324 E 648 F 1,018 PR Sample Question - EOY

81 Question 13/25: Part Topic: rea of a Rectangle 59 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. 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. & E Select from each set of choices to correctly complete the sentence. The perimeter of the screen is approximately inches, and the area of the screen is square inches. D 324 E 648 F 1,018 PR Sample Question - EOY Slide 146 () / 152 Question 15/25 Topic: rea & Perimeter of a Figures in the oordinate Plane 60 The figure shows rectangle D in the coordinate plane with point at (0, 2.76), at (3.87, 2.76), at (3.87, 0) and D at the origin. Rectangle D can be used to approximate the size of the state of olorado with x and y scales representing hundreds of miles. Slide 147 / 152 Part ased on the information given, how many miles is the perimeter of olorado? PR Sample Question - EOY Question 15/25 Topic: rea & Perimeter of a Figures in the oordinate Plane 60 The figure shows rectangle D in the coordinate plane with point at (0, 2.76), at (3.87, 2.76), at (3.87, 0) and D at the origin. Rectangle D can be used to approximate the size of the state of olorado with x and y scales representing hundreds of miles. Slide 147 () / 152 Part ased on the information given, how many miles is the perimeter of olorado? 1326 miles PR Sample Question - EOY

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