{"id":50904,"date":"2022-04-05T11:42:55","date_gmt":"2022-04-05T06:12:55","guid":{"rendered":"https:\/\/ccssmathanswers.com\/?p=50904"},"modified":"2022-04-07T11:12:44","modified_gmt":"2022-04-07T05:42:44","slug":"big-ideas-math-geometry-answers-chapter-11","status":"publish","type":"post","link":"https:\/\/bigideasmathanswers.com\/big-ideas-math-geometry-answers-chapter-11\/","title":{"rendered":"Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume"},"content":{"rendered":"

High School Students who are seeking homework help can refer to this Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume<\/strong> Ultimate Guide. By using this ultimate preparation material, you all can easily understand and learn the concepts covered in chapter 11 Circumference, Area, and Volume. Enhance your subject knowledge and score high in various exams after referring and practicing the questions and solutions provided in the BIM Geometry Ch 11 Circumference, Area, and volume Textbook Answer Key Pdf. In this article, you will find the aligned Lesson-wise Big Ideas Geometry Answers<\/a> Ch 11 in pdf format to download & access offline too for free of cost.<\/p>\n

Big Ideas Math Book Geometry Answer Key Chapter 11 Circumference, Area, and Volume<\/h2>\n

Access the Lesson-wise Big Ideas Math Geometry Chapter 11 Solution Key and solve all easy & complex questions of Circumference, Area, and Volume with ease. Also, you can excel in math concepts by practicing the BIM Geometry Textbook Answer Key<\/strong>.\u00a0Ace up your preparation with BIM Geometry Ch 11 Circumference, Area, and Volume Answer Key and check your knowledge accordingly for better scores and math skills.<\/p>\n

Circumference, Area, and Volume Maintaining Mathematical Proficiency<\/h3>\n

Find the surface area of the prism.<\/p>\n

Question 1.
\n\"Big<\/p>\n

Answer:
\nThe surface area of the prism = 158.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nl = 5 ft, w = 8 ft, and h = 3 ft.
\nwhere l = length, w = width, and h = height.
\nThe surface area of the rectanguler prism = 2(lw + lh + wh).
\nsurface area = 2(5×8 + 5×3 + 8×3).
\nsurface area = 2(40 + 15 + 24).
\nsurface area = 2(79).
\nsurface area = 158.<\/p>\n

Question 2.
\n\"Big<\/p>\n

Answer:
\nThe surface area of the triangular prism = 68m.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nl = 10 m, p = 4 m, and h = 10.
\nthe surface area of the triangular prism = 2B + ph.
\nb = base, p = perimeter, and h = height.
\nsurface area = 2(6 + 8) + 4(10).
\nsurface area = 2(14) + 40.
\nsurface area = 28 + 40.
\nsurface area = 68m.<\/p>\n

Question 3.
\n\"Big<\/p>\n

Answer:
\nThe surface area of the triangular prism = 42m.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nw = 10 cm, p = 4 cm, and h = 5 cm, l = 6 cm.
\nthe surface area of the triangular prism = 2B + ph.
\nb = base, p = perimeter, and h = height.
\nsurface area = 2(6 + 5) + 4(5).
\nsurface area = 2(11) + 20.
\nsurface area = 22 + 20.
\nsurface area = 42cm.<\/p>\n

Find the missing dimension.<\/p>\n

Question 4.
\nA rectangle has a perimeter 0f 28 inches and a width of 5 inches. What is the length of the rectangle?<\/p>\n

Answer:
\nThe length of the rectangle = 9 in.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nA rectangle has a perimeter of 28 inches and a width of 5 inches.
\nlength of the rectangle = p\/2 – w.
\nlength = 28\/2 – 5.
\nwhere perimeter = 28 in, and w = 5 in.
\nlength = 14 – 5.
\nlength = 9.
\nso the length of the rectangle = 9 in.<\/p>\n

Question 5.
\nA triangle has an area of 12 square centimeters and a height of 12 centimeters. What is the base of the triangle?<\/p>\n

Answer:
\nThe base of the triangle = 2 cm.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nA triangle has an area of 12 sq cm and a height of 12 cm.
\nThe base of the triangle = 2(A)\/h.
\nbase = 2(12)\/12.
\nbase = 24\/12.
\nbase = 2cm.
\nso the base of the triangle = 2 cm.<\/p>\n

Question 6.
\nA rectangle has an area of 84 square feet and a width of 7 feet. What is the length of the rectangle?<\/p>\n

Answer:
\nThe length of the rectangle = 12 ft.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nA rectangle has an area of 84 sq ft and a width of 7 feet.
\narea of the rectangle = l x b.
\n84 = l x 7.
\nl = 84\/7.
\nl = 12.
\nso the length of the rectangle = 12 ft.<\/p>\n

Question 7.
\nABSTRACT REASONING<\/strong>
\nWrite an equation for the surface area of a Prism with a length, width, and height of x inches. What solid figure does the prism represent?<\/p>\n

Answer:
\nThe surface area of a prism = 2(lw + wh + lh).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nlength = l, width = w, and height = x inches.
\nthe surface area of the prism = 2(lw + wh + lh).
\nthe solid figure does the prism represent the rectangular prism.<\/p>\n

Circumference, Area, and Volume Monitoring Progress<\/h3>\n

Draw a net of the three-dimensional figure. Label the dimensions.<\/p>\n

Question 1.
\n\"Big<\/p>\n

Answer:
\nThe surface area of the prism = 64 cm.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nl = 2 ft, w = 4 ft, and h = 4 ft.
\nwhere l = length, w = width, and h = height.
\nThe surface area of the rectanguler prism = 2(lw + lh + wh).
\nsurface area = 2(2×4 + 4×4 + 4×2).
\nsurface area = 2(8 + 16 + 8).
\nsurface area = 2(32).
\nsurface area = 64.<\/p>\n

Question 2.
\n\"Big<\/p>\n

Answer:
\nThe surface area of the prism = 392 m.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nl =8 m, w = 12 m, and h = 5 m.
\nwhere l = length, w = width, and h = height.
\nThe surface area of the rectanguler prism = 2(lw + lh + wh).
\nsurface area = 2(8×12 + 12×5 + 5×8).
\nsurface area = 2(96 + 60 + 40).
\nsurface area = 2(196).
\nsurface area = 392.<\/p>\n

Question 3.
\n\"Big<\/p>\n

Answer:
\nThe surface area of the triangular prism = 170 in.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nB = 10 in, p = 15 in, and h = 15 in, l = 10 in.
\nthe surface area of the triangular prism = 2B + ph.
\nb = base, p = perimeter, and h = height.
\nsurface area = 2(10) + 15(10).
\nsurface area = 2(10) + 150.
\nsurface area = 20 + 150.
\nsurface area = 170 in.<\/p>\n

11.1 Circumference and Arc Length<\/p>\n

Exploration 1<\/strong><\/p>\n

Finding the Length of a Circular Arc<\/p>\n

Work with a partner: Find the length of each red circular arc.<\/p>\n

a. entire circle
\n\"Big<\/p>\n

Answer:<\/p>\n

b. one-fourth of a circle
\n\"Big
\nAnswer:<\/p>\n

c. one-third of a circle
\n\"Big
\nAnswer:<\/p>\n

d. five-eights of a circle
\n\"Big
\nAnswer:<\/p>\n

Exploration 2<\/strong><\/p>\n

Using Arc Length<\/p>\n

Work with a partner: The rider is attempting to stop with the front tire of the motorcycle in the painted rectangular box for a skills test. The front tire makes exactly one-half additional revolution before stopping. The diameter of the tire is 25 inches. Is the front tire still in contact with the painted box? Explain.
\n\"Big<\/p>\n

Answer:<\/p>\n

Communicate Your Answer<\/p>\n

Question 3.
\nHow can you find the length of a circular arc?
\nAnswer:
\nThe length of a circular arc = 2
\nLOOKING FOR REGULARITY IN REPEATED REASONING<\/strong>
\nTo be proficient in math, you need to notice if calculations are repeated and look both for general methods and for shortcuts.
\nAnswer:<\/p>\n

Question 4.
\nA motorcycle tire has a diameter of 24 inches. Approximately how many inches does the motorcycle travel when its front tire makes three-fourths of a revolution?
\nAnswer:<\/p>\n

Lesson 11.1 Circumference and Arc Length<\/h3>\n

Monitoring Progress<\/strong><\/p>\n

Question 1.
\nFind the circumference of a circle with a diameter of 5 inches.<\/p>\n

Answer:
\nCircumference C = \u03c0d
\nC = 3.14x 5 = 15.7 in<\/p>\n

Question 2.
\nFind the diameter of a circle with a circumference of 17 feet.<\/p>\n

Answer:
\nDiameter d = C\/\u03c0
\nd = 17\/\u03c0 = 5.41 ft<\/p>\n

Find the indicated measure.<\/p>\n

Question 3.
\narc length of \\(\\widehat{P Q}\\)
\n\"Big<\/p>\n

Answer:
\narc length of \\(\\widehat{P Q}\\) is 5.887<\/p>\n

Explanation:
\n\\(\\widehat{P Q}\\) = \\(\\frac { 75 }{ 360 } \\) . \u03c0(9)
\n= 5.887<\/p>\n

Question 4.
\ncircumference of \u2299N
\n\"Big<\/p>\n

Answer:
\narc length of LM\/C = LM\/360
\n61.26\/C = 270\/360
\nC = 81.68<\/p>\n

Question 5.
\nradius of \u2299G
\n\"Big<\/p>\n

Answer:
\narc length of EF = \\(\\frac { 60 }{ 360 } \\) \u2022 2\u03c0r
\n10.5 = \\(\\frac { 1 }{ 6 } \\) \u2022 2\u03c0r
\nr = 10.02<\/p>\n

Question 6.
\nA car tire has a diameter of 28 inches. How many revolutions does the tire make while traveling 500 feet?<\/p>\n

Answer:
\nThe car tire have to make 69 revolutions to travel 500 ft.<\/p>\n

Explanation:
\nCircumference C = 2\u03c0r = \u03c0d
\nC = 28\u03c0
\nDistance travelled = number of revolutions x C
\n500 x 12 = number of revolutions x 28\u03c0
\nnumber of revolutions = 68.2<\/p>\n

Question 7.
\nIn Example 4. the radius of the arc for a runner on the blue path is 44.02 meters, as shown in the diagram. About how far does this runner travel to go once around the track? Round to the nearest tenth of a meter.<\/p>\n

Answer:<\/p>\n

Question 8.
\nConvert 15\u00b0 to radians.<\/p>\n

Answer:
\n15\u00b0 = 15 . \\(\\frac { \u03c0 radians }{ 180\u00b0 } \\) = \\(\\frac { \u03c0 }{ 12 } \\) radians<\/p>\n

Question 9.
\nConvert \\(\\frac{4 \\pi}{3}\\) radians to degrees.<\/p>\n

Answer:
\n\\(\\frac{4 \\pi}{3}\\) radians = \\(\\frac{4 \\pi}{3}\\) radians . \\(\\frac { 180\u00b0 }{ \u03c0 radians } \\) = 240 degrees<\/p>\n

Exercise 11.1 Circumference and Arc Length<\/h3>\n

Vocabulary and Core Concept Check<\/p>\n

Question 1.
\nCOMPLETE THE SENTENCE<\/strong>
\nThe circumference of a circle with diameter d is C = _______ .
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nWRITING<\/strong>
\nDescribe the difference between an arc measure and an arc length.<\/p>\n

Answer:
\nAn arc measure is measured in degrees while an arc length is the distance along an arc measured in linear units.<\/p>\n

Monitoring Progress and Modeling with Mathematics<\/p>\n

In Exercises 3 – 10, find the indicated measure.<\/p>\n

Question 3.
\ncircumference of a circle with a radius of 6 inches
\nAnswer:
\n\"Big<\/p>\n

Question 4.
\ndiameter of a circle with a circumference of 63 feet<\/p>\n

Answer:
\nC = 63 ft
\n\u03c0d = 63
\nd = 20.05<\/p>\n

Question 5.
\nradius of a circle with a circumference of 28\u03c0
\nAnswer:
\n\"Big<\/p>\n

Question 6.
\nexact circumference of a circle with a diameter of 5 inches<\/p>\n

Answer:
\nC = \u03c0d
\nC = 5\u03c0 = 15.707<\/p>\n

Question 7.
\narc length of \\(\\widehat{A B}\\)
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\nm\\(\\widehat{D E}\\)
\n\"Big<\/p>\n

Answer:
\n\\(\\frac { arc length of DE }{ 2\u03c0r } \\) = \\(\\frac { DE }{ 360 } \\)
\n\\(\\frac { 8.73 }{ 2\u03c0(10) } \\) = \\(\\frac { DE }{ 360 } \\)
\nDE = 50.01\u00b0<\/p>\n

Question 9.
\ncircumference of \u2299C
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\nradius of \u2299R
\n\"Big
\nAnswer:
\n\\(\\frac { arc length of LM }{ 2\u03c0r } \\) = \\(\\frac { LM }{ 360 } \\)
\n\\(\\frac { 38.95 }{ 2\u03c0r } \\) = \\(\\frac { 260 }{ 360 } \\)
\nr = 8.583<\/p>\n

Question 11.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in finding the circumference of \u2299C.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in finding the length of \\(\\widehat{G H}\\).
\n\"Big<\/p>\n

Answer:
\n\\(\\frac { arc length of GH }{ 2\u03c0r } \\) = \\(\\frac { m GH }{ 360 } \\)
\n\\(\\widehat{G H}\\). = \\(\\frac { 5 }{ 24 } \\) . 2\u03c0(10)
\n= 13.08<\/p>\n

Question 13.
\nPROBLEM SOLVING<\/strong>
\nA measuring wheel is used to calculate the length of a path. The diameter of the wheel is 8 inches. The wheel makes 87 complete revolutions along the length of the path. To the nearest foot, how long is the path?
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\nPROBLEM SOLVING<\/strong>
\nYou ride your bicycle 40 meters. How many complete revolutions does the front wheel make?
\n\"Big<\/p>\n

Answer:
\nCircumference of the front wheel = 2\u03c0(32.5)
\n= 65\u03c0 cm
\nDistance covered = 40 m = 40 x 100 = 4000 cm
\nNumber of revolutions = \\(\\frac { 4000 }{ 65\u03c0 } \\) = 19.58<\/p>\n

In Exercises 15-18 find the perimeter of the shaded region.<\/p>\n

Question 15.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\n\"Big
\nAnswer:
\nTwo horizontal edges are 2 . 3 = 6
\nCircumference of circle = 2\u03c0(3) = 6\u03c0
\nThe perimeter of the shaded region = 6 + 6\u03c0<\/p>\n

Question 17.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\n\"Big
\nAnswer:<\/p>\n

In Exercises 19 – 22, convert the angle measure.<\/p>\n

Question 19.
\nConvert 70\u00b0 to radians.
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\nConvert 300\u00b0 to radians.<\/p>\n

Answer:
\n300 \u2022 (\\(\\frac { \u03c0 }{ 180 } \\)) = \\(\\frac { 5\u03c0 }{ 3 } \\) radian<\/p>\n

Question 21.
\nConvert \\(\\frac{11 \\pi}{12}\\) radians to degrees.
\nAnswer:
\n\"Big<\/p>\n

Question 22.
\nConvert \\(\\frac{\\pi}{8}\\) radian to degrees.<\/p>\n

Answer:
\n\\(\\frac { \u03c0 }{ 8 } \\) \u2022 \\(\\frac { 180 }{ \u03c0 } \\)) = 22.5\u00b0<\/p>\n

Question 23.
\nPROBLEM SOLVING<\/strong>
\nThe London Eye is a Ferris wheel in London, England, that travels at a speed of 0.26 meter per second. How many minutes does it take the London Eye to complete one full revolution?
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 24.
\nPROBLEM SOLVING<\/strong>
\nYou are planning to plant a circular garden adjacent to one of the corners of a building, as shown. You can use up to 38 feet of fence to make a border around the garden. What radius (in feet) can the garden have? Choose all that apply. Explain your reasoning.
\n\"Big
\n(A) 7
\n(B) 8
\n(C) 9
\n(D) 10<\/p>\n

Answer:
\nC = 38 ft
\n2\u03c0r = 38
\nr = 6.04<\/p>\n

In Exercises 25 and 26, find the circumference of the circle with the given equation. Write the circumference in terms of \u03c0<\/p>\n

Question 25.
\nx2<\/sup> + y2<\/sup> = 16
\nAnswer:
\n\"Big<\/p>\n

Question 26.
\n(x + 2)2<\/sup> + (y – 3)2<\/sup> = 9<\/p>\n

Answer:
\nThe radius of circle (x + 2)\u00b2 + (y \u2013 3)\u00b2 = 9 is 3
\nC = 2\u03c0r = 2\u03c0(3) = 6\u03c0
\nThe circumference of the circle is 6\u03c0 units.<\/p>\n

Question 27.
\nUSING STRUCTURE<\/strong>
\nA semicircle has endpoints (- 2, 5) and (2, 8). Find the arc length of the semicircle.
\nAnswer:
\n\"Big<\/p>\n

Question 28.
\nREASONING<\/strong>
\n\\(\\widehat{E F}\\) is an arc on a circle with radius r. Let x\u00b0 be the measure of \\(\\widehat{E F}\\). Describe the effect on the length of \\(\\widehat{E F}\\) if you (a) double the radius of the circle, and (b) double the measure of \\(\\widehat{E F}\\).
\nAnswer:<\/p>\n

Question 29.
\nMAKING AN ARGUMENT<\/strong>
\nYour friend claims that it is possible for two arcs with the same measure to have different arc lengths. Is your friend correct? Explain your reasoning.
\nAnswer:
\n\"Big<\/p>\n

Question 30.
\nPROBLEM SOLVING<\/strong>
\nOver 2000 years ago, the Greek scholar Eratosthenes estimated Earth\u2019s circumference by assuming that the Sun\u2019s rays were Parallel. He chose a day when the Sun shone straight down into a well in the city of Syene. At noon, he measured the angle the Sun\u2019s rays made with a vertical stick in the city of Alexandria. Eratosthenes assumed that the distance from Syene to Alexandria was equal to about 575 miles. Explain how Eratosthenes was able to use this information to estimate Earth\u2019s circumference. Then estimate Earth\u2019s circumference.
\n\"Big
\nAnswer:<\/p>\n

Question 31.
\nANALYZING RELATIONSHIPS<\/strong>
\nIn \u2299C the ratio of the length of \\(\\widehat{P Q}\\) to the length of \\(\\widehat{R S}\\) is 2 to 1. What is the ratio of m\u2220PCQ to m\u2220RCS?
\n(A) 4 to 1
\n(B) 2 to 1
\n(C) 1 to 4
\n(D) 1 to 2
\nAnswer:
\n\"Big<\/p>\n

Question 32.
\nANALYZING RELATIONSHIPS<\/strong>
\nA 45\u00b0 arc in \u2299C and a 30\u00b0 arc in \u2299P have the same length. What is the ratio of the radius r1<\/sub> of \u2299C to the radius r2<\/sub> of \u2299P? Explain your reasoning.
\nAnswer:<\/p>\n

Question 33.
\nPROBLEM SOLVING<\/strong>
\nHow many revolutions does the smaller gear complete during a single revolution of the larger gear?
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 34.
\nUSING STRUCTURE<\/strong>
\nFind the circumference of each circle.
\na. a circle circumscribed about a right triangle whose legs are 12 inches and 16 inches long
\nAnswer:
\n\"Big
\nc\u00b2 = a\u00b2 + b\u00b2
\nc\u00b2 = 12\u00b2+ 16\u00b2= 400
\nc = 20 in
\ncircumference c = d\u03c0
\n= 20\u03c0 = 62.83 in<\/p>\n

b. a circle circumscribed about a square with a side length of 6 centimeters
\nAnswer:
\n\"Big
\nd\u00b2 = 6\u00b2+ 6\u00b2 = 72
\nd = 8.49 cm
\nC = d\u03c0
\nC = 8.49\u03c0
\nC = 26.67 cm<\/p>\n

c. a circle inscribed in an equilateral triangle with a side length of 9 inches
\nAnswer:
\n\"Big
\nr = \\(\\frac { a\u221a3 }{ 3 } \\)
\nr = \\(\\frac { 9\u221a3 }{ 3 } \\)
\nr = 3\u221a3 = 5.2
\nC = 2\u03c0r
\nC = 2\u03c0 (5.2) = 32.67 in<\/p>\n

Question 35.
\nREWRITING A FORMULA<\/strong>
\nWrite a formula in terms of the measure \u03b8 (theta) of the central angle in radians) that can he used to find the length of an arc of a circle. Then use this formula to find the length of an arc of a circle with a radius of 4 inches and a central angle of \\(\\frac{3 \\pi}{4}\\) radians.
\nAnswer:
\n\"Big<\/p>\n

Question 36.
\nHOW DO YOU SEE IT?<\/strong>
\nCompare the circumference of \u2299P to the length of \\(\\widehat{D E}\\). Explain your reasoning.
\n\"Big
\nAnswer:<\/p>\n

Question 37.
\nMAKING AN ARGUMENT<\/strong>
\nIn the diagram. the measure of the red shaded angle is 30\u00b0. The arc length a is 2. Your classmate claims that it is possible to find the circumference of the blue circle without finding the radius of either circle. Is your classmate correct? Explain your reasoning.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 38.
\nMODELING WITH MATHEMATICS<\/strong>
\nWhat is the measure (in radians) of the angle formed by the hands of a clock at each time? Explain your reasoning.
\na. 1 : 30 P.M.<\/p>\n

Answer:
\n3\u03c0\/4<\/p>\n

b. 3:15 P.M.<\/p>\n

Answer:
\n\u03c0\/24<\/p>\n

Question 39.
\nMATHEMATICAL CONNECTIONS<\/strong>
\nThe sum of the circumferences of circles A, B, and C is 63\u03c0. Find AC.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 40.
\nTHOUGHT PROVOKING<\/strong>
\nIs \u03c0 a rational number? Compare the rational number \\(\\frac{355}{113}\\) to \u03c0. Find a different rational number that is even closer \u03c0.<\/p>\n

Answer:
\n\u03c0 is not a rational number as it can not be represented as an equivalent fraction. \u03c0 = 3.14 and 355\/113 = 3.14. This fraction resembles that value of \u03c0. Therefore a more accurate fraction will be starting by the value of 7 decimla places of \u03c0, therefore 3.1415926 x x = a.<\/p>\n

Question 41.
\nPROOF<\/strong>
\nThe circles in the diagram are concentric and \\(\\overrightarrow{F G}\\) \u2245 \\(\\overrightarrow{G H}\\) Prove that \\(\\widehat{J K}\\) and \\(\\widehat{N G}\\) have the same length.
\n\"Big
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 42.
\nREPEATED REASONING<\/strong>
\n\\(\\overline{A B}\\) is divided into four congruent segments, and semicircles with radius r are drawn.
\n\"Big
\na. What is the sum of the four arc lengths?
\nAnswer:<\/p>\n

b. What would the sum of the arc lengths be if \\(\\overline{A B}\\) was divided into 8 congruent segments? 16 congruent segments? n congruent segments? Explain your reasoning.
\nAnswer:<\/p>\n

Maintaining Mathematical Proficiency<\/p>\n

Find the area of the polygon with the given vertices.<\/p>\n

Question 43.
\nX(2, 4), Y(8, – 1), Z(2, – 1)
\nAnswer:
\n\"Big<\/p>\n

Question 44.
\nL(- 3, 1), M(4, 1), N(4, – 5), P(- 3, – 5)<\/p>\n

Answer:
\n\"Big
\nLP = \u221a(-3 + 3)\u00b2 + (-5 – 1)\u00b2 = 6
\nPN = \u221a(4 + 3)\u00b2 + (-5 + 5)\u00b2 = 7
\nMN = \u221a(4 – 4)\u00b2 + (-5 – 1)\u00b2 = 6
\nLM = \u221a(4 + 3)\u00b2 + (1 – 1)\u00b2= 7
\nArea = 6 x 7 = 42 units<\/p>\n

11.2 Areas of Circles and Sectors<\/h3>\n

Exploration 1<\/strong><\/p>\n

Finding the Area of a Sector of a Circle<\/p>\n

Work with a partner: A sector of a circle is the region bounded by two radii of the circle and their intercepted arc. Find the area of each shaded circle or sector of a circle.<\/p>\n

a. entire circle
\n\"Big
\nAnswer:<\/p>\n

b. one – fourth of a circle
\n\"Big
\nAnswer:<\/p>\n

c. seven – eighths of a circle.
\n\"Big
\nAnswer:<\/p>\n

d. two – thirds of a circle
\n\"Big
\nAnswer:<\/p>\n

Exploration 2<\/p>\n

Finding the Area of a Circular Sector<\/p>\n

Work with a partner: A center pivot irrigation system consists of 400 meters of sprinkler equipment that rotates around a central pivot point at a rate of once every 3 days to irrigate a circular region with a diameter of 800 meters. Find the area of the sector that is irrigated by this system in one day.
\nREASONING ABSTRACTLY<\/strong>
\nTo be proficient in math, you need to explain to yourself the meaning of a problem and look for entry points to its solution.
\n\"Big
\nAnswer:<\/p>\n

Communicate Your Answer<\/p>\n

Question 3.
\nHow can you find the area of a sector of a circle?
\nAnswer:<\/p>\n

Question 4.
\nIn Exploration 2, find the area of the sector that is irrigated in 2 hours.
\nAnswer:<\/p>\n

Lesson 11.2 Areas of Circles and Sectors<\/h3>\n

Monitoring progress<\/strong><\/p>\n

Question 1.
\nFind the area of a circle with a radius of 4.5 meters.<\/p>\n

Answer:
\nCircle area = \u03c0r\u00b2
\nA = \u03c0(4.5)\u00b2 = 20.25\u03c0<\/p>\n

Question 2.
\nFind the radius of a circle with an area of 176.7 square feet.<\/p>\n

Answer:
\nCircle area = \u03c0r\u00b2
\n176.7 = \u03c0r\u00b2
\nr\u00b2 = 56.24
\nr = 7.499<\/p>\n

Question 3.
\nAbout 58,000 people live in a region with a 2-mile radius. Find the population density in people per square mile.<\/p>\n

Answer:
\nThe population density is about 4615.49 people per square mile.<\/p>\n

Explanation:
\nA = \u03c0r\u00b2 = \u03c0 \u2022 2\u00b2 = 4\u03c0
\nPopulation density = \\(\\frac { number of people }{ area of land } \\)
\n= \\(\\frac { 58000 }{ 4\u03c0 } \\) = 4615.49<\/p>\n

Question 4.
\nA region with a 3-mile radius has a population density of about 1000 people per square mile. Find the number of people who live in the region.<\/p>\n

Answer:
\nThe number of people who live in the region are 28274.<\/p>\n

Explanation:
\nA = \u03c0r\u00b2 = \u03c0 \u2022 3\u00b2 = 9\u03c0
\nPopulation density = \\(\\frac { number of people }{ area of land } \\)
\nNumber of people = 1000 x 9\u03c0 = 28274<\/p>\n

Find the indicated measure<\/p>\n

\"Big<\/p>\n

Question 5.
\narea of red sector<\/p>\n

Answer:
\nThe area of red sector = 205.25<\/p>\n

Explanation:
\nm\u2220FDE = 120\u00b0, FE = 120\u00b0 and FGE = 360\u00b0 – 120\u00b0 = 240\u00b0
\nArea of red sector = \\(\\frac { FE }{ 360\u00b0 } \\) \u2022 \u03c0r\u00b2
\n= \\(\\frac { 120 }{ 360\u00b0 } \\) \u2022 \u03c0(14\u00b2)
\n= 205.25<\/p>\n

Question 6.
\narea of blue sector<\/p>\n

Answer:
\nArea of blue sector = 410.5<\/p>\n

Explanation:
\nArea of blue sector = \\(\\frac { FGE }{ 360\u00b0 } \\) \u2022 \u03c0r\u00b2
\n= \\(\\frac { 240 }{ 360\u00b0 } \\) \u2022 \u03c0(14\u00b2) = 410.5<\/p>\n

Question 7.
\nFind the area of \u2299H.<\/p>\n

Answer:
\nArea of \u2299H = 907.92 sq cm<\/p>\n

Explanation:
\nArea of sector FHG =\\(\\frac { FG }{ 360\u00b0 } \\) \u2022 Area of \u2299H
\n214.37 = \\(\\frac { 85 }{ 360\u00b0 } \\) \u2022 Area of \u2299H
\nArea of \u2299H = 907.92 sq cm<\/p>\n

Question 8.
\nFind the area of the figure.<\/p>\n

Answer:
\nArea of triangle = \\(\\frac { 1 }{ 2 } \\) \u2022 7 \u2022 7
\n= 24.5 sq m
\nArea of semi circle = \u03c0r\u00b2\/2
\n= \u03c0(3.5)\u00b2\/2
\n= 19.242255
\nArea of the figure = 24.5 + 19.24 = 43.74 sq m<\/p>\n

Question 9.
\nIf you know the area and radius of a sector of a circle, can you find the measure of the intercepted arc? Explain.<\/p>\n

Answer:<\/p>\n

Exercise 11.2 Areas of Circles and Sectors<\/h3>\n

Vocabulary and Core Concept Check<\/p>\n

Question 1.
\nVOCABULARY<\/strong>
\nA(n) ____________ of a circle is the region bounded by two radii of the circle and their intercepted arc.
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nWRITING<\/strong>
\nThe arc measure of a sector in a given circle is doubled. will the area of the sector also be doubled? Explain our reasoning.<\/p>\n

Answer:
\nYes<\/p>\n

Explanation:
\nArea of sector with arc measure x and radius r is s = \u03c0\/180(xr)
\nIf x becomes doube, then s1 = \u03c0\/180(2xr) = 2s
\nThis means that if the arc measure doubles, area of the sector also doubles.<\/p>\n

Monitoring Progress and Modeling with Mathematics<\/p>\n

In Exercise 3 – 10, find the indicated measure,<\/p>\n

Question 3.
\narea of \u2299C
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 4.
\narea of \u2299C
\n\"Big<\/p>\n

Answer:
\nArea A = \u03c0r\u00b2
\nA = \u03c0(10)\u00b2 = 100\u03c0 sq in<\/p>\n

Question 5.
\narea of a circle with a radius of 5 inches
\nAnswer:
\n\"Big<\/p>\n

Question 6.
\narea of a circle with a diameter of 16 feet<\/p>\n

Answer:
\nd = 2r
\nCircle area = \u03c0r\u00b2 = (\u03c0\/4)d\u00b2
\n= (\u03c0\/4)16\u00b2 = 64\u03c0<\/p>\n

Question 7.
\nradius of a circle with an area of 89 square feet
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\nradius of a circle with an area of 380 square inches<\/p>\n

Answer:
\nA = \u03c0r\u00b2
\n380 = \u03c0r\u00b2
\nr = 10.99<\/p>\n

Question 9.
\ndiameter of a circle with an area of 12.6 square inches
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\ndiameter of a circle with an area of 676\u03c0 square centimeters<\/p>\n

Answer:
\nArea A = 676\u03c0 square centimeters
\n(\u03c0\/4)d\u00b2 = 676\u03c0
\nd\u00b2 = 2704
\nd = 52<\/p>\n

In Exercises 11 – 14, find the indicated measure.<\/p>\n

Question 11.
\nAbout 210,000 people live in a region with a 12-mile radius. Find the population density in people per square mile.
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\nAbout 650,000 people live in a region with a 6-mile radius. Find the population density in people per square mile.<\/p>\n

Answer:
\nThe population density is about 5747 people per square mile.<\/p>\n

Explanation:
\nArea of region = \u03c0(6)\u00b2 = 36\u03c0
\nPopulation density = \\(\\frac { Number of people }{ area of land } \\)
\n= \\(\\frac { 650,000 }{ 36\u03c0 } \\) = 5747.2<\/p>\n

Question 13.
\nA region with a 4-mile radius has a population density of about 6366 people per square mile. Find the number of people who live in the region.
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\nAbout 79,000 people live in a circular region with a population density of about 513 people per square mile. Find the radius of the region.<\/p>\n

Answer:
\nThe radius of the region is 7<\/p>\n

Explanation:
\nPopulation density = \\(\\frac { Number of people }{ area of land } \\)
\n513 = \\(\\frac { 79,000 }{ \u03c0r\u00b2 } \\)
\n\u03c0r\u00b2 = 153.99
\nr = 7<\/p>\n

In Exercises 15-18 find the areas of the sectors formed by\u2220DFE.<\/p>\n

Question 15.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\n\"Big<\/p>\n

Answer:
\nArea of sector = \\(\\frac { 104\u00b0 }{ 360\u00b0 } \\) \u2022 \u03c0(14)\u00b2
\n= 177.88
\nArea of red region is 177.88 sq cm
\nArea of blue region = \\(\\frac { 256\u00b0 }{ 360\u00b0 } \\) \u2022 \u03c0(14)\u00b2
\n= 437.86 sq cm<\/p>\n

Question 17.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\n\"Big<\/p>\n

Answer:
\nArea of red region is 10.471 sq ft
\nArea of the blue region is 39.79 sq ft<\/p>\n

Explanation:
\nArea of sector = \\(\\frac { 75\u00b0 }{ 360\u00b0 } \\) \u2022 \u03c0(4)\u00b2
\n= 10.471
\nArea of red region is 10.471 sq ft
\nArea of blue region = \\(\\frac { 285\u00b0 }{ 360\u00b0 } \\) \u2022 \u03c0(4)\u00b2
\n= 39.79 sq ft<\/p>\n

Question 19.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in finding. the area of the circle.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in finding the area of sector XZY when the area of \u2299Z is 255 square feet.
\n\"Big<\/p>\n

Answer:
\nArea of \u2299Z is 255 square feet
\n\u03c0r\u00b2 = 255
\nr = 9
\nArea of sector XZY = \\(\\frac { 115 }{ 360 } \\) \u2022 255
\nn = 81.458 sq ft<\/p>\n

In Exercises 21 and 22, the area of the shaded sector is show. Find the indicated measure.<\/p>\n

Question 21.
\narea of \u2299M
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 22.
\nradius of \u2299M
\n\"Big<\/p>\n

Answer:
\nradius of \u2299M = 3.98<\/p>\n

Explanation:
\nArea of region = \\(\\frac { 89 }{ 360 } \\) . Area of \u2299M
\n12.36 = \\(\\frac { 89 }{ 360 } \\) . Area of \u2299M
\nArea of \u2299M = 49.99
\n\u03c0r\u00b2 = 49.99
\nr = 3.98<\/p>\n

In Exercises 23 – 28, find the area of the shaded region.<\/p>\n

Question 23.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 24.
\n\"Big<\/p>\n

Answer:
\nThe area of the shaded region is 85.840 sq in.<\/p>\n

Explanation:
\nArea of square = 20\u00b2 = 400
\nDiameter of one circle = 10
\nradius of one circle = 5 in
\nArea of one circle = \u03c0(5)\u00b2 = 78.53
\nAreas of four circle = 314.159
\nArea of shaded region = 400 – 314.159 = 85.840<\/p>\n

Question 25.
\n\"Big
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 26.
\n\"Big<\/p>\n

Answer:
\nThe area of shaded region is 301.59<\/p>\n

Explanation:
\nThe radius of smaller circle is 8 cm
\nThe radius of bigger circle is 16 cm
\nArea of smaller semicircle = \\(\\frac { 1 }{ 2 } \\)(\u03c0(8)\u00b2) = 100.53
\nArea of lager semicircle = \\(\\frac { 1 }{ 2 } \\)(\u03c0(16)\u00b2) = 402.123
\nArea of shaded region = 402.123 – 100.53 = 301.59<\/p>\n

Question 27.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 28.
\n\"Big
\nAnswer:
\nArea of shaded region = 7.63<\/p>\n

Explanation:
\nc\u00b2 = 3\u00b2 + 4\u00b2 = 25
\nc = 5
\nRadius = 2.5
\nCircle area = \u03c0(2.5)\u00b2 = 19.63
\nArea of triangle = (3 x 4)\/2 = 6
\nArea of shaded region = 19.63 – 12 = 7.63<\/p>\n

Question 29.
\nPROBLEM SOLVING<\/strong>
\nThe diagram shows the shape of a putting green at a miniature golf course. One part of the green is a sector of a circle. Find the area of the putting green.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 30.
\nMAKING AN ARGUMENT<\/strong>
\nYour friend claims that if the radius of a circle is doubled, then its area doubles. Is your friend correct? Explain your reasoning.<\/p>\n

Answer:
\nThe friend is not correct. doubling the radius quadruples the area.<\/p>\n

Explanation:
\nArea of circle with radius r = \u03c0r\u00b2
\nArea of circle with radius 2r = \u03c0(2r)\u00b2 = 4\u03c0r\u00b2<\/p>\n

Question 31.
\nMODELING WITH MATHEMATICS<\/strong>
\nThe diagram shows the area of a lawn covered by a water sprinkler.
\n\"Big
\na. What is the area of the lawn that is covered by the sprinkler?
\nb. The water pressure is weakened so that the radius is 12 feet. What is the area of the lawn that will be covered?
\nAnswer:
\n\"Big<\/p>\n

Question 32.
\nMODELING WITH MATHEMATICS<\/strong>
\nThe diagram shows a projected beam of light from a lighthouse.
\n\"Big
\na. What is the area of water that can be covered by the light from the lighthouse?
\nAnswer:
\nArea = \\(\\frac { 115 }{ 360 } \\) x \u03c0(18)\u00b2
\n= 325.15 sq mi<\/p>\n

b. What is the area of land that can be covered by the light from the lighthouse?
\nAnswer:
\nArea = \\(\\frac { 245 }{ 360 } \\) x \u03c0(18)\u00b2
\n= 692.72 sq mi<\/p>\n

Question 33.
\nANALYZING RELATIONSHIPS<\/strong>
\nLook back at the Perimeters of Similar Polygons Theorem (Theorem 8.1) and the Areas of Similar PoIyons Theorem (Theorem 8.2) in Section 8.1. How would you rewrite these theorems to apply to circles? Explain your reasoning.
\nAnswer:
\n\"Big<\/p>\n

Question 34.
\nANALYZING RELATIONSHIPS<\/strong>
\nA square is inscribed in a circle. The same square is also circumscribed about a smaller circle. Draw a diagram that represents this situation. Then find the ratio of the area of the larger circle to the area of the smaller circle.<\/p>\n

Answer:
\nWe start by assigning a variable to the radius of the inner circle. It is r, therefore the area of the circle is \u03c0r\u00b2
\nIt can be seen that the side length of square is twice this radius. Therefore it can be said that the side length of this square is 2r.
\nNext, it can be seen that the diagonal of the square is diameter of outer circle. Therefore, length of the diagonal of the circle d = 2r\u221a2. outer circle radius = r\u221a2
\nArea of outer circle 2\u03c0r\u00b2
\nThe ratio of the area of larger circle to the smaller circle = 2.<\/p>\n

Question 35.
\nCONSTRUCTION<\/strong>
\nThe table shows how students get to school.
\n\"Big
\na. Explain why a circle graph is appropriate for the data.
\nb. You will represent each method by a sector of a circle graph. Find the central angle to use for each sector. Then construct the graph using a radius of 2 inches.
\nc. Find the area of each sector in your graph.
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 36.
\nHOW DO YOU SEE IT?<\/strong>
\nThe outermost edges of the pattern shown form a square. If you know the dimensions of the other square, is it possible to compute the total colored area? Explain.
\n\"Big
\nAnswer:<\/p>\n

Question 37.
\nABSTRACT REASONING<\/strong>
\nA circular pizza with a 12-inch diameter is enough for you and 2 friends. You want to buy pizzas for yourself and 7 friends. A 10-inch diameter pizza with one topping Costs $6.99 and a 14-inch diameter pizza with one topping Costs $12.99. How many 10-inch and 14-inch pizzas should you buy in each situation? Explain.
\na. You want to spend as little money as possible.
\nb. You want to have three pizzas. each with a different topping, and spend as little money as possible.
\nC. You want to have as much of the thick outer crust as possible.
\nAnswer:
\n\"Big<\/p>\n

Question 38.
\nTHOUGHT PROVOKING<\/strong>
\nYou know that the area of a circle is \u03c0r2<\/sup>. Find the formula for the area of an ellipse, shown below.
\n\"Big<\/p>\n

Answer:
\nEllipse area = \u03c0ab<\/p>\n

Question 39.
\nMULTIPLE REPRESENTATIONS<\/strong>
\nConsider a circle with a radius of 3 inches.
\na. Complete the table, where x is the measure of the arc and is the area of the corresponding sector. Round your answers to the nearest tenth.
\n\"Big
\nb. Graph the data in the table.
\nc. Is the relationship between x and y linear? Explain.
\nd. If parts (a) – (c) were repeated using a circle with a radius of 5 inches, would the areas in the table change? Would your answer to part (c) change? Explain your reasoning.
\nAnswer:
\n\"Big
\n\"Big
\n\"Big<\/p>\n

Question 40.
\nCRITICAL THINKING<\/strong>
\nFind the area between the three congruent tangent circles. The radius of each circle is 6 inches.
\n\"Big
\nAnswer:<\/p>\n

Question 41.
\nPROOF<\/strong>
\nSemicircles with diameters equal to three sides of a right triangle are drawn, as shown. Prove that the sum of the areas of the two shaded crescents equals the area of the triangle.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Maintaining Mathematical proficiency<\/p>\n

Find the area of the figure.<\/p>\n

Question 42.
\n\"Big<\/p>\n

Answer:
\nArea = \\(\\frac { 1 }{ 2 } \\)(base x height)
\nArea = \\(\\frac { 1 }{ 2 } \\)(18 x 6) = 54 sq in<\/p>\n

Question 43.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 44.
\n\"Big
\nAnswer:
\nArea = \\(\\frac { 1 }{ 2 } \\)(base x height)
\nA = \\(\\frac { 1 }{ 2 } \\)(13 x 9) = 58.5 sq in<\/p>\n

Question 45.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

11.3 Areas of Polygons<\/h3>\n

Exploration 1<\/strong><\/p>\n

Finding the Area of a Regular Polygon<\/p>\n

Work with a partner: Use dynamic geometry software to construct each regular polygon with side lengths of 4, as shown. Find the apothem and use it to find the area of the polygon. Describe the steps that you used.
\na.
\n\"Big
\nAnswer:<\/p>\n

b.
\n\"Big
\nAnswer:<\/p>\n

c.
\n\"Big
\nAnswer:<\/p>\n

d.
\n\"Big
\nAnswer:<\/p>\n

Exploration 2<\/strong><\/p>\n

Writing a Formula for Area<\/p>\n

Work with a partner: Generalize the steps you used in Exploration 1 to develop a formula for the area of a regular polygon.
\nREASONING ABSTRACTLY<\/strong>
\nTo be proficient in math, you need to know and flexibly use different properties of operations and objects.
\nAnswer:<\/p>\n

Communicate Your Answer<\/p>\n

Question 3.
\nHow can you find the area of a regular polygon?
\nAnswer:<\/p>\n

Question 4.
\nRegular pentagon ABCDE has side lengths of 6 meters and an apothem of approximately 4.13 meters. Find the area of ABCDE.
\nAnswer:<\/p>\n

Lesson 11.3 Areas of Polygons<\/h3>\n

Monitoring Progress<\/strong><\/p>\n

Question 1.
\nFind the area of a rhombus with diagonals d1<\/sub> = 4 feet and d2<\/sub> = 5 feet.<\/p>\n

Answer:
\nArea of rhombus = \\(\\frac { 1 }{ 4 } \\)(d\u2081d\u2082)
\n= \\(\\frac { 1 }{ 4 } \\)(4 x 5) = 5 sq ft<\/p>\n

Question 2.
\nFind the area of a kite with diagonals d1<\/sub> = 12 inches and d1<\/sub> = 9 inches.<\/p>\n

Answer:
\nArea of kite = \\(\\frac { 1 }{ 4 } \\)(d\u2081d\u2082)
\n= \\(\\frac { 1 }{ 4 } \\)(12 x 9) = 27 sq in<\/p>\n

In the diagram. WXYZ is a square inscribed in \u2299P.<\/p>\n

\"Big<\/p>\n

Question 3.
\nIdentify the center, a radius, an apothem, and a central angle of the polygon.<\/p>\n

Answer:
\nP is the center, PY or PX is the radius, PQ is apothem, \u2220XPY is the central angle.<\/p>\n

Question 4.
\nFind m\u2220XPY, m\u2220XPQ, and m\u2220PXQ.<\/p>\n

Answer:
\nm\u2220XPY = \\(\\frac { 360 }{ 4 } \\) = 90
\nm\u2220XPQ = 90\/2 = 45
\nm\u2220PXQ = 180 – (90 + 45) = 45<\/p>\n

Find the area of regular polygon<\/p>\n

Question 5.
\n\"Big<\/p>\n

Answer:
\nc = \u221a(8\u00b2 + 6.5\u00b2) = 10.3
\na = 20.61
\nArea = 0.25(\u221a5(5+2\u221a5) a\u00b2
\nArea = 0.25(\u221a5(5+2\u221a5) 20.61\u00b2 = 730.8<\/p>\n

Question 6.
\n\"Big
\nAnswer:
\nArea = \\(\\frac { 5a\u00b2 }{ 2 } \\)\u221a(5+2\u221a5)
\n= \\(\\frac { 5(7\u00b2) }{ 2 } \\)\u221a(5+2\u221a5)
\nArea = 55377<\/p>\n

Exercise 11.3 Areas of Polygons<\/h3>\n

Vocabulary and Core Concept Check<\/p>\n

Question 1.
\nWRITING<\/strong>
\nExplain how to find the measure of a central angle of a regular polygon.
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nDIFFERENT WORDS, SAME QUESTION<\/strong>
\nWhich is different? Find \u201cboth\u201d answers.
\n\"Big
\nFind the radius of \u2299F. Answer:<\/p>\n

Answer:
\nEF = radius = 6.8<\/p>\n

Find the apothem of polygon ABCDE.<\/p>\n

Answer:
\nGF = apothem = 5.5<\/p>\n

Find AF.<\/p>\n

Answer:
\nAF = \u221a4\u00b2 + 5.5\u00b2
\nAF = 6.8<\/p>\n

Find the radius of polygon ABCDE.
\nAnswer:
\nAF = radius = 6.8<\/p>\n

Monitoring Progress and Modeling with Mathematics<\/p>\n

In Exercises 3 – 6, find the area of the kite or rhombus.<\/p>\n

Question 3.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 4.
\n\"Big<\/p>\n

Answer:
\nd\u2081 = 6 + 6 = 12
\nd\u2082 = 2 + 10 = 12
\narea A = \\(\\frac { 1 }{ 4 } \\)(d\u2081d\u2082)
\n= \\(\\frac { 1 }{ 4 } \\)(12 x 12)
\n= 36<\/p>\n

Question 5.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 6.
\n\"Big
\nAnswer:
\narea A = \\(\\frac { 1 }{ 4 } \\)(d\u2081d\u2082)
\nA = \\(\\frac { 1 }{ 4 } \\)(5 x 6) = 7.5<\/p>\n

In Exercises 7 – 10, use the diagram<\/p>\n

\"Big<\/p>\n

Question 7.
\nIdentify the center of polygon JKLMN?
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\nIdentify a central angle of polygon JKLMN.<\/p>\n

Answer:
\n\u2220NPM is the central angle of polygon JKLMN<\/p>\n

Question 9.
\nWhat is the radius of polygon JKLMN?
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\nWhat is the apothem of polygon JKLMN?
\nAnswer:
\nQP is the apothem of polygon JKLMN<\/p>\n

In Exercises 11 – 14, find the measure of a central angle of a regular polygon with the given number of sides. Round answers to the nearest tenth of a degree, if necessary.<\/p>\n

\"Big<\/p>\n

Question 11.
\n10 sides
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\n18 sides<\/p>\n

Answer:
\nThe measure of central angle = \\(\\frac { 360 }{ 18 } \\) = 20<\/p>\n

Question 13.
\n24 sides
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\n7 sides<\/p>\n

Answer:
\nThe measure of central angle = \\(\\frac { 360 }{ 7 } \\) = 51.42<\/p>\n

In Exercises 15 – 18, find the given angle measure for regular octagon ABCDEFGH.<\/p>\n

Question 15.
\nm\u2220GJH
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\nm\u2220GJK<\/p>\n

Answer:
\nm\u2220GJK = m\u2220GJH\/2
\nm\u2220GJK = 22.5<\/p>\n

Question 17.
\nm\u2220KGJ
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\nm\u2220EJH
\nAnswer:
\nm\u2220EJH = 3(45) = 135<\/p>\n

In Exercises 19 – 24, find the area of the regular polygon.<\/p>\n

Question 19.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\n\"Big
\nAnswer:
\nArea = \\(\\frac { 5a\u00b2 }{ 2 } \\)\u221a(5+2\u221a5)
\nArea = \\(\\frac { 5(6.84)\u00b2 }{ 2 } \\)\u221a(5+2\u221a5)
\nA = 359.9784<\/p>\n

Question 21.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 22.
\n\"Big
\nAnswer:
\nArea = \\(\\frac { 3\u221a3 a\u00b2}{ 2 } \\)
\nA = \\(\\frac { 3\u221a3 (7)\u00b2}{ 2 } \\)
\nA = 127.30<\/p>\n

Question 23.
\nan octagon with a radius of 11 units
\nAnswer:
\n\"Big<\/p>\n

Question 24.
\na pentagon with an apothem of 5 units<\/p>\n

Answer:
\nA = 90.75<\/p>\n

Explanation:
\n\"Big<\/p>\n

We know apothem a = and it divides pentagon into triangles, the central angle is divided into 360\/5 = 72
\nAfter that, we halved this angle and got 2 right triangles with x = 44 and y = 36. Since we know one side and all three angles of the triangle, we can calculate p with the tangent function.
\ntan y = p\/a
\ntan 36 = p\/5
\np = 3.63
\nSince p is just half of the length of the side, we have to multiply it by 2
\n2 . p = 2 . 3.63 = 7.26 = s
\nArea = \\(\\frac { a . s. n }{ 2 } \\)
\nA = \\(\\frac { 15 x 7.26 x 5 }{ 2 } \\)
\nA = 90.75<\/p>\n

Question 25.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in finding the area of the kite.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 26.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in finding. the area of the regular hexagon.
\n\"Big<\/p>\n

Answer:
\ns = \u221a15\u00b2 – 13\u00b2 = 7.48
\nArea = \\(\\frac { 1 }{ 2 } \\)(a . ns)
\nA = \\(\\frac { 1 }{ 2 } \\)(13 x 6 x 7.48)
\nA = 291.72<\/p>\n

In Exercises 27 – 30, find the area of the shaded region.<\/p>\n

Question 27.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 28.
\n\"Big<\/p>\n

Answer:
\nArea of the shaded region = 223.75<\/p>\n

Explanation:
\nSquare side = diagonal\/\u221a2
\n= 28\/\u221a2 = 19.79
\nArea of square = 19.79\u00b2 = 392
\nCircle area = \u03c0(14)\u00b2 = 615.75
\nArea of the shaded region = 615.75 – 392 = 223.75<\/p>\n

Question 29.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 30.
\n\"Big
\nAnswer:<\/p>\n

Question 31.
\nMODELING WITH MATHEMATICS<\/strong>
\nBasaltic columns arc geological formations that result from rapidly cooling lava. Giant\u2019s Causeway in Ireland contains many hexagonal basaltic columns. Suppose the top of one of the columns is in the shape of a regular hexagon with a radius of 8 inches. Find the area of the top of the column to the nearest square inch.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 32.
\nMODELING WITH MATHEMATICS<\/strong>
\nA watch has a circular surface on a background that is a regular octagon. Find the area of the octagon. Then find the area of the silver border around the circular face.
\n\"Big
\nAnswer:<\/p>\n

CRITICAL THINKING<\/strong>
\nIn Exercises 33 – 35, tell whether the statement is true or false. Explain your reasoning<\/p>\n

Question 33.
\nThe area of a regular n-gon of a fixed radius r increases as n increases.
\nAnswer:
\n\"Big<\/p>\n

Question 34.
\nThe apothem of a regular polygon is always less than the radius.<\/p>\n

Answer:
\ntrue, the radius always reaches the end of the circle but the apothem never does<\/p>\n

Question 35.
\nThe radius of a regular polygon is always less than the side length.
\nAnswer:
\n\"Big<\/p>\n

Question 36.
\nREASONING<\/strong>
\nPredict which figure has the greatest area and which has the least area. Explain your reasoning. Check by finding the area of each figure.
\n(A) \"Big
\n(B) \"Big
\n(C) \"Big
\nAnswer:
\n(B) has the highest area and C has the lowest area.<\/p>\n

Explanation:
\n(A) area = \u03c0(6.5)\u00b2 = 132.73
\n(B) area = 139.25
\n(C) area = \\(\\frac { 1 }{ 2 } \\)(18 x 15) = 135<\/p>\n

Question 37.
\nUSING EQUATIONS<\/strong>
\nFind the area of a regular
\npentagon inscribed in a circle whose equation is given by (x – 4)2<\/sup> + y + 2)2<\/sup> = 25.
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 38.
\nREASONING<\/strong>
\nWhat happens to the area of a kite if you double the length of one of the diagonals? if you double the length of both diagonals? Justify your answer.<\/p>\n

Answer:
\nArea of a kite = \\(\\frac { 1 }{ 4 } \\)(d\u2081d\u2082)
\nIf you double the length of one diagonal, then d\u2081 = 2d\u2081
\nArea of kite = \\(\\frac { 1 }{ 2 } \\)(d\u2081d\u2082)
\nIf you double length of both diagonals
\nArea = \\(\\frac { 1 }{ 4 } \\)(2d\u20812d\u2082) = d\u2081d\u2082
\nIf you double the length of one diagonal, then the area becomes halve. If you double length of both diagonals, then area becomes 4 times.<\/p>\n

MATHEMATICAL CONNECTIONS<\/strong>
\nIn Exercises 39 and 40, write and solve an equation to find the indicated lengths. Round decimal answers to the nearest tenth.<\/p>\n

Question 39.
\nThe area of a kite is 324 square inches. One diagonal is twice as long as the other diagonal. Find the length of each diagonal.
\nAnswer:
\n\"Big<\/p>\n

Question 40.
\nOne diagonal of a rhombus is four times the length of the other diagonal. The area of the rhombus is 98 square feet. Find the length of each diagonal.<\/p>\n

Answer:
\nThe length of each diagonal is 9.89, 2.47.<\/p>\n

Explanation:
\nOne diagonal of a rhombus is four times the length of the other diagonal.
\nd\u2081 = 4d\u2082
\nArea = \\(\\frac { 1 }{ 4 } \\)(d\u2081d\u2082)
\n98 = \\(\\frac { 1 }{ 4 } \\)(d\u2081(4d\u2081))
\nd\u2081 = 9.89
\nd\u2082 = 2.47<\/p>\n

Question 41.
\nREASONING<\/strong>
\nThe perimeter of a regular nonagon. or 9-gon, is 18 inches. Is this enough information to find the area? If so, find the area and explain your reasoning. If not, explain why not.
\nAnswer:
\n\"Big<\/p>\n

Question 42.
\nMAKING AN ARGUMENT<\/strong>
\nYour friend claims that it is possible to find the area of any rhombus if you only know the perimeter of the rhombus. Is your friend correct? Explain your reasoning.<\/p>\n

Answer:
\nNo; A rhombus is not a regular polygon.<\/p>\n

Question 43.
\nPROOF<\/strong>
\nProve that the area of any quadrilateral with perpendicular diagonals is A = \\(\\frac{1}{2}\\)d1<\/sub>d2<\/sub>, where d1<\/sub> and d2<\/sub> are the lengths of the diagonals.
\n\"Big
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 44.
\nHOW DO YOU SEE IT?<\/strong>
\nExplain how to find the area of the regular hexagon by dividing the hexagon into equilateral triangles.
\n\"Big<\/p>\n

Answer:<\/p>\n

Question 45.
\nREWRITING A FORMULA<\/strong>
\nRewrite the formula for the area of a rhombus for the special case of a square with side length s. Show that this is the same as the formula for the area of a square, A = s2<\/sup>.
\nAnswer:
\n\"Big<\/p>\n

Question 46.
\nREWRITING A FORMULA<\/strong>
\nUse the formula for the area of a regular polygon to show that the area of an equilateral triangle can be found by using the formula A = \\(\\frac{1}{4}\\)s2<\/sup>\u221a3 where s is the side length.
\nAnswer:<\/p>\n

Question 47.
\nCRITICAL THINKING<\/strong>
\nThe area of a regular pentagon is 72 square centimeters. Find the length of one side.
\nAnswer:
\n\"Big<\/p>\n

Question 48.
\nCRITICAL THINKING<\/strong>
\nThe area of a dodecagon, or 12-gon, is 140 square inches. Find the apothem of the polygon.<\/p>\n

Answer:
\nLet the side length of dodecagon be 2x. The measure of each interior angle of a regular decagon is 150. This implies that the base angle C and A of the resulting isosceles triangle formed by the red sides is equal to 150\/2 = 75. The adjacent to this angle is the length 2x\/2 = x inches, while the opposite to it is the blue apothem in the right triangle BDC formed. Therefore a = x tan 75. Therefore, area of dodecagon is
\n140 = 1\/2 (x tan75)(12 . 2x)
\n140 = 44.785 x\u00b2
\nx\u00b2 = 3.126
\nx = 1.768<\/p>\n

Question 49.
\nUSING STRUCTURE<\/strong>
\nIn the figure, an equilateral triangle lies inside a square inside a regular pentagon inside a regular hexagon. Find the approximate area of the entire shaded region to the nearest whole number.
\n\"Big
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 50.
\nTHOUGHT PROVOKING<\/strong>
\nThe area of a regular n-gon is given by A = \\(\\frac{1}{2}\\)aP. As n approaches infinity, what does the n-gon approach? What does P approach? What does a approach? What can you conclude from your three answers? Explain your reasoning.
\n\"Big
\nAnswer:<\/p>\n

Question 51.
\nCOMPARING METHODS<\/strong>
\nFind the area of regular pentagon ABCDE by using the formula A = \\(\\frac{1}{2}\\)aP, or A = \\(\\frac{1}{2}\\)a \u2022 ns. Then find the area by adding the areas of smaller polygons. Check that both methods yield the same area. Which method do you prefer? Explain your reasoning.
\n\"Big
\nAnswer:
\n\"Big
\n\"Big
\n\"Big<\/p>\n

Question 52.
\nUSING STRUCTURE<\/strong>
\nTwo regular polygons both have n sides. One of the polygons is inscribed in, and the other is circumscribed about, a circle of radius r. Find the area between the two polygons in terms of n and r.<\/p>\n

Answer:
\nThe radius of the smaller polygon is equal to the apothem of the larger polygon. The central angle is 360\/n, therefore the apothem makes an angle of 180\/n. Use sine and cosine to find the apothem and side length of the smaller polygon.
\nasmall<\/sub> = r sin\\(\\frac { 180 }{ n } \\)
\nssmall<\/sub> = 2r cos\\(\\frac { 180 }{ n } \\)
\nUse tangent to find the side length of the large polygon.
\nSlarge<\/sub> = 2r tan\\(\\frac { 180 }{ n } \\)
\nUse the formula to find the area of the smaller polygon.
\nAsmall<\/sub> = 1\/2 . asmall<\/sub> . n . ssmall<\/sub>
\nAsmall<\/sub> = 1\/2 . r sin\\(\\frac { 180 }{ n } \\) . n . 2r cos\\(\\frac { 180 }{ n } \\)
\nAsmall<\/sub> = nr\u00b2 sin \\(\\frac { 180 }{ n } \\) cos\\(\\frac { 180 }{ n } \\)
\nUse the formula to find the area of the larger polygon.
\nALarge<\/sub> = 1\/2 . alarge<\/sub> . n . slarge
\n= nr\u00b2 tan\\(\\frac { 180 }{ n } \\)
\nThe area between the polygons is equal to the area of the larger polygon minus the area of the smaller polygon. Use some trig identities to simplify the expression.
\nA = Alarge<\/sub> – Asmall<\/sub>
\nA = nr\u00b2 tan\\(\\frac { 180 }{ n } sin\u00b2[latex]\\frac { 180 }{ n }<\/p>\n

Maintaining Mathematical Proficiency<\/p>\n

Determine whether the figure has line symmetry, rotational symmetry, both, or neither. If the
\nfigure has line symmetry. determine the number of lines of symmetry. It the figure has rotational
\nsymmetry, describe any rotations that map the figure onto itself.<\/p>\n

Question 53.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 54.
\n\"Big
\nAnswer:
\nThe figure has rotational symmetry.<\/p>\n

Question 55.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 56.
\n\"Big
\nAnswer:
\nThe figure has one line symmetry.<\/p>\n

11.4 Three-Dimensional Figures<\/h3>\n

Exploration 1<\/strong><\/p>\n

Analyzing a Property of Polyhedra<\/p>\n

Work with a partner: The five Platonic solids are shown below. Each of these solids has congruent regular polygons as faces. Complete the table by listing the numbers of vertices, edges, and faces of each Platonic solid.
\n\"Big
\n\"Big
\nAnswer:<\/p>\n

Communicate Your Answer<\/p>\n

Question 2.
\nWhat is the relationship between the numbers of vertices V, edges E, and faces F of a polyhedron? (Note: Swiss mathematician Leonhard Euler (1707 – 1783) discovered a formula that relates these quantities.)
\nCONSTRUCTING VIABLE ARGUMENTS<\/strong>
\nTo be proficient in math, you need to reason inductively about data.
\nAnswer:<\/p>\n

Question 3.
\nDraw three polyhedra that are different from the Platonic solids given in Exploration 1. Count the numbers of vertices, edges, and faces of each polyhedron Then verify that the relationship you found in Question 2 is valid for each polyhedron.
\nAnswer:<\/p>\n

Lesson 11.4 Three-Dimensional Figures<\/h3>\n

Monitoring progress<\/strong><\/p>\n

Tell whether the solid is a polyhedron. If it is, name the polyhedron.<\/p>\n

Question 1.
\n\"Big<\/p>\n

Answer:
\nThe solid is formed by polygons, so it is a polyhedron. The base is a square, it is a square pyramid.<\/p>\n

Question 2.
\n\"Big
\nAnswer:
\nThe solid have curved faces. So it is not a polyhedron.<\/p>\n

Question 3.
\n\"Big
\nAnswer:
\nThe solid is formed by polygons, so it is a polyhedron. iT has two triangles, one rectangle and two squares.<\/p>\n

Describe the shape formed by the intersection of the plane and the solid.<\/p>\n

Question 4.
\n\"Big
\nAnswer:
\nThe cross-section is a pentagon.<\/p>\n

Question 5.
\n\"Big
\nAnswer:
\nThe cross-section is a hexagon.<\/p>\n

Question 6.
\n\"Big
\nAnswer:
\nThe cross-section is a circle.<\/p>\n

Sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.<\/p>\n

Question 7.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 9.
\n\"Big
\nAnswer:<\/p>\n

Exercise 11.4 Three-Dimensional Figures<\/h3>\n

Vocabulary and Core Concept Check<\/p>\n

Question 1.
\nVOCABULARY<\/strong>
\nA(n) ___________ is a solid that is bounded by polygons.
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nWHICH ONE DOESN\u2019T BELONG?<\/strong>
\nWhich solid does not belong with the other three? Explain your reasoning.
\n\"Big<\/p>\n

Answer:
\nCone does not belong with the other three as it has a curved surface and others not.<\/p>\n

Monitoring Progress and Modeling with Mathematics<\/p>\n

In Exercises 3 – 6, match the polyhedron with its name.<\/p>\n\n\n\n\n\n\n
3. \"Big<\/td>\nA. triangular Prism<\/td>\n<\/tr>\n
4. \"Big<\/td>\nB. rectangular pyramid<\/td>\n<\/tr>\n
5. \"Big<\/td>\nC. hexagonal pyramid<\/td>\n<\/tr>\n
6. \"Big<\/td>\nD. Pentagonal prism<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\n\"Big
\n\"Big<\/p>\n

In Exercises 7 – 10, tell whether the solid is a polyhedron. If it is, name the polyhedron.<\/p>\n

Question 7.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\n\"Big
\nAnswer:
\nYes, it is a polyhedron. It is a hexagonal prism.<\/p>\n

Question 9.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\n\"Big
\nAnswer:
\nYes, it is a polyhedron, truncated square pyramid.<\/p>\n

In Exercises 11 – 14, describe the cross section formed by the intersection of the plane and the solid.<\/p>\n

Question 11.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\n\"Big
\nAnswer:
\nThe cross-section is a square.<\/p>\n

Question 13.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\n\"Big
\nAnswer:
\nThe cross-section is a hexagon<\/p>\n

In Exercises 15 – 18, sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.<\/p>\n

Question 15.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 17.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 19.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in identifying the solid.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\nHOW DO YOU SEE IT?<\/strong>
\nIs the swimming pool shown a polyhedron? If it is, name the polyhedron. If not, explain why not.
\n\"Big
\nAnswer:
\nIt is an octagonal polyhedron.<\/p>\n

In Exercises 21 – 26, sketch the polyhedron.<\/p>\n

Question 21.
\ntriangular prism
\nAnswer:
\n\"Big<\/p>\n

Question 22.
\nrectangular prism<\/p>\n

Answer:
\n\"Big<\/p>\n

Question 23.
\npentagonal prism
\nAnswer:
\n\"Big<\/p>\n

Question 24.
\nhexagonal prism
\nAnswer:
\n\"Big<\/p>\n

Question 25.
\nsquare pyramid
\nAnswer:
\n\"Big<\/p>\n

Question 26.
\npentagonal pyramid<\/p>\n

Answer:
\n\"Big<\/p>\n

Question 27.
\nMAKING AN ARGUMENT<\/strong>
\nYour Friend says that the polyhedron shown is a triangular prism. Your cousin says that it is a triangular pyramid. Who is correct? Explain your reasoning.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 28.
\nATTENDING TO PRECISION<\/strong>
\nThe figure shows a plane intersection a cube through four of its vertices. The edge length of the cube is 6 inches.
\n\"Big
\na. Describe the shape of the cross section.
\nAnswer:
\nThe cross-section is a rectangle.<\/p>\n

b. What is the perimeter of the cross section?
\nAnswer:
\nThe perimeter is 2(l + b)<\/p>\n

c. What is the area of the cross section?
\nAnswer:
\nArea is lb.<\/p>\n

REASONING<\/strong>
\nIn Exercises 29 – 34, tell whether it is possible for a cross section of a cube to have the given shape. If it is, describe or sketch how the plane could intersect the cube.<\/p>\n

Question 29.
\ncircle
\nAnswer:
\n\"Big<\/p>\n

Question 30.
\npentagon
\nAnswer:
\nyes, cross-section of the cube can be a pentagon.<\/p>\n

Question 31.
\nrhombus
\nAnswer:
\n\"Big<\/p>\n

Question 32.
\nisosceles triangle
\nAnswer:
\nYes, the cross-section can be an isosceles triangle.<\/p>\n

Question 33.
\nhexagon
\nAnswer:
\n\"Big<\/p>\n

Question 34.
\nscalene triangle
\nAnswer:
\nYes, the cross-section can be scalene triangle.<\/p>\n

Question 35.
\nREASONING<\/strong>
\nSketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.
\na.
\n\"Big
\nb.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 36.
\nTHOUGHT PROVOKING<\/strong>
\nDescribe how Plato might have argued that there are precisely five Platonic Solids (see page 617). (Hint: Consider the angles that meet at a vertex.)
\nAnswer:<\/p>\n

Maintaining Mathematical proficiency<\/p>\n

Decide whether enough information is given to prove that the triangles are congruent. It so, state the theorem you would use.<\/p>\n

Question 37.
\n\u2206ABD, \u2206 CDB
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 38.
\n\u2206JLK, \u2206JLM
\n\"Big<\/p>\n

Answer:
\n\u2206JLK \u2245 \u2206JLM by SAS congruence theorem.<\/p>\n

Question 39.
\n\u2206RQP, \u2206RTS
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

11.1 – 11.4 Quiz<\/h3>\n

Find the indicated measure.<\/p>\n

Question 1.
\nm[latex]\\widehat{E F}\\)
\n\"Big
\nAnswer:
\n13.7 = \\(\\frac { m[latex]\\widehat{E F}\\) }{ 360 } [\/latex] \u2022 2\u03c0(7)
\nm\\(\\widehat{E F}\\) = 112.13<\/p>\n

Question 2.
\narc length of \\(\\widehat{Q S}\\)
\n\"Big
\nAnswer:
\narc length of \\(\\widehat{Q S}\\) = \\(\\frac { 83 }{ 360 } \\) \u2022 2\u03c0(4)
\n= 5.79<\/p>\n

Question 3.
\ncircumference of \u2299N
\n\"Big
\nAnswer:
\n8 = \\(\\frac { 48 }{ 360 } \\) \u2022 2\u03c0r
\nC = 60<\/p>\n

Question 4.
\nConvert 26\u00b0 to radians and \\(\\frac{5 \\pi}{9}\\) radians to degrees.<\/p>\n

Answer:
\n26\u00b0 = 26 . \\(\\frac { \u03c0 }{ 180 } \\) = \\(\\frac { 13\u03c0 }{ 90 } \\) radians
\n\\(\\frac{5 \\pi}{9}\\) = \\(\\frac{5 \\pi}{9}\\) . \\(\\frac { 180 }{ \u03c0 } \\) = 100\u00b0<\/p>\n

Use the figure to find the indicated measure.<\/p>\n

\"Big<\/p>\n

Question 5.
\narea of red sector<\/p>\n

Answer:
\narea of red sector = \\(\\frac { 100 }{ 360 } \\) . \u03c0(12)\u00b2
\n= 125.66<\/p>\n

Question 6.
\narea of blue sector<\/p>\n

Answer:
\narea of blue sector = \\(\\frac { 260 }{ 360 } \\) . \u03c0(12)\u00b2
\n= 326.72<\/p>\n

In the diagram, RSTUVWXY is a reuIar octagon inscribed in \u2299C.<\/p>\n

\"Big<\/p>\n

Question 7.
\nIdentify the center, a radius, an apothem, and a central angle of the polygon.<\/p>\n

Answer:
\nC is center, CY is radius, CZ is apothem, \u2220YCR is central angle of the polygon<\/p>\n

Question 8.
\nFind m\u2220RCY, m\u2220RCZ, and m\u2220ZRC.<\/p>\n

Answer:
\nm\u2220RCY = 360\/8 = 45
\nm\u2220RCZ = 45\/2 = 22.5
\nm\u2220ZRC = 180 – (22.5 + 90) = 67.5<\/p>\n

Question 9.
\nThe radius of the circle is 8 units. Find the area of the octagon.<\/p>\n

Answer:
\nArea of octagon = 0.5 x 8 x 8 sin 45 = 22.62<\/p>\n

Tell whether the solid is a polyhedron. If it is, name the polyhedron.<\/p>\n

Question 10.
\n\"Big
\nAnswer:
\nIt is not a polyhedron.<\/p>\n

Question 11.
\n\"Big
\nAnswer:
\nThe solid is a polyhedron. It is an octagonal pyramid.<\/p>\n

Question 12.
\n\"Big
\nAnswer:
\nThe solid is a polyhedron. It is a pentagonal prism.<\/p>\n

Question 13.
\nSketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.
\n\"Big<\/p>\n

Answer:
\n\"Big<\/p>\n

Question 14.
\nThe two white congruent circles just fit into the blue circle. What is the area of the blue region?
\n\"Big
\nAnswer:
\nWhite circle diameter = radius of the blue circle.
\n6 = radius of the blue circle.
\nArea of blue circle = \u03c06\u00b2 = 113.09
\nArea of white circle = \u03c03\u00b2 = 28.27
\nArea of blue region = 113.09 – 2(28.27) = 56.541<\/p>\n

Question 15.
\nFind the area of each rhombus tile. Then find the area of the pattern.
\n\"Big<\/p>\n

Answer:
\nArea of yellow tile = \\(\\frac { 1 }{ 4 } \\)(15.7 x 11.4) = 44.745
\narea of red tile = \\(\\frac { 1 }{ 4 } \\)(18.5 x 6) = 27.75
\nArea of pattern = 32(44.745) + 23(27.75) = 2070.09 sq mm<\/p>\n

11.5 Volumes of Prisms and Cylinders<\/h3>\n

Exploration 1<\/strong><\/p>\n

Finding volume<\/p>\n

Work with a partner: Consider a stack of square papers that is in the form of a right prism.<\/p>\n

\"Big<\/p>\n

a. What is the volume the prism?
\nAnswer:<\/p>\n

b. When you twist the slack of papers, as shown at the right, do you change the volume? Explain your reasoning.
\nAnswer:<\/p>\n

c. Write a carefully worded conjecture that describes the conclusion you reached in part (b).
\nATTENDING TO PRECISION<\/strong>
\nTo be proficient in math, you need to communicate precisely to others.
\nAnswer:<\/p>\n

d. Use your conjecture to find the volume of the twisted stack of papers.
\nAnswer:<\/p>\n

Exploration 2<\/strong><\/p>\n

Finding volume<\/p>\n

Work with a partner: Use the conjecture you wrote in Exploration I to find the volume of the cylinder.<\/p>\n

a.
\n\"Big
\nAnswer:<\/p>\n

b.
\n\"Big
\nAnswer:<\/p>\n

Communicate Your Answer<\/p>\n

Question 3.
\nHow can you find the volume of a prism or cylinder that is not a right prism or right cylinder?
\nAnswer:<\/p>\n

Question 4.
\nIn Exploration 1, would the conjecture you wrote change if the papers in each stack were not squares? Explain your reasoning.
\nAnswer:<\/p>\n

Lesson 11.5 Volumes of Prisms and Cylinders<\/h3>\n

Monitoring Progress<\/strong><\/p>\n

Find the volume of the solid.<\/p>\n

Question 1.
\n\"Big
\nAnswer:
\nVolume = Area * height
\nArea = \\(\\frac { 1 }{ 2 } \\)(5 x 9) = 22.5
\nVolume = 22. 5 x 8 = 180 cubic m.<\/p>\n

Question 2.
\n\"Big<\/p>\n

Answer:
\nArea of circle = \u03c0r\u00b2 = \u03c0(8)\u00b2 = 64\u03c0
\nVolume = 64\u03c0 x 14 = 2814.86 cubic ft<\/p>\n

Question 3.
\nThe diagram shows the dimensions of a concrete cylinder. Concrete has a density of 2.3 grams per cubic centimeter. Find the mass of the concrete cylinder to the nearest gram.
\n\"Big
\nAnswer:
\nMass of the concrete cylinder = 32 x \u03c0(24)\u00b2 = 18432\u03c0 cubic in<\/p>\n

Question 4.
\nWHAT IF?<\/strong>
\nIn Example 4, you want the length to be 5 meters, the width to be 3 meters. and the volume to be 60 cubic meters. What should the height be?
\nAnswer:
\nvolume = lbh
\n60 = 5 x 3 x h
\nh = 4 m<\/p>\n

Question 5.
\nWHAT IF?<\/strong>
\nIn Example 5, you want the height to be 5 meters and the volume to be 75 cubic meters. What should the area of the base be? Give a possible length and width.
\nAnswer:
\nvolume V = base x height
\n75 = base x 5
\nBase = 15 sq m<\/p>\n

Question 6.
\nPrism C and prism D are similar. Find the volume of prism D.
\n\"Big
\nAnswer:
\n\\(\\frac { 12 }{ 3 } \\) = \\(\\frac { 1536 }{ v } \\)
\nv = 384
\nVolume of prism D = 384 cubic m<\/p>\n

Question 7.
\nFind the volume of the composite solid.
\n\"Big
\nAnswer:
\nVolume = area x height
\nVolume = \\(\\frac { 1 }{ 2 } \\)(10 x 3 x 6)
\n= 90 cubic ft<\/p>\n

Exercise 11.5 Volumes of Prisms and Cylinders<\/h3>\n

Vocabulary and Core Concept Check<\/p>\n

Question 1.
\nVOCABULARY<\/strong>
\nIn what type of units is the volume of a solid measured?
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nCOMPLETE THE SENTENCE<\/strong>
\nDensity is the amount of _______ that an object has in a given unit of __________ .<\/p>\n

Answer:
\nDensity is the mass of the object divided by its volume.<\/p>\n

Monitoring Progress and Modeling with Mathematics<\/p>\n

In Exercises 3 – 6, find the volume of the prism.<\/p>\n

Question 3.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 4.
\n\"Big
\nAnswer:
\nVolume V = lbh
\nV = 1.5 x 2 x 4 = 12 m\u00b3<\/p>\n

Question 5.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 6.
\n\"Big
\nAnswer:
\nVolume = 6 x 11 x 14
\nV = 924 m\u00b3<\/p>\n

In Exercises 7 – 10. find the volume of the cylinder.<\/p>\n

Question 7.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\n\"Big
\nAnswer:
\nVolume = \u03c0r\u00b2h
\nV = \u03c0(13.4)\u00b2 x 9.8
\nV = 1759.6\u03c0 cm\u00b3<\/p>\n

Question 9.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\n\"Big
\nAnswer:
\nVolume V = \u03c0r\u00b2h
\nShoter side = 18\/2 = 9
\nHeight h = \u221a18\u00b2 – 9\u00b2 = 15.588
\nV = \u03c06\u00b2 x 15.588
\n= 1763 m\u00b3<\/p>\n

In Exercises 11 and 12. make a sketch of the solid and find its volume. Round your answer to the nearest hundredth.<\/p>\n

Question 11.
\nA prism has a height of 11.2 centimeters and an equilateral triangle for a base, where each base edge is 8 centimeters.
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\nA pentagonal prism has a height of 9 feet and each base edge is 3 feet.<\/p>\n

Answer:
\nvolume is 139.32 ft\u00b3<\/p>\n

explanation:
\nPentagon area = 15.48
\nHeight h = 9 ft
\nVolume V = area x height
\n= 15.48 x 9 = 139.32<\/p>\n

Question 13.
\nPROBLEM SOLVING<\/strong>
\nA piece of copper with a volume of 8.25 cubic centimeters has a mass of 73.92 grams. A piece of iron with a volume of 5 cubic centimeters has a mass of 39.35 grams. Which metal has the greater density?
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\nPROBLEM SOLVING<\/strong>
\nThe United States has minted one-dollar silver coins called the American Eagle Silver Bullion Coin since 1986. Each coin has a diameter of 40.6 millimeters and is 2.98 millimeters thick. The density of silver is 10.5 grams per cubic centimeter. What is the mass of an American Eagle Silver Bullion Coin to the nearest grain?
\n\"Big
\nAnswer:
\nV = \u03c0r\u00b2h
\nV =\u03c0(20.3)\u00b2\u00a0 x (2.98)
\nV = 3856 mm\u00b3<\/p>\n

Question 15.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in finding the volume of the cylinder.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in finding the density of an object that has a mass of 24 grams and a volume of 28.3 cubic centimeters.
\n\"Big<\/p>\n

Answer:
\nDensity = mass \/ volume
\nDensity = \\(\\frac { 24 }{ 28.3 } \\)
\nDensity = 0.8480<\/p>\n

In Exercises 17 – 22, find the missing dimension of the prism or cylinder.<\/p>\n

Question 17.
\nVolume = 500 ft3<\/sup>
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\nVolume = 2700 yd3<\/sup>
\n\"Big<\/p>\n

Answer:
\nVolume = 2700 yd\u00b3
\n12 x 5 x v = 2700
\nv = 15 yd<\/p>\n

Question 19.
\nVolume = 80 cm3<\/sup>
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\nVolume = 72.66 in.3<\/sup>
\n\"Big
\nAnswer:
\nVolume = 72.66 in.3<\/sup>
\nArea . x = 72.66
\n10.39 x = 72.66
\nx = 6.9 in<\/p>\n

Question 21.
\nVolume = 3000 ft3<\/sup>
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 22.
\nVolume = 1696.5 m3<\/sup>
\n\"Big
\nAnswer:
\nVolume = 1696.5
\n\u03c0r\u00b2h = 1696.5
\n\u03c0z\u00b2 x 15 = 1696.5
\nz\u00b2 = 36.00
\nz = 6<\/p>\n

In Exercises 23 and 24, find the area of the base of the rectangular prism with the given volume and height. Then give a possible length and width.<\/p>\n

Question 23.
\nV= 154 in.3<\/sup>, h = 11 in.
\nAnswer:
\n\"Big<\/p>\n

Question 24.
\nV = 27 m3<\/sup>,h = 3m<\/p>\n

Answer:
\nV = Bh
\n27 = B x 3
\nB = 9<\/p>\n

In Exercises 25 and 26, the solids are similar. Find the volume of solid B.<\/p>\n

Question 25.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 26.
\n\"Big
\nAnswer:
\n\\(\\frac { 12 }{ 15 } \\) = \\(\\frac { 4608\u03c0 }{ V } \\)
\nV = 5760\u03c0
\nVolume of cylinder B = 5760\u03c0<\/p>\n

In Exercises 27 and 28, the solids are similar. Find the indicated measure.<\/p>\n

Question 27.
\nheight x of the base of prism A
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 28.
\nheight h of cylinder B
\n\"Big
\nAnswer:
\nheight h of cylinder B is 40 ft<\/p>\n

Explanation:
\n\\(\\frac { 7\u03c0 }{ 5 } \\) = \\(\\frac { 56\u03c0 }{ h } \\)
\nh = 40<\/p>\n

In Exercises 29 – 32. find the volume of the composite solid.<\/p>\n

Question 29.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 30.
\n\"Big
\nAnswer:
\nVolume V = 89.32<\/p>\n

Explanation:
\nVolume of square = 4\u00b3 = 64
\nVolume of semicircle = \u03c0(2)\u00b2 x 4 = 8\u03c0<\/p>\n

Question 31.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 32.
\n\"Big<\/p>\n

Answer:
\nThe volume of composite solid is 35 cubic ft<\/p>\n

Explanation:
\nVolume of larger prism = 4 x 2 x 5 = 40
\nVolume of the smaller prism = 1 x 1 x 5 = 5
\nVolume of larger prism – volume of the smaller prism = 40 – 5 = 35 cubic ft<\/p>\n

Question 33.
\nMODELING WITH MATHEMATICS<\/strong>
\nThe Great Blue Hole is a cylindrical trench located off the coast of Belize. It is approximately 1000 feet wide and 400 feet deep. About how many gallons of water does the Great Blue Hole contain? (1 ft3<\/sup> \u2248 7.48 gallons)
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 34.
\nCOMPARING METHODS<\/strong>
\nThe Volume Addition Postulate states that the volume of a solid is the sum of the volumes of all its non overlapping parts. Use this postulate to find the volume of the block of concrete in Example 7 by subtracting the volume of each hole from the volume of the large rectangular prism. Which method do you prefer? Explain your reasoning.
\nAnswer:<\/p>\n

REASONING<\/strong>
\nIn Exercises 35 and 36, you are melting a rectangular block of wax to make candles. how many candles of the given shape can be made using a block that measures 10 centimeters by 9 centimeters by 20 centimeters?<\/p>\n

Question 35.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 36.
\n\"Big<\/p>\n

Answer:
\n7 triangular prism candles with the given measures can be made.<\/p>\n

Explanation:
\nVolume of block = 1800
\nThe volume of triangular prism = 4 x 6 x 10 = 240
\n1800\/240 = 7.5<\/p>\n

Question 37.
\nPROBLEM SOLVING<\/strong>
\nAn aquarium shaped like a rectangular prism has a length of 30 inches, a width of 10 inches, and a height of 20 inches. You fill the aquarium \\(\\frac{3}{4}\\) fill with water. When you submerge a rock in the aquarium, the water level rises 0.25 inch.
\na. Find the volume of the rock.
\nb. How many rocks of this size can you place in the aquarium before water spills out?
\nAnswer:
\n\"Big<\/p>\n

Question 38.
\nPROBLEM SOLVING<\/strong>
\nYou drop an irregular piece of metal into a container partially filled with water and measure that the waler level rises 4.8 centimeters. The square base of the container has a side length of 8 centimeters. You measure the mass of the metal to be 450 grams. What is the density of the metal?<\/p>\n

Answer:
\nThe density of metal is 1.4648<\/p>\n

Explanation:
\nDensity = \\(\\frac { Mass }{ Volume } \\)
\nVolume V = 4.8 x 64 = 307.2
\nDensity = \\(\\frac { 450 }{ 307.2 } \\) = 1.4648<\/p>\n

Question 39.
\nWRITING<\/strong>
\nBoth of the figures shown arc made up of the same number of congruent rectangles. Explain how Cavalieri\u2019s Principle can be adapted to compare the areas of these figures.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 40.
\nHOW DO YOU SEE IT?<\/strong>
\nEach stack of memo papers contains 500 equally-sized sheets of paper. Compare their volumes. Explain your reasoning.
\n\"Big
\nAnswer:<\/p>\n

Question 41.
\nUSING STRUCTURE<\/strong>
\nSketch the solid formed by the net. Then find the volume of the solid.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 42.
\nUSING STRUCTURE<\/strong>
\nSketch the solid with the given views. Then find the volume of the solid.
\n\"Big<\/p>\n

Answer:
\nVolume = 2.5 x 3.5 x 6
\nVolume = 52.5<\/p>\n

Question 43.
\nOPEN-ENDED<\/strong>
\nSketch two rectangular prisms that have volumes of 1000 cubic inches hut different surface areas. Include dimensions in your sketches.
\nAnswer:
\n\"Big<\/p>\n

Question 44.
\nMODELING WITH MATHEMATICS<\/strong>
\nWhich box gives you more cereal for your money? Explain.
\n\"Big<\/p>\n

Answer:
\nFirst one gives more cerel for your money.<\/p>\n

Explanation:
\nBigger one volume = 16 x 4 x 10 = 640
\nSmaller one volume = 2 x 8 x 10 = 160
\n6 – 640 means 1 – 106.66
\n2 – 160 means 1 – 80<\/p>\n

Question 45.
\nCRITICAL THINKING<\/strong>
\nA 3-inch by 5-inch index card is rotated around a horizontal line and a vertical line to produce two different solids. Which solid has a greater volume? Explain your reasoning.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 46.
\nCRITICAL THINKING<\/strong>
\nThe height of cylinder X is twice the height of cylinder Y. The radius of cylinder X is half the radius of cylinder Y. Compare the volumes of cylinder X and cylinder Y. Justify your answer.<\/p>\n

Answer:
\nLet the height of cylinder X be h, radius be r and its volume is \u03c0r\u00b2h
\nSo, the height of cylinder Y is h\/2 and radius is 2r, then the volume is 2\u03c0r\u00b2h
\nFrom both expressions, it can be seen that the volume of cylinder y is twice that of cylinder X.<\/p>\n

Question 47.
\nUSING STRUCTURE<\/strong>
\nFind the volume of the solid shown. The bases of the solid are sectors of circles.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 48.
\nMATHEMATICAL CONNECTIONS<\/strong>
\nYou drill a circular hole of radius r through the base of a cylinder of radius R. Assume the hole is drilled completely through to the other base. You want the volume of the hole to be half the volume of the cylinder. Express r as a function of R.<\/p>\n

Answer:
\nr = \u221aR\u00b2\/2<\/p>\n

Explanation:
\nThe radius of a solid cylinder without a hole is R. So its volume is \u03c0R\u00b2h
\nAs per the given condition, the volume of the hole must be half of that of the solid cylinder, hole volume is \u03c0R\u00b2h\/2
\nVolume of cylinder V = \u03c0r\u00b2h
\n\u03c0R\u00b2h\/2 = \u03c0r\u00b2h
\nR\u00b2\/2 = r\u00b2
\nr = \u221aR\u00b2\/2
\nr = \\(\\frac { R\u221a2 }{ 2 } \\)<\/p>\n

Question 49.
\nANALYZING RELATIONSHIPS<\/strong>
\nHow can you change the height of a cylinder so that the volume is increased by 25% but the radius remains the same?
\nAnswer:
\n\"Big<\/p>\n

Question 50.
\nANALYZING RELATIONSHIPS<\/strong>
\nHow can you change the edge length of a cube so that the volume is reduced by 40%?<\/p>\n

Answer:
\nWrite the equation of volume of rectangular prism which can be used to evaluate the cube volume
\nVolume = s x s x s
\nThe above equation shows that the volume of a cube is directly proportional to one of its side length, therefore, if the volume is to be reduced by 40%, then its the length of one of its side must be reduced by 40%, without changing the 2 other of its sides.<\/p>\n

Question 51.
\nMAKING AN ARGUMENT<\/strong>
\nYou have two objects of equal V0lume. Your friend says you can compare the densities of the objects by comparing their mass, because the heavier object will have a greater density. Is your friend correct? Explain your reasoning.
\nAnswer:
\n\"Big<\/p>\n

Question 52.
\nTHOUGHT PROVOKING<\/strong>
\nCavalieri\u2019s Principle states that the two solids shown below have the same volume. Do they also have the same surface area? Explain your reasoning.
\n\"Big
\nAnswer:<\/p>\n

Question 53.
\nPROBLEM SOLVING<\/strong>
\nA barn is in the shape of a pentagonal prism with the dimensions shown, The volume of the barn is 9072 cubic feel. Find the dimensions of each half of the root.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 54.
\nPROBLEM SOLVING<\/strong>
\nA wooden box is in the shape of a regular pentagonal prism. The sides, top, and bottom of the box are 1 centimeter thick. Approximate the volume of wood used to construct the box. Round your answer to the nearest tenth.
\n\"Big
\nAnswer:
\nPentagon area = \\(\\frac { 5 }{ 2 } \\)(4)\u00b2 sin 72
\nA = 38
\nVolume = Area x height
\nVolume = 38 x 6 = 228<\/p>\n

Maintaining Mathematical Proficiency<\/p>\n

Find the surface area of the regular pyramid.<\/p>\n

Question 55.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 56.
\n\"Big
\nAnswer:
\nSurface area = base area + 3bs
\nSurface area = 166.3 + 3(8)(10)
\n= 406.3<\/p>\n

Question 57.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

11.6 Volumes of Pyramids<\/h3>\n

Exploration 1<\/strong><\/p>\n

Finding the Volume of a Pyramid<\/p>\n

Work with a partner: The pyramid and the prism have the same height and the same square base.
\n\"Big
\nWhen the pyramid is filled with sand and poured into the prism, it takes three pyramids to fill the prism.
\n\"Big
\nUse this information to write a formula for the volume V of a pyramid.
\nLOOKING FOR STRUCTURE<\/strong>
\nTo be proficient in math, you need to look closely to discern a pattern or structure.
\nAnswer:<\/p>\n

Exploration 2<\/strong><\/p>\n

Finding the Volume of a Pyramid
\nWork with a partner: Use the formula you wrote in Exploration 1 to find the volume of the hexagonal pyramid.
\n\"Big
\nAnswer:<\/p>\n

Communicate Your Answer<\/p>\n

Question 3.
\nHow can you find the volume of a pyramid?
\nAnswer:<\/p>\n

Question 4.
\nIn Section 11 .7, you will study volumes of cones. How do you think you could use a method similar to the one presented in Exploration 1 to write a formula for the volume of a cone? Explain your reasoning.
\nAnswer:<\/p>\n

Lesson 11.6 Volumes of Pyramids<\/h3>\n

Monitoring Progress<\/strong><\/p>\n

Find the volume of the pyramid.<\/p>\n

Question 1.
\n\"Big
\nAnswer:
\nThe volume of the pyramid is 400 cm\u00b3<\/p>\n

Explanation:
\nVolume V = \\(\\frac { 1 }{ 3 } \\)Bh
\nV = \\(\\frac { 1 }{ 3 } \\)(10 x 10 x 12)
\nV = 400<\/p>\n

Question 2.
\n\"Big<\/p>\n

Answer:
\nThe volume of the pyramid is 2494.13 cm\u00b3<\/p>\n

Explanation:
\nVolume V = \\(\\frac { 1 }{ 3 } \\)Bh
\nV = \\(\\frac { 1 }{ 3 } \\)(374.12 x 20)
\nV = 2494.13<\/p>\n

Question 3.
\nThe volume of a square pyramid is 75 cubic meters and the height is 9 meters. Find the side length of the square base.<\/p>\n

Answer:
\nThe side length of the square base is 5 m<\/p>\n

Explanation:
\nVolume V = \\(\\frac { 1 }{ 3 } \\)Bh = 75
\n\\(\\frac { 1 }{ 3 } \\)B(9) = 75
\nB = 25
\ns = 5<\/p>\n

Question 4.
\nFind the height of the triangular pyramid at the left.
\n\"Big<\/p>\n

Answer:
\nThe height of the triangular pyramid is 8 m<\/p>\n

Explanation:
\nV = 24
\n\\(\\frac { 1 }{ 3 } \\)Bh = 24
\nB = \\(\\frac { 1 }{ 2 } \\)(3 x 6) = 9
\n\\(\\frac { 1 }{ 3 } \\)(9)h = 24
\nh = 8<\/p>\n

Question 5.
\nPyramid C and pyramid D are similar. Find the volume of pyramid D.
\n\"Big<\/p>\n

Answer:
\nThe volume of pyramid D is 12 m\u00b3<\/p>\n

Explanation:
\n\\(\\frac { volume of pyramid C }{ volume of pyramid D } \\) = (\\(\\frac { pyramid C base }{ pyramid D base } \\))\u00b3
\n\\(\\frac { 324 }{ V } \\) = (\\(\\frac { 9 }{ 3 } \\))\u00b3
\nV = 12<\/p>\n

Question 6.
\nFind the volume of the composite solid.
\n\"Big<\/p>\n

Answer:
\nthe volume of solid = 96<\/p>\n

Explanation:
\nVolume of prism = Bh
\nB = 8 x 2 = 16
\nV = 16 x 5 = 80
\n\\(\\frac { 1 }{ 3 } \\)Bh
\n= \\(\\frac { 1 }{ 3 } \\)(16 x 3) = 16
\nthe volume of solid = 16 + 80 = 96<\/p>\n

Exercise 11.6 Volumes of Pyramids<\/h3>\n

Vocabulary and Core Concept Check<\/p>\n

Question 1.
\nVOCABULARY<\/strong>
\nExplain the difference between a triangular prism and a triangular pyramid.
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nREASONING<\/strong>
\nA square pyramid and a cube have the same base and height. Compare the volume of the square pyramid to the volume of the cube.<\/p>\n

Answer:
\nSquare pyramid = 1\/3 Bh
\nCube = BH
\nSo, the volume of the square pyramid is 1\/3 of the volume of the cube.<\/p>\n

Monitoring Progress and Modeling with Mathematics<\/p>\n

In Exercises 3 and 4, find the volume of the pyramid.<\/p>\n

Question 3.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 4.
\n\"Big<\/p>\n

Answer:
\nV = 6 in\u00b3<\/p>\n

Explanation:
\nV = \u00a0\\(\\frac { 1 }{ 3 } \\)Bh
\nB = 2 x 3 = 6
\nV = \\(\\frac { 1 }{ 3 } \\)(6 x 3)<\/p>\n

In Exercises 5 – 8, find the indicated measure.<\/p>\n

Question 5.
\nA pyramid with a square base has a volume of 120 cubic meters and a height of 10 meters. Find the side length of the square base.
\nAnswer:
\n\"Big<\/p>\n

Question 6.
\nA pyramid with a square base has a volume of 912 cubic feet and a height of 19 feet. Find the side length of the square base.<\/p>\n

Answer:
\nThe side length of the square base is 12 ft<\/p>\n

Explanation:
\nA pyramid with a square base has a volume of 912 cubic feet
\nh = 19
\n\\(\\frac { 1 }{ 3 } \\)Bh = 912
\n\\(\\frac { 1 }{ 3 } \\)B(19) = 912
\nB = 144
\ns = 12<\/p>\n

Question 7.
\nA pyramid with a rectangular base has a volume of 480 cubic inches and a height of 10 inches. The width of the rectangular base is 9 inches. Find the length of the rectangular base.
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\nA pyramid with a rectangular base has a volume of 105 cubic centimeters and a height of 15 centimeters. The length of the rectangular base is 7 centimeters. Find the width of the rectangular base.<\/p>\n

Answer:
\nThe width of the rectangular base is 3 cm<\/p>\n

Explanation:
\nA pyramid with a rectangular base has a volume of 105 cubic centimeters
\nh = 15
\nl = 7
\n\\(\\frac { 1 }{ 3 } \\)Bh = 105
\n\\(\\frac { 1 }{ 3 } \\)lbh = 105
\n\\(\\frac { 1 }{ 3 } \\)(7 x 15 x b) = 105
\nb = 3<\/p>\n

Question 9.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in finding the volume of the pyramid.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\nOPEN-ENDED<\/strong>
\nGive an example of a pyramid and a prism that have the same base and the same volume. Explain your reasoning.
\nAnswer:
\nLet the rectangular prism have the base dimensions 4 x 2 nad a height of 5 so its volume is 4 x 2 x 5 = 40 cubic units
\nTherefore the base of the rectangular prism also have the dimensions of 4 x 2 and a height of 5 x 3 = 15 units so its volume V = 1\/3 x 4 x 2 x 15 = 40 cubic units<\/p>\n

In Exercises 11 – 14, find the height of the pyramid.<\/p>\n

Question 11.
\nVolume = 15 ft3<\/sup>
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\nVolume = 224 in.3<\/sup>
\n\"Big<\/p>\n

Answer:
\nThe height of the pyramid is 10.5 in<\/p>\n

Explanation:
\nVolume = 224
\n\\(\\frac { 1 }{ 3 } \\)Bh = 224
\nB = 8\u00b2 = 64
\n\\(\\frac { 1 }{ 3 } \\)(64)h = 224
\nh = 10.5<\/p>\n

Question 13.
\nVolume = 198 yd3<\/sup>
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\nVolume = 392 cm3<\/sup>
\n\"Big<\/p>\n

Answer:
\nThe height of the pyramid is 12 cm<\/p>\n

Explanation:
\nVolume = 392
\n\\(\\frac { 1 }{ 3 } \\)Bh = 392
\nB = 14 x 7 = 98
\n\\(\\frac { 1 }{ 3 } \\)(98)h = 392
\nh = 12<\/p>\n

In Exercises 15 and 16, the pyramids are similar. Find the volume of pyramid B.<\/p>\n

Question 15.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\n\"Big<\/p>\n

Answer:
\nVolume of A = 80<\/p>\n

Explanation:
\n\\(\\frac { Volume of B }{ Volume of A } \\) = (\\(\\frac { Side of B }{ side of A } \\))\u00b3
\n\\(\\frac { V }{ 10 } \\) = (\\(\\frac { 6 }{ 3 } \\))\u00b3
\nV = 8 x 10
\nVolume of A = 80<\/p>\n

In Exercises 17 – 20, find the volume of the composite solid.<\/p>\n

Question 17.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\n\"Big<\/p>\n

Answer:
\nComposite solid volume = 306<\/p>\n

Explanation:
\nBase area = \\(\\frac { 1 }{ 2 } \\)bh = \\(\\frac { 1 }{ 2 } \\)(12 x 9) = 54
\nBottom solid volume V = \\(\\frac { 1 }{ 3 } \\)Bh = \\(\\frac { 1 }{ 3 } \\)(54 x 10)
\nV = 180
\nTop solid volume v = \\(\\frac { 1 }{ 3 } \\)(54 x 7) = 126
\nComposite solid volume = 180 + 126 = 306<\/p>\n

Question 19.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\n\"Big<\/p>\n

Answer:
\nComposite solid volume = 1152<\/p>\n

Explanation:
\nVolume of Box = 12 x 12 x 12 = 1728
\nSquare pyramid volume = \\(\\frac { 1 }{ 3 } \\)Bh = \\(\\frac { 1 }{ 3 } \\)(144 x 12)
\n= 576
\nComposite solid volume = 1728 – 576 = 1152<\/p>\n

Question 21.
\nABSTRACT REASONING<\/strong>
\nA pyramid has a height of 8 feet and a square base with a side length of 6 feet.<\/p>\n

a. How does the volume of the pyramid change when the base slays the same and the height is doubled?
\nb. How does the volume of the pyramid change when the height stays the same and the side length of the base is doubled?
\nC. Are your answers 10 parts (a) and (b) true for any square pyramid? Explain your reasoning.
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 22.
\nHOW DO YOU SEE IT?<\/strong>
\nThe cube shown is formed by three pyramids. each with the same square base and the same height. How could you use this to verify the formula for the volume of a pyramid?
\n\"Big
\nAnswer:<\/p>\n

Question 23.
\nCRITICAL THINKING<\/strong>
\nFind the volume of the regular pentagonal pyramid. Round your answer to the nearest hundredth. In the diagram. m\u2220ABC = 35\u00b0
\n\"Big
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 24.
\nTHOUGHT PROVOKING<\/strong>
\nA frustum of a pyramid is the part of the pyramid that lies between the base and a plane parallel to the base, as shown. Write a formula for the volume of the frustum of a square pyramid in terms of a, b, and h. (Hint: Consider the \u201cmissing\u201d top of the pyramid and use similar triangles.)
\n\"Big
\nAnswer:<\/p>\n

Question 25.
\nMODELING WITH MATHEMATICS<\/strong>
\nNautical deck prisms were used as a safe way to illuminate decks on ships. The deck prism shown here is composed of the following three solids: a regular hexagonal prism with an edge length of 3.5 inches and a height of 1.5 inches, a regular hexagonal prism with an edge length of 3.25 inches arid a height of 0.25 inch, and a regular hexagonal pyramid with an edge length of 3 inches and a height of 3 inches. Find the volume of the deck prism.
\n\"Big
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Maintaining Mathematical Proficiency<\/p>\n

Find the value of X. Round your answer to the nearest tenth.<\/p>\n

Question 26.
\n\"Big
\nAnswer:
\ntan 35 = \\(\\frac { 9 }{ x } \\)
\n0.7 = \\(\\frac { 9 }{ x } \\)
\nx = 12.8<\/p>\n

Question 27.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 28.
\n\"Big
\nAnswer:
\ntan 30 = \\(\\frac { x }{ 10 } \\)
\n0.577 = \\(\\frac { x }{ 10 } \\)
\nx = 5.77<\/p>\n

Question 29.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

11.7 Surface Areas and Volumes of Cones<\/h3>\n

Exploration 1<\/strong><\/p>\n

Finding the Surface Area of a Cone<\/p>\n

Work with a partner: Construct a circle with a radius of 3 inches. Mark the circumference of the circle into six equal parts, and label the length of each part. Then cut out one sector of the circle and make a cone.
\n\"Big
\na. Explain why the base of the cone is a circle. What are the circumference and radius
\nof the base?
\nAnswer:<\/p>\n

b. What is the area of the original circle? What is the area with one sector missing?
\nAnswer:<\/p>\n

c. Describe the surface area of the cone, including the base. Use your description to find the surface area.
\nAnswer:<\/p>\n

Exploration 2<\/strong><\/p>\n

Finding the Volume of a Cone<\/p>\n

Work with a partner: The cone and the cylinder have the same height and the same circular base.
\n\"Big
\nWhen the cone is filled with sand and poured into the cylinder. it takes three cones to fill the cylinder.
\n\"Big
\nUse this information to write a formula for the volume V of a cone.
\nCONSTRUCTING VIABLE ARGUMENTS<\/strong>
\nTo be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results in constructing arguments.
\nAnswer:<\/p>\n

Communicate Your Answer<\/p>\n

Question 3.
\nHow can you find the surface area and the volume of a cone?
\nAnswer:<\/p>\n

Question 4.
\nIn Exploration 1, cut another sector from the circle and make a cone. Find the radius of the base and the surface area of the cone. Repeat this three times, recording your results in a table. Describe the pattern.
\nAnswer:<\/p>\n

Lesson 11.7 Surface Areas and Volumes of Cones<\/h3>\n

Monitoring progress<\/p>\n

Question 1.
\nFind the surface area of the right cone.
\n\"Big<\/p>\n

Answer:
\nThe surface area of the right cone is 436.17 m\u00b2<\/p>\n

Explanation:
\nr = 7.8
\nl = 10
\nS = \u03c0r\u00b2 + \u03c0rl
\nS = \u03c0(7.8)\u00b2 + \u03c0(7.8 x 10)
\nS = 436.17<\/p>\n

Find the volume of the cone.<\/p>\n

Question 2.
\n\"Big<\/p>\n

Answer:
\nThe volume of the cone is 2206.44 in\u00b3<\/p>\n

Explanation:
\nr = 7, h = 13
\nl = \u221a13\u00b2 – 7\u00b2= 10.95
\nS = \u03c0r\u00b2 + \u03c0rl
\nS = \u03c07\u00b2 + \u03c0(7 x 10.95)
\nS = 394.74
\nVolume V = \\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2h)
\nV = \\(\\frac { 1 }{ 3 } \\)(\u03c0 x 7\u00b2 x 13)
\nV = 2206.44<\/p>\n

Question 3.
\n\"Big<\/p>\n

Answer:
\nThe volume of the cone is 163.4 m\u00b3<\/p>\n

Explanation:
\nh = \u221a8\u00b2 – 5\u00b2 = 6.24
\nVolume V = \\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2h)
\nV = \\(\\frac { 1 }{ 3 } \\)(\u03c0 x 5\u00b2 x 6.24)
\nV = 163.4<\/p>\n

Question 4.
\nCone C and cone D are similar. Find the volume of cone D.
\n\"Big<\/p>\n

Answer:
\nVolume of cone D = 18.84 cm\u00b3<\/p>\n

Explanation:
\n\\(\\frac { Volumeof cone C }{ Volume of cone D } \\) = (\\(\\frac { height of C }{ height of D } \\))\u00b3
\n\\(\\frac { 384\u03c0 }{ Volume of cone D } \\)= (\\(\\frac { 8 }{ 2 } \\))\u00b3
\nVolume of cone D = 18.84<\/p>\n

Question 5.
\nFind the volume of the composite solid.
\n\"Big<\/p>\n

Answer:
\nComposite solid volume = 329.86 cm\u00b3<\/p>\n

Explanation:
\nVolume of cylinder = \u03c0r\u00b2h = \u03c0(3)\u00b2 x 10 = 90\u03c0
\nVolume of cone = \\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2h)
\n= \\(\\frac { 1 }{ 3 } \\)(\u03c0 x 3\u00b2 x 5)
\n= 15\u03c0
\nComposite solid volume = 15\u03c0 + 90\u03c0 = 105\u03c0<\/p>\n

Exercise 11.7 Surface Areas and Volumes of Cones<\/h3>\n

Vocabulary and Core Concept Check<\/p>\n

Question 1.
\nWRITING<\/strong>
\nDescribe the differences between pyramids and cones. Describe their similarities.
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nCOMPLETE THE SENTENCE<\/strong>
\nThe volume of a cone with radius r and height h is \\(\\frac{1}{3}\\) the volume of a(n) __________ with radius r and height h.
\nAnswer:<\/p>\n

Monitoring Progress and Modeling with Mathematics<\/p>\n

In Exercises 3 – 6, find the surface area of the right cone.<\/p>\n

Question 3.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 4.
\n\"Big<\/p>\n

Answer:
\nThe surface area of cone is 219.44 sq cm.<\/p>\n

Explanation:
\nS = \u03c0r\u00b2 + \u03c0rl
\nS = \u03c0(5.5)\u00b2 + \u03c0(5.5 x 7.2)
\nS = 219.44<\/p>\n

Question 5.
\nA right cone has a radius of 9 inches and a height of 12 inches.
\nAnswer:
\n\"Big<\/p>\n

Question 6.
\nA right cone has a diameter of 11.2 feet and a height of 19.2 feet.<\/p>\n

Answer:
\nThe surface area is 421.52 sq ft.<\/p>\n

Explanation:
\nr = 5.6
\nh = 19.2
\nl = \u221a19.2\u00b2 – 5.6\u00b2 = 18.36
\nSurface area S = \u03c0r\u00b2 + \u03c0rl
\nS = \u03c0(5.6)\u00b2 + \u03c0(5.6 x 18.36)
\nS = 421.52<\/p>\n

In Exercises 7 – 10, find the volume of the cone.<\/p>\n

Question 7.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\n\"Big<\/p>\n

Answer:
\nThe volume is 2.09 cubic meter<\/p>\n

Explanation:
\nVolume of cone = \\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2h)
\nV = \\(\\frac { 1 }{ 3 } \\)(\u03c0(1)\u00b2 x 2)
\nV = 2.09<\/p>\n

Question 9.
\nA cone has a diameter of 11.5 inches and a height of 15.2 inches.
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\nA right cone has a radius of 3 feet and a slant height of 6 feet.<\/p>\n

Answer:
\nThe volume is 56.54 cubic ft<\/p>\n

Explanation:
\nVolume of cone = \\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2h)
\nV = \\(\\frac { 1 }{ 3 } \\)(\u03c0 x 3\u00b2 x 6)
\nV = 56.54<\/p>\n

In Exercises 11 and 12, find the missing dimension(s).<\/p>\n

Question 11.
\nSurface area = 75.4 cm2<\/sup>
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\nVolume = 216\u03c0 in.3<\/sup><\/p>\n

Answer:
\nThe radius is 6.13 in<\/p>\n

Explanation:
\nVolume = 216\u03c0 in.3<\/sup>
\n\\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2h) = 216
\n\\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2 x 18) = 216
\nr = 6.13<\/p>\n

In Exercises 13 and 14, the cones are similar. Find the volume of cone B.<\/p>\n

Question 13.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\n\"Big<\/p>\n

Answer:
\nVolume of cone B = 24.127<\/p>\n

Explanation:
\n\\(\\frac { Volumeof cone A }{ Volume of cone B } \\) = (\\(\\frac { height of A }{ height of B } \\))\u00b3
\n\\(\\frac { 120\u03c0 }{ Volume of cone B } \\) = (\\(\\frac { 10 }{ 4 } \\))\u00b3
\nVolume of cone B = 24.127<\/p>\n

In Exercises 15 and 16, find the volume of the composite solid.<\/p>\n

Question 15.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\n\"Big<\/p>\n

Answer:
\nVolume of the composite solid = 97.93 cubic m<\/p>\n

Explanation:
\nVolume of box = lbh
\nV = 51 x 5.1 x 5.1 = 132.651
\nCone volume = \\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2h)
\nv = \\(\\frac { 1 }{ 3 } \\)(\u03c0 x 2.55\u00b2 x 5.1)
\nv = 34.72
\nVolume of the composite solid = 132.651 – 34.72 = 97.93<\/p>\n

Question 17.
\nANALYZING RELATIONSHIPS<\/strong>
\nA cone has height h and a base with radius r. You warn to change the cone so its volume is doubled. What is the new height if you change only the height? What is the new radius if you change only the radius? Explain.
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\nHOW DO YOU SEE IT<\/strong>
\nA snack stand serves a small order of popcorn in a cone-shaped container and a large order of popcorn in a cylindrical container. Do not perform any calculations.
\n\"Big
\na. How many small containers of popcorn do you have to buy to equal the amount of popcorn in a large container? Explain.<\/p>\n

Answer:
\nVolume of cone = \\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2h)
\nV = \\(\\frac { 1 }{ 3 } \\)(\u03c0 x 3\u00b2 x 8) = 75.39
\nVolume of cylinder = \u03c0r\u00b2h = \u03c0 x 3\u00b2 x 8 = 226.19
\nVolume of cylinder \/ Volume of cone = \\(\\frac { 226.19 }{ 75.39 } \\) = 3
\nYou have to buy 3 small containers of popcorn to equal the amount of popcorn in a large containe.<\/p>\n

b. Which container gives you more popcorn for your money? Explain.
\nAnswer:
\n$1.25 -> 75.39 i.e $1 = 60.312
\n$2.50 -> 226.19 i.e $1 = 90.47
\nSo, large containers gives you more popcorn for your money<\/p>\n

In Exercises 19 and 20. find the volume of the right cone.<\/p>\n

Question 19.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\n\"Big<\/p>\n

Answer:
\nVolume of cone is 575.62 cubic yd<\/p>\n

Explanation:
\ntan 32 = \\(\\frac { 7 }{ h } \\)
\nh = 11.21
\nVolume of cone = \\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2h)
\nV = \\(\\frac { 1 }{ 3 } \\)(\u03c0 x 7\u00b2 x 11.21)
\nV = 575.62<\/p>\n

Question 21.
\nMODELING WITH MATHEMATICS<\/strong>
\nA cat eats hail a cup of food, twice per day. Will the automatic pet feeder hold enough food for 10 days? Explain your reasoning. (1 cup \u2248 14.4 in.3<\/sup>)
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 22.
\nMODELING WITH MATHEMATICS<\/strong>
\nDuring a chemistry lab, you use a funnel to pour a solvent into a flask. The radius of the funnel is 5 centimeters and its height is 10 centimeters. You pour the solvent into the funnel at a rate of 80 milliliters per second and the solvent flows out of the funnel at a rate of 65 milliliters per second. How long will it be before the funnel overflows? (1 mL = 1 cm3<\/sup>)
\nAnswer:
\n17.45 seconds<\/p>\n

Explanation:
\nVolume of cone = \\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2h)
\nV = \\(\\frac { 1 }{ 3 } \\)(\u03c0 x 5\u00b2 x 10)
\nV = 261.8
\n\\(\\frac { 261.8 }{ 15 } \\) = 17.45<\/p>\n

Question 23.
\nREASONING<\/strong>
\nTo make a paper drinking cup, start with a circular piece of paper that has a 3-inch radius, then follow the given steps. How does the surface area of the cup compare to the original paper circle? Find m\u2220ABC.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 24.
\nTHOUGHT PROVOKING<\/strong>
\nA frustum of a cone is the part of the cone that lies between the base and a plane parallel to the base, as shown. Write a formula for the volume of the frustum of a cone in terms of a, b, and h. (Hint: Consider the \u201cmissing\u201d top of the cone and use similar triangles.)
\n\"Big
\nAnswer:
\nVolume V \u00a0= (1\/3) * \u03c0 * h * (r1\u00b2 + r2\u00b2 + (r1 * r2))<\/p>\n

Question 25.
\nMAKING AN ARGUMENT<\/strong>
\nIn the figure, the two cylinders are congruent The combined height of the two smaller cones equals the height of the larger cone. Your friend claims that this means the total volume of the two smaller cones is equal to the volume of the larger cone. Is your friend correct? Justify your answer.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 26.
\nCRITICAL THINKING<\/strong>
\nWhen the given triangle is rotated around each of its sides. solids of revolution are formed. Describe the three solids and find their volumes. Give your answers in terms of \u03c0.
\n\"Big
\nAnswer:<\/p>\n

Maintaining Mathematical Proficiency<\/p>\n

Find the indicated measure.<\/p>\n

Question 27.
\narea of a circle with a radius of 7 feet
\nAnswer:
\n\"Big<\/p>\n

Question 28.
\narea of a circle with a diameter of 22 centimeters<\/p>\n

Answer:
\nd = 11
\nA = \u03c0r\u00b2
\nA = 121\u03c0<\/p>\n

Question 29.
\ndiameter of a circle with an area of 256 square meters
\nAnswer:
\n\"Big<\/p>\n

Question 30.
\nradius of a circle with an area of 529 \u03c0 square inches
\nAnswer:
\nA = \u03c0r\u00b2
\n529\u03c0 = \u03c0r\u00b2
\nr = 23<\/p>\n

11.8 Surface Areas and Volumes of Spheres<\/h3>\n

Exploration 1<\/strong><\/p>\n

Finding the Surface Area of a Sphere<\/p>\n

Work with a partner: Remove the covering from a baseball or softball.
\n\"Big
\nYou will end up with two \u201cfigure 8\u201d pieces of material, as shown above. From the amount of material it takes to cover the ball, what would you estimate the surface area S of the ball to be? Express your answer in terms of the radius r of the ball.
\n\"Big
\nUse the Internet or some other resource to confirm that the formula you wrote for the surface area of a sphere is correct.
\nUSING TOOLS STRATEGICALLY<\/strong>
\nTo be proficient in math, you need to identify relevant external mathematical resources, such as content located on a website.
\nAnswer:<\/p>\n

Exploration 2<\/strong><\/p>\n

Finding the volume of a sphere<\/p>\n

Work with a partner: A cylinder is circumscribed about a sphere, as shown. Write a formula for the volume V of the cylinder in terms of the radius r.
\n\"Big
\n\"Big
\nWhen half of the sphere (a hemisphere) is filled with sand and poured into the cylinder, it takes three hemispheres to till the cylinder. Use this information to write a formula for the volume V of a sphere in terms of the radius r.
\n\"Big
\n\"Big
\nAnswer:<\/p>\n

Communicate Your Answer<\/p>\n

Question 3.
\nHow can you find the surface area and the volume of a sphere?
\nAnswer:<\/p>\n

Question 4.
\nUse the results of Explorations 1 and 2 to find the surface area and the volume of a sphere with a radius of(a) 3 inches and (b) 2 centimeters.
\nAnswer:<\/p>\n

Lesson 11.8 Surface Areas and Volumes of Spheres<\/h3>\n

Monitoring Progress<\/strong><\/p>\n

Find the surface area of the sphere.<\/p>\n

Question 1.
\n\"Big<\/p>\n

Answer:
\nThe surface area of the sphere is 5026.54 ft\u00b2<\/p>\n

Explanation:
\nD = 40
\nr = 20
\nThe surface area of the sphere = 4\u03c0r\u00b2
\nS = 4 x \u03c0 x (20)\u00b2
\nS = 5026.54 ft\u00b2<\/p>\n

Question 2.
\n\"Big<\/p>\n

Answer:
\nThe surface area of the sphere is 113.09 ft\u00b2<\/p>\n

Explanation:
\nCircumference C = 6\u03c0
\n2\u03c0r = 6\u03c0
\nr = 3
\nThe surface area of the sphere = 4\u03c0r\u00b2
\nS = 4\u03c0 x 3\u00b2
\nS = 113.09<\/p>\n

Question 3.
\nFind the radius of the sphere.
\n\"Big<\/p>\n

Answer:
\nThe radius of the sphere is 2.73 m<\/p>\n

Explanation:
\nThe surface area of the sphere = 4\u03c0r\u00b2
\n30\u03c0 = 4\u03c0r\u00b2
\nr = 2.73<\/p>\n

Question 4.
\nThe radius of a sphere is 5 yards. Find the volume of the sphere.<\/p>\n

Answer:
\nThe volume of the sphere is 523.59 yards\u00b3<\/p>\n

Explanation:
\nr = 5
\nThe volume of the sphere V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 5\u00b3
\nV = 523.59 yards\u00b3<\/p>\n

Question 5.
\nThe diameter of a sphere is 36 inches. Find the volume of the sphere.<\/p>\n

Answer:
\nThe volume of the sphere is 24429.02 in\u00b3<\/p>\n

Explanation:
\nD = 36
\nr = 18
\nThe volume of the sphere V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 18\u00b3
\nV = 24429.02<\/p>\n

Question 6.
\nThe surface area of a sphere is 576\u03c0 square centimeters. Find the volume of the sphere.<\/p>\n

Answer:
\nThe volume of the sphere is 2304\u03c0 cm\u00b3<\/p>\n

Explanation:
\nThe surface area of the sphere = 4\u03c0r\u00b2
\n576\u03c0 = 4\u03c0r\u00b2
\nr = 12
\nThe volume of the sphere V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 12\u00b3
\nV = 2304\u03c0<\/p>\n

Question 7.
\nFind the volume of the composite solid at the left.
\n\"Big<\/p>\n

Answer:
\nThe volume of the composite solid is 7.324 m\u00b3<\/p>\n

Explanation:
\nThe volume of cone = \u03c0r\u00b2\\(\\frac { h }{ 3 } \\)
\n= \u03c0 x 1\u00b2 x \\(\\frac { 5 }{ 3 } \\) = 5.23
\nThe volume of sphere = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\n= \\(\\frac { 4 }{ 3 } \\)\u03c0 x 1\u00b3 = 4.188
\nThe volume of the composite solid = The volume of cone + The volume of sphere\/2
\n= 5.23 + 4.188\/2
\n= 7.324 m\u00b3<\/p>\n

Exercise 11.8 Surface Areas and Volumes of Spheres<\/h3>\n

Question 1.
\nVOCABULARY<\/strong>
\nWhen a plane intersects a sphere. what must be true for the intersection to be a great circle?
\nAnswer:
\n\"Big<\/p>\n

Question 2.
\nWRITING<\/strong>
\nExplain the difference between a sphere and a hemisphere.<\/p>\n

Answer:
\nHemisphere is a related term of the sphere. Sphere and hemisphere are three-dimensional solids. The volume of sphere is \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3 and hemisphere volume is \\(\\frac { 2 }{ 3 } \\)\u03c0r\u00b3. The surface area of the sphere is 4\u03c0r\u00b2 and hemisphere surface area is 3\u03c0r\u00b2.<\/p>\n

Monitoring progress and Modeling with Mathematics<\/p>\n

In Exercises 3 – 6, find the surface area of the sphere.<\/p>\n

Question 3.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 4.
\n\"Big<\/p>\n

Answer:
\nThe surface area of the sphere is 225\u03c0 cm\u00b2<\/p>\n

Explanation:
\nThe surface area of the sphere = 4\u03c0r\u00b2
\nS = 4\u03c0 x 7.5\u00b2
\nS = 225\u03c0<\/p>\n

Question 5.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 6.
\n\"Big<\/p>\n

Answer:
\nThe surface area of the sphere is 8\u03c0 ft\u00b2<\/p>\n

Explanation:
\nC = 4\u03c0
\n2\u03c0r = 4\u03c0
\nr = 2
\nThe surface area of the sphere = 4\u03c0r\u00b2
\nS = 4\u03c0 x 2\u00b2
\nS = 8\u03c0<\/p>\n

In Exercises 7 – 10. find the indicated measure.<\/p>\n

Question 7.
\nFind the radius of a sphere with a surface area of 4\u03c0 square feet.
\nAnswer:
\n\"Big<\/p>\n

Question 8.
\nFind the radius of a sphere with a surface area of 1024\u03c0 square inches.<\/p>\n

Answer:
\nThe radius of a sphere is 16 in<\/p>\n

Explanation:
\nThe surface area of the sphere = 1024\u03c0
\n4\u03c0r\u00b2 = 1024\u03c0
\nr = 16<\/p>\n

Question 9.
\nFind the diameter of a sphere with a surface area of 900\u03c0 square meters.
\nAnswer:
\n\"Big<\/p>\n

Question 10.
\nFind the diameter of a sphere with a surface area of 196\u03c0 square centimeters.<\/p>\n

Answer:
\nThe diameter of a sphere is 14 cm<\/p>\n

Explanation:
\nThe surface area of the sphere = 196\u03c0
\n4\u03c0r\u00b2 = 196\u03c0
\nr = 7
\nD = 2(7) = 14<\/p>\n

In Exercises 11 and 12, find the surface area of the hemisphere.<\/p>\n

Question 11.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 12.
\n\"Big<\/p>\n

Answer:
\nThe surface area of the hemisphere is 108\u03c0 in\u00b2<\/p>\n

Explanation:
\nD = 12, r = 6
\nThe surface area of the sphere = 3\u03c0r\u00b2
\nS = 3\u03c0 x 6\u00b2
\nS = 108\u03c0<\/p>\n

In Exercises 13 – 18. find the volume of the sphere.<\/p>\n

Question 13.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 14.
\n\"Big<\/p>\n

Answer:
\nThe volume of the sphere is 268.08 ft\u00b3<\/p>\n

Explanation:
\nr = 4 ft
\nVolume of the sphere V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 4\u00b3
\nV = 268.08 ft<\/p>\n

Question 15.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\n\"Big<\/p>\n

Answer:
\nThe volume of the sphere is 1436.75 ft\u00b3<\/p>\n

Explanation:
\nD = 14 ft
\nr = 7 ft
\nVolume of the sphere V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 7\u00b3
\nV = 1436.75 ft<\/p>\n

Question 17.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 18.
\n\"Big<\/p>\n

Answer:
\nThe volume of the sphere is 179.89 in\u00b3<\/p>\n

Explanation:
\nC = 7\u03c0
\n2\u03c0r = 7\u03c0
\nr = 3.5
\nVolume of the sphere V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 3.5\u00b3
\nV = 179.89 in<\/p>\n

In Exercises 19 and 20, find the volume of the sphere with the given surface area.<\/p>\n

Question 19.
\nSurface area = 16\u03c0 ft2<\/sup>
\nAnswer:
\n\"Big<\/p>\n

Question 20.
\nSurface area = 484\u03c0 cm2<\/sup><\/p>\n

Answer:
\nThe volume of the sphere is 5575.27 cm\u00b3<\/p>\n

Explanation:
\nSurface area = 484\u03c0
\n4\u03c0r\u00b2 = 484\u03c0
\nr = 11
\nVolume of the sphere V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 11\u00b3
\nV = 5575.27<\/p>\n

Question 21.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in finding the volume of the sphere.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 22.
\nERROR ANALYSIS<\/strong>
\nDescribe and correct the error in finding the volume of the sphere.
\n\"Big<\/p>\n

Answer:
\nDiameter = 3
\nradius = 1.5
\nVolume of the sphere V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x (1.5)\u00b3
\nV = 14.137 cubic in<\/p>\n

In Exercises 23 – 26, find the volume of the composite solid.<\/p>\n

Question 23.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 24.
\n\"Big<\/p>\n

Answer:
\nVolume is 288\u03c0 ft\u00b3<\/p>\n

Explanation:
\nVolume of hemipshere = \\(\\frac { 2 }{ 3 } \\)\u03c0r\u00b3
\n= \\(\\frac { 2 }{ 3 } \\)\u03c0 x 6\u00b3 = 144\u03c0
\nvolume of the cone = \u03c0r\u00b2\\(\\frac { h }{ 3 } \\)
\n= \u03c0 x 6\u00b2 x \\(\\frac { 12 }{ 3 } \\) = 144\u03c0
\nArea of circle = \u03c0r\u00b2 = \u03c0 x 6\u00b2 = 36\u03c0
\nVolume of hemipshere + volume of the cone = 144\u03c0 + 144\u03c0 = 288\u03c0<\/p>\n

Question 25.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 26.
\n\"Big<\/p>\n

Answer:
\nThe volume of solid is 296\u03c0 m\u00b3<\/p>\n

Explanation:
\nVolume of hemipshere = \\(\\frac { 2 }{ 3 } \\)\u03c0r\u00b3
\n= \\(\\frac { 2 }{ 3 } \\)\u03c0 x 6\u00b3 = 144\u03c0
\nVolume of cylinder = \u03c0r\u00b2h
\n= \u03c0 x 6\u00b2 x 14 = 504\u03c0
\nVolume of solid = 504\u03c0 – 2(144\u03c0) = 296\u03c0<\/p>\n

In Exercises 27 – 32, find the surface area and volume of the ball.<\/p>\n

Question 27.
\nbowling ball
\n\"Big
\nd = 8.5 in.
\nAnswer:
\n\"Big<\/p>\n

Question 28.
\nbasketball
\n\"Big
\nC = 29.5 in.<\/p>\n

Answer:
\nThe surface area is 277 in\u00b2, volume is 43212.27 in\u00b3<\/p>\n

Explanation:
\nC = 29.5
\n2\u03c0r = 29.5
\nr = 4.69
\nSurface area = 4\u03c0r\u00b2
\nS = 4\u03c0 x 4.69\u00b2 = 277
\nVolume V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 4.69\u00b3
\nV = 43212.27<\/p>\n

Question 29.
\nsoftball
\n\"Big
\nC = 12 in.
\nAnswer:
\n\"Big<\/p>\n

Question 30.
\ngolf ball
\n\"Big
\nd = 1.7 in.<\/p>\n

Answer:
\nThe surface area is 9.07 in\u00b2, volume is 2.57 in\u00b3<\/p>\n

Explanation:
\nd = 1.7
\nr = 0.85
\nSurface area = 4\u03c0r\u00b2
\nS = 4\u03c0 x 0.85\u00b2 = 9.07
\nVolume V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 0.85\u00b3
\nV = 2.57<\/p>\n

Question 31.
\nvolleyball
\n\"Big
\nC = 26 in.
\nAnswer:
\n\"Big<\/p>\n

Question 32.
\nbaseball
\n\"Big
\nC = 9 in.<\/p>\n

Answer:
\nThe surface area is 25.78 in\u00b2, volume is 12.24 in\u00b3<\/p>\n

Explanation:
\nC = 9
\n2\u03c0r = 9
\nr = 1.43
\nSurface area = 4\u03c0r\u00b2
\nS = 4\u03c0 x 1.43\u00b2
\nS = 25.78
\nVolume V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 1.43\u00b3
\nV = 12.24<\/p>\n

Question 33.
\nMAKING AN ARGUMENT<\/strong>
\nYou friend claims that if the radius of a sphere is doubled, then the surface area of the sphere will also be doubled. Is our friend correct? Explain your reasoning.
\nAnswer:
\n\"Big<\/p>\n

Question 34.
\nREASONING<\/strong>
\nA semicircle with a diameter of 18 inches is rotated about its diameter. Find the surface area and the volume of the solid formed.<\/p>\n

Answer:
\nThe surface area is 1018 in\u00b2, volume is 3054.02 in\u00b3<\/p>\n

Explanation:
\nDiameter = 18
\nradius r = 9
\nVolume V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 9\u00b3
\nV = 3054.02
\nSurface area = 4\u03c0r\u00b2
\nS = 4\u03c0 x 9\u00b2
\nS = 1018<\/p>\n

Question 35.
\nMODELING WITH MATHEMATICS<\/strong>
\nA silo has the dimensions shown. The top of the silo is a hemispherical shape. Find the volume of the silo.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 36.
\nMODELING WITH MATHEMATICS<\/strong>
\nThree tennis balls are stored in a cylindrical container with a height of 8 inches and a radius of 1.43 inches. The circumference of a tennis ball is 8 inches.
\n\"Big
\na. Find the volume of a tennis ball.<\/p>\n

Answer:
\nC = 8 in
\n2\u03c0r = 8
\nr = 1.27
\nVolume V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 1.27\u00b3 = 8.64
\nThe volume of tennis ball = 8.64 in\u00b3<\/p>\n

b. Find the amount of space within the cylinder not taken up by the tennis balls.
\nAnswer:
\nThe surface area of tennis ball S = 4\u03c0r\u00b2
\nS = 4\u03c0 x 1.27\u00b2 = 20.26
\nArea of cylinder s = 2\u03c0rh+2\u03c0r\u00b2
\ns = 2\u03c0 x 1.43 x 8+2\u03c0 x 1.43\u00b2
\ns = 84.72
\nRemaining space = 84.72 – 20.26 = 64.46 in\u00b2<\/p>\n

Question 37.
\nANALYZING RELATIONSHIPS<\/strong>
\nUse the table shown for a sphere.
\n\"Big
\na. Copy and complete the table. Leave your answers in terms of \u03c0.
\nb. What happens to the surface area of the sphere when the radius is doubled? tripled? quadrupled?
\nc. What happens to the volume of the sphere when the radius is doubled? tripled? quadrupled?
\nAnswer:
\n\"Big<\/p>\n

Question 38.
\nMATHEMATICAL CONNECTIONS<\/strong>
\nA sphere has a diameter of 4(x + 3) centimeters and a surface area of 784 \u03c0 square centimeters. Find the value of x.<\/p>\n

Answer:
\nx =11<\/p>\n

Explanation:
\nSurface area = 4\u03c0r\u00b2
\n784\u03c0 = \u03c0r\u00b2
\nr = 28
\n2r = diameter = 4(x + 3)
\nr = 2(x + 3)
\n28 = 2(x + 3)
\nx = 11<\/p>\n

Question 39.
\nMODELING WITH MATHEMATICS<\/strong>
\nThe radius of Earth is about 3960 miles. The radius of the moon is about 1080 miles.
\na. Find the surface area of Earth and the moon.
\nb. Compare the surface areas of Earth and the moon.
\nc. About 70% of the surface of Earth is water. How many square miles of water are on Earth\u2019s surface?
\nAnswer:
\n\"Big<\/p>\n

Question 40.
\nMODELING WITH MATHEMATICS<\/strong>
\nThe Torrid Zone on Earth is the area between the Tropic of Cancer and the Tropic of Capricorn. The distance between these two tropics is about 3250 miles. You can estimate the distance as the height of a cylindrical belt around the Earth at the equator.
\n\"Big
\na. Estimate the surface area of the Torrid Zone. (The radius of Earth is about 3960 miles.)
\nAnswer:
\nSurface area of cylinder = 2\u03c0rh
\nS = 2\u03c0 x 3960 x 3250 = 80875080
\nsurface area of earth = 4\u03c0r\u00b2
\n= 4\u03c0 x 3960\u00b2 = 197086348.8<\/p>\n

b. A meteorite is equally likely to hit anywhere on Earth. Estimate the probability that a meteorite will land in the Torrid Zone.
\nAnswer:
\nProbability of meteorites hitting the torrid zone = 80875080\/197086348.8 = 0.4104<\/p>\n

Question 41.
\nABSTRACT REASONING<\/strong>
\nA sphere is inscribed in a cube with a volume of 64 cubic inches. What is the surface area of the sphere? Explain your reasoning.
\nAnswer:
\n\"Big<\/p>\n

Question 42.
\nHOW DO YOU SEE IT?<\/strong>
\nThe formula for the volume of a hemisphere and a Cone are shown. If each solid has the same radius and r = h, which solid will have a greater volume? Explain your reasoning.
\n\"Big
\nAnswer:
\nThe hemisphere has the highest volume.<\/p>\n

Explanation:
\nVolume of hemisphere v = \\(\\frac { 2 }{ 3 } \\)\u03c0r\u00b3
\nVolume of cone V = \\(\\frac { 1 }{ 3 } \\)\u03c0r\u00b2h
\nIf r = h
\nVolume of cone V = \\(\\frac { 1 }{ 3 } \\)\u03c0r\u00b2 x r = \\(\\frac { 1 }{ 3 } \\)\u03c0r\u00b3
\nSo, the hemisphere has the highest volume.<\/p>\n

Question 43.
\nCRITICAL THINKING<\/strong>
\nLet V be the volume of a sphere. S be the surface area of the sphere, and r be the radius of the sphere. Write an equation for V in terms of r and S. (Hint: Start with the ratio \\(\\frac{V}{S}\\).)
\nAnswer:
\n\"Big<\/p>\n

Question 44.
\nTHOUGHT PROVOKING<\/strong>
\nA spherical lune is the region between two great circles of a sphere. Find the formula for the area of a lune.
\n\"Big
\nAnswer:<\/p>\n

Question 45.
\nCRITICAL THINKING<\/strong>
\nThe volume of a right cylinder is the same as the volume of a sphere. The radius of the sphere is 1 inch. Give three possibilities for the dimensions of the cylinder.
\nAnswer:
\n\"Big
\n\"Big<\/p>\n

Question 46.
\nPROBLEM SOLVING<\/strong>
\nA spherical cap is a portion of a sphere cut off by a plane. The formula for the volume of a spherical cap is V = \\(\\frac{\\pi h}{6}\\) (3a2<\/sup> + h2<\/sup>), where a is the radius of the base of the cap and h is the height of the cap. Use the diagram and given information to find the volume of each spherical cap.
\n\"Big
\na. r = 5ft, a = 4ft
\nAnswer:<\/p>\n

b. r = 34 cm, a = 30 cm
\nAnswer:<\/p>\n

c. r = 13 m, h = 8 m
\nAnswer:<\/p>\n

d. r=75 in., h = 54in.
\nAnswer:<\/p>\n

Question 47.
\nCRITICAL THINKING<\/strong>
\nA sphere with a radius of 2 inches is inscribed in a right cone with a height of 6 inches. Find the surface area and the volume of the cone.
\nAnswer:
\n\"Big
\n\"Big
\n\"Big<\/p>\n

Maintaining Mathematical Proficiency<\/p>\n

Solve the triangle. Round decimal answers to the nearest tenth.<\/p>\n

Question 48.
\nA = 26\u00b0, C = 35\u00b0, b = 13<\/p>\n

Answer:
\nB = 119\u00b0, a = 7.16, c = 9.5<\/p>\n

Explanation:
\nB = 180 – (26 + 35) = 119
\n\\(\\frac { sin A }{ a } \\) = \\(\\frac { sin B }{ b } \\)
\n\\(\\frac { sin 26 }{ a } \\) = \\(\\frac { sin 119 }{ 13 } \\)
\na = 7.16
\n\\(\\frac { sin C }{ c } \\) = \\(\\frac { sin B }{ b } \\)
\n\\(\\frac { sin 35 }{ c } \\) = \\(\\frac { sin 119 }{ 13 } \\)
\nc = 9.5<\/p>\n

Question 49.
\nB = 102\u00b0, C = 43\u00b0, b = 21
\nAnswer:
\n\"Big<\/p>\n

Question 50.
\na = 23, b = 24, c = 20<\/p>\n

Answer:
\nA = 62.2, B = 65.5, C = 49.4<\/p>\n

Explanation:
\na\u00b2 = b\u00b2 + c\u00b2 – 2bc cos A
\n23\u00b2 = 24\u00b2+ 20\u00b2 – 2(24 x 20) cos A
\nA = 62.2
\n\\(\\frac { sin 62.2 }{ 23 } \\) = \\(\\frac { sin B }{ 24 } \\)
\nB = 65.5
\n\\(\\frac { sin 62.2 }{ 23 } \\) = \\(\\frac { sin C }{ 20 } \\)
\nC = 49.4<\/p>\n

Question 51.
\nA = 103\u00b0, b = 15, c = 24
\nAnswer:
\n\"Big<\/p>\n

Circumference, Area, and Volume Review<\/h3>\n

11.1 Circumference and Arc Length<\/h3>\n

Find the indicated measure.<\/p>\n

Question 1.
\ndiameter of \u2299P
\n\"Big<\/p>\n

Answer:
\ndiameter of \u2299P is 29.99<\/p>\n

Explanation:
\nCircumference = 94.24
\n\u03c0d = 94.24
\nd = 29.99<\/p>\n

Question 2.
\ncircumference of \u2299F
\n\"Big<\/p>\n

Answer:
\ncircumference of \u2299F = 56.57<\/p>\n

Explanation:
\n5.5 = \\(\\frac { 35 }{ 360 } \\) . C
\nC = 56.57<\/p>\n

Question 3.
\narc length of \\(\\widehat{A B}\\)
\n\"Big<\/p>\n

Answer:
\narc length of \\(\\widehat{A B}\\) = 26.09<\/p>\n

Explanation:
\narc length of \\(\\widehat{A B}\\) = \\(\\frac { 115 }{ 360 } \\) . 2\u03c0(13)
\n= 26.09<\/p>\n

Question 4.
\nA mountain bike tire has a diameter of 26 inches. To the nearest foot, how far does the tire travel when it makes 32 revolutions?<\/p>\n

Answer:
\nThe tire travels 2613.80 inches.<\/p>\n

Explanation:
\nD = 26 in
\nr = 13 in
\nCircumference C = 2\u03c0(13) = 81.68
\n32 revolutions = 32 x 81.68 = 2613.80<\/p>\n

11.2 Areas of Circles and Sectors<\/h3>\n

Find the area of the blue shaded region.<\/p>\n

Question 5.
\n\"Big<\/p>\n

Answer:
\nArea = \\(\\frac { 240 }{ 360 } \\) . \u03c0(9)\u00b2
\n= 169.64<\/p>\n

Question 6.
\n\"Big<\/p>\n

Answer:
\nArea of shaded region = 11.43<\/p>\n

Explanation:
\nArea of rectangle = 6 x 4 = 24
\nArea of semicircle = \u03c0(2)\u00b2 = 4\u03c0
\nArea of shaded region = 24 – 4\u03c0 = 11.43<\/p>\n

Question 7.
\n\"Big<\/p>\n

Answer:
\nArea of shaded region = 173.13<\/p>\n

Explanation:
\nArea of small region = 27.93 = \\(\\frac { 50 }{ 360 } \\) . \u03c0r\u00b2
\n\u03c0r\u00b2 = 201.096
\nr = 8
\nArea of shaded region = \\(\\frac { 310 }{ 360 } \\) . \u03c0(8)\u00b2
\n= 173.13<\/p>\n

11.3 Areas of Polygons<\/h3>\n

Find the area of the kite or rhombus.<\/p>\n

Question 8.
\n\"Big
\nAnswer:
\nArea = 65<\/p>\n

Explanation:
\nArea = \\(\\frac { 1 }{ 4 } \\)(d\u2081d\u2082)
\nA = \\(\\frac { 1 }{ 4 } \\)(13 x 20)
\nA = 65<\/p>\n

Question 9.
\n\"Big
\nAnswer:
\nArea = 48<\/p>\n

Explanation:
\nArea = \\(\\frac { 1 }{ 4 } \\)(d\u2081d\u2082)
\nA = \\(\\frac { 1 }{ 4 } \\)(16 x 12)
\nA = 48<\/p>\n

Question 10.
\n\"Big
\nAnswer:
\nArea = 52.5<\/p>\n

Explanation:
\nArea = \\(\\frac { 1 }{ 4 } \\)(d\u2081d\u2082)
\nA = \\(\\frac { 1 }{ 4 } \\)(14 x 15)
\nA = 52.5<\/p>\n

Find the area of the regular polygon.<\/p>\n

Question 11.
\n\"Big
\nAnswer:
\nArea = \\(\\frac { 3\u221a3 }{ 2 } \\)a\u00b2
\nA = \\(\\frac { 3\u221a3 }{ 2 } \\)(8.8)\u00b2
\nA = 201.195<\/p>\n

Question 12.
\n\"Big
\nAnswer:
\nArea = \\(\\frac { 1 }{ 2 } \\)(n . a. s)
\nA = \\(\\frac { 1 }{ 2 } \\)(9 . 5.2 . 7.6)
\nA = 117.84<\/p>\n

Question 13.
\n\"Big
\nAnswer:
\nArea = \\(\\frac { 1 }{ 2 } \\)(n . a. s)
\nA = \\(\\frac { 1 }{ 2 } \\)(5 . 4 .\u00a0 3.3)
\nA = 33<\/p>\n

Question 14.
\nA platter is in the shape of a regular octagon with an apothem of 6 inches. Find the area of the platter.<\/p>\n

Answer:
\nArea = \\(\\frac { 1 }{ 2 } \\)(n . a. s)
\nA = \\(\\frac { 1 }{ 2 } \\)(8 . 6 . sin 45)
\nA = 16.97<\/p>\n

11.4 Three-Dimensional Figures<\/h3>\n

Sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.<\/p>\n

Question 15.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 16.
\n\"Big
\nAnswer:
\n\"Big<\/p>\n

Question 17.
\n\"Big
\nAnswer:<\/p>\n

Describe the cross section formed by the intersection of the plane and the solid.<\/p>\n

Question 18.
\n\"Big
\nAnswer:
\nThe cross section is a rectangle.<\/p>\n

Question 19.
\n\"Big
\nAnswer:
\nThe cross-section is a square.<\/p>\n

Question 20.
\n\"Big
\nAnswer:
\nthe cross-section is a triangle.<\/p>\n

11.5 Volumes of Prisms and Cylinders<\/h3>\n

Find the volume of the solid.<\/p>\n

Question 21.
\n\"Big<\/p>\n

Answer:
\nVolume = lbh
\nV = 3.6 x 2.1 x 1.5 = 113.4 m\u00b3<\/p>\n

Question 22.
\n\"Big<\/p>\n

Answer:
\nVolume = \u03c0r\u00b2h
\nV = \u03c0(2)\u00b2 x 8 = 100.53 mm\u00b3<\/p>\n

Question 23.
\n\"Big<\/p>\n

Answer:
\nPentagon area = 6.88
\nVolume = Area x height
\nV = 6.88 x 4 = 27.52 yd\u00b3<\/p>\n

11.6 Volumes of Pyramids<\/h3>\n

Find the volume of the pyramid.<\/p>\n

Question 24.
\n\"Big<\/p>\n

Answer:
\nVolume V = Base area x height\/3
\nBase Area = 9\u00b2 = 81
\nV = 81 x 7\/3 = 189 ft\u00b3<\/p>\n

Question 25.
\n\"Big
\nAnswer:
\nVolume V = Base area x height\/3
\nBase Area = 4 x 15 = 60
\nV = 60x 20\/3 = 400 yd\u00b3<\/p>\n

Question 26.
\n\"Big<\/p>\n

Answer:
\nVolume V = Base area x height\/3
\nbase area = 18 x 10 = 180
\nV = 180 x 5\/3 = 300 m\u00b3<\/p>\n

Question 27.
\nThe volume of a square pyramid is 60 cubic inches and the height is 15 inches. Find the side length of the square base.<\/p>\n

Answer:
\nThe side length of the square base is 3.46 in<\/p>\n

Explanation:
\nThe volume of a square pyramid is 60 cubic inches
\nV = 60
\ns\u00b2h\/3 = 60
\ns\u00b2 x 15\/3 = 60
\ns\u00b2 = 12
\ns = 3.46<\/p>\n

Question 28.
\nThe volume of a square pyramid is 1024 cubic inches. The base has a side length of 16 inches. Find the height of the pyramid
\nAnswer:
\nThe volume of a square pyramid is 1024 cubic inches
\ns\u00b2h\/3 = 1024
\n16\u00b2h = 3072
\nh = 12<\/p>\n

11.7 Surface Areas and Volumes of Cones<\/h3>\n

Find the surface area and the volume of the cone.<\/p>\n

Question 29.
\n\"Big
\nAnswer:
\nSurface area is 678.58 cm\u00b2
\nvolume is 1017.87 cm\u00b3<\/p>\n

Explanation:
\nSurface area of cone S = \u03c0r\u00b2 + \u03c0rl
\nS = \u03c0 x 9\u00b2 + \u03c0 x 9 x 15
\nS = 678.58
\nVolume of cone = \\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2h)
\nV = \\(\\frac { 1 }{ 3 } \\)(\u03c0 x 9\u00b2 x 12)
\nV = 1017.87<\/p>\n

Question 30.
\n\"Big<\/p>\n

Answer:
\nSurface area is 2513.27 cm\u00b2
\nvolume is 8042.47 cm\u00b3<\/p>\n

Explanation:
\nSurface area of cone S = \u03c0r\u00b2 + \u03c0rl
\nS = \u03c0 x 16\u00b2 + \u03c0 x 16 x 34
\nS = 2513.27
\nVolume of cone = \\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2h)
\nV = \\(\\frac { 1 }{ 3 } \\)(\u03c0 x 16\u00b2 x 30)
\nV = 8042.47<\/p>\n

Question 31.
\n\"Big<\/p>\n

Answer:
\nSurface area is 439.82 m\u00b2
\nvolume is 562.102 m\u00b3<\/p>\n

Explanation:
\nSurface area of cone S = \u03c0r\u00b2 + \u03c0rl
\nS = \u03c0 x 7\u00b2 + \u03c0 x 7 x 13
\nS = 439.82
\nh = \u221a13\u00b2 – 7\u00b2 = 10.95
\nVolume of cone = \\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2h)
\nV = \\(\\frac { 1 }{ 3 } \\)(\u03c0 x 7\u00b2 x 10.95)
\nV = 562.102<\/p>\n

Question 32.
\nA cone with a diameter of 16 centimeters has a volume of 320\u03c0 cubic centimeters. Find the height of the cone.<\/p>\n

Answer:
\nThe height of the cone = 15 cm.<\/p>\n

Explanation:
\nr = 8
\nVolume V = 320\u03c0
\n\\(\\frac { 1 }{ 3 } \\)(\u03c0r\u00b2h) = 320\u03c0
\n\\(\\frac { 1 }{ 3 } \\)(\u03c0 x 8\u00b2 x h) = 320\u03c0
\nh = 15<\/p>\n

11.8 Surface Areas and Volumes of Spheres<\/h3>\n

Find the surface area and the volume of the sphere.<\/p>\n

Question 33.
\n\"Big<\/p>\n

Answer:
\nThe surface area is 615.75 in\u00b2, volume is 1436.75 in\u00b3<\/p>\n

Explanation:
\nSurface area S = 4\u03c0r\u00b2
\nS = 4\u03c0 x 7\u00b2
\nS = 615.75
\nVolume V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 7\u00b3
\nV = 1436.75<\/p>\n

Question 34.
\n\"Big<\/p>\n

Answer:
\nThe surface area is 907.92 ft\u00b2, volume is 2572.44 ft\u00b3<\/p>\n

Explanation:
\nd = 17
\nr = 8.5
\nSurface area S = 4\u03c0r\u00b2
\nS = 4\u03c0 x 8.5\u00b2
\nS = 907.92
\nVolume V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 7\u00b3
\nV = 2572.44<\/p>\n

Question 35.
\n\"Big<\/p>\n

Answer:
\nThe surface area is 2827.43 ft\u00b2, volume is 14137.16 ft\u00b3<\/p>\n

Explanation:
\nC = 30\u03c0
\n2\u03c0r = 30\u03c0
\nr = 15
\nSurface area S = 4\u03c0r\u00b2
\nS = 4\u03c0 x 15\u00b2
\nS = 2827.43
\nVolume V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 15\u00b3
\nV = 14137.16<\/p>\n

Question 36.
\nThe shape of Mercury can be approximated by a sphere with a diameter of 4880 kilometers. Find the surface area and the volume of Mercury.<\/p>\n

Answer:
\nThe surface area and the volume of Mercury is 23814400\u03c0, 19369045330\u03c0<\/p>\n

Explanation:
\nd = 4880
\nr = 2440
\nSurface area S = 4\u03c0r\u00b2
\nS = 4\u03c0 x 2440\u00b2 = 23814400\u03c0
\nVolume V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 2440\u00b3
\nV = 19369045330\u03c0<\/p>\n

Question 37.
\nA solid is composed of a cube with a side length of 6 meters and a hemisphere with a diameter of 6 meters. Find the volume of the composite solid.<\/p>\n

Answer:
\nVolume of the composite solid = 272.52<\/p>\n

Explanation:
\nVolume of cube = a\u00b3
\n= 6\u00b3 = 216
\nVolume of hemisphere = \\(\\frac { 4 }{ 6 } \\)\u03c0r\u00b3
\n= \\(\\frac { 4 }{ 6 } \\)\u03c0 x 3\u00b3 = 18\u03c0
\nVolume of the composite solid = 216 + 18\u03c0 = 272.52<\/p>\n

Circumference, Area, and Volume Test<\/h3>\n

Find the volume of the solid.<\/p>\n

Question 1.
\n\"Big<\/p>\n

Answer:
\nVolume = 2577.29 m\u00b3<\/p>\n

Explanation:
\nVolume = \\(\\frac { 3\u221a3 }{ 2 } \\)a\u00b2h
\n= \\(\\frac { 3\u221a3 }{ 2 } \\) x 8\u00b2 x 15.5
\n= 2577.29<\/p>\n

Question 2.
\n\"Big<\/p>\n

Answer:
\nVolume is 17.157 ft\u00b3<\/p>\n

Explanation:
\nd = 3.2
\nr = 1.6
\nVolume V = \\(\\frac { 4 }{ 3 } \\)\u03c0r\u00b3
\nV = \\(\\frac { 4 }{ 3 } \\)\u03c0 x 1.6\u00b3
\nV = 17.157<\/p>\n

Question 3.
\n\"Big<\/p>\n

Answer:
\nVolume of sloid = 402.11 m\u00b3<\/p>\n

Explanation:
\nVolume of cone = \\(\\frac { 1 }{ 3 } \\)\u03c0r\u00b2h
\n= \\(\\frac { 1 }{ 3 } \\)\u03c0 x 4\u00b2 x 3
\n= 50.26
\nVolume of cylinder = \u03c0r\u00b2h
\n= \u03c0 x 4\u00b2 x 6 = 301.59
\nVolume of sloid = 2(50.26) + 301.59 = 402.11<\/p>\n

Question 4.
\n\"Big<\/p>\n

Answer:
\nVolume of solid = 106.66<\/p>\n

Explanation:
\nVolume of rectangular box = 5 x 2 x 8 = 80
\nVolume of pyramid = 80\/3 = 26.66
\nVolume of solid = 80 + 26.66 = 106.66<\/p>\n

Find the indicated measure.<\/p>\n

Question 5.
\ncircumference of \u2299F
\n\"Big<\/p>\n

Answer:
\ncircumference of \u2299F is 109.71 in<\/p>\n

Explanation:
\n64 = \\(\\frac { 210 }{ 360 } \\) \u2022 C
\nC = 109.7<\/p>\n

Question 6.
\nm\\(\\widehat{G H}\\)
\n\"Big<\/p>\n

Answer:
\nm\\(\\widehat{G H}\\) = 74.27<\/p>\n

Explanation:
\n35 = \\(\\frac { x }{ 360 } \\) \u2022 2\u03c0 x 27
\nx = 74.27<\/p>\n

Question 7.
\narea of shaded sector
\n\"Big<\/p>\n

Answer:
\nArea is 142.41 in\u00b2<\/p>\n

Explanation:
\nArea = \\(\\frac { 360 – 105 }{ 360 } \\) \u2022 \u03c0 x 8\u00b2
\nArea = 142.41<\/p>\n

Question 8.
\nSketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.
\n\"Big
\nAnswer:<\/p>\n

Question 9.
\nFind the surface area of a right cone with a diameter of 10 feet and a height of 12 feet.<\/p>\n

Answer:
\nThe surface area is 486.7 sq ft<\/p>\n

Explanation:
\nl\u00b2 = r\u00b2 + h\u00b2
\nl\u00b2 = 5\u00b2 + 12\u00b2
\nl = 13
\nSurface area S = \u03c0r\u00b2 + 2\u03c0rl
\nS = \u03c0 x 5\u00b2 + 2\u03c0 x 5 x 13
\nS = 486.7<\/p>\n

Question 10.
\nYou have a funnel with the dimensions shown.
\n\"Big
\na. Find the approximate volume of the funnel.<\/p>\n

Answer:
\nVolume = \\(\\frac { 1 }{ 3 } \\)\u03c0r\u00b2h
\n= \\(\\frac { 1 }{ 3 } \\)\u03c0 x 6\u00b2 x 10
\n= 376.99<\/p>\n

b. You use the funnel to put oil in a ear. Oil flows out of the funnel at a rate of 45 milliliters per second. How long will it take to empty the funnel when it is full of oil? (1 mL = 1 cm3<\/sup>)
\nAnswer:<\/p>\n

c. How long would it take to empty a funnel with a radius of 10 centimeters and a height of 6 centimeters if oil flows out of the funnel at a rate of 45 milliliters per second?
\nAnswer:<\/p>\n

d. Explain why you can claim that the time calculated in part (c) is greater than the time calculated in part (b) without doing any calculations.
\nAnswer:<\/p>\n

Question 11.
\nA water bottle in the shape of a cylinder has a volume of 500 cubic centimeters. The diameter of a base is 7.5 centimeters. What is the height of the bottle? Justify your answer.<\/p>\n

Answer:
\nThe height of the bottle is 11.3 cm<\/p>\n

Explanation:
\nVolume of cylinder = 500
\n\u03c0r\u00b2h = 500
\n\u03c0(3.75)\u00b2h = 500
\nh = 11.3 cm<\/p>\n

Question 12.
\nFind the area of a dodecagon (12 sides) with a side length of 9 inches.<\/p>\n

Answer:
\nArea is 237.31<\/p>\n

Explanation:
\nArea = \\(\\frac { 1 }{ 4 } \\)\u03c0a\u00b2cot(\u03c0\/n)
\n= \\(\\frac { 1 }{ 4 } \\)\u03c0 x 9\u00b2 x cot(\u03c0\/12)
\n= 237.31<\/p>\n

Question 13.
\nIn general, a cardboard fan with a greater area does a better job of moving air and cooling you. The fan shown is a sector of a cardboard circle. Another fan has a radius of 6 centimeters and an intercepted are of 150\u00b0. Which fan does a better job of cooling you?
\n\"Big<\/p>\n

Answer:<\/p>\n

Explanation:<\/p>\n

Circumference, Area, and Volume Cumulative Assessment<\/h3>\n

Question 1.
\nIdentify the shape of the cross section formed by the intersection of the plane and the
\nsolid.<\/p>\n

a.
\n\"Big
\nAnswer:
\nThe cross-section is a trapezoid.<\/p>\n

b.
\n\"Big
\nAnswer:
\nThe cross-section is a pentagon.<\/p>\n

c.
\n\"Big
\nAnswer:
\nThe cross-section is a rectangle.<\/p>\n

Question 2.
\nIn the diagram, \"Big is tangent to \u2299P at Q and \\(\\overline{P Q}\\) is a radius of \u2299P? What must be true about \"Big and \\(\\overline{P Q}\\)? Select all that apply.
\n\"Big
\n\"Big
\nAnswer:
\nPQ is perpendicular to RS.<\/p>\n

Question 3.
\nA crayon can be approximated by a composite solid made from a cylinder and a cone.
\nA crayon box is a rectangular prism. The dimensions of a crayon and a crayon box containing 24 crayons are shown.
\n\"Big
\na. Find the volume of a crayon.
\nAnswer:
\nThe volume of a crayon = \u03c0r\u00b2h + \\(\\frac { 1 }{ 3 } \\)\u03c0r\u00b2h
\n= \u03c0 x 4.25\u00b2 x 80 + \\(\\frac { 1 }{ 3 } \\)\u03c0 x 3.25\u00b2 x 10
\n= 4539.6 + 110.61
\n= 4650.21 mm\u00b3<\/p>\n

b. Find the amount of space within the crayon box not taken up by the crayons.
\nAnswer:
\nVolume of box = 94 x 28 x 71 = 186872
\nThe volume of a crayon = 4650.21
\nRemaining space = 186872 – 24 x 4650.21
\n= 75266.96<\/p>\n

Question 4.
\nWhat is the equation ol the line passing through the point (2, 5) that is parallel to the line x + \\(\\frac{1}{2}\\)y = – 1?
\n(A) y = – 2x + 9
\n(B) y = 2x + 1
\n(C) y = \\(\\frac{1}{2}\\)x + 4
\n(D) y = –\\(\\frac{1}{2}\\)x + 6
\nAnswer:
\n(A) y = – 2x + 9<\/p>\n

Explanation:
\nx + \\(\\frac{1}{2}\\)y = – 1
\ny = -2 – 2x
\nThe slope of the line is -2
\nThe euation of line is y – 5 = -2(x – 2)
\ny – 5 = -2x + 4
\ny = -2x + 9<\/p>\n

Question 5.
\nThe top of the Washington Monument in Washington, D.C., is a square pyramid, called a pyramidion. What is the volume of the pyramidion?
\n\"Big
\n(A) 22,019.63 ft3<\/sup>
\n(B) 172,006.91 ft3<\/sup>
\n(C) 66,058.88 ft3<\/sup>
\n(D) 207,530.08 ft3<\/sup><\/p>\n

Answer:
\n(A) 22,019.63 ft3<\/sup><\/p>\n

Explanation:
\nVolume = a\u00b2\\(\\frac { h }{ 3 } \\)
\n= 34.5\u00b2 x \\(\\frac { 55.5 }{ 3 } \\)
\n= 22019.62<\/p>\n

Question 6.
\nProve or disprove that the point (1, \u221a3 ) lies on the circle centered at the origin and containing the point (0, 2).
\nAnswer:
\nWe consider the circle centered at the origin and containing the point (0, 2).
\nTherefore, we canconclude that rdaius is 2 and points be (0, 0), (1, \u221a3)
\ndistance = \u221a(1 – 0)\u00b2 + (\u221a3 – 0)\u00b2 = 2
\nAs radius and distance are same. The point B(1, \u221a3) lies on the circle.<\/p>\n

Question 7.
\nYour friend claims that the house shown can be described as a composite solid made from a rectangular prism and a triangular prism. Do you support your friend\u2019s claim? Explain your reasoning.
\n\"Big
\nAnswer:
\nYes<\/p>\n

Question 8.
\nThe diagram shows a square pyramid and a cone. Both solids have the same height, h, and the base of the cone has radius r. According to Cavalieri\u2019s Principle, the solids will have the same volume if the square base has sides of length ______ .
\n\"Big
\nAnswer:
\n2r\/\u221a2<\/p>\n

Explanation:
\nVolume of square pyramid = a\u00b2\\(\\frac { h }{ 3 } \\)
\nSquare diagonal = \u221a2a
\nradius = \u221a2a\/2
\na = 2r\/\u221a2
\nVolume of cone = \\(\\frac { 1 }{ 3 } \\)\u03c0r\u00b2h<\/p>\n

Question 9.
\nAbout 19,400 people live in a region with a 5-mile radius. Find the population density in people per square mile.
\n\"Big<\/p>\n

Answer:
\nThe number of people per square mile is 247<\/p>\n

Explanation:
\nS = \u03c0r\u00b2
\n= \u03c0 x 5\u00b2 = 78.5
\nNumber of people per square mile = 19400\/78.5 = 247<\/p>\n","protected":false},"excerpt":{"rendered":"

High School Students who are seeking homework help can refer to this Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume Ultimate Guide. By using this ultimate preparation material, you all can easily understand and learn the concepts covered in chapter 11 Circumference, Area, and Volume. Enhance your subject knowledge and score high … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":89440,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[4],"tags":[],"jetpack_featured_media_url":"https:\/\/bigideasmathanswers.com\/wp-content\/uploads\/2021\/04\/Big-Ideas-Math-Geometry-Answers-Chapter-11-Circumference-Area-and-Volume.png","_links":{"self":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/50904"}],"collection":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/comments?post=50904"}],"version-history":[{"count":1,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/50904\/revisions"}],"predecessor-version":[{"id":90297,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/50904\/revisions\/90297"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media\/89440"}],"wp:attachment":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media?parent=50904"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/categories?post=50904"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/tags?post=50904"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}