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Math 1332 Section 10A

Math 1332 Section 10A

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Matthew Sievers

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17 Slides • 12 Questions

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Math 1332 Section 10A

Fundamentals of Geometry

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45. Find the perimeter and area.


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45.Find the perimeter.

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45.Find the area.

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 47.

A picture window has a length of 8 feet and a height of 6 feet, with a semicircular cap on each end. How much metal trim is needed for the perimeter of the entire window, and how much glass is needed for the opening of the window?

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47. How much metal trim is needed for the perimeter of the entire window?

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47. How much glass is needed for the opening of the window?

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49.

Refer to Figure 10.14, showing the region to be covered with plywood under a set of stairs. Suppose that the stairs rise at a steeper angle and are 11 feet tall. What is the area of the region to be covered in that case?

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49.

  • Refer to Figure 10.14, showing the region to be covered with plywood under a set of stairs. Suppose that the stairs rise at a steeper angle and are 11 feet tall. What is the area of the region to be covered in that case?

  • A = 1/2 (12) (11)

  • = 66 square feet = 66 ft266\ ft^2  

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51.

 A parking lot is shaped like a parallelogram and bounded on four sides by streets, as shown in Figure 10.23. How much asphalt (in square yards) is needed to pave the parking lot?

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51. A parking lot is shaped like a parallelogram and bounded on four sides by streets, as shown in Figure 10.23. How much asphalt (in square yards) is needed to pave the parking lot?

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59. Water Reservoir. The water reservoir for a city is shaped like a rectangular prism 250 meters long, 60 meters wide, and 12 ­meters deep. At the end of the day, the reservoir is 70% full. How much water must be added overnight to fill the reservoir?

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59.

  • Find the area of the whole figure.

  • 12 meters * 250 meters * 60 meters

  • = 180,000 cubic meters

  • Find 30% of 180,000 cubic meters.

  • (0.30)(180,000) = 54,000 cubic meters =

     54,000 m354,000\ m^3  

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Suppose you build an architectural model of a new concert hall using a scale factor of 30.


63. How will the surface area of the actual concert hall compare to the surface area of the scale model?


Response: The surface area of the actual concert hall will be __ times larger than the surface area of the scale model.

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Suppose you magically quadrupled in size—that is, your height, width, and depth increased by a factor of 4.

71. By what factor has your weight increased?

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Consider a person named Sam, who is 20% taller than you but proportioned in exactly the same way. (That is, Sam looks like a larger version of you.)

73. What size is your waist? (Assume your waist is 39.7 in.) What size is Sam’s waist?

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Open Ended

Squirrels and humans are both mammals that stay warm through metabolism that takes place in the body volume. Mammals must constantly generate internal heat to replace the heat that they lose through the surface area of their skin.


75. In general terms, how does the surface-area-to-volume ratio of a squirrel compare to that of a human being?

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Open Ended

Squirrels and humans are both mammals that stay warm through metabolism that takes place in the body volume. Mammals must constantly generate internal heat to replace the heat that they lose through the surface area of their skin.


76. Which animal must maintain a higher rate of metabolism to replace the heat lost through the skin: squirrels or humans? Based on your answer, which animal would you expect to eat more food in proportion to its body weight each day? Explain.

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Fill in the Blank

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84. Backyard. Figure 10.26 shows the layout of a backyard that is to be seeded with grass except for the patio and flower garden. What is the area of the region that is to be seeded with grass? (two decimal places)

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84.

  • Full rectangle 20*10 = 200

  • Garden = 16

  • Half circle = (3.14)(5^2) (1/2) = (3.14)(25)(1/2) = 39.25

  • Full rectangle - garden - patio = 200 - 16 - 39.25

  • = 144.75 square meters

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85.

  • Human Lung. The human lung has approximately 300 million nearly spherical air sacs (alveoli), each with a diameter of about 1/3 millimeter. The key feature of the air sacs is their surface area, because on their surfaces gas is exchanged between the bloodstream and the air.

  • a. What is the total surface area of the air sacs? What is the total volume of the air sacs?

  •  SA = 4πr2SA\ =\ 4\pi r^2  

  •  SA = 4 ( 3.14)(16)20.3489 mm2SA\ =\ 4\ \left(\ 3.14\right)\left(\frac{1}{6}\right)^2\approx0.3489\ mm^2  

  • Multiply by 300 million    1.05 ×108=105,000,000 mm2\approx\ 1.05\ \times10^8=105,000,000\ mm^2 

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85.

  • Human Lung. The human lung has approximately 300 million nearly spherical air sacs (alveoli), each with a diameter of about 1/3 millimeter. The key feature of the air sacs is their surface area, because on their surfaces gas is exchanged between the bloodstream and the air.

  • a. What is the total surface area of the air sacs? What is the total volume of the air sacs?

  •  V=43πr3V=\frac{4}{3}\pi r^3  

  •  V=43(3.14)(16)30.0193827V=\frac{4}{3}\left(3.14\right)\left(\frac{1}{6}\right)^3\approx0.0193827  

  • Multiply by 300 million:  5.81×106 mm3\approx5.81\times10^6\ mm^3  

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85.

  • The human lung has approximately 300 million nearly spherical air sacs (alveoli), each with a diameter of about 1/3 millimeter. The key feature of the air sacs is their surface area, because on their surfaces gas is exchanged between the bloodstream and the air.

  • b. Suppose a single sphere were made that had the same volume as the total volume of the air sacs. What would be the radius and surface area of such a sphere? How would this surface area compare to that of the air sacs?

  •  V=43πr3V=\frac{4}{3}\pi r^3  

  •  5.8×106=43πr35.8\times10^6=\frac{4}{3}\pi r^3 

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85. b.)

  •  5.8×106÷(43π) = r35.8\times10^6\div\left(\frac{4}{3}\pi\right)\ =\ r^3  

  •  r3=1.3846×106r^3=1.3846\times10^6  

  •  r=31.3846×106r=^3\sqrt{1.3846\times10^6}  

  •  =111.46 mm  112 mm=111.46\ mm\ \approx\ 112\ mm  is the radius of big sphere

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85. b.)

  •  SA=4πr2SA=4\pi r^2  

  •  SA=4(3.14)(112)2SA=4\left(3.14\right)\left(112\right)^2  

  •  =157552.64 1.58×105 mm2=157552.64\ \approx1.58\times10^5\ mm^2  is the surface area of big sphere

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85. b.)

  •  Surface Area of Big SphereSurface Area of Little Sphere=1.05×1081.58×105\frac{Surface\ Area\ of\ Big\ Sphere}{Surface\ Area\ of\ Little\ Sphere}=\frac{1.05\times10^8}{1.58\times10^5}  

  •  664.56 670 times greater\approx664.56\ \approx670\ times\ greater  

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85. c.)

  • If a single sphere had the same surface area as the total surface area of the air sacs, what would be its radius? Based on your results, comment on the design of the human lung.

  •  SA=4πr2SA=4\pi r^2  

  •  1.05×108=4πr21.05\times10^8=4\pi r^2  

  •  r2=(1.05×108)4π8.36×106r^2=\frac{\left(1.05\times10^8\right)}{4\pi}\approx8.36\times10^6  

  •  r=8.36×1062891.34 mmr=\sqrt{8.36\times10^6}\approx2891.34\ mm  

  • 2.89 meters / for their relatively small volume, the air sacs have a very large surface area

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87.

  • The English Channel Tunnel, or “Chunnel,” runs a distance of 50.45 kilometers from Dover, England, to Calais, France. The Chunnel actually consists of three individual, adjacent tunnels, each shaped like a half-cylinder with a radius of 4 meters (the height of the tunnel). How much earth (volume) was removed to build the entire Chunnel?

  •  V=πr2hV=\pi r^2h      (convert 50.45 km to 50450 m)

  •  V=(3.14)(4)2(50450)2,543,608V=\left(3.14\right)\left(4\right)^2\left(50450\right)\approx2,543,608  

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87.

  • There are 3 tunnels. Multiply by 3 to get 7,603,824 cubic meters.

  • Each tunnels is a half, so divide by 2. (Alternatively, one could multiply by 3/2)

  • 3,801,912 = 3.8 x 106 cubic meters

Math 1332 Section 10A

Fundamentals of Geometry

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