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Surface Areas and Volumes of Spheres

For Exercises 1–3, find the surface area. Leave your answer in terms of . 1. a sphere whose diameter is 13 cm 2. a ball whose circumference is 19 mm 3. a sphere whose volume is 288 in. 3 For Exercises 4 and 5, find the volume. Round your answer to the nearest whole number.

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Surface Areas and Volumes of Spheres

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  1. For Exercises 1–3, find the surface area. Leave your answer in terms of . 1. a sphere whose diameter is 13 cm 2. a ball whose circumference is 19 mm 3. a sphere whose volume is 288 in.3 For Exercises 4 and 5, find the volume. Round your answer to the nearest whole number. 4. sphere with radius 5 m 5. sphere with diameter 7 yd 169 cm2 361 mm2 144 in.2 Surface Areas and Volumes of Spheres Lesson 11-6 Lesson Quiz 524 m3 180 yd3 11-7

  2. Areas and Volumes of Similar Solids Lesson 11-7 Check Skills You’ll Need (For help, go to Lessons 8-2 and 11-4.) Are the figures similar? Explain. Include the similarity ratio. 1. two squares, one with 3-in. sides and the other with 1-in. sides 2. two isosceles right triangles, one with a 3-cm hypotenuse and the other a 1-cm leg Find the volume of each space figure. 3. cube with a 3-in. edge 4. 3-m by 5-m by 9-m rectangular prism 5. cylinder with radius 4 cm and height 8 cm Check Skills You’ll Need 11-7

  3. 1. Yes, four pairs of angles are congruent and the corresponding sides are proportional with a similarity ratio of . 2. Yes, three pairs of angles are congruent and the corresponding sides are proportional with a similarity ratio of . 3. To find the volume, use the formula V = s3: V = (3)3 = (3)(3)(3) = 27 in.3 4. To find the volume, use the formula V = bhl: V = (3)(5)(9) = 135 m3. 3 1 3 2 3 2 2 = 5. The volume of the cylinder is V = Bh, where B is the area of the base. The base is a circle whose area is 16 .So, V = 16 (8) = 128  402.1 cm3. Areas and Volumes of Similar Solids Lesson 11-7 Check Skills You’ll Need Solutions 11-7

  4. Areas and Volumes of Similar Solids Lesson 11-7 Notes Similar solids have the same shape, and all their corresponding dimensions are proportional. The ratio of corresponding linear dimensions of two similar solids is the similarity ratio. Any two cubes are similar, as are any two spheres. 11-7

  5. Areas and Volumes of Similar Solids Lesson 11-7 Notes 11-7

  6. 8 26 8 26 3 9 The ratio of the radii is , and the ratio of the height is . / 3 9 The cones are not similar because = . Areas and Volumes of Similar Solids Lesson 11-7 Additional Examples Identifying Similar Solids Are the two solids similar? If so, give the similarity ratio. Both solid figures have the same shape. Check that the ratios of the corresponding dimensions are equal. Quick Check 11-7

  7. Find the similarity ratio of two similar cylinders with surface areas of 98 ft2 and 2 ft2. 98 2 = The ratio of the surface areas is a2 : b2. 49 1 a2 b2 a2 b2 = Simplify. a b 7 1 = Take the square root of each side. Areas and Volumes of Similar Solids Lesson 11-7 Additional Examples Finding the Similarity Ratio Use the ratio of the surface areas to find the similarity ratio. The similarity ratio is 7 : 1. Quick Check 11-7

  8. 48 162 = The ratio of the volumes is a3 : b3. 8 27 = Simplify. a b a3 b3 a3 b3 2 3 = Take the cube root of each side. Areas and Volumes of Similar Solids Lesson 11-7 Additional Examples Using a Similarity Ratio Two similar square pyramids have volumes of 48 cm3 and 162 cm3. The surface area of the larger pyramid is 135 cm2. Find the surface area of the smaller pyramid. Step 1: Use the ratio of the volumes to find the similarity ratio. 11-7

  9. 22 32 = The ratio of the surface areas is a2 : b2. S1 S2 S1 S2 4 9 = Simplify. S1 135 4 9 = Substitute 135 for S2, the surface area of the larger pyramid. Quick Check 4 9 S1= • 135 Solve for S1. Areas and Volumes of Similar Solids Lesson 11-7 Additional Examples (continued) Step 2: Use the similarity ratio to find the surface area S1 of the smaller pyramid. S1= 60 Simplify. The surface area of the smaller pyramid is 60 cm2. 11-7

  10. Areas and Volumes of Similar Solids Lesson 11-7 Additional Examples Real-World Connection A box of detergent shaped like a rectangular prism is 6 in. high and holds 3.25 lb of detergent. How much detergent would a similar box that is 8 in. tall hold? Round your answer to the nearest tenth. The ratio of the heights of the boxes is 6 : 8, so the similarity ratio is 6 : 8, or 3 : 4 in simplest terms. Because the weights are proportional to the volumes, the ratio of the weights equals the cube of their similarity ratio: 33 : 43, or 27 : 64. 11-7

  11. 27 64 3.25x = Let x  the weight of the larger box of detergent. x 7.7037037 Divide each side by 27. Areas and Volumes of Similar Solids Lesson 11-7 Additional Examples (continued) 27x= 64  3.25 Cross-Product Property. 27x= 208 Simplify. The box that is 8 in. tall would hold about 7.7 lb of detergent. Quick Check 11-7

  12. For Exercises 1 and 2, are the two solids similar? If so, give the similarity ratio of the smaller figure to the larger figure. 1.2. 3. Find the similarity ratio of two spheres with volumes of 20 m3 and 160 m3. 4. The volumes of two similar solids are 54 ft3 and 250 ft3. The surface area of the smaller solid is 45 ft2. Find the surface area of the larger solid. 5. A solid chocolate rabbit is 6 in. high and weighs 0.25 lb. A similar chocolate rabbit is 12 in. high. How much does it weigh? Areas and Volumes of Similar Solids Lesson 11-7 Lesson Quiz yes; 2 : 3 no 1 : 2 125 ft2 2 lb 11-7

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