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Holt Geometry 10-8 Spheres Learn and apply the formula for the volume of a sphere. Learn and apply the formula for the surface area of a sphere. Objectives.

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Presentation on theme: "Holt Geometry 10-8 Spheres Learn and apply the formula for the volume of a sphere. Learn and apply the formula for the surface area of a sphere. Objectives."— Presentation transcript:

1 Holt Geometry 10-8 Spheres Learn and apply the formula for the volume of a sphere. Learn and apply the formula for the surface area of a sphere. Objectives

2 Holt Geometry 10-8 Spheres sphere center of a sphere radius of a sphere hemisphere great circle Vocabulary

3 Holt Geometry 10-8 Spheres A sphere is the locus of points in space that are a fixed distance from a given point called the center of a sphere. A radius of a sphere connects the center of the sphere to any point on the sphere. A hemisphere is half of a sphere. A great circle divides a sphere into two hemispheres

4 Holt Geometry 10-8 Spheres

5 Holt Geometry 10-8 Spheres Example 1A: Finding Volumes of Spheres Find the volume of the sphere. Give your answer in terms of . = 2304 in 3 Simplify. Volume of a sphere.

6 Holt Geometry 10-8 Spheres Example 1B: Finding Volumes of Spheres Find the diameter of a sphere with volume 36,000 cm 3. Substitute 36,000  for V. 27,000 = r 3 r = 30 d = 60 cm d = 2r Take the cube root of both sides. Volume of a sphere.

7 Holt Geometry 10-8 Spheres Example 1C: Finding Volumes of Spheres Find the volume of the hemisphere. Volume of a hemisphere Substitute 15 for r. = 2250 m 3 Simplify.

8 Holt Geometry 10-8 Spheres Check It Out! Example 1 Find the radius of a sphere with volume 2304 ft 3. Volume of a sphere Substitute for V. r = 12 ftSimplify.

9 Holt Geometry 10-8 Spheres Example 2: Sports Application A sporting goods store sells exercise balls in two sizes, standard (22-in. diameter) and jumbo (34- in. diameter). How many times as great is the volume of a jumbo ball as the volume of a standard ball? standard ball: jumbo ball: A jumbo ball is about 3.7 times as great in volume as a standard ball.

10 Holt Geometry 10-8 Spheres

11 Holt Geometry 10-8 Spheres Example 3A: Finding Surface Area of Spheres Find the surface area of a sphere with diameter 76 cm. Give your answers in terms of . S = 4r 2 S = 4(38) 2 = 5776 cm 2 Surface area of a sphere

12 Holt Geometry 10-8 Spheres Example 3B: Finding Surface Area of Spheres Find the volume of a sphere with surface area 324 in 2. Give your answers in terms of . Substitute 324 for S. 324 = 4r 2 r = 9Solve for r. Substitute 9 for r. The volume of the sphere is 972 in 2. S = 4r 2 Surface area of a sphere

13 Holt Geometry 10-8 Spheres Example 3C: Finding Surface Area of Spheres Find the surface area of a sphere with a great circle that has an area of 49 mi 2. Substitute 49  for A. 49 = r 2 r = 7 Solve for r. S = 4r 2 = 4(7) 2 = 196 mi 2 Substitute 7 for r. A = r 2 Area of a circle

14 Holt Geometry 10-8 Spheres Check It Out! Example 3 Find the surface area of the sphere. Substitute 25 for r. S = 2500 cm 2 S = 4r 2 S = 4(25) 2 Surface area of a sphere

15 Holt Geometry 10-8 Spheres Example 4: Exploring Effects of Changing Dimensions The radius of the sphere is multiplied by. Describe the effect on the volume. original dimensions: radius multiplied by : Notice that. If the radius is multiplied by, the volume is multiplied by, or.

16 Holt Geometry 10-8 Spheres Check It Out! Example 4 The radius of the sphere is divided by 3. Describe the effect on the surface area. original dimensions:dimensions divided by 3: The surface area is divided by 9. S = 4r 2 = 4(3) 2 = 36 m 3 S = 4r 2 = 4(1) 2 = 4 m 3

17 Holt Geometry 10-8 Spheres Example 5: Finding Surface Areas and Volumes of Composite Figures Find the surface area and volume of the composite figure. Give your answer in terms of . Step 1 Find the surface area of the composite figure. The surface area of the composite figure is the sum of the curved surface area of the hemisphere, the lateral area of the cylinder, and the base area of the cylinder.

18 Holt Geometry 10-8 Spheres Example 5 Continued The surface area of the composite figure is L(cylinder) = 2rh = 2(6)(9) = 108 in 2 B(cylinder) = r 2 = (6) 2 = 36 in 2 72 + 108 + 36 = 216 in 2. Find the surface area and volume of the composite figure. Give your answer in terms of .

19 Holt Geometry 10-8 Spheres Step 2 Find the volume of the composite figure. Example 5 Continued Find the surface area and volume of the composite figure. Give your answer in terms of . The volume of the composite figure is the sum of the volume of the hemisphere and the volume of the cylinder. The volume of the composite figure is 144 + 324 = 468 in 3.

20 Holt Geometry 10-8 Spheres Check It Out! Example 5 Find the surface area and volume of the composite figure. Step 1 Find the surface area of the composite figure. The surface area of the composite figure is the sum of the curved surface area of the hemisphere, the lateral area of the cylinder, and the base area of the cylinder.

21 Holt Geometry 10-8 Spheres Check It Out! Example 5 Continued The surface area of the composite figure is Find the surface area and volume of the composite figure. L(cylinder) = 2rh = 2(3)(5) = 30 ft 2 B(cylinder) = r 2 = (3) 2 = 9 ft 2 18 + 30 + 9 = 57 ft 2.

22 Holt Geometry 10-8 Spheres Step 2 Find the volume of the composite figure. Find the surface area and volume of the composite figure. Check It Out! Example 5 Continued The volume of the composite figure is the volume of the cylinder minus the volume of the hemisphere. V = 45 – 18 = 27 ft 3


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