Surface Areas and Volumes of Spheres LESSON 12–6 Surface Areas and Volumes of Spheres

Five-Minute Check (over Lesson 12–5) TEKS Then/Now New Vocabulary Key Concept: Surface Area of a Sphere Example 1: Surface Area of a Sphere Example 2: Use Great Circles to Find Surface Area Key Concept: Volume of a Sphere Example 3: Volumes of Spheres and Hemispheres Example 4: Real-World Example: Solve Problems Involving Solids Lesson Menu

Find the volume of the cone. Round to the nearest tenth if necessary. A. 134.0 mm3 B. 157.0 mm3 C. 201.1 mm3 D. 402.1 mm3 5-Minute Check 1

Find the volume of the pyramid. Round to the nearest tenth if necessary. A. 36 ft3 B. 125 ft3 C. 180 ft3 D. 270 ft3 5-Minute Check 2

Find the volume of the cone. Round to the nearest tenth if necessary. A. 323.6 ft3 B. 358.1 ft3 C. 382.5 ft3 D. 428.1 ft3 5-Minute Check 3

Find the volume of the pyramid. Round to the nearest tenth if necessary. A. 1314.3 in3 B. 1177.0 in3 C. 1009.4 in3 D. 987.5 in3 5-Minute Check 4

Find the volume of a cone with a diameter of 8 Find the volume of a cone with a diameter of 8.4 meters and a height of 14.6 meters. A. 192.6 m3 B. 237.5 m3 C. 269.7 m3 D. 385.2 m3 5-Minute Check 5

Find the height of a hexagonal pyramid with a base area of 130 square meters and a volume of 650 cubic meters. A. 12 m B. 15 m C. 17 m D. 22 m 5-Minute Check 6

Mathematical Processes G.1(A), G.1(E) Targeted TEKS G.11(C) Apply the formulas for the total and lateral surface area of three-dimensional figures, including prisms, pyramids, cones, cylinders, spheres, and composite figures, to solve problems using appropriate units of measure. G.11(D) Apply the formulas for the volume of three-dimensional figures, including prisms, pyramids, cones, cylinders, spheres, and composite figures, to solve problems using appropriate units of measure. Also addresses G.10(B). Mathematical Processes G.1(A), G.1(E) TEKS

You found surface areas of prisms and cylinders. Find surface areas of spheres. Find volumes of spheres. Then/Now

great circle pole hemisphere Vocabulary

Concept

Find the surface area of the sphere. Round to the nearest tenth. Surface Area of a Sphere Find the surface area of the sphere. Round to the nearest tenth. S = 4r2 Surface area of a sphere = 4(4.5)2 Replace r with 4.5. ≈ 254.5 Simplify. Answer: 254.5 in2 Example 1

Find the surface area of the sphere. Round to the nearest tenth. A. 462.7 in2 B. 473.1 in2 C. 482.6 in2 D. 490.9 in2 Example 1

A. Find the surface area of the hemisphere. Use Great Circles to Find Surface Area A. Find the surface area of the hemisphere. Find half the area of a sphere with the radius of 3.7 millimeters. Then add the area of the great circle. Example 2A

Surface area of a hemisphere Use Great Circles to Find Surface Area Surface area of a hemisphere Replace r with 3.7. ≈ 129.0 Use a calculator. Answer: about 129.0 mm2 Example 2A

Use Great Circles to Find Surface Area B. Find the surface area of a sphere if the circumference of the great circle is 10 feet. First, find the radius. The circumference of a great circle is 2r. So, 2r = 10 or r = 5. Example 2B

S = 4r2 Surface area of a sphere = 4(5)2 Replace r with 5. Use Great Circles to Find Surface Area S = 4r2 Surface area of a sphere = 4(5)2 Replace r with 5. ≈ 314.2 Use a calculator. Answer: about 314.2 ft2 Example 2B

Use Great Circles to Find Surface Area C. Find the surface area of a sphere if the area of the great circle is approximately 220 square meters. First, find the radius. The area of a great circle is r2. So, r2 = 220 or r ≈ 8.4. Example 2C

S = 4r2 Surface area of a sphere ≈ 4(8.4)2 Replace r with 5. Use Great Circles to Find Surface Area S = 4r2 Surface area of a sphere ≈ 4(8.4)2 Replace r with 5. ≈ 886.7 Use a calculator. Answer: about 886.7 m2 Example 2C

A. Find the surface area of the hemisphere. A. 110.8 m2 B. 166.3 m2 C. 169.5 m2 D. 172.8 m2 Example 2A

B. Find the surface area of a sphere if the circumference of the great circle is 8 feet. A. 100.5 ft2 B. 201.1 ft2 C. 402.2 ft2 D. 804.3 ft2 Example 2B

C. Find the surface area of the sphere if the area of the great circle is approximately 160 square meters. A. 320 ft2 B. 440 ft2 C. 640 ft2 D. 720 ft2 Example 2C

Concept

Volumes of Spheres and Hemispheres A. Find the volume a sphere with a great circle circumference of 30 centimeters. Round to the nearest tenth. Find the radius of the sphere. The circumference of a great circle is 2r. So, 2r = 30 or r = 15. Volume of a sphere (15)3 r = 15 ≈ 14,137.2 cm3 Use a calculator. Example 3A

Answer: The volume of the sphere is approximately 14,137.2 cm3. Volumes of Spheres and Hemispheres Answer: The volume of the sphere is approximately 14,137.2 cm3. Example 3A

The volume of a hemisphere is one-half the volume of the sphere. Volumes of Spheres and Hemispheres B. Find the volume of the hemisphere with a diameter of 6 feet. Round to the nearest tenth. The volume of a hemisphere is one-half the volume of the sphere. Volume of a hemisphere r 3 Use a calculator. Answer: The volume of the hemisphere is approximately 56.5 cubic feet. Example 3B

A. Find the volume of the sphere to the nearest tenth. A. 268.1 cm3 B. 1608.5 cm3 C. 2144.7 cm3 D. 6434 cm3 Example 3A

B. Find the volume of the hemisphere to the nearest tenth. A. 3351.0 m3 B. 6702.1 m3 C. 268,082.6 m3 D. 134,041.3 m3 Example 3B

Analyze You know that the volume of the stone is 36,000 cubic inches. Solve Problems Involving Solids ARCHEOLOGY The stone spheres of Costa Rica were made by forming granodiorite boulders into spheres. One of the stone spheres has a volume of about 36,000 cubic inches. What is the diameter of the stone sphere? Analyze You know that the volume of the stone is 36,000 cubic inches. Formulate First use the volume formula to find the radius. Then find the diameter. Example 4

Determine Volume of a sphere Solve Problems Involving Solids Determine Volume of a sphere Replace V with 36,000. Divide each side by Use a calculator to find 2700 ( 1 ÷ 3 ) ENTER 30 The radius of the stone is 30 inches. So, the diameter is 2(30) or 60 inches. Example 4

Justify You can work backward to check the solution. Solve Problems Involving Solids Answer: 60 inches Justify You can work backward to check the solution. If the diameter is 60, then r = 30. If r = 30, then V = cubic inches. The solution is correct.  Evaluate Since the stone was a sphere allows use of the formula for the volume of a sphere to find a solution that closely approximates the actual diameter. Example 4

RECESS The jungle gym outside of Jada’s school is a perfect hemisphere RECESS The jungle gym outside of Jada’s school is a perfect hemisphere. It has a volume of 4,000 cubic feet. What is the diameter of the jungle gym? A. 10.7 feet B. 12.6 feet C. 14.4 feet D. 36.3 feet Example 4

Surface Areas and Volumes of Spheres LESSON 12–6 Surface Areas and Volumes of Spheres