Geometry Section 11-4
- 2. Essential Questions How do you find surface areas of spheres? How do you find volumes of spheres?
- 3. Vocabulary 1. Great Circle: 2. Pole: 3. Hemisphere:
- 4. Vocabulary 1. Great Circle: 2. Pole: 3. Hemisphere: A circle formed when a plane intersects a sphere and the circle has the same center as the sphere
- 5. Vocabulary 1. Great Circle: 2. Pole: 3. Hemisphere: A circle formed when a plane intersects a sphere and the circle has the same center as the sphere The endpoints of the diameter of a great circle
- 6. Vocabulary 1. Great Circle: 2. Pole: 3. Hemisphere: A circle formed when a plane intersects a sphere and the circle has the same center as the sphere The endpoints of the diameter of a great circle One of the two congruent halves of a sphere created by a great circle
- 7. Formulas Surface Area of a Sphere: Volume of a Sphere:
- 8. Formulas Surface Area of a Sphere: SA = 4πr2 Volume of a Sphere:
- 9. Formulas Surface Area of a Sphere: SA = 4πr2 Volume of a Sphere: V = 4 3 πr3
- 10. Example 1 Find the surface area of the sphere.
- 11. Example 1 Find the surface area of the sphere. SA = 4πr2
- 12. Example 1 Find the surface area of the sphere. SA = 4πr2 SA = 4π(4.5)2
- 13. Example 1 Find the surface area of the sphere. SA = 4πr2 SA = 4π(4.5)2 SA ≈ 254.47 in2
- 14. Example 2 Find the surface area of a hemisphere with a diameter of 8 mm to the nearest hundredth.
- 15. Example 2 Find the surface area of a hemisphere with a diameter of 8 mm to the nearest hundredth. SA = 1 2 i 4πr2 + πr2
- 16. Example 2 Find the surface area of a hemisphere with a diameter of 8 mm to the nearest hundredth. SA = 1 2 i 4πr2 + πr2 r = 1 2 d
- 17. Example 2 Find the surface area of a hemisphere with a diameter of 8 mm to the nearest hundredth. SA = 1 2 i 4πr2 + πr2 r = 1 2 d r = 1 2 (8)
- 18. Example 2 Find the surface area of a hemisphere with a diameter of 8 mm to the nearest hundredth. SA = 1 2 i 4πr2 + πr2 r = 1 2 d r = 1 2 (8) r = 4 mm
- 19. Example 2 Find the surface area of a hemisphere with a diameter of 8 mm to the nearest hundredth. SA = 1 2 i 4πr2 + πr2 r = 1 2 d r = 1 2 (8) r = 4 mm SA = 2π(4)2 + π(4)2
- 20. Example 2 Find the surface area of a hemisphere with a diameter of 8 mm to the nearest hundredth. SA = 1 2 i 4πr2 + πr2 r = 1 2 d r = 1 2 (8) r = 4 mm SA = 2π(4)2 + π(4)2 SA ≈ 150.8 mm2
- 21. Example 3 Find the surface area of a sphere if the circumference of the great circle is 14π inches.
- 22. Example 3 Find the surface area of a sphere if the circumference of the great circle is 14π inches. SA = 4πr2
- 23. Example 3 Find the surface area of a sphere if the circumference of the great circle is 14π inches. SA = 4πr2 C = 2πr
- 24. Example 3 Find the surface area of a sphere if the circumference of the great circle is 14π inches. SA = 4πr2 C = 2πr 14π = 2πr
- 25. Example 3 Find the surface area of a sphere if the circumference of the great circle is 14π inches. SA = 4πr2 C = 2πr 14π = 2πr 2π 2π
- 26. Example 3 Find the surface area of a sphere if the circumference of the great circle is 14π inches. SA = 4πr2 C = 2πr 14π = 2πr 2π 2π r = 7 in.
- 27. Example 3 Find the surface area of a sphere if the circumference of the great circle is 14π inches. SA = 4πr2 C = 2πr 14π = 2πr 2π 2π r = 7 in. SA = 4π(7)2
- 28. Example 3 Find the surface area of a sphere if the circumference of the great circle is 14π inches. SA = 4πr2 C = 2πr 14π = 2πr 2π 2π r = 7 in. SA = 4π(7)2 SA ≈ 615.75 in2
- 29. Example 4 Find the volume of the sphere rounded to the nearest hundredth.
- 30. Example 4 Find the volume of the sphere rounded to the nearest hundredth. V = 4 3 πr3
- 31. Example 4 Find the volume of the sphere rounded to the nearest hundredth. V = 4 3 πr3 V = 4 3 π(4.5)3
- 32. Example 4 Find the volume of the sphere rounded to the nearest hundredth. V = 4 3 πr3 V = 4 3 π(4.5)3 V ≈ 381.70 in3
- 33. Example 5 Find the volume of a hemisphere with a diameter of 6 feet to the nearest hundredth.
- 34. Example 5 Find the volume of a hemisphere with a diameter of 6 feet to the nearest hundredth. V = 1 2 i 4 3 πr3
- 35. Example 5 Find the volume of a hemisphere with a diameter of 6 feet to the nearest hundredth. V = 1 2 i 4 3 πr3 r = 1 2 d
- 36. Example 5 Find the volume of a hemisphere with a diameter of 6 feet to the nearest hundredth. V = 1 2 i 4 3 πr3 r = 1 2 d r = 1 2 (6)
- 37. Example 5 Find the volume of a hemisphere with a diameter of 6 feet to the nearest hundredth. V = 1 2 i 4 3 πr3 r = 1 2 d r = 1 2 (6) r = 3 ft
- 38. Example 5 Find the volume of a hemisphere with a diameter of 6 feet to the nearest hundredth. V = 1 2 i 4 3 πr3 r = 1 2 d r = 1 2 (6) r = 3 ft V = 2 3 π(3)3
- 39. Example 5 Find the volume of a hemisphere with a diameter of 6 feet to the nearest hundredth. V = 1 2 i 4 3 πr3 r = 1 2 d r = 1 2 (6) r = 3 ft V = 2 3 π(3)3 V ≈ 56.55 ft3
- 40. Example 6 Find the volume of a sphere that has a great circle with an area of 9π square feet.
- 41. Example 6 Find the volume of a sphere that has a great circle with an area of 9π square feet. V = 4 3 πr3
- 42. Example 6 Find the volume of a sphere that has a great circle with an area of 9π square feet. V = 4 3 πr3 A = πr2
- 43. Example 6 Find the volume of a sphere that has a great circle with an area of 9π square feet. V = 4 3 πr3 A = πr2 9π = πr2
- 44. Example 6 Find the volume of a sphere that has a great circle with an area of 9π square feet. V = 4 3 πr3 A = πr2 9π = πr2 π π
- 45. Example 6 Find the volume of a sphere that has a great circle with an area of 9π square feet. V = 4 3 πr3 A = πr2 9π = πr2 π π 9 = r2
- 46. Example 6 Find the volume of a sphere that has a great circle with an area of 9π square feet. V = 4 3 πr3 A = πr2 9π = πr2 π π r = 3 ft. 9 = r2
- 47. Example 6 Find the volume of a sphere that has a great circle with an area of 9π square feet. V = 4 3 πr3 A = πr2 9π = πr2 π π r = 3 ft. 9 = r2 V = 4 3 π(3)3
- 48. Example 6 Find the volume of a sphere that has a great circle with an area of 9π square feet. V = 4 3 πr3 A = πr2 9π = πr2 π π r = 3 ft. 9 = r2 V = 4 3 π(3)3 V ≈ 113.10 ft3
- 49. Problem Set
- 50. Problem Set p. 822 #1-9 all “With self-discipline most anything is possible.” - Theodore Roosevelt