Warm Up #9: The figure shows a hemisphere and a cylinder ...
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5/23/12 1 Holt Geometry 10-8 Spheres Warm Up #9: The figure shows a hemisphere and a cylinder with a cone removed from its interior. The cross section of the hemisphere is a circle with radius: so its area is: r 2 − x 2 A = π ( r 2 − x 2 ) The cross section of the cylinder with the cone removed has an outer radius of: and an inner radius of: so its area is: A = πr 2 − πx 2 = π ( r 2 − x 2 ) r x Holt Geometry 10-8 Spheres I will learn and apply the formulas for the surface area and volume of a sphere. CN#7 Objectives Holt Geometry 10-8 Spheres sphere center of a sphere radius of a sphere hemisphere great circle Vocabulary 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
Transcript of Warm Up #9: The figure shows a hemisphere and a cylinder ...
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Holt Geometry
10-8 Spheres
Warm Up #9: The figure shows a hemisphere and a cylinder with a cone removed from its interior.
The cross section of the hemisphere is a circle!with radius: !so its area is:!
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r2 − x 2
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A = π(r2 − x 2)
The cross section of the cylinder with the cone!removed has an outer radius of: !and an inner radius of: !so its area is:!
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A = πr2 −πx 2 = π(r2 − x 2)
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r
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x
Holt Geometry
10-8 Spheres
I will learn and apply the formulas for the surface area and volume of a sphere.
CN#7 Objectives
Holt Geometry
10-8 Spheres
sphere center of a sphere radius of a sphere hemisphere great circle
Vocabulary
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
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Holt Geometry
10-8 Spheres
The figure shows a hemisphere and a cylinder with a cone removed from its interior. The cross sections have the same area at every level, so the volumes are equal by Cavalieri’s Principle.
The height of the hemisphere is equal to the radius.
Holt Geometry
10-8 Spheres V(hemisphere) = V(cylinder) – V(cone)
The volume of a sphere with radius r is twice the volume of the hemisphere, or .
Holt Geometry
10-8 Spheres
Holt Geometry
10-8 Spheres Example 1A: Finding Volumes of Spheres
Find the volume of the sphere. Give your answer in terms of π.
= 2304π in3 Simplify.
Volume of a sphere.
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Holt Geometry
10-8 Spheres Example 1B: Finding Volumes of Spheres
Find the diameter of a sphere with volume 36,000π cm3.
Substitute 36,000π for V.
27,000 = r3
r = 30
d = 60 cm d = 2r
Take the cube root of both sides.
Volume of a sphere.
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.
Holt Geometry
10-8 Spheres
In the figure, the vertex of the pyramid is at the center of the sphere. The height of the pyramid is approximately the radius r of the sphere. Suppose the entire sphere is filled with n pyramids that each have base area B and height r.
Holt Geometry
10-8 Spheres
4πr2 ≈ nB
If the pyramids fill the sphere, the total area of the bases is approximately equal to the surface area of the sphere S, so 4πr2 ≈ S.
As the number of pyramids increases, the approximation gets closer to the actual surface area.
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Holt Geometry
10-8 Spheres
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 π.
S = 4πr2
S = 4π(38)2 = 5776π cm2
Surface area of a sphere
Holt Geometry
10-8 Spheres Example 3B: Finding Surface Area of Spheres
Find the volume of a sphere with surface area 324π in2. Give your answers in terms of π.
Substitute 324π for S. 324π = 4πr2
r = 9 Solve for r.
Substitute 9 for r.
The volume of the sphere is 972π in3.
S = 4πr2 Surface area of a sphere
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=43π (9)3 = 972π .in3
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 .
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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 π.
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.
Holt Geometry
10-8 Spheres Example 5 Continued
The surface area of the composite figure is
L(cylinder) = 2πrh = 2π(6)(9) = 108π in2
B(cylinder) = πr2 = π(6)2 = 36π in2
72π + 108π + 36π = 216π in2.
Find the surface area and volume of the composite figure. Give your answer in terms of π.
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 π.
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 144π + 324π = 468π in3.
