Practice
Cylinders and Spheres Practice
Fifty original questions on cylinder and sphere formulas, great circles, hemispheres, composites, holes, scaling, nets, units, and water transfer.
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Question 1
Explanation
A right cylinder has two congruent parallel circles and a perpendicular height between their centers.
- Geometry and units
Identify the bases and the perpendicular altitude.
Question 2
Explanation
A sphere is the set of points in three-dimensional space equidistant from a center.
- Geometry and units
Distinguish the sphere's surface definition from a solid cylinder or cone.
Question 3
Explanation
A circular section through the center is a great circle and has the sphere's full radius.
- Geometry and units
Check whether the cutting plane contains the center.
Question 4
Explanation
Cylinder volume is circular base area πr² times perpendicular height h.
- Geometry and units
Use area of a circular base, not its circumference.
Question 5
Explanation
A full sphere has volume (4/3)πr³; (2/3)πr³ is a hemisphere.
- Geometry and units
Separate sphere surface area from sphere volume.
Question 6
Explanation
V=πr²h=π(3²)(8)=72π cm³.
- Geometry and units
Square the radius, then multiply by the perpendicular height.
Question 7
Explanation
V=πr²h=π(4²)(11)=176π cm³.
- Geometry and units
Square the radius, then multiply by the perpendicular height.
Question 8
Explanation
V=πr²h=π(5²)(7)=175π cm³.
- Geometry and units
Square the radius, then multiply by the perpendicular height.
Question 9
Explanation
V=πr²h=π(6²)(9)=324π cm³.
- Geometry and units
Square the radius, then multiply by the perpendicular height.
Question 10
Explanation
V=πr²h=π(8²)(5)=320π cm³.
- Geometry and units
Square the radius, then multiply by the perpendicular height.
Question 11
Explanation
LA=2πrh=2π(2)(10)=40π m².
- Geometry and units
Use only the curved rectangle.
Question 12
Explanation
TA=2πrh+2πr²=42π+18π=60π m².
- Geometry and units
Include the curved surface and both circular bases.
Question 13
Explanation
LA=2πrh=2π(4)(6)=48π m².
- Geometry and units
Use only the curved rectangle.
Question 14
Explanation
TA=2πrh+2πr²=90π+50π=140π m².
- Geometry and units
Include the curved surface and both circular bases.
Question 15
Explanation
TA=2πrh+2πr²=96π+72π=168π m².
- Geometry and units
Include the curved surface and both circular bases.
Question 16
Explanation
From πr²(10)=200π, r²=20. The positive physical radius is r=2.
- Geometry and units
Cancel pi, isolate r squared, and retain the positive square root.
Question 17
Explanation
From πr²(12)=432π, r²=36. The positive physical radius is r=3.
- Geometry and units
Cancel pi, isolate r squared, and retain the positive square root.
Question 18
Explanation
From πr²(8)=320π, r²=40. The positive physical radius is r=5.
- Geometry and units
Cancel pi, isolate r squared, and retain the positive square root.
Question 19
Explanation
From πr²(15)=1125π, r²=75. The positive physical radius is r=5.
- Geometry and units
Cancel pi, isolate r squared, and retain the positive square root.
Question 20
Explanation
From πr²(18)=1458π, r²=81. The positive physical radius is r=4.5.
- Geometry and units
Cancel pi, isolate r squared, and retain the positive square root.
Question 21
Explanation
SA=4πr²=4π(2²)=16π cm².
- Geometry and units
Use square units with 4πr².
Question 22
Explanation
V=(4/3)πr³=(4/3)π(3³)=36π cm³.
- Geometry and units
Use cubic units and retain the factor 4/3.
Question 23
Explanation
SA=4πr²=4π(4²)=64π cm².
- Geometry and units
Use square units with 4πr².
Question 24
Explanation
V=(4/3)πr³=(4/3)π(6³)=288π cm³.
- Geometry and units
Use cubic units and retain the factor 4/3.
Question 25
Explanation
SA=4πr²=4π(9²)=324π cm².
- Geometry and units
Use square units with 4πr².
Question 26
Explanation
Cylinder volume is 72π ft³. Hemisphere volume is 18π ft³. Their nonoverlapping volumes add to 90π ft³.
- Geometry and units
Add cylinder and half-sphere volumes; the shared base has no volume to subtract.
Question 27
Explanation
Cylinder volume is 720π ft³. Hemisphere volume is 1152π ft³. Their nonoverlapping volumes add to 1872π ft³.
- Geometry and units
Add cylinder and half-sphere volumes; the shared base has no volume to subtract.
Question 28
Explanation
Cylinder volume is 675π ft³. Hemisphere volume is 2250π ft³. Their nonoverlapping volumes add to 2925π ft³.
- Geometry and units
Add cylinder and half-sphere volumes; the shared base has no volume to subtract.
Question 29
Explanation
Cylinder volume is 252π ft³. Hemisphere volume is 144π ft³. Their nonoverlapping volumes add to 396π ft³.
- Geometry and units
Add cylinder and half-sphere volumes; the shared base has no volume to subtract.
Question 30
Explanation
Cylinder volume is 324π ft³. Hemisphere volume is 486π ft³. Their nonoverlapping volumes add to 810π ft³.
- Geometry and units
Add cylinder and half-sphere volumes; the shared base has no volume to subtract.
Question 31
Explanation
Remaining volume is πh(R²-r²)=π(8)(5²-2²)=168π cm³.
- Geometry and units
Subtract cross-sectional areas before multiplying by the shared height.
Question 32
Explanation
Remaining volume is πh(R²-r²)=π(7)(6²-3²)=189π cm³.
- Geometry and units
Subtract cross-sectional areas before multiplying by the shared height.
Question 33
Explanation
Remaining volume is πh(R²-r²)=π(10)(7²-2²)=450π cm³.
- Geometry and units
Subtract cross-sectional areas before multiplying by the shared height.
Question 34
Explanation
Remaining volume is πh(R²-r²)=π(6)(8²-4²)=288π cm³.
- Geometry and units
Subtract cross-sectional areas before multiplying by the shared height.
Question 35
Explanation
Remaining volume is πh(R²-r²)=π(5)(9²-3²)=360π cm³.
- Geometry and units
Subtract cross-sectional areas before multiplying by the shared height.
Question 36
Explanation
Cylinder volume scales by a²b. Here (2)²(0.5)=2.
- Geometry and units
Square the radius scale factor but use the height scale factor once.
Question 37
Explanation
Cylinder volume scales by a²b. Here (3)²(2)=18.
- Geometry and units
Square the radius scale factor but use the height scale factor once.
Question 38
Explanation
Cylinder volume scales by a²b. Here (0.5)²(4)=1.
- Geometry and units
Square the radius scale factor but use the height scale factor once.
Question 39
Explanation
Cylinder volume scales by a²b. Here (2)²(3)=12.
- Geometry and units
Square the radius scale factor but use the height scale factor once.
Question 40
Explanation
Cylinder volume scales by a²b. Here (4)²(0.25)=4.
- Geometry and units
Square the radius scale factor but use the height scale factor once.
Question 41
Explanation
The net rectangle has sides 24 and 32. Its diagonal is √(24²+32²)=40 cm.
- Geometry and units
Use cylinder height and circumference as the two legs of the net rectangle.
Question 42
Explanation
The net rectangle has sides 30 and 40. Its diagonal is √(30²+40²)=50 cm.
- Geometry and units
Use cylinder height and circumference as the two legs of the net rectangle.
Question 43
Explanation
The net rectangle has sides 18 and 24. Its diagonal is √(18²+24²)=30 cm.
- Geometry and units
Use cylinder height and circumference as the two legs of the net rectangle.
Question 44
Explanation
The net rectangle has sides 20 and 21. Its diagonal is √(20²+21²)=29 cm.
- Geometry and units
Use cylinder height and circumference as the two legs of the net rectangle.
Question 45
Explanation
The net rectangle has sides 12 and 35. Its diagonal is √(12²+35²)=37 cm.
- Geometry and units
Use cylinder height and circumference as the two legs of the net rectangle.
Question 46
Explanation
Conservation gives π(3²)Δh=180, so Δh=180/(9π)≈6.366 cm.
- Geometry and units
Divide transferred volume by the cylinder's circular base area.
Question 47
Explanation
Conservation gives π(4²)Δh=256, so Δh=256/(16π)≈5.093 cm.
- Geometry and units
Divide transferred volume by the cylinder's circular base area.
Question 48
Explanation
Conservation gives π(5²)Δh=375, so Δh=375/(25π)≈4.775 cm.
- Geometry and units
Divide transferred volume by the cylinder's circular base area.
Question 49
Explanation
Conservation gives π(6²)Δh=324, so Δh=324/(36π)≈2.865 cm.
- Geometry and units
Divide transferred volume by the cylinder's circular base area.
Question 50
Explanation
Conservation gives π(8²)Δh=512, so Δh=512/(64π)≈2.546 cm.
- Geometry and units
Divide transferred volume by the cylinder's circular base area.
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Questions to review
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- Question 1Cylinder elementsEasy
- Question 2Sphere elementsEasy
- Question 3Great circlesEasy
- Question 4Cylinder volumeEasy
- Question 5Sphere volumeEasy
- Question 6Cylinder volumeEasy
- Question 7Cylinder volumeEasy
- Question 8Cylinder volumeEasy
- Question 9Cylinder volumeEasy
- Question 10Cylinder volumeEasy
- Question 11Cylinder lateral areaEasy
- Question 12Cylinder total areaEasy
- Question 13Cylinder lateral areaEasy
- Question 14Cylinder total areaEasy
- Question 15Cylinder total areaEasy
- Question 16Reverse cylinder volumeMedium
- Question 17Reverse cylinder volumeMedium
- Question 18Reverse cylinder volumeMedium
- Question 19Reverse cylinder volumeMedium
- Question 20Reverse cylinder volumeMedium
- Question 21Sphere surface areaMedium
- Question 22Sphere volumeMedium
- Question 23Sphere surface areaMedium
- Question 24Sphere volumeMedium
- Question 25Sphere surface areaMedium
- Question 26Composite volumeMedium
- Question 27Composite volumeMedium
- Question 28Composite volumeMedium
- Question 29Composite volumeMedium
- Question 30Composite volumeMedium
- Question 31Hollow cylindersMedium
- Question 32Hollow cylindersMedium
- Question 33Hollow cylindersMedium
- Question 34Hollow cylindersMedium
- Question 35Hollow cylindersMedium
- Question 36Cylinder scalingMedium
- Question 37Cylinder scalingMedium
- Question 38Cylinder scalingMedium
- Question 39Cylinder scalingMedium
- Question 40Cylinder scalingMedium
- Question 41Cylinder netsHard
- Question 42Cylinder netsHard
- Question 43Cylinder netsHard
- Question 44Cylinder netsHard
- Question 45Cylinder netsHard
- Question 46Water transferHard
- Question 47Water transferHard
- Question 48Water transferHard
- Question 49Water transferHard
- Question 50Water transferHard