Practice
Pyramids and Cones Practice
Fifty original questions on pyramid and cone formulas, slant height, reverse information, similar cuts, composites, scaling, and modeling.
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Question 1
Explanation
A pyramid has exactly one polygonal base, and its triangular lateral faces share a vertex.
- Geometry and units
Classify the solid by its bases and lateral faces.
Question 2
Explanation
A cone occupies one third of a cylinder with the same circular base and perpendicular height.
- Geometry and units
Use h for volume and include the factor 1/3.
Question 3
Explanation
The triangular lateral faces combine to one half times base perimeter times slant height.
- Geometry and units
Surface area uses slant height, not perpendicular height.
Question 4
Explanation
Volume uses perpendicular height h; lateral area πrℓ uses slant height ℓ.
- Geometry and units
Label h and ℓ before selecting a formula.
Question 5
Explanation
The pyramid volume is one third of the matching prism volume: 270/3=90.
- Geometry and units
Use the same-base, same-height one-third relationship.
Question 6
Explanation
V=(1/3)Bh=(1/3)(48)(9)=144 cubic units.
- Geometry and units
Multiply base area by perpendicular height, then divide by 3.
Question 7
Explanation
V=(1/3)Bh=(1/3)(75)(12)=300 cubic units.
- Geometry and units
Multiply base area by perpendicular height, then divide by 3.
Question 8
Explanation
V=(1/3)Bh=(1/3)(96)(15)=480 cubic units.
- Geometry and units
Multiply base area by perpendicular height, then divide by 3.
Question 9
Explanation
V=(1/3)Bh=(1/3)(126)(8)=336 cubic units.
- Geometry and units
Multiply base area by perpendicular height, then divide by 3.
Question 10
Explanation
V=(1/3)Bh=(1/3)(150)(18)=900 cubic units.
- Geometry and units
Multiply base area by perpendicular height, then divide by 3.
Question 11
Explanation
P=4(6)=24. LA=(1/2)Pℓ=60 cm², and adding B=36 gives SA=96 cm².
- Geometry and units
Find lateral area, then add one square base.
Question 12
Explanation
P=4(8)=32. LA=(1/2)Pℓ=112 cm².
- Geometry and units
Use half the base perimeter times slant height.
Question 13
Explanation
P=4(10)=40. LA=(1/2)Pℓ=180 cm², and adding B=100 gives SA=280 cm².
- Geometry and units
Find lateral area, then add one square base.
Question 14
Explanation
P=4(12)=48. LA=(1/2)Pℓ=240 cm².
- Geometry and units
Use half the base perimeter times slant height.
Question 15
Explanation
P=4(14)=56. LA=(1/2)Pℓ=364 cm², and adding B=196 gives SA=560 cm².
- Geometry and units
Find lateral area, then add one square base.
Question 16
Explanation
V=(1/3)πr²h=(1/3)π(3²)(12)=36π cm³.
- Geometry and units
Square the radius, multiply by h, and divide by 3.
Question 17
Explanation
V=(1/3)πr²h=(1/3)π(4²)(15)=80π cm³.
- Geometry and units
Square the radius, multiply by h, and divide by 3.
Question 18
Explanation
V=(1/3)πr²h=(1/3)π(5²)(18)=150π cm³.
- Geometry and units
Square the radius, multiply by h, and divide by 3.
Question 19
Explanation
V=(1/3)πr²h=(1/3)π(6²)(9)=108π cm³.
- Geometry and units
Square the radius, multiply by h, and divide by 3.
Question 20
Explanation
V=(1/3)πr²h=(1/3)π(8²)(12)=256π cm³.
- Geometry and units
Square the radius, multiply by h, and divide by 3.
Question 21
Explanation
LA=πrℓ=15π and base area is 9π, so SA=24π square units.
- Geometry and units
Use slant height for the curved area, then add one base.
Question 22
Explanation
LA=πrℓ=65π square units.
- Geometry and units
The lateral-area formula is πrℓ.
Question 23
Explanation
LA=πrℓ=60π and base area is 36π, so SA=96π square units.
- Geometry and units
Use slant height for the curved area, then add one base.
Question 24
Explanation
LA=πrℓ=136π square units.
- Geometry and units
The lateral-area formula is πrℓ.
Question 25
Explanation
LA=πrℓ=135π and base area is 81π, so SA=216π square units.
- Geometry and units
Use slant height for the curved area, then add one base.
Question 26
Explanation
An axial cross-section gives ℓ²=r²+h²: 5²=3²+4². Therefore the missing slant height is 5.
- Geometry and units
Use the radius and perpendicular height as legs; slant height is the hypotenuse.
Question 27
Explanation
An axial cross-section gives ℓ²=r²+h²: 13²=5²+12². Therefore the missing perpendicular height is 12.
- Geometry and units
Use the radius and perpendicular height as legs; slant height is the hypotenuse.
Question 28
Explanation
An axial cross-section gives ℓ²=r²+h²: 17²=8²+15². Therefore the missing radius is 8.
- Geometry and units
Use the radius and perpendicular height as legs; slant height is the hypotenuse.
Question 29
Explanation
An axial cross-section gives ℓ²=r²+h²: 25²=7²+24². Therefore the missing slant height is 25.
- Geometry and units
Use the radius and perpendicular height as legs; slant height is the hypotenuse.
Question 30
Explanation
An axial cross-section gives ℓ²=r²+h²: 41²=9²+40². Therefore the missing perpendicular height is 40.
- Geometry and units
Use the radius and perpendicular height as legs; slant height is the hypotenuse.
Question 31
Explanation
2πr=12π gives r=6. Then V=(1/3)π(6²)(9)=108π cm³.
- Geometry and units
Convert circumference to radius before using volume.
Question 32
Explanation
196π=(1/3)π(7²)h, so 196=(49/3)h and h=12.
- Geometry and units
Cancel π, then isolate h.
Question 33
Explanation
The vessel volume is (1/3)π(6²)(10)=120π ft³. Time=120π/(15π)=8 minutes.
- Geometry and units
Find total volume, then divide by the constant volumetric rate.
Question 34
Explanation
560=(1/3)(105)h=35h, so h=16 cm.
- Geometry and units
Multiply volume by 3, then divide by base area.
Question 35
Explanation
πrℓ=65π gives 5ℓ=65, so ℓ=13 cm.
- Geometry and units
Use lateral area πrℓ and cancel π.
Question 36
Explanation
The cones are similar. The height scale is 6/18=1/3, so the radius is 9(1/3)=3.
- Geometry and units
Use one common linear scale factor for height and radius.
Question 37
Explanation
Similar-solid volume scales by the cube of the linear factor: (1/2)³=1/8.
- Geometry and units
Cube a linear scale factor when comparing volumes.
Question 38
Explanation
Full volume is 256π and removed volume is 4π. The difference is 252π cubic units.
- Geometry and units
Find both cone volumes with their own dimensions, then subtract.
Question 39
Explanation
Full volume is 405π and removed volume is 15π. The difference is 390π cubic units.
- Geometry and units
Find both cone volumes with their own dimensions, then subtract.
Question 40
Explanation
Full volume is 864π and removed volume is 32π. The difference is 832π cubic units.
- Geometry and units
Find both cone volumes with their own dimensions, then subtract.
Question 41
Explanation
Cylinder volume is 108π. The two cones total 36π. The difference is 72π cm³.
- Geometry and units
Subtract both cone volumes from the enclosing cylinder.
Question 42
Explanation
Cylinder volume is 648π. The two cones total 216π. The difference is 432π cm³.
- Geometry and units
Subtract both cone volumes from the enclosing cylinder.
Question 43
Explanation
Cylinder volume is π(4²)(7)=112π. Cone volume is (1/3)π(4²)(9)=48π. Total=160π cm³.
- Geometry and units
Add nonoverlapping component volumes; the shared circle has no volume.
Question 44
Explanation
The prism volume is 81(15)=1,215. The pyramid occupies one third, 405. Empty space is 1,215−405=810 cm³.
- Geometry and units
The pyramid occupies one third, so two thirds of the matching prism remains.
Question 45
Explanation
Combined volume is (1/3)π(6²)(5+7)=(1/3)π(36)(12)=144π.
- Geometry and units
For equal radii, combine the heights before applying the cone formula.
Question 46
Explanation
Cone volume scales as r²h, so the factor is 2²·3=12.
- Geometry and units
Square the radius scale and multiply by the height scale.
Question 47
Explanation
The cone has one third the volume of the matching cylinder, which is 33⅓%.
- Geometry and units
Use the one-third relationship before calculating dimensions.
Question 48
Explanation
The factor 1/3 applies to cone volume, not to its curved lateral area.
- Geometry and units
Separate volume formulas from surface-area formulas.
Question 49
Explanation
An open cone has no circular base, so use lateral area only: πrℓ=π(5)(12)=60π cm².
- Geometry and units
Decide which surfaces are physically present before adding a base.
Question 50
Explanation
Area scales by 3²=9 and volume by 3³=27. Thus 40(9)=360 and 70(27)=1,890.
- Geometry and units
Square a uniform linear factor for area and cube it for volume.
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Questions to review
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- Question 1Solid classificationEasy
- Question 2Cone volumeEasy
- Question 3Pyramid lateral areaEasy
- Question 4Height versus slant heightEasy
- Question 5Pyramid-prism comparisonEasy
- Question 6Pyramid volumeEasy
- Question 7Pyramid volumeEasy
- Question 8Pyramid volumeEasy
- Question 9Pyramid volumeEasy
- Question 10Pyramid volumeEasy
- Question 11Pyramid surface areaEasy
- Question 12Pyramid surface areaEasy
- Question 13Pyramid surface areaEasy
- Question 14Pyramid surface areaEasy
- Question 15Pyramid surface areaEasy
- Question 16Cone volumeMedium
- Question 17Cone volumeMedium
- Question 18Cone volumeMedium
- Question 19Cone volumeMedium
- Question 20Cone volumeMedium
- Question 21Cone surface areaMedium
- Question 22Cone surface areaMedium
- Question 23Cone surface areaMedium
- Question 24Cone surface areaMedium
- Question 25Cone surface areaMedium
- Question 26Cone slant heightMedium
- Question 27Cone slant heightMedium
- Question 28Cone slant heightMedium
- Question 29Cone slant heightMedium
- Question 30Cone slant heightMedium
- Question 31Reverse cone informationMedium
- Question 32Reverse cone informationMedium
- Question 33Volume ratesMedium
- Question 34Reverse pyramid informationMedium
- Question 35Reverse cone informationMedium
- Question 36Similar conesMedium
- Question 37Similar-cone scalingMedium
- Question 38Frustum volumeMedium
- Question 39Frustum volumeMedium
- Question 40Frustum volumeMedium
- Question 41Composite empty volumeHard
- Question 42Composite empty volumeHard
- Question 43Composite volumeHard
- Question 44Pyramid-prism comparisonHard
- Question 45Composite volumeHard
- Question 46Cone scalingHard
- Question 47Cone-cylinder comparisonHard
- Question 48Error analysisHard
- Question 49Exposed surface areaHard
- Question 50Three-dimensional scalingHard