NeokaalStudy

Surface Areas and Volumes: Practice Worksheet

Surface area, volume, capacity, and displacement across distinct combined-solid models.

18 problems·35–45 min·★★★★☆
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  1. Problem 1
    surface-area·★★☆☆☆
    Two cubes of edge 7 cm are joined face-to-face. Find the surface area of the resulting cuboid.
    ▶Answer
    $490$ cm$^2$
    ▶Step-by-step solution
    1. The new cuboid is $14\times7\times7$ cm.
    2. Its surface area is $2(lw+lh+wh)=490$ cm$^2$.
  2. Problem 2
    surface-area·★★★☆☆
    A hollow cylindrical cup of radius 5 cm and height 12 cm has a hollow hemispherical bowl of the same radius fixed at one end. Find its inner surface area.
    ▶Answer
    $170\pi$ cm$^2$
    ▶Step-by-step solution
    1. Add only the inner curved surfaces.
    2. $2\pi rh+2\pi r^2=170\pi$ cm$^2$.
  3. Problem 3
    surface-area·★★★★☆
    A cone of radius 6 cm and height 8 cm is joined to a hemisphere of the same radius. Find the exposed surface area.
    ▶Answer
    $132\pi$ cm$^2$
    ▶Step-by-step solution
    1. The cone's slant height is $\sqrt{6^2+8^2}=10$ cm.
    2. Add $\pi rl+2\pi r^2=132\pi$ cm$^2$.
  4. Problem 4
    surface-area·★★★★★
    A hemisphere of the greatest possible radius is mounted on a cube of edge 12 cm. Find the exposed surface area.
    ▶Answer
    $864+36\pi$ cm$^2$
    ▶Step-by-step solution
    1. The greatest radius is 6 cm.
    2. Subtract the covered base circle from the cube and add the hemisphere's curved area, leaving a net $+\pi r^2$.
  5. Problem 5
    surface-area·★★☆☆☆
    A hemispherical depression of greatest possible radius is cut into one face of a cube of edge 12 cm. Find the remaining surface area.
    ▶Answer
    $864+36\pi$ cm$^2$
    ▶Step-by-step solution
    1. Remove the circular opening and add the curved surface of the depression.
    2. The net change is $-\pi r^2+2\pi r^2=+\pi r^2$.
  6. Problem 6
    surface-area·★★★☆☆
    A capsule has radius 3 cm and total length 14 cm, including two hemispherical ends. Find its surface area.
    ▶Answer
    $84\pi$ cm$^2$
    ▶Step-by-step solution
    1. The cylindrical length is $14-2(3)=8$ cm.
    2. Add $2\pi rh+4\pi r^2=84\pi$ cm$^2$.
  7. Problem 7
    surface-area·★★★★☆
    An open-bottom tent consists of a cylinder of radius 12 m and height 9 m topped by a cone of height 16 m. Find the canvas area.
    ▶Answer
    $456\pi$ m$^2$
    ▶Step-by-step solution
    1. Cone slant height is 20 m.
    2. Canvas area $=2\pi rh+\pi rl=456\pi$ m$^2$.
  8. Problem 8
    surface-area·★★★★★
    A cone of radius 3 cm and depth 4 cm is hollowed from one end of a cylinder with the same radius and height 4 cm. Find the total exposed surface area.
    ▶Answer
    $48\pi$ cm$^2$
    ▶Step-by-step solution
    1. Add the cylinder's curved surface, the untouched circular base, and the cavity's curved surface.
    2. The cone slant height is 5 cm, giving $48\pi$ cm$^2$.
  9. Problem 9
    surface-area·★★☆☆☆
    A hemisphere is scooped from each flat end of a cylinder of radius 4 cm and height 13 cm. Find the exposed surface area.
    ▶Answer
    $168\pi$ cm$^2$
    ▶Step-by-step solution
    1. The exposed parts are the cylinder's curved surface and two hemispherical cavities.
    2. $2\pi rh+4\pi r^2=168\pi$ cm$^2$.
  10. Problem 10
    volume·★★★☆☆
    A cone of radius 9 cm and height 10 cm stands on a hemisphere of the same radius. Find its volume.
    ▶Answer
    $\frac{2268}3\pi$ cm$^3$
    ▶Step-by-step solution
    1. Add cone and hemisphere volumes.
    2. $\frac13\pi r^2h+\frac23\pi r^3=\frac{2268}3\pi$ cm$^3$.
  11. Problem 11
    volume·★★★★☆
    A cylinder of radius 9 cm and length 5 cm has a cone of height 9 cm attached at each end. Find the total volume.
    ▶Answer
    $\frac{2673}3\pi$ cm$^3$
    ▶Step-by-step solution
    1. Add one cylinder and two cone volumes.
    2. $\pi r^2h+2(\frac13\pi r^2H)=\frac{2673}3\pi$ cm$^3$.
  12. Problem 12
    volume·★★★★★
    A capsule has a cylindrical middle of radius 6 cm and length 11 cm, plus two hemispherical ends. Find its volume.
    ▶Answer
    $\frac{2052}3\pi$ cm$^3$
    ▶Step-by-step solution
    1. The hemispheres form one sphere.
    2. $\pi r^2h+\frac43\pi r^3=\frac{2052}3\pi$ cm$^3$.
  13. Problem 13
    volume·★★☆☆☆
    A 20 cm by 10 cm by 3 cm wooden block has 6 conical depressions, each radius 1 cm and depth 3 cm. Find the remaining volume.
    ▶Answer
    $600-6\pi$ cm$^3$
    ▶Step-by-step solution
    1. Block volume is 600 cm$^3$ and each cone has volume $\pi$ cm$^3$.
    2. Subtract the volume of all depressions.
  14. Problem 14
    volume·★★★☆☆
    Small spheres of radius 2 cm are dropped into a full inverted cone of radius 6 cm and height 16 cm. If their combined volume equals the displaced liquid, how many spheres are needed?
    ▶Answer
    $18$
    ▶Step-by-step solution
    1. Equate the cone volume to the total sphere volume.
    2. $n(4\pi(2)^3/3)=\pi(6)^2(16)/3$, so $n=18$.
  15. Problem 15
    volume·★★★★☆
    A stepped solid consists of cylinders $(r,h)=(4,8)$ cm and $(2,5)$ cm. Its material density is 3 g/cm$^3$. Find its mass.
    ▶Answer
    $444\pi$ g
    ▶Step-by-step solution
    1. Total volume is $\pi(4^28+2^25)=148\pi$ cm$^3$.
    2. Multiply by density.
  16. Problem 16
    volume·★★★★★
    A cylinder of radius 6 cm and height 15 cm is full of water. A solid cone of height 8 cm joined to a hemisphere, both radius 3 cm, is immersed completely. Find the water remaining.
    ▶Answer
    $498\pi$ cm$^3$
    ▶Step-by-step solution
    1. Cylinder volume is $540\pi$ cm$^3$.
    2. The immersed solid displaces $42\pi$ cm$^3$; subtract.
  17. Problem 17
    volume·★★☆☆☆
    A vessel consists of a spherical body of radius 6 cm and a cylindrical neck of radius 2 cm and height 5 cm. Find its capacity.
    ▶Answer
    $\frac{924}3\pi$ cm$^3$
    ▶Step-by-step solution
    1. Add the sphere and neck volumes.
    2. $4\pi r^3/3+\pi R^2h=924\pi/3$ cm$^3$.
  18. Problem 18
    volume·★★★☆☆
    A cone of radius 9 cm and height 12 cm is joined to a hemisphere and placed snugly inside a cylinder of radius 9 cm and height 21 cm. Find the volume inside the cylinder but outside the solid.
    ▶Answer
    $891\pi$ cm$^3$
    ▶Step-by-step solution
    1. Subtract the cone-plus-hemisphere volume from the circumscribing cylinder.
    2. The difference is $(1701-810)\pi=891\pi$ cm$^3$.