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IB Mathematics: Applications and Interpretation SL · 12.10: Geometry & Trigonometry · 12.10.9

Volumes and surface areas of solids Problem set A One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. A cone has base radius 5.4 cm and perpendicular height 10.2 cm.

    10.2 cm5.4 cm
    1. (a) Find the slant height.

      • \(l=\sqrt{5.4^2+10.2^2}=11.541\) cm
      • 11.5 cm
    2. (b) Find the volume.

      • \(V=\frac13\pi r^2h=\frac13\pi(5.4)^2(10.2)=311.47\)
      • 311 cm³
  2. A cone has base radius 7 cm and slant height 18.5 cm.

    1. (a) Find the curved surface area.

      • \(\pi rl=\pi(7)(18.5)=406.84\)
      • 407 cm²
    2. (b) Find the total surface area, including the base.

      • \(\pi r^2=153.94\), so total \(=406.84+153.94=560.77\)
      • 561 cm²
  3. A sphere has radius 8.5 cm.

    1. (a) Find its volume.

      • \(\frac43\pi(8.5)^3=2572.44\)
      • 2570 cm³
    2. (b) Find its surface area.

      • \(4\pi(8.5)^2=907.92\)
      • 908 cm²
    3. (c) A solid hemisphere has the same radius. Find its total surface area, including the flat face.

      • Curved \(2\pi r^2=453.96\), flat face \(\pi r^2=226.98\)
      • Total \(3\pi r^2=680.94\), so 681 cm²
  4. A sphere has volume 2500 cm³.

    1. (a) Find the radius of the sphere.

      • \(\frac43\pi r^3=2500\)
      • Solve for \(r\): \(r=8.419\)
      • 8.42 cm
    2. (b) A cone has volume 330 cm³ and base radius 5 cm. Find its perpendicular height.

      • \(\frac13\pi(5)^2h=330\)
      • \(h=\frac{3\times330}{\pi\times25}=12.605\)
      • 12.6 cm

In context

  1. A hemispherical bowl has internal diameter 36 cm.

    1. (a) Find the capacity in litres.

      • \(r=18\) cm, \(V=\frac23\pi(18)^3=12214.5\) cm³
      • \(12214.5\div1000=12.215\) litres, so 12.2 litres
    2. (b) How many 0.25 litre cups can be filled completely from a full bowl?

      • \(12.215\div0.25=48.86\)
      • 48 complete cups
  2. A water tower holds its water in a cylinder of radius 3.2 m and height 7.5 m, with a hemispherical top of the same radius.

    6.4 m7.5 m
    1. (a) Find the volume.

      • Cylinder \(\pi(3.2)^2(7.5)=241.27\), hemisphere \(\frac23\pi(3.2)^3=68.63\)
      • Total 309.9
    2. (b) Find the area of its curved outside surface (wall and top, not the floor).

      • Wall \(2\pi(3.2)(7.5)=150.80\), curved top \(2\pi(3.2)^2=64.34\)
      • Total 215.1
  3. A garden tower is a cuboid with a square base of side 2.8 m and height 4.5 m, with a square-based pyramid of height 1.5 m on top, with the same base.

    2.8 m4.5 m
    1. (a) Find the total volume.

      • Cuboid \(2.8\times2.8\times4.5=35.28\), pyramid \(\frac13(2.8)^2(1.5)=3.920\)
      • Total 39.20
    2. (b) Find the total area of the four sloping triangular faces.

      • Slant height \(\sqrt{1.5^2+1.4^2}=2.052\)
      • Four faces: \(4\times\frac12\times2.8\times2.052=11.49\)
  4. A solid wax sphere of radius 7.5 cm is melted and recast into 15 identical cones. Each cone has base radius 3 cm. No wax is lost. Find the perpendicular height of each cone.

    • Sphere: \(\frac43\pi(7.5)^3=1767.15\) cm³
    • 15 cones: \(15\times\frac13\pi(3)^2h=141.372\,h\)
    • \(h=\frac{1767.15}{141.372}=12.500\), so 12.5 cm

Stretch

  1. A toy is a solid hemisphere of radius 5 cm fixed to the base of a cone of the same base radius. The volume of the whole toy is 450 cm³. Find the perpendicular height of the cone.

    • Hemisphere: \(\frac23\pi(5)^3=261.80\)
    • Cone: \(\frac13\pi(5)^2h\). Total: \(261.80+\frac13\pi(5)^2h=450\)
    • \(h=\frac{3(450-261.80)}{\pi(5)^2}=7.189\), so 7.19 cm
  2. A conical filter has internal radius 6 cm at the top and internal depth 15 cm. It stands on its point and holds water to a depth of 10 cm.

    12 cm10 cm
    1. (a) Find the radius of the liquid surface.

      • \(\frac{r}{6}=\frac{10}{15}\)
      • \(r=4.000\), so 4.00 cm
    2. (b) Find the volume of liquid.

      • \(\frac13\pi(4.000)^2(10)=167.55\)
      • 168 cm³
    3. (c) Find the volume of liquid as a percentage of the volume of the full container.

      • Full container \(\frac13\pi(6)^2(15)=565.5\)
      • \(\frac{167.55}{565.5}\times100=29.6\)%

jmaths.xyz · AI SL · 12.10.9 Volumes and surface areas of solids