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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 B 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 4.6 cm and perpendicular height 8.3 cm.

    8.3 cm4.6 cm
    1. (a) Find the slant height.

      • \(l=\sqrt{4.6^2+8.3^2}=9.489\) cm
      • 9.49 cm
    2. (b) Find the volume.

      • \(V=\frac13\pi r^2h=\frac13\pi(4.6)^2(8.3)=183.92\)
      • 184 cm³
  2. A cone has base radius 5.5 cm and slant height 14.2 cm.

    1. (a) Find the curved surface area.

      • \(\pi rl=\pi(5.5)(14.2)=245.36\)
      • 245 cm²
    2. (b) Find the total surface area, including the base.

      • \(\pi r^2=95.03\), so total \(=245.36+95.03=340.39\)
      • 340 cm²
  3. A sphere has radius 6.4 cm.

    1. (a) Find its volume.

      • \(\frac43\pi(6.4)^3=1098.07\)
      • 1100 cm³
    2. (b) Find its surface area.

      • \(4\pi(6.4)^2=514.72\)
      • 515 cm²
    3. (c) A solid hemisphere has the same radius. Find its total surface area, including the flat face.

      • Curved \(2\pi r^2=257.36\), flat face \(\pi r^2=128.68\)
      • Total \(3\pi r^2=386.04\), so 386 cm²
  4. A sphere has volume 900 cm³.

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

      • \(\frac43\pi r^3=900\)
      • Solve for \(r\): \(r=5.989\)
      • 5.99 cm
    2. (b) A cone has volume 450 cm³ and base radius 6 cm. Find its perpendicular height.

      • \(\frac13\pi(6)^2h=450\)
      • \(h=\frac{3\times450}{\pi\times36}=11.937\)
      • 11.9 cm

In context

  1. A hemispherical fish tank has internal diameter 42 cm.

    1. (a) Find the capacity in litres.

      • \(r=21\) cm, \(V=\frac23\pi(21)^3=19396.2\) cm³
      • \(19396.2\div1000=19.396\) litres, so 19.4 litres
    2. (b) How many 0.5 litre bottles can be filled completely from the full tank?

      • \(19.396\div0.5=38.79\)
      • 38 complete bottles
  2. A drinks flask is a cylinder of radius 4.5 cm and height 20 cm, closed by a hemispherical cap of the same radius.

    9 cm20 cm
    1. (a) Find the volume.

      • Cylinder \(\pi(4.5)^2(20)=1272.35\), hemisphere \(\frac23\pi(4.5)^3=190.85\)
      • Total 1463.2
    2. (b) Find the area of its curved outside surface (wall and cap, not the base).

      • Wall \(2\pi(4.5)(20)=565.49\), curved top \(2\pi(4.5)^2=127.23\)
      • Total 692.7
  3. A tent is a cuboid with a square base of side 3.6 m and height 1.6 m, with a square-based pyramid roof of height 1.2 m, with the same base.

    3.6 m1.6 m
    1. (a) Find the total volume.

      • Cuboid \(3.6\times3.6\times1.6=20.74\), pyramid \(\frac13(3.6)^2(1.2)=5.184\)
      • Total 25.92
    2. (b) Find the area of canvas needed for the four sloping triangular faces of the roof.

      • Slant height \(\sqrt{1.2^2+1.8^2}=2.163\)
      • Four faces: \(4\times\frac12\times3.6\times2.163=15.58\)
  4. A solid steel ball of radius 5.4 cm is melted and made into 9 identical cones, each with base radius 2.7 cm. No steel is lost. Find the perpendicular height of each cone.

    • Sphere: \(\frac43\pi(5.4)^3=659.58\) cm³
    • 9 cones: \(9\times\frac13\pi(2.7)^2h=68.707\,h\)
    • \(h=\frac{659.58}{68.707}=9.600\), so 9.60 cm

Stretch

  1. A pencil-top ornament is a solid hemisphere of radius 4 cm joined to the base of a cone of the same base radius. The volume of the whole ornament is 300 cm³. Find the perpendicular height of the cone.

    • Hemisphere: \(\frac23\pi(4)^3=134.04\)
    • Cone: \(\frac13\pi(4)^2h\). Total: \(134.04+\frac13\pi(4)^2h=300\)
    • \(h=\frac{3(300-134.04)}{\pi(4)^2}=9.905\), so 9.90 cm
  2. A conical paper cup has internal radius 9 cm at the top and internal depth 12 cm. It stands on its point and holds juice to a depth of 9 cm.

    18 cm9 cm
    1. (a) Find the radius of the liquid surface.

      • \(\frac{r}{9}=\frac{9}{12}\)
      • \(r=6.750\), so 6.75 cm
    2. (b) Find the volume of liquid.

      • \(\frac13\pi(6.750)^2(9)=429.42\)
      • 429 cm³
    3. (c) Find the volume of liquid as a percentage of the volume of the full container.

      • Full container \(\frac13\pi(9)^2(12)=1017.9\)
      • \(\frac{429.42}{1017.9}\times100=42.2\)%

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