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Area, Surface Area and Volume: Question 2

Syllabus C5.2, E5.2, C5.4, E5.4

Structured Extended 8 marks

A solid ornament is made by fixing a cone on top of a cylinder, so that the flat circular face of the cone exactly covers the top circular face of the cylinder.

The cylinder has radius 6 cm6\text{ cm} and height 15 cm15\text{ cm}. The cone has the same radius, 6 cm6\text{ cm}, and a vertical (perpendicular) height of 8 cm8\text{ cm}.

The ornament stands upright with the cone sitting point-upwards on top of the cylinder, so the single solid looks like a sharpened pencil.

(a) Calculate the total volume of the ornament. Give your answer in terms of π\pi and also correct to the nearest cubic centimetre. [4]

(b) The entire outer surface of the ornament is painted. Calculate the total area that is painted. Give your answer in terms of π\pi and also correct to 3 significant figures. [4]

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Worked solution

Part (a): Total volume

The ornament is a composite solid = cylinder + cone, so add the two volumes.

Volume of the cylinder (radius r=6r=6, height h=15h=15): Vcyl=πr2h=π(6)2(15)=π×36×15=540π cm3.V_{\text{cyl}} = \pi r^2 h = \pi (6)^2 (15) = \pi \times 36 \times 15 = 540\pi \text{ cm}^3.

Volume of the cone (radius r=6r=6, vertical height h=8h=8): Vcone=13πr2h=13π(6)2(8)=13π×36×8=13(288π)=96π cm3.V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi (6)^2 (8) = \tfrac{1}{3}\pi \times 36 \times 8 = \tfrac{1}{3}(288\pi) = 96\pi \text{ cm}^3.

Total volume: V=540π+96π=636π cm3.V = 540\pi + 96\pi = 636\pi \text{ cm}^3.

As a decimal: 636π=636×3.14159…=1998.05…≈1998 cm3.636\pi = 636 \times 3.14159\ldots = 1998.05\ldots \approx 1998 \text{ cm}^3.

V=636π cm3≈1998 cm3\boxed{V = 636\pi \text{ cm}^3 \approx 1998 \text{ cm}^3}

Part (b): Total painted (surface) area

Because the cone’s flat base exactly covers the top of the cylinder, the join is hidden. So neither the top circle of the cylinder nor the base circle of the cone is painted. The painted surfaces are:

  1. the bottom circular face of the cylinder,
  2. the curved surface of the cylinder,
  3. the curved surface of the cone.

1. Bottom face of cylinder: Abase=πr2=π(6)2=36π cm2.A_{\text{base}} = \pi r^2 = \pi (6)^2 = 36\pi \text{ cm}^2.

2. Curved surface of cylinder: Acyl=2πrh=2π(6)(15)=180π cm2.A_{\text{cyl}} = 2\pi r h = 2\pi (6)(15) = 180\pi \text{ cm}^2.

3. Curved surface of cone: first find the slant height ll using Pythagoras’ theorem on the right triangle formed by the radius, the vertical height and the slant: l=r2+h2=62+82=36+64=100=10 cm.l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ cm}.

Then: Acone=πrl=π(6)(10)=60π cm2.A_{\text{cone}} = \pi r l = \pi (6)(10) = 60\pi \text{ cm}^2.

Total painted area: A=36π+180π+60π=276π cm2.A = 36\pi + 180\pi + 60\pi = 276\pi \text{ cm}^2.

As a decimal: 276π=276×3.14159…=867.08…≈867 cm2 (3 s.f.).276\pi = 276 \times 3.14159\ldots = 867.08\ldots \approx 867 \text{ cm}^2 \ (\text{3 s.f.}).

A=276π cm2≈867 cm2\boxed{A = 276\pi \text{ cm}^2 \approx 867 \text{ cm}^2}

Method note

Keeping every term as a multiple of π\pi (e.g. 540π540\pi, 96π96\pi, 180π180\pi) and only converting to a decimal at the very end avoids rounding errors and keeps the working tidy. The key “trick” in part (b) is recognising that the matching radii make the join surface internal, so it must not be counted, and that the cone needs the slant height l=10l=10, not the vertical height 88.