Year 10A Surface Area and Volume of Pyramids, Cones and Spheres Worksheets & Practice

Worked examples

Year 10A surface area and volume of pyramids, cones and spheres: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

1. Example 1 · Slant height and surface area of a square pyramid

   A right pyramid has a square base with side length 8 m and a perpendicular height of 3 m. Find the total surface area of the pyramid.

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   Key idea: The slant height of each triangular face is found using Pythagoras' theorem with the perpendicular height and the distance from the centre of the base to the midpoint of a side.

   1. Distance from centre of base to midpoint of a side: $\frac{8}{2} = 4$ m
   2. Slant height: $l = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ m
   3. Area of one triangular face: $\frac{1}{2} \times 8 \times 5 = 20$ m²
   4. Total area of 4 triangular faces: $4 \times 20 = 80$ m²
   5. Base area: $8^2 = 64$ m²
   6. Total surface area: $64 + 80 = 144$ m²

   Answer: $144$ m²

2. Example 2 · Surface area of a right cone

   A right cone has a base radius of 5 cm and a perpendicular height of 12 cm. Find the total surface area of the cone. Give your answer correct to one decimal place.

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   Key idea: Find the slant height using Pythagoras' theorem, then apply the cone surface area formula $\pi r l + \pi r^2$.

   1. Slant height: $l = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$ cm
   2. Curved surface area: $\pi \times 5 \times 13 = 65\pi$ cm²
   3. Base area: $\pi \times 5^2 = 25\pi$ cm²
   4. Total surface area: $65\pi + 25\pi = 90\pi \approx 282.7$ cm²

   Answer: $282.7$ cm²

3. Example 3 · Volume of a composite solid (cone on a cylinder)

   A grain silo is made up of a cylinder topped with a right cone. The cylinder has a radius of 3 m and a height of 10 m. The cone sits on top of the cylinder, has the same radius of 3 m, and has a perpendicular height of 4 m. Find the total volume of the silo. Give your answer correct to one decimal place.

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   Key idea: The total volume is the volume of the cylinder plus the volume of the cone, using $V_{\text{cone}} = \frac{1}{3}\pi r^2 h$.

   1. Volume of cylinder: $\pi \times 3^2 \times 10 = 90\pi$ m³
   2. Volume of cone: $\frac{1}{3} \times \pi \times 3^2 \times 4 = \frac{1}{3} \times 36\pi = 12\pi$ m³
   3. Total volume: $90\pi + 12\pi = 102\pi \approx 320.4$ m³

   Answer: $320.4$ m³

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