Mensuration | Surface Area & Volume of 3D Solids

Synopsis

This article covers the essential skills of surface area and volume of 3D solids – a core topic in DSE Paper 1 Section A(2) and Section B worth 4–6 marks. You will learn the formulas for volume and surface area of common solids: prisms, cylinders, cones, spheres, and pyramids. You will also learn the length, area, and volume ratios for similar solids, and how to solve problems involving composite solids. The article includes step-by-step worked examples, DSE exam techniques, practice questions. These are essential skills that appear regularly in DSE papers.


Learning Objectives

By the end of this article, you should be able to:

  • Recall and apply the formulas for volume of prisms, cylinders, cones, spheres, and pyramids.
  • Recall and apply the formulas for total surface area and lateral surface area of these solids.
  • Apply the length, area, and volume ratios for similar solids.
  • Solve problems involving composite solids (combinations of two or more solids).
  • Determine the effect of changing dimensions on volume and surface area.
  • Solve DSE-style problems involving 3D mensuration.

1. Introduction:

3D mensuration appears every year in DSE Paper 1 Section A(2) and Section B. Questions may ask you to:

  • Calculate the volume or surface area of a given solid
  • Find an unknown dimension given volume or surface area
  • Apply similar solids ratios to find missing measurements
  • Solve composite solid problems (e.g., a cylinder with a hemisphere on top)

These are essential skills that also appear in real-world contexts like capacity, packaging, and construction.

DSE Exam Tip

3D mensuration questions often appear in Section A(2) as short-answer questions worth 3–4 marks, and in Section B as longer problems worth 5–6 marks.

2. Volume and Surface Area Formulas

Prisms

  • Volume = Area of base × Height
  • Total Surface Area = 2 × (Area of base) + (Perimeter of base) × Height (lateral area)

Cylinders

  • Volume = \( \pi r^2 h \)
  • Lateral Surface Area = \( 2\pi r h \)
  • Total Surface Area = \( 2\pi r^2 + 2\pi r h \)

Cones

  • Volume = \( \frac{1}{3} \pi r^2 h \)
  • Slant height \( l = \sqrt{r^2 + h^2} \)
  • Lateral Surface Area = \( \pi r l \)
  • Total Surface Area = \( \pi r^2 + \pi r l \)

Spheres

  • Volume = \( \frac{4}{3} \pi r^3 \)
  • Surface Area = \( 4\pi r^2 \)

Pyramids

  • Volume = \( \frac{1}{3} \times \text{Area of base} \times \text{Height} \)
  • Surface area = sum of areas of all faces (base + triangular faces)
DSE Memory Aid

For cones and pyramids, the volume is one-third of the corresponding prism/cylinder. Remember: \( V_{\text{cone}} = \frac{1}{3} V_{\text{cylinder}} \).

3. Similar Solids

If two solids are similar, their corresponding lengths are proportional. Let the linear scale factor be \( k \) (ratio of corresponding lengths). Then:

  • Area ratio = \( k^2 \)
  • Volume ratio = \( k^3 \)

This means if you double the dimensions (scale factor 2), the surface area becomes 4 times larger, and the volume becomes 8 times larger.

Worked Example

Question: Two similar cylinders have heights 5 cm and 15 cm. If the smaller cylinder has volume 200 cm³, find the volume of the larger cylinder.

Solution

Scale factor \( k = \frac{15}{5} = 3 \).

Volume ratio = \( k^3 = 27 \).

Volume of larger = \( 200 \times 27 = 5400 \) cm³.

Answer: 5400 cm³.

4. Composite Solids

A composite solid is made up of two or more simple solids. To find the total volume or surface area:

  1. Identify the individual solids (e.g., cylinder + hemisphere).
  2. Calculate the volume or surface area of each part.
  3. For volume: add the volumes.
  4. For surface area: add the exposed areas (do not count internal faces where solids meet).

Worked Example

Question: A solid is made of a cylinder of radius 4 cm and height 6 cm, with a hemisphere of the same radius on top. Find the total volume.

Solution

Volume of cylinder = \( \pi (4)^2 (6) = 96\pi \) cm³.

Volume of hemisphere = \( \frac{1}{2} \times \frac{4}{3}\pi (4)^3 = \frac{2}{3}\pi (64) = \frac{128\pi}{3} \) cm³.

Total volume = \( 96\pi + \frac{128\pi}{3} = \frac{288\pi + 128\pi}{3} = \frac{416\pi}{3} \) cm³.

Answer: \( \frac{416\pi}{3} \) cm³.

5. Worked Examples

Example 1: Volume of a Cone

Question: A cone has radius 6 cm and height 8 cm. Find its volume.

Solution

\( V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (6)^2 (8) = \frac{1}{3} \pi \times 36 \times 8 = 96\pi \) cm³.

Answer: \( 96\pi \) cm³.

Example 2: Surface Area of a Sphere

Question: Find the surface area of a sphere with radius 7 cm.

Solution

\( A = 4\pi r^2 = 4\pi (7)^2 = 4\pi \times 49 = 196\pi \) cm².

Answer: \( 196\pi \) cm².

Example 3: Similar Solids

Question: Two similar pyramids have volumes 54 cm³ and 432 cm³. Find the ratio of their corresponding heights.

Solution

Volume ratio = \( \frac{432}{54} = 8 \).

Since volume ratio = \( k^3 \), \( k = \sqrt[3]{8} = 2 \).

So the height ratio is 2:1.

Answer: 2:1.

6. DSE-Style Practice Questions

Section A(2) & Section B Style

Question 1 MC
Find the volume of a cylinder with radius 3 cm and height 10 cm.
A. \( 30\pi \)     B. \( 60\pi \)     C. \( 90\pi \)     D. \( 120\pi \)

Question 2 MC
The surface area of a sphere is \( 144\pi \) cm². Find its radius.
A. \( 4 \) cm     B. \( 6 \) cm     C. \( 8 \) cm     D. \( 12 \) cm

Question 3 Short Answer
A cone has radius 5 cm and slant height 13 cm. Find its total surface area.

Question 4 Short Answer
Two similar cylinders have heights 4 cm and 10 cm. If the smaller has volume 80 cm³, find the volume of the larger.

Question 5 MC
A solid is formed by a hemisphere on top of a cylinder of the same radius. If the radius is 3 cm and the cylinder height is 5 cm, what is the total volume?
A. \( 45\pi + 18\pi \)     B. \( 45\pi + 9\pi \)     C. \( 45\pi + 12\pi \)     D. \( 45\pi + 6\pi \)

7. Solutions with Explanations

Question 1: C. \( 90\pi \)
\( V = \pi r^2 h = \pi (3)^2 (10) = 90\pi \).
Question 2: B. \( 6 \) cm
\( 4\pi r^2 = 144\pi \)\( r^2 = 36 \)\( r = 6 \).
Question 3: \( 90\pi \) cm²
\( r = 5 \), \( l = 13 \). \( A = \pi r^2 + \pi r l = \pi (25) + \pi (5)(13) = 25\pi + 65\pi = 90\pi \).
Question 4: \( 1250 \) cm³
Scale factor \( k = \frac{10}{4} = 2.5 \). Volume ratio = \( k^3 = 15.625 \). Volume = \( 80 \times 15.625 = 1250 \).
Question 5: A. \( 45\pi + 18\pi \)
Cylinder volume = \( \pi (3)^2 (5) = 45\pi \). Hemisphere volume = \( \frac{1}{2} \times \frac{4}{3}\pi (3)^3 = \frac{2}{3}\pi (27) = 18\pi \). Total = \( 45\pi + 18\pi \).

9. Exercise

Click the following link to have
 An Exercise on Mensuration | Surface Area & Volume of 3D Solids

Key Takeaways

What You Should Remember
  • Volume formulas: Prisms/cylinders (\( \text{Base} \times \text{Height} \)), cones/pyramids (\( \frac{1}{3} \text{Base} \times \text{Height} \)), spheres (\( \frac{4}{3}\pi r^3 \)).
  • Surface area: Sum of areas of all faces. For cylinders, \( 2\pi r^2 + 2\pi r h \); for cones, \( \pi r^2 + \pi r l \); for spheres, \( 4\pi r^2 \).
  • Similar solids: Area ratio = \( k^2 \), Volume ratio = \( k^3 \).
  • Composite solids: Break into parts, add volumes, add exposed surface areas.
  • This topic guarantees 4–6 marks in DSE Paper 1 Section A(2) and Section B – master these skills!

Summary Checklist for Revision

  • Volume of prism/cylinder: base area × height
  • Volume of cone/pyramid: \( \frac{1}{3} \) base area × height
  • Volume of sphere: \( \frac{4}{3}\pi r^3 \)
  • Surface area: sum of all faces
  • Slant height of cone: \( l = \sqrt{r^2 + h^2} \)
  • Similar solids: area ratio = \( k^2 \), volume ratio = \( k^3 \)
  • Composite solids: add/subtract parts
  • Check units (cm³, m³, etc.)