Surface Area and Volume of Solid Shapes – SS1 Mathematics Lesson Note (Third Term)

A solid shape is a three-dimensional figure with length, breadth and height, and its surface area and volume are found with standard formulae.

Behavioural objectives

By the end of the lesson, students should be able to:

1. State the number of faces, edges and vertices of a cube, cuboid, triangular prism, cylinder, cone and rectangular-based pyramid, and describe their nets.
2. Use the formulae for the volume and surface area of a cube, cuboid and cylinder.
3. Show that the curved surface area of a cone is πrl and its total surface area is πr(l + r).
4. Use Pythagoras’ theorem to find the slant height or height of a cone or pyramid.
5. Calculate the volume of a pyramid and of a cone.

What are the key terms used for solid shapes?

These are the words used to describe any solid.

• Face: a flat or curved surface of a solid.
• Edge: the line where two faces meet.
• Vertex: the point where edges meet (plural: vertices).
• Net: a flat pattern that can be folded to make the solid.
• Prism: a solid with the same cross-section all through its length.
• Slant height (l): the distance from the vertex of a cone, down the curved surface, to the edge of the base.

What are the properties of solid shapes?

Each common solid has a fixed number of faces, edges and vertices, and its own net.

Cube

• Faces: 6 square faces.
• Edges: 12 straight edges.
• Vertices: 8.
• Net: 6 square faces joined together.

Cuboid

• Faces: 6 rectangular faces.
• Edges: 12 straight edges.
• Vertices: 8.
• Net: 6 rectangular faces joined together.

Triangular prism

• Faces: 3 rectangular faces and 2 triangular end faces.
• Edges: 9 straight edges.
• Vertices: 6.
• Net: 3 rectangles and 2 triangles joined together.

Cylinder

• Faces: 2 circular faces and 1 curved surface.
• Edges: 2 curved edges.
• Vertices: none.
• Net: 2 circles and 1 rectangle.

Cone

• Faces: 1 circular face and 1 curved surface.
• Edges: 1 curved edge.
• Vertices: 1.
• Net: a sector of a circle and a circle.

Rectangular-based pyramid

• Faces: 4 triangular faces and 1 rectangular face.
• Edges: 8 straight edges.
• Vertices: 5.
• Net: 4 triangles and 1 rectangle.

Summary table of properties

What are the surface area and volume formulae for common solids?

The volume of a prism is the area of its uniform cross-section multiplied by its perpendicular height.

• A prism has the same cross-section all through its length.
• Cubes, cuboids and cylinders are examples of prisms.
• Volume of prism: area of cross-section × height = area of base × height.

Worked example: cube

Find the volume and total surface area of a cube of side 5 cm.

1. Volume = l³ = 5 × 5 × 5 = 125 cm³.
2. Surface area = 6l² = 6 × 25 = 150 cm².

Worked example: cuboid

Find the volume and total surface area of a cuboid 6 cm by 4 cm by 3 cm.

1. Volume = lbh = 6 × 4 × 3 = 72 cm³.
2. Surface area = 2(lb + lh + bh) = 2(24 + 18 + 12) = 2 × 54 = 108 cm².

Worked example: cylinder

Find the volume and total surface area of a cylinder of radius 7 cm and height 10 cm. Use π = 22/7.

1. Volume = πr²h = 22/7 × 7 × 7 × 10 = 1540 cm³.
2. Total surface area = 2πr(h + r) = 2 × 22/7 × 7 × (10 + 7).
3. = 44 × 17 = 748 cm².

How do you find the surface area of a cone?

The curved surface area of a cone is πrl, where r is the base radius and l is the slant height.

• A sector of a circle can be bent to form the curved surface of an open cone.
• In the figure, the sector OAXB has radius l, and the arc AXB subtends angle θ at O.
• The sector is bent to form a cone of base radius r and slant height l.
• The area of the sector equals the curved surface area of the cone.
• The length of arc AXB equals the circumference of the circular base of the cone.

How do you derive the formulae for a cone?

1. Curved surface area of cone = (θ/360°) × πl² . . . (i)
2. The arc length equals the base circumference: (θ/360°) × 2πl = 2πr.
3. Divide both sides by 2π: (θ/360°) × l = r.
4. Divide both sides by l: θ/360° = r/l.
5. Substitute r/l for θ/360° in (i): curved surface area = (r/l) × πl² = πrl.
6. Total surface area = curved surface area + area of circular base = πrl + πr² = πr(l + r).

Worked example: a paper cone

A paper cone has a diameter of 8 cm and a height of 3 cm.

• (a) Sketch the cone and use Pythagoras’ theorem to calculate its slant height.
• (b) Calculate the curved surface area of the cone in terms of π.
• (c) If the cone is cut and opened out into the sector of a circle, find the angle of the sector.
• (d) Assuming the paper cone is closed at its base, find the total surface area.

Solution to (a)

1. Radius = diameter ÷ 2 = 8 ÷ 2 = 4 cm.
2. In the right-angled triangle OBC: l² = |OB|² + |BC|².
3. l² = 3² + 4² = 9 + 16 = 25.
4. l = √25 = 5 cm.

Solution to (b)

1. Curved surface area = πrl = π × 4 × 5.
2. = 20π cm².

Solution to (c)

1. The area of the sector equals the curved surface area: (θ/360) × π × 5² = 20π.
2. (θ/360) × 25 = 20.
3. 5θ/72 = 20 (dividing 360 and 25 by 5).
4. 5θ = 72 × 20 = 1440.
5. θ = 288°.

Solution to (d)

1. Area of circular base = πr² = π × 4² = 16π cm².
2. Total surface area = 20π + 16π = 36π cm².

How do you find the volume of a pyramid and a cone?

The volume of a pyramid or a cone is one third of the base area multiplied by the height.

• Square-based pyramid: ⅓ × b² × h.
• Rectangular-based pyramid: ⅓ × l × b × h.
• Cone: ⅓ × πr² × h.

Example 1: rectangular-based pyramid

A pyramid 8 cm high stands on a rectangular base 6 cm by 4 cm. Calculate the volume of the pyramid.

1. Volume = ⅓ × l × b × h.
2. = ⅓ × 6 × 4 × 8 cm³.
3. = 8 × 8 cm³.
4. = 64 cm³.

Example 2: right pyramid with a slanted edge

A right pyramid on a base 4 cm square has a slanted edge of 6 cm. Calculate the volume of the pyramid.

Let ABCD be the square base, O the vertex of the pyramid and E the midpoint of DB.

1. Find the diagonal DB: |DB|² = |DC|² + |CB|² (Pythagoras’ theorem).
2. |DB|² = 4² + 4² = 16 + 16 = 32.
3. |DB| = √32 = 4√2 cm.
4. E is the midpoint of DB, so |EB| = ½ × 4√2 = 2√2 cm.
5. In the right-angled triangle OEB: OE² + EB² = OB².
6. OE² + (2√2)² = 6².
7. OE² + 8 = 36.
8. OE² = 36 – 8 = 28.
9. OE = √28 = √(4 × 7) = 2√7 cm. This is the height of the pyramid.
10. Volume = ⅓ × b² × h = ⅓ × 4² × 2√7 cm³.
11. = (32 × √7)/3 cm³.
12. = (32 × 2.646)/3 = 32 × 0.882 = 28.224 cm³.
13. = 28.2 cm³ to 1 d.p.

Example 3: cone

Calculate the volume of a cone 14 cm in base diameter and 24 cm high. Use π = 22/7.

1. Radius = 14 ÷ 2 = 7 cm.
2. Volume = ⅓πr²h.
3. = ⅓ × 22/7 × 7² × 24 cm³.
4. = ⅓ × 22/7 × 49 × 24 cm³.
5. = 22 × 56 cm³.
6. = 1232 cm³.

Evaluation

1. State the number of faces, edges and vertices of a triangular prism.
2. Write down the formula for the total surface area of a cylinder.
3. Describe the net of a cone.
4. A cone has a base radius of 3 cm and a slant height of 5 cm. Calculate its curved surface area in terms of π.
5. A pyramid 9 cm high stands on a rectangular base 5 cm by 3 cm. Calculate its volume.

Assignment

Answer all the questions below. Use π = 22/7 where needed.

1. Calculate the volume of a cylinder which has a radius of 21 cm and a height of 6 cm.
   A. 8500 cm³; B. 8316 cm³; C. 7632 cm³; D. 7500 cm³; E. 8000 cm³
2. Calculate the total surface area of the cylinder in question 1.
   A. 5346 cm²; B. 4653 cm²; C. 3000 cm²; D. 3564 cm²; E. 3800 cm²
3. Calculate the volume of a cone which has a base diameter of 7 cm and a height of 6 cm.
   A. 77 cm³; B. 70 cm³; C. 88 cm³; D. 90 cm³; E. 65 cm³
4. Calculate the curved surface area of the cone in question 3 above.
   A. 152 cm²; B. 150 cm²; C. 132 cm²; D. 142 cm²; E. 160 cm²
5. Calculate the total surface area of a cuboid which is 8 cm by 5 cm by 3 cm.
   A. 198 cm²; B. 178 cm²; C. 188 cm²; D. 168 cm²; E. 158 cm²
6. A water tank is 1.2 m square and 1.35 m deep. It is half full of water. How many times can a 9-litre bucket be filled from the tank?
7. A measuring cylinder of radius 3 cm contains water to a height of 49 cm. If this water is poured into a similar cylinder of radius 7 cm, what will be the height of the water column?
8. A solid cone has a circular base of radius 7 cm. The vertical height of the cone is 15 cm. The cone is melted and recast into a metal cube of side x cm. Calculate, correct to 3 significant figures, the value of x.
9. A cylindrical container with a diameter of 80 cm and a height of 50 cm is full of liquid. The liquid is then poured into another cylinder with a diameter of 90 cm. Calculate the depth of the liquid in the second cylinder.