Measurement

Measurement deals with length, perimeter, area, surface area, volume and capacity. In examination questions, learners must select the correct formula, use consistent units, substitute accurately and round only the final answer. Diagrams are not always drawn to scale, so use the given measurements rather than estimating from the drawing.

Two dimensional measurement

Two dimensional measurement concerns flat figures. Length is measured in units such as millimetres, centimetres and metres. Perimeter is measured in linear units, while area is measured in square units.

Important formulas

• Rectangle:

P=2(l+w)

A=lw

• Triangle:

A=\frac{1}{2}bh

• Parallelogram:

A=bh

• Trapezium:

A=\frac{1}{2}(a+b)h

• Circle:

C=2\pi r

A=\pi r^2

Remember

• The perpendicular height must be used in an area formula.
• The radius is half the diameter.
• Convert all measurements to the same unit before calculating.

Example

Question: A trapezium has parallel sides of length 8\text{ cm} and 14\text{ cm}. Its perpendicular height is 6\text{ cm}. Calculate its area.

Step 1: Identify the given information.

a=8\text{ cm},\quad b=14\text{ cm},\quad h=6\text{ cm}

Step 2: Write the formula.

A=\frac{1}{2}(a+b)h

Step 3: Substitute the values.

A=\frac{1}{2}(8+14)(6)

Step 4: Simplify.

A=\frac{1}{2}(22)(6)

A=66

Step 5: State the final answer.

A=66\text{ cm}^2

Common Mistake

Do not add all the side lengths when area is required. Adding side lengths gives the perimeter.

Exam Tip

A complete examination solution should show the required length, the correct area formula, substitution and the final answer with square units. This step-by-step approach is reflected in the allocation of marks for area calculations.

Three dimensional measurement

Three dimensional objects have length, width and height. They occupy space and may have flat faces, curved surfaces, edges and vertices.

Important facts

• A prism has a constant cross-section throughout its length.
• A cylinder has two congruent circular bases.
• A pyramid has one base and triangular faces meeting at an apex.
• A cone has a circular base and a curved surface meeting at an apex.
• A sphere has one continuous curved surface.
• The perpendicular height is used for volume.
• The slant height of a cone or pyramid may be needed for surface area.

Example

Question: A cone has a perpendicular height of 12\text{ cm} and a radius of 5\text{ cm}. Calculate its slant height.

Step 1: Identify the given information.

h=12\text{ cm},\quad r=5\text{ cm}

Step 2: Apply the theorem of Pythagoras.

s^2=h^2+r^2

Step 3: Substitute the values.

s^2=12^2+5^2

Step 4: Simplify.

s^2=144+25

s^2=169

s=\sqrt{169}

Step 5: State the final answer.

s=13\text{ cm}

Common Mistake

Do not use the slant height in the volume formula. Volume always uses the perpendicular height.

Surface area

Surface area is the total area of all the outside surfaces of a three-dimensional object. A net can help identify every face that must be included.

Important formulas

• Rectangular prism:

SA=2(lw+lh+wh)

• Closed cylinder:

SA=2\pi r^2+2\pi rh

• Closed cone:

SA=\pi r^2+\pi rs

• Sphere:

SA=4\pi r^2

For an open container, exclude the missing face or base.

Example

Question: Calculate the total surface area of a closed cylinder with radius 3\text{ cm} and height 10\text{ cm}.

Step 1: Identify the given information.

r=3\text{ cm},\quad h=10\text{ cm}

Step 2: Write the formula.

SA=2\pi r^2+2\pi rh

Step 3: Substitute the values.

SA=2\pi(3)^2+2\pi(3)(10)

Step 4: Simplify.

SA=18\pi+60\pi

SA=78\pi

SA\approx245{,}04

Step 5: State the final answer.

SA\approx245{,}04\text{ cm}^2

Common Mistake

Forgetting one or both circular ends of a closed cylinder leads to an incomplete surface area.

Volume

Volume measures the amount of space inside a three-dimensional object. It is measured in cubic units. Capacity is commonly measured in millilitres or litres.

Important conversions

1\text{ cm}^3=1\text{ mL}

1\,000\text{ cm}^3=1\text{ L}

1\text{ m}^3=1\,000\text{ L}

Important formulas

• Prism:

V=A_{\text{base}}h

• Cylinder:

V=\pi r^2h

• Pyramid:

V=\frac{1}{3}A_{\text{base}}h

• Cone:

V=\frac{1}{3}\pi r^2h

• Sphere:

V=\frac{4}{3}\pi r^3

Example

Question: Calculate the volume of a sphere with radius 6\text{ cm}.

Step 1: Identify the given information.

r=6\text{ cm}

Step 2: Write the formula.

V=\frac{4}{3}\pi r^3

Step 3: Substitute the value.

V=\frac{4}{3}\pi(6)^3

Step 4: Simplify.

V=\frac{4}{3}\pi(216)

V=288\pi

V\approx904{,}78

Step 5: State the final answer.

V\approx904{,}78\text{ cm}^3

Exam Tip

Keep the value involving pi on the calculator and round only the final answer. Do not assume missing measurements, because assumed values are not accepted in examination solutions.

Composite solids

A composite solid consists of two or more basic solids joined together or with one solid removed from another. Break the object into familiar parts before calculating.

Method

• Identify each basic solid.
• Decide whether volumes must be added or subtracted.
• Identify any shared surfaces that must not be included in surface area.
• Write a separate formula for each part.
• Use consistent units.

Example

Question: A solid consists of a cylinder with a hemisphere attached to its top. Both parts have radius 3\text{ cm}. The cylinder has height 8\text{ cm}. Calculate the total volume.

Step 1: Write the volume of the cylinder.

V_{\text{cylinder}}=\pi r^2h

Step 2: Substitute the values.

V_{\text{cylinder}}=\pi(3)^2(8)

V_{\text{cylinder}}=72\pi

Step 3: Write the volume of the hemisphere.

V_{\text{hemisphere}}=\frac{1}{2}\left(\frac{4}{3}\pi r^3\right)

Step 4: Substitute the radius.

V_{\text{hemisphere}}=\frac{2}{3}\pi(3)^3

V_{\text{hemisphere}}=18\pi

Step 5: Add the volumes.

V_{\text{total}}=72\pi+18\pi

V_{\text{total}}=90\pi

V_{\text{total}}\approx282{,}74

Step 6: State the final answer.

V_{\text{total}}\approx282{,}74\text{ cm}^3

Common Mistake

When calculating the surface area of joined solids, do not include the circular surface where the cylinder and hemisphere meet because it is inside the solid.

Applications involving prisms, cylinders, cones, pyramids and spheres

Measurement is used in construction, packaging, storage, transport, manufacturing and design.

Prisms

Prisms model objects such as boxes, buildings, swimming pools and channels. Their volume is the area of the constant cross-section multiplied by the length.

Cylinders

Cylinders model tanks, pipes, cans and silos. Questions may involve capacity, material cost or the rate at which a tank fills.

Cones

Cones model funnels, tents and conical containers. Read carefully whether the perpendicular height or slant height is required.

Pyramids

Pyramids appear in roof structures, ornaments and packaging. The area of the base depends on whether the base is a square, rectangle or another polygon.

Spheres

Spheres model balls, globes and spherical tanks. A hemisphere is half a sphere.

Example

Question: A cylindrical water tank has radius 1{,}5\text{ m} and height 2\text{ m}. Calculate its capacity in litres.

Step 1: Identify the given information.

r=1{,}5\text{ m},\quad h=2\text{ m}

Step 2: Write the formula.

V=\pi r^2h

Step 3: Substitute the values.

V=\pi(1{,}5)^2(2)

Step 4: Simplify.

V=4{,}5\pi

V\approx14{,}137\text{ m}^3

Step 5: Convert to litres.

14{,}137\times1\,000=14\,137

Step 6: State the final answer.

\text{Capacity}\approx14\,137\text{ L}

Exam Tip

• Draw and label a simplified diagram.
• Distinguish between radius and diameter.
• Check whether the object is open or closed.
• Use square units for area and cubic units for volume.
• Show all formulas and substitutions.
• Do not round intermediate answers.
• Check whether the final answer is reasonable in the given context.