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Mathematics Teaching Self-evaluation Tools

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Here you can see a summary of the areas in which you are confident and those in which you are less confident; there are some ideas and suggestions which may help you in your professional learning.

Showing all next steps for the selected topic.Click on a question to show more information.

the distinction between mass and weight?(show/hide all)

What this might look like in the classroom

Question 1:
Suggest suitable units for measuring the mass of:

a car

a person

an egg

Answer 1:

tonnes

kilogrammes

grammes

Taking this mathematics further

Physicists in particular work with the distinct difference between mass and weight. Why is this? How does this connect to an understanding of the force acting on an object?

Newton's second law is F = ma
where F is the force m the mass and a the acceleration
So if we pushed a mass of 1kg with a horizontal force of 10 newtons. The mass would move with an acceleration of 10 m/s^{2}.

Making connections

Colloquially we often use “weight” when we actually mean “mass”. However, it is important to distinguish between mass and weight and to use them in the correct contexts.

the distinction between volume and capacity?(show/hide all)

What this might look like in the classroom

Question 1:
Suggest suitable units for measuring:

The volume of a computer mouse

The volume of water in a swimming pool

The amount of drink in a can

The volume of the rubber on the end of a pencil

The volume of rock removed from a quarry in one day

The amount of water a swimming pool can hold

Answer 1:

Cubic centimetres

Cubic metres

Millilitres

Cubic millimetres

Cubic metres

Litres

Taking this mathematics further

Research the history of the metric system:

When was it developed?

Who developed it?

Why did they develop it?

What is a metre?

Where is the metre?

In addition to the standard prefixes of milli, centi and kilo, find out what these prefixes mean in the metric system:

Deci

Deca

Hecto

Making connections

The metric system also interrelates volume and capacity with mass:

1 millilitre of water has a mass of 1 gram

1 litre of water has a mass of 1 kilogram

1 cubic metre of water has a mass of 1 tonne.

As such, in the metric system, length, area, volume, capacity and mass are all related.

Using the connection between volume and capacity means that irregular shaped objects can have their volume measured by considering the amount of water that they displace

Question 1: A piece of tape is measured and found to be 3.2 cm long to 1 decimal place. Which of the following are possible lengths for the tape?

3.22 cm

3.19 cm

3.25 cm

3.15 cm

Answer 1:
a, b and d

Question 2:
The world record for the 100 m sprint is held by Usain Bolt at 9.72 seconds. What are the fastest and slowest times this could be? If another runner runs 100 m in 9.719 seconds is this a new world record?

Answer 2:
Fastest time 9.715 seconds
Slowest time 9^{.}.7249
9.719 seconds might or might not be faster as it is 9.72 seconds to the nearest hundredth and we don’t know Usain Bolt’s time to that level of accuracy to compare.

Question 3:
If the mass of an object is 60 kg to the nearest whole number, why could its mass be more than that of another object that has a mass of 60 kg to the nearest whole number?

Answer 3:
A mass of 60 kg has its minimum possible mass as 59.5 kg. Another mass could have mass 60.2 kg, for example. They would then both be 60 kg to the nearest whole number but one would have a greater mass than another.

Taking this mathematics further

Can you think of some practical examples where you would always want to work with the upper bound rather than the approximate value?

All branches of practical science rely on a good understanding of approximation in measurement. Quality control, for example in the checking of materials in manufacture, requires an understanding of what is an acceptable level of variation in a measurement. Surveying, architecture and other disciplines relying on measurement also require an understanding of the approximate nature of measurement.

This area of mathematics is developed further into tolerance intervals and the calculation of absolute and relative errors.

Making connections

Any length can always be measured to a greater degree of accuracy and can never be measured exactly. This is true of other units of measure using a continuous scale, such as mass, capacity and time. This means that any stated measure is always an approximation.

The measurement can be inaccurate by up to half a unit larger or half a unit smaller. The greatest possible value for the measurement for the given degree of accuracy is called the upper bound and the smallest possible value is called the lower bound. All values in the interval given by the two bounds round to the stated measurement given the degree of accuracy.

Work on accuracy relies on a sound understanding of rounding and approximation and the notion of giving a solution to an appropriate degree of accuracy.

Higher tier GCSE examinations commonly contain problems where students are required to calculate with approximate measurements.

Question 1:
A motorbike travels 80 kilometres in 2 hours. What is its average speed?

Answer 1:
As average speed can be calculated using
speed =

distance

time

The motorbike’s average speed is 80 ¸ 2 = 40 (km/h). As the original units are ‘kilometres’ and ‘hours’, the correct ‘compound measure’ is kilometres per hour (km/h). Of course, it is straightforward to find 40 km/h without using the formula in this case.

Question 2:
A car travels 29 miles in 45 minutes. What is its average speed?

Answer 2:
Applying the formula speed =

distance

time

gives an average speed of 0.644…, miles per minutes. This now needs to be converted into a more sensible choice of measure of speed − miles per hour. Alternatively, the minutes could be converted into hours first (45 minutes = 0.75 hours). The formula then yields the result 38.7 mph (to 1 decimal place).

Taking this mathematics further

Research the densities of precious metals and what different ‘carats’ of gold refer to.

Use opportunities to establish cross−curricular links where appropriate; e.g. density and pressure measurements in science and population density in geography.

Average speed = ‘total distance’ ¸ ‘total time’ is an example of a harmonic mean. Establish a definition of this and its possible implications.

Making connections

Learners will need to be confident at rounding, and at converting between units − especially of time.

Challenge the understanding of average speed by considering an example such as:
Joshua walks to school at 4 km/h. His average speed for the journey to and from school was 2 km/h. What was his speed on the return journey?
The answer cannot be 0 km/h, and this demonstrates that you do not find average speeds by averaging two (or more) speeds. Average speed is actually ‘total distance’ ¸ total time.
Take his journey to school to be 1 km in length and work out the average speed. Repeat for a 2 km journey.

Create opportunities to bring together knowledge of conversions, approximations and calculation to solve problems such as:
An explorer walks 16 000 miles in 18 months. What is his average speed?

Point C is on a bearing of 032° from A, and a bearing of 289° from B. Mark the position of point C on the diagram.

Answer 1

The bearing of A from B is 63^{o} + 180^{o} = 243^{o}.
The bearing of B from A is 63^{o}.
360^{o} − 289^{o} = 71^{o}. So:

Taking this mathematics further

Some learners (especially those involved in the Duke of Edinburgh Award Scheme for example) may be familiar with using a compass and finding bearings; solving problems that can have a real impact on safety. Find out about:

Setting the map

Following a bearing taken from a map

Magnetic variation

Back bearings

Resection

Making connections

Questions on bearings will require learners to work accurately with pencil and paper constructions.

It is important to think about the language of bearings. Learners need to be clear that the ‘bearing of A from B’, implies that they are standing at B.

Knowledge of angle properties with parallel lines is likely to be needed in order to solve problems involving bearings

Simple problems involving bearings might be solved using accurate scale drawing

Some problems involving bearings will involve trigonometry.

the area and perimeter of sectors and segments of a circle?(show/hide all)

What this might look like in the classroom

Question 1
A circle has a radius of 3 cm.
If I cut it into 3 sectors of 120^{o}, 90^{o} and 150^{o}, what will be the area and perimeter of each sector?

Answer 1
Each sector is a fraction of the whole circle The 120^{o} sector is

120

360

=

1

3

of the circle
The 90^{o} sector is

90

360

=

1

4

of the circle
The 150^{o} sector is

150

360

=

5

12

of the circle

The area of the whole circle is A = π x 3^{2} = 28.27433388… cm^{2}.

So the area of each sector is:

1

3

of 28.27433388… cm^{2} = 9.42 cm^{2} (3sf)

1

4

of 28.27433388… cm^{2} = 7.07 cm^{2} (3sf)

5

12

of 28.27433388… cm^{2} = 11.8 cm^{2} (3sf)

The circumference of the whole circle is C = π x d = π x 6 = 18.84955592… cm

So the arc length of each sector is:

1

3

of 18.84955592… cm = 6.28 cm (3sf)

1

4

of 18.84955592… cm = 4.71 cm (3sf)

5

12

of 18.84955592… cm = 7.85 cm (3sf)

The perimeter of each sector would include the length of the radius added to these figures twice:
12.28 cm (3sf)
10.71 cm (3sf)
13.85 cm (3sf)

Taking this mathematics further

What would the angle at the centre have to be if the sector had an area twice that of the segment?

Making connections

Learners need to be adept at calculating areas and circumferences of circles, and also at using the formula in reverse; for example, calculating the diameter of a circle given the area.

Ensure that learners do not round answers in the middle of a calculation.

Trigonometrical methods will be required in order to calculate the area of a segment.

the properties of simple solids to solve problems involving surface area and volume?(show/hide all)

What this might look like in the classroom

Question
Work out the exact volume and surface area of a cylinder with height 2cm, and diameter of the cross−section 20cm.

Answer
The radius of the circular cross−section is 10 cm, so the area of the circle is:
A = πr^{2} = π x 10 x 10 = 100πcm^{2}

The volume is the area of the circle multiplied by the height:
V = 100π x 2 = 200πcm^{3}

The surface area of the cylinder can be found by adding the area of the circle (twice) to the area of a rectangle with dimensions 2cm and 20πcm (the circumference of the circle):
SA = 100πcm^{2} + 100πcm^{2} + 40 cm^{2} = 240 cm^{2}.

Taking this mathematics further

Explore ways of finding the surface area of the Platonic solids

Find out about Euler's theorem.

What is meant by an equable shape in three dimensions? What examples of equable 3D shapes can you find?

Making connections

Learners need to be adept at calculating areas and circumferences of circles, and also at using the formula in reverse; for example, calculating the diameter of a circle given the area.

Likewise, fluency with the use of the formulae for areas of rectangles, parallelograms, triangles and trapezia may be required.

When working with shapes such as a triangular prism, Pythagoras’ theorem may be required.

the use of the sine rule and cosine rule?(show/hide all)

What this might look like in the classroom

We can use the cosine formula to find the length of a side or size of an angle in a non−right−angled triangle.

Question
Find the missing side of this triangle Redraw triangle with the side opposite angle A marked ‘a’, the one opposite B marked ‘b = 7 cm’, and the one opposite C marked ‘c = 3 cm’

Answer Using the cosine rule:
a^{2} = b^{2} + c^{2} − 2bc cos A
a^{2} = 7^{2} + 3^{2} − 2 × 3 × 7 × cos 35^{o}
a^{2} = 58 − 42 × cos 35^{o}
a^{2} = 23.59561414…
a = 4.86 cm (3sf)

Taking this mathematics further

Research how the use of the cosine and sine rules has contributed to scientific development; for example, surveys to calculate the distance from the earth to the moon.

Making connections

The sine and cosine rules can be used to solve problems involving missing sides and angles in triangles which are not right−angled. Although it is unnecessarily complicated, the cosine rule can be used in right angled triangles as the fact that cos 90^{o} = 0 means that a^{2} = b^{2} + c^{2} − 2bc cos A reduces to a^{2} = b^{2} + c^{2}.

These triangles may be generated by problems involving bearings.

why caution is sometimes needed in interpreting results from the sine rule?(show/hide all)

What this might look like in the classroom

If you know that a triangle has sides of length 4 cm and 6 cm, it must be true that the third side is smaller than 10 (since 6 + 4 = 10) and bigger than 2 (since 6 − 4 = 2).

Problem Sketch two triangles with sides of 4 cm and 8 cm. If the angle between these sides is 40(, where is the biggest angle in the triangle?
Answer
The biggest angle is always opposite the longest side

Taking this mathematics further

Think about the conditions for two triangles to be congruent. Is it enough that two sides and an angle are the same? If not, why not?

Making connections

Use compasses ruler and protractor to construct the triangles.

Use the graph of y = sin x to explore why it is that there are two possible solutions in some cases.

Problem
The area of a triangle is 8.2cm^{2} and two of its sides are 3.2 cm and 7.1 cm. Find the angle between these sides.

Answer Using the formula
Area =

1

2

ab sin C:
8.2 =

1

2

x 3.2 x 7.1 x sin C
8.2 = 11.36 x sin C
0.7218… = sin C
C = 46.2^{o} (3sf)

Taking this mathematics further

Another formula for finding the area of a triangle is Heron's formula.
What information is required in order to calculate the area of a triangle using his formula?
How does it complement the two methods for calculating the area of a triangle already known?
Find out more about this mathematician and the contributions he made.

Making connections

Learners first think of the area of a triangle as the area of half of a rectangle. They then explore obtuse−angled triangles where the dissection of a rectangle is not so obvious, but the derived formula; half the base multiplied by the perpendicular height, still holds.

Learners can now find the area of a triangle if they know two sides of a triangle and the angle between them.

It is also possible to find the area of a triangle if the three sides are known (see ‘Taking this mathematics further’)

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