physic

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forces and movement

speed,velocity and accerelation-

Average speed

When an object moves in a straight line at a steady speed, you can calculate its average speed if you know how far it travels and how long it takes. The following equation shows the relationship between average speed, distance moved and time taken:

Equation: average~speed = frac{distance~moved}{time~taken} (http://equation-chef.files.bbci.co.uk/content/4a0acb2cae12474a61e1ed626a67266e/18)

where:

average speed is measured in metres per second, m/s

distance moved is measured in metres, m

time taken is measured in seconds, s

For example, a car travels 300 m in 20 s. Its average speed is:

300 ÷ 20 = 15 m/s

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speed, velocity and accrerlation

Distance-time graphs

A distance-time graph shows how far something travels over a period of time. The vertical axis of a distance-time graph is the distance travelled from the start. The horizontal axis is the time from the start.

Features of the graphs

When an object is stationary, the line on the graph is horizontal. When an object is moving at a steady speed in a straight line, the line on the graph is straight but sloped.

Red line extends 0 to 10 ms over 2s, steady speed. Green line extends 0 to 8 ms over 4s, steady speed. Stays at 8s for 3s, stationary, returns to 0 over 3s, steady speed returning to start. (http://a.files.bbci.co.uk/bam/live/content/zwydxnb/small)Red line extends 0 to 10 ms over 2s, steady speed. Green line extends 0 to 8 ms over 4s, steady speed. Stays at 8s for 3s, stationary, returns to 0 over 3s, steady speed returning to start. (http://a.files.bbci.co.uk/bam/live/content/zwydxnb/large)

Note that the steeper the line, the faster the object is travelling. The purple line is steeper than the green line because the purple line represents an object which is moving more quickly.

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electricity

Electrical symbols

Standard symbols

The diagram shows the standard circuit symbols.

Circuit symbols of open switch, closed switch, lamp, voltmeter, resistor, variable resistor, diode, light emitting diode, battery, cell, ammeter, fuse, thermistor and light dependant resistor. (http://a.files.bbci.co.uk/bam/live/content/zp8h34j/small)Circuit symbols of open switch, closed switch, lamp, voltmeter, resistor, variable resistor, diode, light emitting diode, battery, cell, ammeter, fuse, thermistor and light dependant resistor. (http://a.files.bbci.co.uk/bam/live/content/zp8h34j/large) < >

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  2. 2
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1 of 3 Glossary

  1. circuitA closed loop through which charge moves - from an energy source, through a series of components, and back into the energy source.
  2. currentMoving electric charges, for example, electrons moving through a metal wire.
  3. electrical componentA device in an electric circuit, such as a battery, switch or lamp.
  4. fuseAn electrical component that protects circuits and electrical devices from overload by melting when the current becomes too high.
  5. oscilloscopeA device used to measure and observe electrical signals over time.
  6. potential differenceThe potential difference (or voltage) of a supply is a measure of the energy given to the charge carriers in a circuit, units volts (V).

More Guides

  1. Electrical circuits, AC and DC
  2. Electrical safety
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circuits

Properties of circuits

The components in electrical circuits can be connected in series or in parallel.

Series connections

Components that are connected one after another on the same loop of the circuit are connected in series. The current that flows through each component connected in series is the same.

Two lamps connected in series

The circuit diagram shows a circuit with two lamps connected in series. If one lamp breaks, the other lamp will not light.

Two lamps connected in series with an open switch and a cell. (http://a.files.bbci.co.uk/bam/live/content/zndpb9q/small)Two lamps connected in series with an open switch and a cell. (http://a.files.bbci.co.uk/bam/live/content/zndpb9q/small)

Two lamps connected in series with an open switch and a cell

Series circuits are useful if you want a warning that one of the components in the circuit has failed. For example, a circuit breaker or fuse must be connected in series in order for it to work. If Christmas tree lights all go out when one bulb breaks, they are connected in series.

The sum of all the potential differences across the components in a series circuit is equal to the total potential difference across the power supply.

Parallel connections

Components that are connected on separate loops are connected in parallel. The current is shared between each component connected in parallel. The total amount of current flowing into the junction, or split, is equal to the total current flowing out. The current is described as being conserved.

Two lamps connected in parallel

The circuit diagram shows a circuit with two lamps connected in parallel. If one lamp breaks, the other lamp will still light.

Two lamps connected in parallel with an open switch and a cell. (http://a.files.bbci.co.uk/bam/live/content/zswngk7/small)Two lamps connected in parallel with an open switch and a cell. (http://a.files.bbci.co.uk/bam/live/content/zswngk7/small)

Two lamps connected in parallel with an open switch and a cell

The lights in most houses are connected in parallel. This means that they all receive the full voltage and if one bulb breaks the others remain on.

For a parallel circuit the current from the electrical supply is greater than the current in each branch. The sum of all the current in every branch is equal to the current from the electrical supply.

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ac dc current

AC and DC

AC electricity

Alternating current

If the current constantly changes direction it is called alternating current, or AC. Mains electricity is an AC supply. The UK mains supply is about 230 V. It has a frequency of 50 Hz, which means that it changes direction and back again 50 times a second. The diagram shows an oscilloscope screen displaying the signal from an AC supply.

An oscilloscope screen displaying the signal from an alternating current (AC) supply. It is a 50Hz sine wave that peaks at 230V. (http://a.files.bbci.co.uk/bam/live/content/z3m3kqt/small)

Oscilloscope trace illustrating alternating current

DC electricity

Direct current

If the current flows in only one direction it is called direct current, or DC. Batteries and solar cells supply DC electricity. A typical battery may supply 1.5 V. The diagram shows an oscilloscope screen displaying the signal from a DC supply.

An oscilloscope screen displaying the signal from a direct current (DC) supply. It is a horizontal straight line at 1.5V. (http://a.files.bbci.co.uk/bam/live/content/zv2v9j6/small)

Oscilloscope trace illustrating direct current

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atoms

Atomic structure

This section describes the nuclear model of an atom.

The nuclear model

An atom with the nucleus in the centre and five electrons around it. (http://a.files.bbci.co.uk/bam/live/content/z6ktvcw/small)An atom with the nucleus in the centre and five electrons around it. (http://a.files.bbci.co.uk/bam/live/content/z6ktvcw/large)

The structure of the atom

Atoms contain three sub-atomic particles called protons, neutrons and electrons. The protons and neutrons are found in the nucleus at the centre of the atom. The nucleus is very much smaller than the atom as a whole. The electrons are arranged in energy levels around the nucleus.

The table shows the properties of these three sub-atomic particles:

ParticleRelative massRelative charge Proton 1 +1 Neutron 1 0 Electron Almost zero –1

The number of electrons in an atom is always the same as the number of protons, so atoms are electrically neutral overall. Atoms can lose or gain electrons. When they do, they form charged particles called ions:

  • if an atom loses one or more electrons, it becomes a positively charged ion
  • if an atom gains one or more electrons, it becomes a negatively charge
6 of 12

maths-numbers

7 of 12

math numbers

Prime factors and decomposition

Prime numbers

You have most likely heard the term factor before. A factor is a number that goes into another. The factors of 10 for example are 1, 2, 5 and 10.

Prime numbers are a special set of numbers that only have two factors: themselves and 1.

An example of a prime number is 13 as it only has two factors: 13 and 1, whereas 9 is not a prime number as it has three factors: 9, 3 and 1.

The first 10 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.

It is interesting to note that 2 is the only even prime number. The number 1 is not prime as it only has a single factor (1 itself), and as we previously mentioned prime numbers must have two factors exactly.

Expressing numbers in prime factor form

Every whole number (with only one exception – the number 1) can be expressed as a product of prime numbers.

Examples

8 = 2 × 2 × 2 = 23

9 = 3 × 3 = 32

10 = 2 × 5

39 = 3 × 13

The process of breaking a number down into its prime factors is sometimes called prime factor decomposition.

Example

Express 300 in prime factor form.

First we start with the lowest prime number, 2. Because 2 is a factor of 300, we make a note of the '2' and then divide 300 by 2, leaving 150.

We can use a table to make this easier to see:

A two row table labelled Number and Prime Factors. The prime factor of 300 is 2. The prime factor of 150 is left blank. (http://a.files.bbci.co.uk/bam/live/content/zcpr7hv/small)A two row table labelled Number and Prime Factors. The prime factor of 300 is 2. The prime factor of 150 is left blank. (http://a.files.bbci.co.uk/bam/live/content/zcpr7hv/small)

Now we can divide by 2 again, leaving 75:

A three row table labelled Number and Prime Factors. The prime factor of 300 is 2. The prime factor of 150 is 2. The prime factor of 75 is left blank. (http://a.files.bbci.co.uk/bam/live/content/zgpr7hv/small)A three row table labelled Number and Prime Factors. The prime factor of 300 is 2. The prime factor of 150 is 2. The prime factor of 75 is left blank. (http://a.files.bbci.co.uk/bam/live/content/zgpr7hv/small)

We can no longer divide by 2, as 2 is not a factor of 75. We now try to divide by the next largest prime number which is 3:

A four row table labelled Number and Prime Factors. The prime factor of 300 is 2. The prime factor of 150 is 2. The prime factor of 75 is 3. The prime factor of 25 is left blank. (http://a.files.bbci.co.uk/bam/live/content/z242ycw/small)A four row table labelled Number and Prime Factors. The prime factor of 300 is 2. The prime factor of 150 is 2. The prime factor of 75 is 3. The prime factor of 25 is left blank. (http://a.files.bbci.co.uk/bam/live/content/z242ycw/small)

We can no longer divide by 3, as 3 is not a factor of 25. We must again look for a larger prime number to use. The next prime number in the list is 5:

A five row table labelled Number and Prime Factors. The prime factor of 300 is 2, 150 is 2, 75 is 3, 25 is 5. The prime factor of 5 is left blank. (http://a.files.bbci.co.uk/bam/live/content/zyxvdmn/small)A five row table labelled Number and Prime Factors. The prime factor of 300 is 2, 150 is 2, 75 is 3, 25 is 5. The prime factor of 5 is left blank. (http://a.files.bbci.co.uk/bam/live/content/zyxvdmn/small)

Finally we can divide by 5 again, leaving 1:

A six row table labelled Number and Prime Factors. The prime factor of 300 is 2, 150 is 2, 75 is 3, 25 is 5 and 5 is 5 . The prime factor of 1 is left blank. (http://a.files.bbci.co.uk/bam/live/content/zcyk6fr/small)A six row table labelled Number and Prime Factors. The prime factor of 300 is 2, 150 is 2, 75 is 3, 25 is 5 and 5 is 5 . The prime factor of 1 is left blank. (http://a.files.bbci.co.uk/bam/live/content/zcyk6fr/small)

When we have a 1 in the left-hand column, we have finished the process.

From the table we can see that 300 = 2 × 2 × 3 × 5 × 5 = 22 × 3 × 52. You can check this by doing the multiplication on a calculator.

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number index

Positive indices

Indices are a way of writing numbers in a more convenient form. The index or power is the small, raised number next to a normal letter or number. It represents the number of times that normal letter or number has been multiplied by itself, for example:

Equation: text{a}^{2}~text{= a times~a} (http://equation-chef.files.bbci.co.uk/content/655d53fac9e83fff1a6638357873596c/18)

Equation: text{6}^{4}~text{= 6 times~6 times~6 times~6} (http://equation-chef.files.bbci.co.uk/content/603b207cabb9a83a6c1663be34dac647/18)

Equation: text{b}^{5}~text{= b times~b times~b times~b times~b} (http://equation-chef.files.bbci.co.uk/content/4ecd45871f7acd49fc1f61a0fdc1ad5a/18)

For Equation: text{b}^{5} (http://equation-chef.files.bbci.co.uk/content/43f9675dc098340be4570ec943869def/18), Equation: text{b} (http://equation-chef.files.bbci.co.uk/content/9c40af9cbfd0d960d5e8d6b356333784/18) is the ‘base number’ and Equation: text{5} (http://equation-chef.files.bbci.co.uk/content/96c20eeb0907773de26b4531d116ae27/18) is the ‘index’.

The rules we are about to look at only work when the base numbers in the question are the same.

Multiplying indices

To multiply indices, add the powers together.

Example

24 × 22 = (2 × 2 × 2 × 2) × (2 × 2)

= 2 × 2 × 2 × 2 × 2 × 2

= 26

QQuestion

Evaluate 35 × 37

AAnswerReveal answerHide answer

312, because 5 + 7 = 12

Dividing indices

To divide indices, subtract the powers.

Example

A worked example of how to divide indices by subtracting the powers. (http://a.files.bbci.co.uk/bam/live/content/zccmv4j/small)A worked example of how to divide indices by subtracting the powers. (http://a.files.bbci.co.uk/bam/live/content/zccmv4j/small)

QQuestion

Evaluate y9 ÷ y6

AAnswerReveal answerHide answer

y3, because 9 - 6 = 3

Raising a power to a power

When a power is raised to a power, multiply the powers

Example

(53)2 = 53 × 53.

= 56, using the condition for multiplying indices

QQuestion

Evaluate (86)4

AAnswerReveal answerHide answer

824, because 6 × 4 = 24

Whole numbers and indices

We need to deal with the numbers and the powers separately. First multiply the numbers in front of the letters together, and then use the rule for multiplying indices to deal with the letters and powers.

Example

2a3 × 3a4

2 × 3 = 6

a3 × a4 = a7

So, 2a3 × 3a4 = 6a7

QQuestion

Evaluate 8b7 × 4b2

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ratio

Understanding ratio

A ratio shows the relationship between two quantities.

It is written in the form a:b which is read as ‘a to b’.

Example

Two teenagers in cinema seats eating sweets (http://a.files.bbci.co.uk/bam/live/content/z3j2rdm/small)

John has a pack of sweets. He shares them with a friend in the ratio 3:2. This means that for every 3 sweets John eats, he gives his friend 2.

If John eats 15 sweets (5 × 3), his friend will receive 10 sweets (5 × 2).

In the example, this means that John has 5 lots of 3 sweets so his friend will have 5 lots of 2 sweets.

Inverse proportion - Intermediate and Higher tier

If two values are inversely proportional, this means that as one value increases the other decreases.

Speed and time can be inversely proportional; as the speed increases, time taken to complete the journey will decrease.

C is inversely proportional to D.

We can write this as Equation: {C}�frac{1}{D} (http://equation-chef.files.bbci.co.uk/content/0805a3c1fdb62611f751ec4a6074e587/18)

This can be converted into a formula:

Equation: {C}={k}timesfrac{1}{D} (http://equation-chef.files.bbci.co.uk/content/5a53b28abe5b4bdf60d5da9c90121ad0/18)

where Equation: text{k} (http://equation-chef.files.bbci.co.uk/content/0378f182909bdab8d85f3dafbab72d93/18) is the constant of proportionality. This can also be written as:

Equation: {C}=frac{k}{D} (http://equation-chef.files.bbci.co.uk/content/347a9c8a0d1dd7660bf52a5d032afe55/18)

The number of plumbers is inversely proportional to the number of days of work needed. 32 plumbers can complete a job in 15 days.

This can be written as an equation:

15 days = Equation: frac{k}{32} (http://equation-chef.files.bbci.co.uk/content/77cb7c0b8367645bee2a2f9b6da7443a/18) plumbers.

To find the value of Equation: text{k} (http://equation-chef.files.bbci.co.uk/content/0378f182909bdab8d85f3dafbab72d93/18) multiply both sides of the equation by 32:

15 × 32 = Equation: text{k} (http://equation-chef.files.bbci.co.uk/content/0378f182909bdab8d85f3dafbab72d93/18) = 480

We can rewrite the equation as Equation: days~=frac{480}{plumbers} (http://equation-chef.files.bbci.co.uk/content/ac75c8a2d6e8c82d5256d303818bb2ea/18) and use this to calculate one value given the other.

Example

If there are 20 plumbers, how many days will it take to complete the job?

Days = Equation: frac{480}{plumbers} (http://equation-chef.files.bbci.co.uk/content/2086258cbadf3f01f8ed8e342573a110/18)

Days = Equation: frac{480}{20} (http://equation-chef.files.bbci.co.uk/content/6063a0cdd9be437471297f399497860c/18)

Days = 24

Illustrating direct and inverse proportion

How do we know if a graph represents either direct or inverse proportionality? We look for certain features of the graph and perform some test calculations to help us decide.

Look at this graph:

A straight-line graph labelled "Time (s)" on the x-axis and "Distance (m)" on the y-axis (http://a.files.bbci.co.uk/bam/live/content/ztcksg8/small)A straight-line graph labelled "Time (s)" on the x-axis and "Distance (m)" on the y-axis (http://a.files.bbci.co.uk/bam/live/content/ztcksg8/large)

The graph shows that the distance travelled and the time taken are proportional, but how do we know that?

Notice that the graph is a straight line starting from the origin. When both of these features are present we know that the two quantities on the graph must be directly proportional. However, take note of the scales on the graph, if they do not start at (0,0) or if they are not linear take care.

Look at the values on both of the axes - when the distance axis is 4 the time axis is 2, when the distance axis shows 8 the time axis shows 4. This means that when one of the variables doubles the other variable also doubles, this is the test for proportionality. If this condition is true and the graph is a straight line then we must have a directly proportional relationship.

The values on the axes also serve another important purpose as they allow us to discover the constant of proportionality from the graph so that we can describe the relationship using an equation. Look at the time value when the distance is 40, you should be able to see that it is 20. This means that the constant of proportionality which is linking distance to time is 40 ÷ 20 = 2.

Equation: {distance}�{time} (http://equation-chef.files.bbci.co.uk/content/838ef3fa87d8544033083f2c661ad0f2/18)

Removing the proportional sign and adding the constant of proportionality gives:

Equation: {distance}={k}times{time} (http://equation-chef.files.bbci.co.uk/content/93deb9d800fd8a0fae5df0cecb3e0223/18)

Rearranging we get:

Equation: {k}=frac{distance}{time}=frac{40}{20}={2} (http://equation-chef.files.bbci.co.uk/content/47835ca510185117690bcee811fad972/18)

So finally we can write:

Equation: {distance}={2}times{time} (http://equation-chef.files.bbci.co.uk/content/059befd23adf75beb4ea33ff2f5e46a8/18) or Equation: {d}={2t} (http://equation-chef.files.bbci.co.uk/content/75e67d566613fca457b45c365f8d9fde/18)

If we want to make Equation: {time} (http://equation-chef.files.bbci.co.uk/content/9a70046ce050b4719bf214812b7fd4ac/18) the subject of the equation, we would have to divide both sides of the equation by 2. This would give:

Equation: {t}=frac{d}{2} (http://equation-chef.files.bbci.co.uk/content/38b7935c79b14dfd862cf6d23d4bf3e1/18)

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geometry- sector

Parts of a circle

Do you remember some key parts of a circle?

  • an arc is a section of the circumference of the circle
  • a sector is an area enclosed by two radii and an arc
  • a chord is a straight line connecting two points on the circumference of a circle
  • a segment is a section formed between an arc and a chord

A diagram showing key parts of circle, labelled chord, segment, sector and arc. (http://a.files.bbci.co.uk/bam/live/content/zgcx39q/small)A diagram showing key parts of circle, labelled chord, segment, sector and arc. (http://a.files.bbci.co.uk/bam/live/content/zgcx39q/small)

Length of an arc

We can find the length of an arc by using the formula:

Equation: frac{texttheta}{360} times pi~text{d} (http://equation-chef.files.bbci.co.uk/content/e7160821e9669b0a11e974e3a0cf9c7b/18)

Equation: texttheta (http://equation-chef.files.bbci.co.uk/content/2ed42a7e3c652dfb69b0575a7ade3274/18) is the angle of the sector and Equation: text{d} (http://equation-chef.files.bbci.co.uk/content/13fbebcbae10fb52668c48da27b4c919/18) is the diameter of the circle.

Example

A sector with an angle of 80° and a radius of 12 cm. (http://a.files.bbci.co.uk/bam/live/content/zpfdhv4/small)A sector with an angle of 80° and a radius of 12 cm. (http://a.files.bbci.co.uk/bam/live/content/zpfdhv4/small)

This sector has a minor arc, because the angle is less than 180⁰.

We are given the radius of the sector so we need to double this to find the diameter.

Here, Equation: text{d} (http://equation-chef.files.bbci.co.uk/content/13fbebcbae10fb52668c48da27b4c919/18) = 24 and Equation: texttheta (http://equation-chef.files.bbci.co.uk/content/2ed42a7e3c652dfb69b0575a7ade3274/18) = 80⁰.

Substituting these values into the formula, we get:

Equation: text{Arc length =}~frac{80}{360} times pi times {24} (http://equation-chef.files.bbci.co.uk/content/83c4e55ca8bca8af1694a949b7b5a7b0/18)

Equation: text{= 16.755...} (http://equation-chef.files.bbci.co.uk/content/f3e6198680068a64a76b9947f20e8ea0/18)

Equation: text{= 16.8 cm (to one decimal place)} (http://equation-chef.files.bbci.co.uk/content/dd51716dd1d89c51133eedbb4d21e7da/18)

QQuestion

Find the length of the minor arc AB.

A sector with an angle of 60° and a radius of 7.3 cm. (http://a.files.bbci.co.uk/bam/live/content/zxh4mnb/small)A sector with an angle of 60° and a radius of 7.3 cm. (http://a.files.bbci.co.uk/bam/live/content/zxh4mnb/small) AAnswerReveal answerHide answer

Equation: text{Arc length =}~frac{60}{360} times pi times {14.6} (http://equation-chef.files.bbci.co.uk/content/f4d34dc1476a5d3fd7fca5e01b97ca57/18)

Equation: text{= 7.644...} (http://equation-chef.files.bbci.co.uk/content/396f8e2734dac14c33f70430a09167a8/18)

Equation: text{= 7.6 cm (to one decimal place)} (http://equation-chef.files.bbci.co.uk/content/4ba3ef274d12fe032e1de682e4dc9b5b/18)

Example

A sector with an angle of 210° and a radius of 60 mm. (http://a.files.bbci.co.uk/bam/live/content/z9ws9qt/small)A sector with an angle of 210° and a radius of 60 mm. (http://a.files.bbci.co.uk/bam/live/content/z9ws9qt/small)

This sector contains a major arc, because the angle is greater than 180⁰.

We still follow the same process for this question. Don’t forget to double the radius to get the diameter.

Equation: text{Arc length =}~frac{210}{360} times pi times {120} (http://equation-chef.files.bbci.co.uk/content/982f29bb4a0137b465a03aef5a7508ec/18)

Equation: text{= 219.91...} (http://equation-chef.files.bbci.co.uk/content/aa40c265bb65e477f94b9cb01a3cbdb5/18)

Equation: text{= 219.9 mm (to one decimal place)} (http://equation-chef.files.bbci.co.uk/content/e40aca3c66c0172986e4a9f9601163ce/18)

QQuestion

Find the length of the major arc AB.

A sector with an angle of 330° and a radius of 4.60 cm. (http://a.files.bbci.co.uk/bam/live/content/zts8tyc/small)A sector with an angle of 330° and a radius of 4.60 cm. (http://a.files.bbci.co.uk/bam/live/content/zts8tyc/small) AAnswerReveal answerHide answer

Equation: text{Arc length =}~frac{330}{360} times pi times {9.2} (http://equation-chef.files.bbci.co.uk/content/bd5742bf7fcb8de8578994fbd71b3e0d/18)

Equation: text{= 26.49...} (http://equation-chef.files.bbci.co.uk/content/c79dab7d80a56febfe880116abde13df/18)

Equation: text{= 26.5 cm (to one decimal place)} (http://equation-chef.files.bbci.co.uk/content/40c2546d551d7fdec67efa689d53db7f/18)

Perimeter of a sector

The perimeter is the distance all around the outside of a shape. We can find the perimeter of a sector using what we know about finding the length of an arc.

A sector is formed between two radii and an arc. To find the perimeter, we need to add these values together.

Equation: text{Perimeter = Arc length + 2r} (http://equation-chef.files.bbci.co.uk/content/8346730ebc64ced3c374060f37721561/18)

A sector with a radius of 45 cm and arc length 27.5 cm. (http://a.files.bbci.co.uk/bam/live/content/zgkhrdm/small)A sector with a radius of 45 cm and arc length 27.5 cm. (http://a.files.bbci.co.uk/bam/live/content/zgkhrdm/small)

Here, we are given the arc length and the radius.

Substituting these values, we get:

Equation: text{Perimeter = 27.5 + (2 times 45)} (http://equation-chef.files.bbci.co.uk/content/a8cf21e99b223d4ae1193f658c70fa2e/18)

Equation: text{= 27.5 + 90} (http://equation-chef.files.bbci.co.uk/content/1279ddabc9e7326a4cf3cdf3cb5ed4ef/18)

Equation: text{= 117.5 cm} (http://equation-chef.files.bbci.co.uk/content/5887aaa420d02b054de230beadf32867/18)

Area of a segment

A segment is the section between a chord and an arc. It is essentially a sector with the triangle cut out, so we need to use our knowledge of triangles here as well.

A diagram showing that we can find the area of a segment by subtracting the area of the triangle within the sector from the area of the whole sector. (http://a.files.bbci.co.uk/bam/live/content/zgdn4j6/small)A diagram showing that we can find the area of a segment by subtracting the area of the triangle within the sector from the area of the whole sector. (http://a.files.bbci.co.uk/bam/live/content/zgdn4j6/small)

To calculate the area of a segment, we will need to do three things:

  1. find the area of the whole sector
  2. find the area of the triangle within the sector
  3. subtract the area of the triangle from the area of the sector

Example

An example showing how to find the area of a segment with an angle of 40° and a radius of 8 cm. The segment is labelled P, R, S. (http://a.files.bbci.co.uk/bam/live/content/zpb7frd/small)An example showing how to find the area of a segment with an angle of 40° and a radius of 8 cm. The segment is labelled P, R, S. (http://a.files.bbci.co.uk/bam/live/content/zpb7frd/small)

1. Equation: text{Area of sector =}~frac{40}{360} times pi times {8}^{2} (http://equation-chef.files.bbci.co.uk/content/f8d415f91fe4cea1caad6a4b43b9ac0c/18)

Equation: text{= 22.340...} (http://equation-chef.files.bbci.co.uk/content/c98f752453bd6924a3ccd868fe02d9b0/18)

Equation: text{= 22.34 cm}^{2}~text{(to two decimal places)} (http://equation-chef.files.bbci.co.uk/content/55094cffe242742edeacd593ee769436/18)

2. This is a non right-angled triangle, so we will need to use the formula:

Equation: text{Area of triangle =}~frac{1}{2}~text{ab}~text{sin C} (http://equation-chef.files.bbci.co.uk/content/9e8754cd04c548ab9f2c8e05c3551e0c/18)

In this formula, Equation: text{a} (http://equation-chef.files.bbci.co.uk/content/c39635983cb91a83aa38e2734bc7aae9/18) and Equation: text{b} (http://equation-chef.files.bbci.co.uk/content/9c40af9cbfd0d960d5e8d6b356333784/18) are the two sides which form the angle Equation: text{C} (http://equation-chef.files.bbci.co.uk/content/94f931c273525fbff508990a78505368/18). So Equation: text{a} (http://equation-chef.files.bbci.co.uk/content/c39635983cb91a83aa38e2734bc7aae9/18) and Equation: text{b} (http://equation-chef.files.bbci.co.uk/content/9c40af9cbfd0d960d5e8d6b356333784/18) are both 8cm, and Equation: text{C} (http://equation-chef.files.bbci.co.uk/content/94f931c273525fbff508990a78505368/18) is 40⁰.

Equation: text{Area of triangle =}~frac{1}{2} times {8} times {8} times text{sin 40} (http://equation-chef.files.bbci.co.uk/content/df25da8d63192798348d744ead083bdc/18)

Equation: text{= 20.569...} (http://equation-chef.files.bbci.co.uk/content/665e4a5b6a93ae5ba682cd8f75d74997/18)

Equation: text{= 20.57 cm}^{2}~text{(to two decimal places)} (http://equation-chef.files.bbci.co.uk/content/e096586890580c4c154bda571774261d/18)

3. To find the area of the shaded segment, we need to subtract the area of the triangle from the area of the sector.

Equation: text{Area of segment = Area of sector - Area of triangle} (http://equation-chef.files.bbci.co.uk/content/2d5377b26b1e25480bdc203af9df1171/18)

Equation: text{= 22.34 - 20.57} (http://equation-chef.files.bbci.co.uk/content/2a372a1ea5f1fbe116d9ba5ad54d42fd/18)

Equation: text{= 1.77 cm}^{2} (http://equation-chef.files.bbci.co.uk/content/3264d6dcd0066eb6d20444d474eb5d87/18)

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loci

Constructing circles

To construct a circle you will need a compass, a pencil and a ruler. You need to know the radius or diameter of the circle you wish to construct.

Open your compass to a distance equal to the radius of the circle (or half the diameter). Place the point of the compass at the centre of the circle and lightly draw the shape.

Circle with the diameter and radius marked. The raduis is labelled 7 cm. Below the circle is a compass with a pencil, and a ruler. The compass is opened opened to 7 cm. (http://a.files.bbci.co.uk/bam/live/content/zq9dycw/small)

Constructing an angle

Constructing a 60° angle

To construct a 60° angle, we first draw a line of any length. Then using a compass, we use this as a base to construct an equilateral triangle.

Two lines forming angle RPQ which is labelled 60°. There is an arc at point R. (http://a.files.bbci.co.uk/bam/live/content/zgjmcwx/small)Two lines forming angle RPQ which is labelled 60°. There is an arc at point R. (http://a.files.bbci.co.uk/bam/live/content/zgjmcwx/small)

To construct the equilateral triangle, we:

  • open the compass to the same dimensions as our original line
  • place the point of the compass on one end of the line and draw an arc
  • repeat this at the other end and the arcs should intersect where the tip of the triangle should be

Connect the tip of the triangle to one end of the base, and you will have a 60° angle.

Constructing a 30° angle

To construct a 30° angle, you must first construct a 60° angle as above and then bisect the angle.

Angle RPQ bisected to form two angles labelled 30° (http://a.files.bbci.co.uk/bam/live/content/zxvrtyc/small)Angle RPQ bisected to form two angles labelled 30° (http://a.files.bbci.co.uk/bam/live/content/zxvrtyc/small)

Constructing a 90° angle

To construct a 90° angle, you draw a base line and then construct a perpendicular bisector to leave a 90° angle.

A 90° angle formed by a base line and a perpendicular bisector (http://a.files.bbci.co.uk/bam/live/content/zs4dycw/small)A 90° angle formed by a base line and a perpendicular bisector (http://a.files.bbci.co.uk/bam/live/content/zs4dycw/small)

Constructing a 45° angle

To construct a 45° angle, first construct a perpendicular bisector of a line to leave a 90° angle. Secondly, following the steps above, bisect the angle. This should leave you with a 45° angle.

A 90° angle bisected to to form two angles labelled 45° (http://a.files.bbci.co.uk/bam/live/content/zwpy7hv/small)A 90° angle bisected to to form two angles labelled 45° (http://a.files.bbci.co.uk/bam/live/content/zwpy7hv/small)Circle with the diameter and radius marked. The raduis is labelled 7 cm. Below the circle is a compass with a pencil, and a ruler. The compass is opened opened to 7 cm. (http://a.files.bbci.co.uk/bam/live/content/zq9dycw/large)

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