These views are shown for some basic shapes below.

]]>{jatex options:inline}(2 + \sqrt{5}) + (2 - \sqrt{5}) = 4{/jatex}

Note that the {jatex options:inline}{} + \sqrt{5}{/jatex} cancels the {jatex options:inline}- \sqrt{5}{/jatex}.

{jatex options:inline} \sqrt{5}{/jatex} and {jatex options:inline}2 \sqrt{5}{/jatex} multiply to give a rational number.

{jatex options:inline} \sqrt{5}+ \times 2 \sqrt{5} =2 \times \sqrt{25} = 2 \times 5 = 10{/jatex}

{jatex options:inline} \sqrt{5}{/jatex} and {jatex options:inline}2 \sqrt{5}{/jatex} divide to give a rational number.

{jatex options:inline} \frac{\sqrt{5}}{ 2 \sqrt{5}} =\frac{1}{2}{/jatex}

{jatex options:inline}3+ \sqrt{5}{/jatex} andhj {jatex options:inline}2 + \sqrt{5}{/jatex} subtract to give a rational number.

{jatex options:inline}(3 + \sqrt{5}) - (2 + \sqrt{5}) =3 + \sqrt{5} - 2 - \sqrt{5} =1 {/jatex}

Note that the {jatex options:inline}{} + \sqrt{5}{/jatex} cancels the {jatex options:inline}{}+ \sqrt{5}{/jatex}.]]>

The expressioncan be factorised, but only when it is recognised that the first two terms can be factorised as a difference of squares:

The last two terms can also be factorised asso thatis a common factor.

Some expressions with four terms, involving four different variables, can also be factorised.

The simplest is

Notice thatis a common factor of the first two terms, andis a common factor of the last two terms. We have

Nowis a common factor, so we can write

Hence

]]>Ifarrange in order of magnitude

Ifthenincreases with increasingso a bigger value ofmeans a bigger value of

Note thatand(1)

Sincewe have

Ifarrange in order of magnitude

Ifthendecreases with increasingso a bigger value ofmeans a smaller value of

Using (1) and thatand

Sincewe have

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is a difference of squares and factorises asand 7 factorises asorsoorororWe may equate factors on either side to give either

andorand

The first of these gives the simultaneous equations

The solution isand

The second of these gives the simultaneous equation

The solution is

The first of these gives the simultaneous equations

The solution isand

The first of these gives the simultaneous equations

The solution isand

Ifandis not a prime number, a great many solutions exist since m+n and m-n must be matched with each factor in turn, either both negative or positive.

]]>We can construct the table

Sibling | Age |

Alf | |

Ben | |

Colin |

Sibling | Age |

Alf | {jatex options:inline}2x{/jatex} |

Ben | {jatex options:inline}2x+3{/jatex} |

Colin | {jatex options:inline}x{/jatex} |

For example, we want to solve the quadratic equation

We could use the quadraticformulaIn this caseso Henceor

The above solution however was not obtained by the use of an algorithm. If we wanted to use an algorithm to solve the equation we could complete the square and makethe subject.

We can start by factorising with 3.

Use the identityNow write

We can collect the last two termsso

Now make x the subject. Add 13 over 36 to both sides.

Square root both sides, remembering that there are two square roots, one + and one -.

Subtractfrom both sides.

Both methods give the same answer, but only the second method is the result of an algorithm.

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Opposite angles are equal

Alternate angles when a line is drawn to intersect parallel lines are equal. To recognise when this rule can be used look for 'Z' shapes.

Supplementary or complementary angles when a line is drawn to intersect parallel lines add to 180 Degrees, soAgain look for 'Z' shapes. The supplementary angles are on the same side of the 'Z'.

Corresponding angles can be 'slid' along a line between parallel lines. Corresponding angles are equal.

Angles on a straight line add up to 180 Degrees.

]]>All the sides in a regular polygon are the same length and the interior angles are all the same. For a triangle they are all 60 degrees, for a square 90 degrees, for a pentagon 108 degrees... The external angles – the angle between a side and an adjacent side extended – are all 180 minus the internal angle, because we can extend each side so that both angles are on a straight line so add to 180 degrees. This is shown below.

As the perimeter is traced out, each time the end of an edge is reached, an angle b is turned through. If the polygon has n sides, there will be n turnings. When the starting point is reached again, there has been one complete turn – an angle of 360 degrees, so thatdegrees. Dividing bygivesdegrees. The interior anglecan then be found by subtracting this from 180 degrees:Adding up all n interior angles gives

]]>The base is a regular pyramid and all the sides are the same length, 2cm. The pyramid is 5 cm high.

We want to find the anglefor which we need to find the lengthThe base of the pyramid is shown below.

We may cut the base up into six equilateral triangles, so that each angle is 60 and each length is 2 cm The length(O is the centre of the base so FOC is a straight line) is then 4 cm and we may construct the triangle below.

The angleis twice the angleThe triangleis right angled soand

Finding the angleis trickier. We have to find the slant heightand the distance

From the triangle above, Pythagoras Theorem givesW can draw the triangle below to find

We can find the distance AB using the cosine rule

The triangleis then

We can findusing the cosine rule.

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