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The angle in a Angles at the 0 B circumference semi-circle is 90 They must come from the same arc. are equal. A Look out for a 2x diameter. A Cyclic Quadrilateral They must come from the same arc. x B is twice the angle at the Circle Theorems Opposite angles The angle at the centre circumference. From any point you can only draw two tangents... equal add up to 1800 The angle between a tangent and a equal 0 radius is 90 Alternate Segment Theorem. Look out for radii. @lhmaths ... and they’ll be equal. Lesley Hall @ Soar Valley College 0 180 straight lines round a point 360 triangles allied angles 0 quadrilaterals parallel opposite lines alternate angle sum = (n – 2) x 1800 polygons corresponding exterior angle Equal …add up to interior angle @lhmaths 3600 Lesley Hall @ Soar Valley College height height rectangle height a triangle is half the area of a rectangle base parallelogram height AREA height Area = base x height 2 Area = base x height Always use the perpendicular base Area = base x height Area = (a + b) x h 2 height a height trapezium @lhmaths base base base circle radius Area = πr2 b Lesley Hall @ Soar Valley College Y=3x+2 Y = 4x + 3 Quadratic Graphs Y=3x-4 y = x2 intercept 3 gradient 3 U shaped – parabola. y = mx + c Y=3x+2 y = -x + 5 3 Parallel lines have the same gradient. GRAPHS Perpendicular lines y = tanx have gradients with a product of -1. y = sinx Sine Curve @lhmaths Trigonometric Graphs y = cosx Cosine Curve Tangent Reciprocal y = ax2 + bx + c Cubic Graphs Linear Graphs Square numbers. 4 3 y = x2 – 3x - 2 y= 1 x Cube numbers. y = x3 Lesley Hall @ Soar Valley College Angle Sum 3 (n – 2) x 1800 triangle 5 pentagon number of Polygons 8 octagon 9 - nonagon 10 - decagon exterior angle interior angle angle sum number of sides 3600 number of sides OR OR exterior 180 – angle 0 @lhmaths 6 hexagon 7 - heptagon triangles 4 x 1800 = 7200 4 quadrilateral interior 1800 – angle Lesley Hall @ Soar Valley College Completing the square: Solving: Factorising Formula Completing the square Drawing a graph x 2 x x2 2x 2 2x 4 half of 4x x2 + 4x - 3 = 0 (x + 2)2 – 4 – 3 = 0 2 x + 4x subtract 22 (x + 2)2 – 7 = 0 = (x + 2)2 - 4 x + 2 = ±7 x = ±7 - 2 complete the square Factorising: The formula: Quadratic Equations x2 + 9x + 20 = 0 x x 5 x2 5x 4 4x 20 ax2 + bx + c fill in the blue squares first work out the factors (red numbers) (x + 5)(x + 4) = 0 x = -5 x = -4 3x2 - 10x + 8 = 0 x -2 the white squares make the x term Difference of Two Squares: x2 - 16 (x – 4)(x + 4) x squared subtract 4 squared 3x 3x 2 -4 -4x -6x Graphs: draw lines to find solutions 8 (3x - 4)(x - 2) = 0 x = x=2 @lhmaths x = -b ±b2 – 4ac 2a Parabola – u shaped graph Lesley Hall @ Soar Valley College y = -fx y = fx + a y = kfx plus a - up reflection minus a – down in x-axis k 0 a a y=sinx y=x 2 y = f(x + a) plus a - left minus a – right a @lhmaths scale factor k stretch in y = f(-x) x-axis reflection 0 in y-axis might think! y-axis y = f(Kx) -a opposite to what u stretch in y=x3 scale factor 1/k Lesley Hall @ Soar Valley College Rotation Translation Rotation of 900, Describe with clockwise, about To describe a rotation a vector centre (2,-1) you need: 3 4 3 4 squares right squares up the angle of rotation the direction the coordinates of anticlockwise the centre Transformations Reflection clockwise Centre Centre of rotation Negative enlargements 3 – HIGHER only! 1 Enlargement Enlargement, scale factor 3, centre (0,7) x=2 2 Describe by naming the line of symmetry Reflection in the line x = 2. @lhmaths Always use TRACING PAPER for translation, reflection & rotation. To describe an enlargement you need: Enlargement of the scale factor coordinates of the centre scale factor -2. Lesley Hall @ Soar Valley College Non-Prisms Prisms Cones Prisms have a uniform cross-section Pyramids 2 Crosssection height V = πr h 3 length Volume = area of cross-section radius Volume × length a cone is one third of a cylinder Frustrums cylinders cuboids Volume = area of base x height 3 πr2 Spheres a frustrum is a pyramid with the top cut off. length Volume = length x width x height height height width You need to find Volume = πr2h radius the volume of both pyramids. Often you need to use similar V = 4πr3 3 shapes in frustrum problems. @lhmaths Lesley Hall @ Soar Valley College @lhmaths Lesley Hall @ Soar Valley College Square 4 equal sides opposite sides are parallel rotational symmetry of order 4 diagonals meet at 900 diagonals of equal length 4 lines of symmetry Rhombus 4 equal sides opposite sides are parallel rotational symmetry of order 2 kite diagonals meet at 900 rotational symmetry of order 1 2 pairs of equal sides Parallelogram 1 line of symmetry rotational symmetry of order 2 no line symmetry Quadrilaterals Trapezium Rectangle @lhmaths opposite sides are equal & parallel one pair of parallel sides rotational symmetry of order 1 rotational symmetry of order 2 diagonals meet at 900 2 lines of symmetry opposite sides are equal & parallel an isosceles trapezium has a line of symmetry 2 lines of symmetry diagonals of equal length angles in a quadrilateral add up to 3600 Lesley Hall @ Soar Valley College 10 millimetres = 1 centimetre units 100 centimetre = 1 METRE inches yards feet MILES Imperial units ounces pounds (lbs) 1000 METRES = 1 kilometre Metric units 1000 grams = 1 kilogram Imperial Metric units STONES Units 1 inch = 2.5 cm 1 kg = 2.2 pounds units 1000 millilitres = 1 litre (6 foot) 1 gallon = 4.5 litres 4 litres = 7 pints 1 litre = 1¾ pints Metric units metres tall. 1 mile = 1.6 km Imperial gallons man is about 1.7 or 1.8 5 miles = 8 km pints An average ‘A litre of water is a pint and three quarters’ @lhmaths 1 foot = 12 inches That’s 30cm – the length of a ruler! 3 feet = 1 yard A yard is almost 1 metre (its 90cm). Lesley Hall @ Soar Valley College Brackets The Power Zero division, so subtract the powers ... but any number divided by itself Multiplying equals 1 ... Indices add the powers Negative Indices Dividing power zero equals 1. Fractional Indices 25 5 5–3=2 x 25 = 25 x 5 1 add powers = 25 must be equal = subtract the powers Any number to the equal 25 2 – 5 = -3 8 @lhmaths 27 Lesley Hall @ Soar Valley College increasing on a calculator 39% of 82 0.39 x 82 fraction to % 15 = 75 = 75% 20 100 x 5 5 Change to a decimal and multiply increase £60 by 12% 12% of 60 = 0.12 x 60 = £7.20 New amount = £60 + £7.20 = £67.20 Percentages OR 15 ÷ 20 x 100 = 75% without a calculator 50% - half 25% - half and half 75% - 50% + 25% % ADD decreasing decrease £60 by 12% 12% of 60 = 0.12 x 60 = £7.20 New amount = £60 - £7.20 = £52.80 10% - divide by 10 5% - half 10% 20% - double 10% SUBTRACT Lesley Hall @ Soar Valley College @lhmaths