Montfort Sec O:N Level E.Math Formula Sheet 2025

Mathematical Formulae Compound interest: Total amount = 1100 n rP Mensuration Curved surface area of a cone = rl Surface area of a sphere = 4 2r Volume of a cone = hr2 3 1  Volume of a sphere = 3 3 4 r Area of triangle ABC = 1sin2ab C Arc length =r, where θ is in radians Sector area = 2 2 1 r , where θ is in radian Trigonometry C c B b A a sinsinsin  Abccba cos2222  Statistics Mean = 12 ...nxx xfx fn   Standard deviation = 22          f fx f fx Simple Interest: 100 PRTI  total amount, A = P + I Curved SA of cylinder 2rh Volume of a pyramid 1base area height3  Cosine Rule 222 cos 2 bcaA bc  __ :s um of :f r e quency :me a n , f fx xf    Indices       0 1 1 if , then mn m n mn m n n n mnmmm n nn n nn m n n n n aaa a aaa a a aa a a ab a b a a m n aa b b            Coordinate Geometry     22 12 21 12 12 12 21 Length of line segment Gradient of line = = yy y yxx yy xx x x     Equation of line x-intercept : 0 y-intercept : 0 y ym x c x     Parallel lines must always have the same gradient and the lines never meet or intersect one another Area of Trapezium 1sum of parallel sides height2  Area of Parallelogram base height Area of Circle 2r Circumference of Circle2 or rd Arc length of a sector 2 360 r   Area of a sector 2 360 r   :d e grees Trigonometry TOA: tan opp adj  opp, b hyp, c CAH: cos adj hyp  SOH: sin opp hyp  Pythagoras’ Theorem: 22 2ca b  adj, a Geometric reasons for lines/angles 1)( on a st. line) 2)( at a point) 3)( sum of ) 4)(alt , // sides) 5)(corr. ,// sides) 6)(base s of isos ) 7)(equilateral ) 8)( at centre = twice at circumference) 9)( in a semicircle = 90 ) 10)(            in the same segment) 11)( in opp. segment) 12)(tangent radius) 13)(tangents from an ext. point) 14)( from centre bisects chord) 15)(equal chords are equidistant from centre)     Polygons Sum of interior angles in a polygon: (2 ) 1 8 0n  For a regular polygon, each interior angle: ( 2) 180n n  Sum of exterior angles: 360 For a regular polygon, each exterior angle: 360 n  General 2 Qu a d r a t i c f o r mu l a 4 2 di stspeed tim e ma s sdensi t y volum e bb a cx a     Probability P(heads) = 1 – P(tails) P(A or B) =P(A) + P(B) P(A and B) =P(A) X P(B) Special Algebraic Identities:     22 2 22 2 22 44 2222 22 () 2 () 2 () () Additional ( )( ) ab a a bb ab a a bb ab ab a b xy xyxy xy xyx y              Inequalities: 4 4 : sign changes when multiply or divide by a negative number 4 4 if a k ka if a k ka     Prefixes: micro: 610 nano: 910 pico: 1210 million: 610 billion: 910 trillion:1210 Mode: highest frequency Where grouped data is given, find the mid-value, x

Box-and-whisker plots Conversion of units: For length 1cm = 10mm 1m = 100cm 1km =1000m = 100000 cm For area 22 22 1m = 10000cm 1km = 1000 000m Cumulative Frequency plots 1: 2: 3: lower quartile (25th percentile) median (50th percentile) upper quartile (75th percentile) Q Q Q 31 Range = Max - Min IQR Q Q Perpendicular Bisector To construct the perpendicular bisector of a line segment PQ: 1) Given or draw PQ 2) Using a pair of compasses, choose a radius more than half the length of PQ 3) Using P as the centre, draw arc 1 above PQ and arc 2 below PQ 4) Using Q as the centre, draw arc 3 above PQ and arc 4 below PQ 5) Join the two points on intersection, X and Y, to get the perpendicular bisector of PQ P Q arc 1 arc 3 arc 2 arc 4 Angle Bisector To construct the angle bisector of angle BAC: 1) Using a pair of compasses, choose a radius less than or equal to the length of AB and AC 2) Using A as the centre, draw an arc to cut AB at X and an arc to cut AC at Y 3) Using X as the centre, draw arc 1 4) Using Y as the centre, draw arc 2, such that arc 1 & 2 meet at R 5) Join A to R and that forms the bisector of angle BAC B A C arc 1 arc 2 Distance-time graph: Gradient = speed

Number Patterns 2 5 8 11 14 Level 1: General Formula Common difference (d) = +3 First term (a) – 2 General term, Tn = ()dn a d (e.g. 3( 2 3 ) 31nn Level 2: Typical follow up question Does 28 lie in the sequence? 31 2 8n 29 s i n c e n i s n o t a n i n t e g e r , 3 28 does not lie in the sequence n  Level 3: Level increment *additional info 3 7 13 21 2 2 2 use Tn an bn c  Then create simultaneous equations to find the value of a, b, c Timeline to O levels Jun – Complete topical revision (TYS) & notes Jul – Kickstart Prelim Papers Aug – Prepare for Prelims Sep – Complete any outstanding yearly TYS + other schools’ Prelim Papers Oct – Final Revision – aim to ACE O’s 4 6 8 Additional Info about Further Trigonometry Tests for CONGRUENT triangles 1) SSS – 3 pairs of sides equal 2) SAS – 2 sides and 1 included  3) ASA – 2s and 1 included side 4) RHS – Right-Hypothenuse Side Tests for SIMILAR triangles 1) AAA – Show any 2 angles equal 2) SSS – Show 3 pairs of equal ratios (i.e. same scale factor) 3) SAS – Show 2 ratios and 1 angle equal (this angle must be an included angle) Ratios of Areas: 22 1212::AA l l Ratios of Volumes: 33 12 1 2::VV l l Completing the Square: 22 22 22 22 22 bbxb x c xb x c bbxc                