YEAR 11 REVISION BLASTER UNIT 2 STAGE 2

GCSE MATHS REVISION UNIT 2 (stage 2) – all about: SHAPE & ALGEBRA & NUMBER Circle theorems Angles & lines & shapes Simplifying & factorising Linear graphs (y=mx±c) Percentages

CIRCLE THEOREMS

The angle made at the centre of a circle (by an arc or chord) is double the angle made at the circumference by the same arc or chord. 2x o xoxo xoxo xoxo xoxo xoxo xoxo xoxo xoxo o o o o o o o o Angle at centre is a reflex angle Another Theorem The Angle at Centre (Arrow) TheoremTheorem 1

o Diameter 90 o angle in a semi-circle 20 o angle sum triangle 90 o angle in a semi-circle e o a b c 70 o d 30 o Find the unknown angles below stating a reason. angle a = angle b = angle c = angle d = angle e =60 o angle sum triangle Angle in Semi-Circle TheoremTheorem 2 This is a special case of Theorem 1: angle in a semi-circle is always 90° Th2

Angles subtended by an arc or chord in the same segment are equal. xoxo xoxo xoxo xoxo xoxo yoyo yoyo Th3 The Chord (Bow) TheoremTheorem 3

The angle between a tangent and a radius is 90 o. (Tan/rad) o The Tangent TheoremTheorem 4

The Alternate Segment TheoremTheorem 5 The angle between a tangent and a chord through the point of contact is equal to the angle made by that chord in the alternate segment. xoxo xoxo yoyo yoyo 45 o (Alt Seg) 60 o (Alt Seg) 75 o angle sum triangle 45 o xoxo yoyo 60 o zozo Find the missing angles below giving reasons in each case. angle x = angle y = angle z = Th5

Cyclic Quadrilateral TheoremTheorem 6 The opposite angles of a cyclic quadrilateral are supplementary. (They sum to 180 o ) w y x z Angles x + z = 180 o Angles y + w = 180 o r p s q Angles p + r = 180 o Angles q + s = 180 o All corners touch the circle circumference

The Two Tangent TheoremTheorem 7 From any point outside a circle only two tangents can be drawn and they are equal in length. P T U Q R PT = PQ P T U Q R Th7

Circle Theorem Answers 1.(a)58° (b)2 angles are the same (=58°) 2.(a)(i)70° (opposite angles of) cyclic quadrilateral (ii)140° angle at centre is twice angle at circumference 3.(a)(i)40° (ii)140° (b)Right angle at A / C (Tangent Theorem): at B 2 x 50° angles 4.x = 52°; y = 76° and z = 58°

Angles

Which angles are equal to each other? a b c d e f g h

Alternate angles are equal a b a = b Look for an F-shape Look for a Z-shape Corresponding angles are equal a b a = b Look for a C- or U-shape Interior angles add up to 180° a b a + b = 180°

a b c Any exterior angle in a triangle is equal to the sum of the two opposite interior angles. a = b + c We can prove this by constructing a line parallel to this side. These alternate angles are equal. These corresponding angles are equal. b c

Calculate the size of angle a. a 28º 45º Hint: Add another parallel line. a = 28º + 45º =73º

Angles & Lines & Shapes Answers 1.(a)x = 60.5 (b)Not parallel because angle A is not equal to x, or because angle B is not equal to x, or because alternate (or Z) angles are not equal, or because corresponding angles are not equal. 2.(a)(i)x = 74 (ii)y = 36° (b)No. Because triangle ABD is not isosceles. or Because y is not equal to 38°. 3.(a)30° (b)40° 4.a = 50° and b = 110°

AD BREAK Year 11 Unit 2 Exam(s). Monday 15 th November- NEXT WEEK!! Don’t Forget: CALCULATORS PEN PENCIL RULER REVISE

Algebra

Expanding and simplifying (x+2)(x+3) x +3 x +2 x²+3x +2x+6 = x² = x² + 5x x+ 2x+6

Factorising x² + 7x + 12 x² +12 x x x +2x = x² + 8x +12 = x² + 6x +2x x +3x = x² + 4x +3x +12 = x² + 7x +12 So, x² + 7x + 12 = ( )( ) x +4 x +3

Which simplifies to: 20a³b³ 4ab x 5a²b² Simplifying 10ab² First, focus on the 4a³b x 5a²b² Rearranging gives: 4x5 x axa² x bxb² 20a³b³ 10ab² = 20 a³ b³ 10 a b² = 2a²b

Simplifying & Factorising Answers 1.(a)3x(x – 2y) (b) (y – 7)(y – 2) 2.(a)(x – 7)(x – 3) 3.(a)24abc 4.a 4 c –1 or 5.(b)5xy(x + 3y2) 6.(b)2a2 b2 c

STRAIGHT LINE GRAPHS

General equation of straight line. y = mx + c

Gradient/ Steepness of the line. Y-intercept. (where the graph crosses the y axis)

Plotting a straight line…y = 2x -1 Pick 3 values of x. (e.g -2, 0, 2) SUBSTITUTE these into the equation (y=2x-1) x y when x = -2, Y = (2x-2) – 1 Y = - 4 – 1 Y = -5 When x = 0, Y = (2x0) – 1 Y = 0 – 1 Y = -1 When x = 2 Y = (2x2) – 1 Y = 4 – 1 Y =

Find the gradient: Draw a triangle under the graph. Make sure the height and width of the triangle are WHOLE numbers Rise = 8 Run = 4 Gradient = Rise = 8 Run = 4 = 2 Plot these pairs of co- ordinates. Join up the points, and extend across the axes.

y = Gradient/ Steepness of the line. Y-intercept. (where the graph crosses the y axis) m x + c 2

Key Facts Parallel lines have the SAME gradient Graphs with a negative gradient slope DOWNHILL Graphs with a positive gradient slope UPHILL The bigger the gradient the steeper the slope

Linear Graphs Answers 1.0.8, 4/5, or equivalent 2.(a)y = 0.4x + 18 (b)58 3.(a)(– 6, 0) (b)½ (c)Line drawn parallel to 2y = x (i)(–½, 0) (ii)reasonably parallel line crossing y axis below origin (iii)parallel lines do not meet

Percentage Increase and Decrease

Increase £20 by 15% Method 1 Find 15% 10% = £2 5% = £1 so 15% = £3 Now add it on to £20 £20 + £3 = £23 Increase by 15% so 1.15 is the multiplier Method 2 I need to find 115% OF original amount 115% of £20 is 1.15 x £20 = £23

Decrease £60 by 5% Method 1 Find 5% of £60 10% = £6 5% = £3 Subtract from £60 £60 - £3 = £57 Method 2 A decrease of 5% is same as 95% of original amount So 0.95 x £60 = £57 95% of original amount means 0.95 is the multiplier

Compound Interest £1,000 in bank earns 5% interest per year. How much will you have in 3 years? After 1 year 1000 x 1.05 = 1050 After 2 years 1050 x 1.05= After 3 years x 1.05 = OR 1000 x 1.05 x 1.05 x1.05= 1000 x 1.05³=£ is the multiplier!

Percentages Answers 1.£2140 (allow £140) 2.(a)1.029 (b) (allow ) 3.1st year is £20; 2nd year is £20.80 Interest is £40.80 so Amir is wrong 4.(a)2.04 (b)6 (windmills)