Geometry and Measures

Midpoint of a line segment What are the coordinates of the midpoint of this line? Add the coordinates Halve them (6, 7) (2, 3) (8, 10) (4,5) + ½ (2,3) (6,7)

Area of 2D shapes Triangle Rectangle Parallelogram Trapezium a l h h h b w A = length x width b A = base x height b h a b b A = ½ bh h A = h (a + b) 2

Area of compound shapes Split the shape in to two (or more) smaller shapes Find any missing lengths Find the area of each shape Add them up Remember the units

Area of a trapezium Add the parallel sides Multiply by the height Halve height height

Real-life area problems The diagram shows the plan of a swimming pool. Harry wants to tile the bottom of the swimming pool. Each pack of tiles cover an area of 7m². Each pack of tiles costs 6.99 How much will it cost Harry to tile the bottom of the pool? 2m Area of A = 2m x 5m = 10m 2 Area of B = 8m x 6m = 48m 2 11m A B 8m 5m 6m Total area of the floor = 10m 2 + 48m 2 1 pack covers 7m 2 2 packs cover 14m 2 4 packs cover 28m 2 8 packs cover 56m 2 9 packs cover 63m 2 9 packs are needed = 58m 2 Cost of tiles = 9 x 6.99 = 62.91

Surface area of a prism Count how many faces there are List them Find the area of each face Add up all of the areas Remember the units

Surface area of a prism Find the surface area of the cuboid 6 faces Front Back Top Bottom Side Side 8 x 5 = 40cm 2 8 x 5 = 40cm 2 8 x 3 = 24cm 2 8 x 3 = 24cm 2 5 x 3 = 15cm 2 5 x 3 = 15cm 2 Total surface area = 158cm 2

Surface area of a prism Find the surface area of the triangular prism 5 faces 5 Front Back Bottom 3 x 5 = 15cm 2 4 x 3 = 12cm 2 3 x 3 = 9cm 2 Total surface area = 48cm 2 Side 3 4 2 = 6cm2 Side 3 4 2 = 6cm2

Surface area of a prism Find the surface area of the prism 8 faces Front 64cm 2 Total surface area = 48cm 2 Back 64cm 2 Bottom 12 x 10 = 120cm 2 Top 1 8 x 10 = 80cm 2 Top 2 4 x 10 = 40cm 2 8 cm 4 cm Area = 48cm 2 Area = 16cm 2 4 cm Side 1 Side 2 Side 3 6 x 10 = 60cm 2 2 x 10 = 20cm 2 4 x 10 = 40cm 2

Parts of a circle

Area and circumference of circles Work out the area of a circle with a radius of 7m. Give your answer in terms of π. Work out the area of a circle with a radius of 12cm. Give your answer in terms of π. A = πr 2 = π x 7 2 = π x 49 = 49π m 2 r = 12cm C = πd = π x 24 = 24π cm d = 24cm

Area of a sector Find the area of the sector in terms of π Area of sector = x 360 A = 40 360 π 62 π radius2 = 1 9 π 36 = 36π 9 = 4π cm 2

Length of an arc Find the perimeter of the sector in terms of π Length of arc = x 360 L = 40 360 2 π r 2 π 18 = 1 9 2 π 18 = 36π 9 = 4π cm Perimeter of sector = 4π + 18 + 18 = 4π + 36

Area of a sector Find the shaded area correct to 2 decimal places. Area of large sector = 35 360 x π x 92 = 24.7400421 cm 2 Find the area of the large sector Find the area of the small sector Subtract to get the shaded area Area of small sector = 35 360 x π x 52 = 7.63581548 cm 2 Shaded area = 24.7400421 cm 2-7.63581548 cm 2 = 17.10422662cm 2 = 17.10cm 2 (2dp)

Area of non-right angled triangles Use Area = 0.5ab sin C Area = 0.5 x length 1 x length 2 x sine of the included angle HIGHER ONLY

Area of non-right angled triangles A = 0.5 x 5 x 5 x sin 60 5cm 60 0 5cm = 23.8156986 60 0 5cm 60 0 = 23.8 cm 2 (1dp) HIGHER ONLY

Area of non-right angled triangles Area of triangle 1 = 0.5 x 80 x 80 x sin 71 = 3025.6594419 m 2 1 2 Area of triangle 2 = 0.5 x 70 x 50 x sin 80 = 1723.41356777 m 2 Area of quadrilateral = 4749.07300967 = 4749 m 2 (nearest m 2 ) HIGHER ONLY

Area of segments Find the area of the segment. Give your answer to 3 significant figures. Area of segment = Area of sector Area of triangle Area of sector = 25 360 π 62 = 7.85398163397 Area of triangle = 0. 5 6 6 sin 25 0 = 7.60712871133 Area of segment = 7.85398163397 7.60712871133 = 0.24685292264 = 0.247 cm 2 (3 sf) HIGHER ONLY

Volume of prism Volume of prism = area of cross section x length

Volume of prism Find the volume of the triangular prism below. Volume prism = Area of cross-section x Length Area of cross-section = 8 3 2 V = 12 x 12 = 144cm 3 = 12cm2

Volume of prism Find the volume of the prism below. Volume of prism = Area of cross-section x Length Area of cross section: 10 + 6 = 16 4 x 16 = 64 64 2 = 32cm 2 V = 32 x 5 = 160cm 3

Volume of prism Find the volume of the prism Volume of prism = Area of cross section x Length = 64cm 2 x 10cm = 640cm 2 8 cm A 4 cm B 4 cm Area of A = 48 cm 2 Area of B = 16cm 2 Area of cross section = 64cm 2

Volume of a cylinder Calculate the volume of the cylinder. Leave your answer in terms of p 6 cm 8 cm V = pr 2 h V = p x 6 2 x 8 = 288pcm 3

Volume of a sphere Find the volume of the hemisphere in terms of π 3 cm Volume of sphere = 4 π r3 3 Volume of hemisphere = 2 π r3 3 V = 2 3 π 33 = 2 3 π 27 = 18πcm 3

Surface area of a cone Find the total surface area of the cone. Give your answer to 1 decimal place. Curved surface area = πrl l 2 = 4 2 + 3 2 = 16 + 9 = 25 l = 5 cm

Surface area of a cone 5 cm Total surface area = curved surface area + area of base = πrl + πr 2 = (π x 3 x 5) + (π x 3 2 ) = 75.3982237 = 75.4 cm 2 (1 dp) HIGHER ONLY

Volume problems The shape below is composed of a solid metal cylinder capped with a solid metal hemi-sphere as shown. Find the volume of the shape. (to 3 sig fig) Volume of hemi-sphere = 4 3 πr3 2 = 2 3 pr3 = 2 3 x p x 33 6 m 4m = 56.5486678m 3 Volume of cylinder = pr 2 h = p x 3 2 x 4 = 113.097336 m 3 Total volume = 169.646003 m 3 = 170 m 3

Volume problems 12 cm A fuel pod consists of cylinder with a hemi-spherical base and a conical top as shown in the diagram. Calculate the surface area of the pod. (answer to 2 sig fig) Curved surface area of cone = prl = p x 5 x 12 = 188.495559 cm 2 40 cm Curved surface area of cylinder = 2prh = 2 x p x 5 x 40 = 1256.63706 cm 2 10 cm Surface area of hemi-sphere = 2pr 2 = 2 x p x 5 2 = 157.079633 cm 2 Total surface area = 1602.21225 cm 2 = 1600 cm 2

Volume of a pyramid Volume of pyramid = 1 / 3 x area of base x h Area of base = 8 x 8 = 64 cm 2 Volume of pyramid = 1 / 3 x 64 x 9 = 192 cm 3

Volume of a frustum Volume of frustum = Volume of big cone Volume of small cone HIGHER ONLY

Volume of a frustum A cone of height 6cm and base radius of 8cm is cut to form a smaller cone and a frustum. Find the volume of the frustum. Give your answer to 3 significant figures. Volume of frustum = Volume of big cone Volume of small cone Volume of big cone = 1 x π x 82 x 6 3 = 402.1238597 cm 3 Volume of small cone = 1 x π x 3 42 x 3 = 50.26548246 cm 3 Volume of frustum = 402.1238597-50.26548246 = 351.85837724 cm 3 = 352 cm 3 (3sf) HIGHER ONLY

Density, mass and volume MOVE DOWN VERA

Density, mass and volume A cuboid has dimensions of 2 m x 90 cm x 140 cm. It has a density of 0.8kg/m 3. Find the mass of the cuboid. Mass = Density x Volume Volume of cuboid = 2.52 m 3 Mass = 0.8 x 2.52 Mass = 2.016 kg 90cm 2m 140cm

Plans and elevations Front The plan of a 3D shape is the two dimensional view that you see from above i.e. the bird s eye view. The front elevation is the two dimensional view of the 3D shape from the front. The side elevation is the two dimensional view of the 3D shape from the side.

Alternate angles Z shape Alternate angles are equal

Corresponding angles F shape Same place but further up or down (or left or right) Corresponding angles are equal

Angles on parallel lines 70 0 70 0 Angle FBC = 70 0 Alternate angles are equal. Angle BFC = 70 0 Base angles in an isosceles triangle are equal. x = 40 0 Angles in a triangle add up to 180 0.

Angles of regular polygons Exterior angle = 360 0 no of sides Interior angle + exterior angle = 180 0 Interior angle = 180 0 exterior angle No. of sides = 360 0 exterior angle Sum of interior angle of any polygon = (number of sides 2) x 180 0

Angles of polygons Sum of interior angles = (5 2) x 180 0 = 540 0 85 0 + 82 0 + 124 0 + 100 0 = 391 0 x = 540 0 391 0 = 149 0

Angles of regular polygons The diagram shows a regular polygon with 8 sides. Work out the value of x. Interior angle of octagon: Exterior angle = 360 0 8 = 45 0 Interior angle = 180 0 45 0 = 135 0 The triangle is isosceles 180 0 135 0 = 45 0 x = 45 0 2 = 22.5 0

Interior angle of decagon: Exterior angle = 360 0 10 = 36 0 Interior angle = 180 0 36 0 = 144 0 The triangle is isosceles 180 0 144 0 = 36 0 36 0 2 = 18 0 x = 144 0 18 0 18 0 = 108 0

Angles of regular polygons Angle DCF Interior angle of a regular pentagon Exterior angle of regular pentagon = 360 0 5 = 72 0 Interior angle of regular pentagon = 180 0 72 0 = 108 0 DCF = 108 0 90 0 = 18 0 108 0

Bearings A bearing is used to represent the direction of one point relative to another point. Bearings are a way of giving directions more accurately. 1. Bearings are always measured from North, which is 000 0 (or 360 0 ). 2. They are always measured clockwise. 3. They must have three figures. So East is 090 0, South is 180 0 and West is 270 0.

Bearings x C Bearings must be written as 3 figures 066

Bearings Find the bearing of B from A 360 o - 30 0 = 330 0

Bearings

Constructing a perpendicular bisector 1. Draw a line connecting the points if their isn t one there. 2. Place compass at A, set over halfway and draw 2 arcs. 3. Place compass at B, with same distance set and draw 2 arcs to intersect first two. 4. Draw the perpendicular bisector through the points of intersection. A B

Constructing an angle bisector 1. Place compass at A, and draw an arc crossing AB and AC. 2. Place compass at intersections and (with the same distance set) draw 2 arcs that intersect. B 3. Draw the angle bisector from A through the point of intersection. A C Angle

Construct a triangle of sides 8 cm, 7cm and 6 cm. 1. Draw a line 8cm long and use it as the base of triangle. 2. Set compass to 7 cm, place at A and draw an arc. 3. Set compass to 6 cm, place at B and draw an arc to intersect the first one. 4. Draw straight lines from A and B to point of intersection. Tri 3 sides 7 cm 6 cm A 8 cm B

Constructing an equilateral triangle 1. Draw base line AB of any length. 2. Place compass at A, set to distance AB and draw arc. 3. Place compass at B, with same distance set, draw an arc to intersect first one. 4. Join intersection point to A and B to form an equilateral triangle. Eq Tri A B

Loci Two lines = angle bisector

Reflection Draw the line x = 1 Reflect the shape Remember to label it

Rotation Rotate the triangle, 90 0 clockwise, about the centre of rotation. centre of rotation

Rotate the triangle, 90 0 clockwise, about the centre of rotation.

Rotate the flag 180 0 about the centre of rotation. 180 0

Enlargement from a centre Enlarge the green shape by scale factor 2 using the dot as the centre of enlargement When enlarging from a centre the image must be twice as big AND twice as far away from the centre

Enlargement from a centre Enlarge shape P by a scale factor of 2 from the origin. Origin = (0, 0) The new shape needs to be twice as big and twice as far away from the centre.

Translation y Translate Hexagon A by the vector 9 8 7 6 5 4 3 2 1 3 left 4 down 4 down B A 3 left æ çç è - - 3 4 ö ø Label the image B 0 1 2 3 4 5 6 7 8 9 x

Describing a single transformation

Translation

Enlargement To find the centre of enlargement draw lines through corresponding points on the image and object and find where they cross

Rotation

Reflection

Similar shapes The two rectangles below are similar. Find x. Scale factor = enlarged length original length = 30 6 x = 5 x = 16 x 5 = 80 cm

Similar shapes Find DE C C 8cm 10cm D E B 12cm A STEP 1 Draw out the two triangles separately. STEP 2 STEP 3 Find the scale factor. Find the missing side. Scale factor = 10 8 = 1.25 DE = 12 1.25 = 9.6cm

The relationship between the lengths, areas and volumes of similar shapes If the scale factor of the lengths is k, then: the scale factors of the areas will be k 2 (i.e. the areas will be k 2 times as big) the scale factor of the volumes will be k 3 (i.e. the volumes will be k 3 times as big) HIGHER ONLY

Similar shapes - area Scale factor of lengths = 18 6 = 3 Scale factor of areas = 3 2 = 9 Area of triangle A = 36 9 = 4cm 2 HIGHER ONLY

Similar shapes - volumes Scale factor of volumes = 54 16 = 3.375 Scale factor of lengths = N = 1.5 Length of cylinder B = 6 x 1.5 = 9cm 3. 375 HIGHER ONLY

Congruent shapes If two or more shapes are described as 'congruent', it means that they are identical in shape and size. Shapes can be congruent even if one of the shapes has been reflected or rotated. Reflect

Congruent triangles Consequently, when two or more triangles are described as congruent, it means that they have exactly the same three sides and exactly the same three angles. These triangles can be congruent even if one of them has been reflected or rotated. For example, the triangles below are congruent because they have exactly the same sides and angles: HIGHER ONLY

Congruent triangles However, it is important to note that triangles can have exactly the same angles but not be congruent. Triangles can only be congruent if all of their sides are exactly the same length. Their angles may be the same also but it is not necessary. For example, the triangles displayed below have exactly the same angles but are not congruent because their sides are different lengths: HIGHER ONLY

Proving that triangles are congruent There are four ways of proving that two triangles are congruent: SSS (Side-Side-Side) SAS (Side-Angle-Side) AAS (Angle-Angle-Side) RHS (Right Angle-Hypotenuse-Side) HIGHER ONLY

Proving that triangles are congruent SSS (Side Side Side) Often referred to as the SSS rule, the Side-Side-Side rule can be used to prove two triangles are congruent. If the three sides of the first triangle are equal to the three sides of the second triangle, then the triangles are congruent to one another. You can mark these SSS triangles in the following manner: HIGHER ONLY

Proving that triangles are congruent SAS (Side Angle Side) Often referred to as the SAS rule, the Side-Angle-Side rule can be used to prove two triangles are congruent. If two sides of the first triangle are equal to two sides of the second triangle, and the included angle is equal, then the two triangles are congruent. You can mark these SAS triangles in the following manner: HIGHER ONLY

Proving that triangles are congruent AAS (Angle Angle Side) Often referred to as the AAS rule, the Angle-Angle-Side rule can be used to prove two triangles are congruent. If two angles of the first triangle are equal to two angles of the second triangle, and one similarly located side is equal, then the two triangles are congruent. You can mark these AAS triangles in the following manner: HIGHER ONLY

Proving that triangles are congruent RHS (Right Angle Hypotenuse - Side) Often referred to as the RHS rule, the Right Angle-Hypotenuse-Side rule can be used to prove two triangles are congruent. If the hypotenuse and one other side of the first right-angled triangle are equal to the hypotenuse and corresponding side of the second right-angled triangle, then the two triangles are congruent. You can mark these RHS triangles in the following manner: HIGHER ONLY

Angle AEB = Angle DEC (Vertically opposite angles are equal) Angle BAE = Angle DCE (Alternate angles are equal) Triangle ABE is congruent to EDC (ASA) HIGHER ONLY

AX = AX AB = AC (Common side) (Given Hypotenuse) Angle AXB = Angle AXC = 90 0 The triangles are congruent (RHS) HIGHER ONLY

AB = AD BC = CD AC = AC (Given - SIDE) (Given SIDE) (Common side SIDE) The triangles are congruent (SSS) HIGHER ONLY

Pythagoras theorem Finding the hypotenuse: square, add, square root Finding a shorter side: square, subtract, square root

Pythagoras theorem Find the length of the line joining (2, 5) and (9, 14) Make a sketch. Draw in a right angled triangle. Find the length of the base and height of the triangle Use Pythagoras theorem x 2 = 7 2 + 9 2 14 5 y 2 x 7 9 9 x = 49 + 81 = 130 x = 130 = 11.401754251 = 11.4 cm (1dp)

Pythagoras theorem Is this a right angled triangle? If it is a right angled triangle, then 8 2 = 5 2 + 6 2 8 2 = 64 5 2 = 25 6 2 = 36 25 + 36 = 61 As 61 is not equal to 64, this is NOT a right angled triangle.

Pythagoras theorem 3D Look at the cuboid below. Calculate the length of the diagonal AG which runs through the space inside the cuboid. E H F G 16cm (1) Start by considering the triangle ABC. D 28cm A B AC = 30.463092423cm or Ö 928 C 12cm A (2) Calculate the length of AC. 28 C 12

Pythagoras theorem 3D Now look at the cuboid again. H E F G 16cm (3) Now consider triangle ACG. (4) Calculate the length of AG. G 16 D C A Ö 928 AG 2 = (Ö 928) 2 + 16 2 Ö 928 12cm AG 2 = 1184 A 28cm B AG = 34.4093011 AG = 34.4cm (1dp)

Trigonometry SOH CAH TOA sin x! = cos x! = tan x! = opposite hypotenuse adjacent hypotenuse opposite adjacent

Trigonometry finding a side SOH CAH TOA HYP OPP H = A C ADJ x = 4 cos 78 = 19.23893738 = 19.2 cm (1dp)

Trigonometry finding an angle HYP to one decimal place X ADJ OPP SOH CAH TOA sin x = opp hyp sin x = 10 11 x = sin -1 (10 11) x = 65.3800267 x = 65.4 0 (1 dp)

Trigonometry ABC is an isosceles triangle. Work out the area of the triangle. Give your answer correct to 1 decimal place. SOH CAH TOA Opp Height Hyp O = T A Height = tan 54 0 x 6 = 8.25829152 cm Area of triangle = ½ bh 54 0 6 cm Adj 12 8.25829152 A = 2 = 49.5497491 = 49.5cm 2

Trigonometry A 6 metre ladder is used to clean windows. It must stand 1.5 metres from the base of a wall on horizontal ground. In order for the window cleaner to be able to safely use the ladder, the angle between the ground and the ladder cannot exceed 75 o. Is the ladder safe to use? Wall Ground 1.5m Ladder 6m x 0 X Opp 1.5m Adj Hyp 6m x 0 Use CAH cos x = adj hyp cos x = 1.5 6 x = cos -1 (1.5 6) x = 75.5224878 x = 75.5 0 (1 dp) The ladder is not safe to use as the angle exceeds 75 0.

Exact trig ratios

Trigonometry using exact ratios Find x. SOH CAH TOA OPP HYP ADJ X O = S x H x = sin 30 0 11 = ½ x 11 = 5.5 cm

Sine and cosine rules Missing sides and angles of non right angled triangles Finding a side Finding an angle Sine rule Cosine rule Sine rule Cosine rule a sin A = Need to know two angles and a side b sin B = c sin C Need to know two sides and an included angle a 2 = b 2 + c 2 2bcCosA Need to know two sides and a non included angle sina a = sinb b = sinc c Need to know all three sides CosA = b2 + c 2 a 2 2bc HIGHER ONLY

Sine and cosine rules HIGHER ONLY A fishing boat leaves a harbour (H) and travels due East for 40 miles to a marker buoy (B). At B the boat sails on a bearing of 035 o to a lighthouse (L) 24 miles away. It then returns to harbour. Find the total distance travelled by the boat to the nearest mile. HL 2 = 40 2 + 24 2 (2 x 40 x 24 x Cos 125 0 ) L HL 2 = 1600 + 576-1101.266758 HL 2 = 3277.26678 H 40 miles 125 o 24 miles B HL = Ö(3277.26678) = 57.247417 miles a 2 = b 2 + c 2 2bcCosA = 57 miles (to the nearest mile) Total distance = 57 + 40 + 24 = 121 miles

Sine and cosine rules Find p to 3 significant figures B c a 15 o p 145 o C A 45 m b a b c = = SinA SinB SinC ü X = ] YZ[ \ YZ[ ^ ü ü a = 45 sin 145 sin 15 p = bc YZ[ dc x sin 145 p = 99.72581278 p = 99.7 m (3sf) HIGHER ONLY

Sine and cosine rules Find y to 1 decimal place. B a 12.7cm 11.4cm c C YZ[ \ 63 o X b y A sin A 12.7 sin A = = YZ[ e f = sin 63 11.4 SinA SinB SinC = = a b c ü YZ[ gh dd.b x 12.7 ü ü Sin A = 0.99261253133 A = sin -1 (0.99261253133) = 83.0312773471 = 83.0 0 (to 1dp) HIGHER ONLY

Sine and cosine rules ABC is a triangle in which a = 8.2 cm, b = 10.5 cm and c = 14.6 cm. Calculate the largest angle in the triangle. Give your answer to one decimal place. A c a b C The largest angle is opposite the largest side Cos A = b2 ic 2 ja 2 2bc Cos A = 10.52 i8.2 2 j14.6 2 2 10.5 8.2 Cos A = j35.67 172.2 A = cos j1 j35.67 172.2 A = 101.954969 = 102.0 0 (1dp) HIGHER ONLY

Circle theorems a = 55 0 The angle at the centre is twice the angle at the circumference b = 98 0 Opposite angles in a cyclic quadrilateral add up to 180 0 d = 35 0 Angles in the same segment are equal HIGHER ONLY

Circle theorems B O 3 cm x o 3 cm A 5 cm PB = 5cm P e = 90 0 The angle in a semicircle is a right angle x = 90 0 A tangent and a radius always meet at right angles Tangents from the same point are equal in length HIGHER ONLY

Circle theorems HIGHER ONLY

Circle theorems Angle ABO = x Base angles in an isosceles triangle are equal 180 2x Angle AOB = 180 2x Angles in a triangle add up to 180 0 x 90 0 180 (180-2x) = 180 180 + 2x = 2x 180 (90 + 2x) = 180 90-2x = 90-2x Angle OBC = 90 0 A tangent and a radius always meet at right angles Angle BOC = 2x Angles on a straight line add up to 180 0 Angle ACB = 90 2x Angles in a triangle add up to 180 0 HIGHER ONLY

Circle theorems 90 - x ODB = x Base angles in an isosceles triangle are equal. BOD = 180 2x Angles in a triangle add up to 180 0. BAD = ½ (180 2x) = 90 - x The angle at the centre is twice the angle at the circumference. BCD = 180 (90 x) = 180 90 + x = 90 + x Opposite angles in a cyclic quadrilateral add up to 180 0 x 180 2x HIGHER ONLY

Vectors a + b - c

Vectors a = 3 2 b = 5 6 a + b = = 2 4 3 + 5 2 + 6 2a 3b = 2 3 2 2 3 5 3 6 = 6 4 15 18 = 6 15 4 18 = 21 22

Vectors Write AB as a column vector. AB = 6 5 B A

Vectors Write AB as a column vector. AB = 5 1 B A

Vectors Write AB as a column vector. AB = 4 2 A B

Vectors Write AB as a column vector. AB = 6 0 A B

Vectors PQ = PS + SQ = -2a + b = b 2a

Vectors CM = ½CB CB = CA + AB = -s + r CM = ½(-s + r) = -½s + ½r HIGHER ONLY

Vectors AD = AB + BD = a + ½ BC BC = BA + AC = -a + b AD = a + ½ (-a + b) = a ½ a + ½ b = ½ a + ½ b HIGHER ONLY

Vectors 2a 2b 6a HIGHER ONLY

Vectors MN = MP + PO + ON = -a 2b+ 3a = 2a 2b

Vectors OP = OA + AP = 6a + 2 3 AC AC = AO + OC = -6a + 6c OP = 6a + 2 (-6a + 6c) 3 = 6a 4a + 4c = 2a + 4c HIGHER ONLY

Vectors 1 2 AB a 3 4 BC 3b 1 4 BC 1 2 AB HIGHER ONLY

Vectors XY = XB + BY = a + 3b YD = YC + CD = b 2a DX = DA + AX = -4b + a HIGHER ONLY

Distance-time graphs

Speed, distance & time 1 S = D T S = 240 miles 2. 5hrs S = 96 mph

Speed, distance & time 2 S = D T 310 miles S = 4. 25 hrs S = 72. 9411765 mph S = D T 310 miles S = 255 mins S = 1. 21568627 miles/min S = 72. 9411765 mph

Speed, distance & time 3 The distance time graph shows a cyclist's journey. How much faster was the cyclist's speed between 20 and 30 minutes compared to 0 to 10 minutes? 0 to 10 minutes Distance = 1.5km Time = 10 mins 10 mins = 1. 5 km x 6 x 6 60 mins = 9 km Speed = 9 km/h 20 to 30 minutes Distance = 2.5km Time = 10 mins 10 mins = 2. 5 km x 6 x 6 60 mins = 15 km The cyclists speed was faster by 6 km/h. Speed = 15 km/h

Velocity-time graphs For a velocity-time graph: The gradient is the acceleration The area under the graph is the distance travelled a) What was the acceleration in the first 4 seconds? b) How far was travelled in the first 10 seconds? HIGHER ONLY

Velocity-time graphs a) What was the acceleration in the first 4 seconds? To find the acceleration find the gradient of the line. Gradient = change in y change in x = 8 4 = 2m/s2 HIGHER ONLY

HIGHER ONLY Velocity-time graphs b) How far was travelled in the first 10 seconds? We need to find the area of the graph between 0 and 10 seconds. Area of triangle = 4 8 = 16 2 Area of rectangle = 6 x 8 = 48 Total area = 64 Distance = 64m

Gradient of curves The gradient of a curve is constantly changing. Gradient is negative and approaching zero Gradient is positive and increasing HIGHER ONLY

Gradient of curves To find the gradient of a curve at a given point, you need to draw a tangent this is a line that only touches the curve at this point. You should then calculate the gradient of the tangent. Example: For the curve opposite, find the gradient when x = 2. 2 4 Gradient = 4 2 = 2 HIGHER ONLY

Gradient of curves Find the gradient of the curve at x = 1-4 Gradient = j4 1 = -4 1 HIGHER ONLY

Gradient of curves Find the gradient of the curve at x = 1 1.5 2 Gradient = 2 1.5 = 4 3 HIGHER ONLY

The graph below shows the first 10 seconds of a cyclists journey. Calculate an estimate for the distance travelled by the cyclist in these 10 seconds. Use 5 strips of equal widths. Areas: Triangle = 2 5 2 = 5 Trapezium 1: 5 + 7 = 12 12 x 2 = 24 24 2 = 12 Trapezium 2: 7 + 8 = 15 15 x 2 = 30 30 2 = 15 Trapezium 3: 8 + 8.5 = 16.5 16.5 x 2 = 33 33 2 = 16.5 Trapezium 4: 8.5 + 9 = 17.5 17.5 x 2 = 35 35 2 = 17.5 Total area = 5 + 12 + 15 + 16.5 + 17.5 = 66 units 2 Distance = 66m HIGHER ONLY

The graph opposite shows the first 8 seconds of a ball s movement. a) Calculate the acceleration of the ball at time t = 6 b) Calculate the distance travelled by the ball. Use 4 strips of equal widths. a) Gradient = 2.1 1 = 2.1 Acceleration = 2.1m/s 2 1 2.1 b) Area = d r 2 0.2 + d r 0.2 + 1 2 + 1 2 1 + 4 2 + 1 2 4 + 10 2 = 0.2 + 1.2 + 5 + 14 = 20.4 units 2 Distance = 20.4m HIGHER ONLY