Mathematics Geometry Revision Notes for Higher Tier Thomas Whitham Sixth Form S J Cooper Pythagoras Theorem Right-angled trigonometry Trigonometry for the general triangle rea & Perimeter Volume of Prisms, Pyramids & Spheres Surface rea Similar shapes and ratios ngles Vectors
. Pythagoras Theorem Using Pythagoras Theorem c 5 5 44 69 c 69 cm If it is a calculator paper there is no need for this line as the above can be typed into the calculator. Using Pythagoras Theorem y 4 0 96 y 96 9.80cm Here we have a difference of the squares as we are finding one of the smaller sides. Thomas Whitham Sixth Form
Remember to square and add when finding the hypotenuse. Square and subtract when finding one of the smaller sides. Using Pythagoras Theorem PQ 5 666 PQ 666 5.8m x. 0.44 x.m Thomas Whitham Sixth Form
y 5 4 0 4 5 6 x Find the exact length of X distance from to = 6 Y distance from to = 6 4 6 40 40 0. The right angled triangle Learn to identify by name the sides of a triangle in relation to a given angle. Opposite Hypotenuse First The hypotenuse (faces the right angle always). djacent Next The side Opposite to your angle Then The djacent side (the last side) Thomas Whitham Sixth Form
. The three trig ratios Learn the following mnemonic: Which enables you to write down TO / SOH / CH Leading to the three trig ratios: tan Opposite djacent sin Opposite Hypotenuse cos djacent Hypotenuse 4. To calculate a given angle in a right angled triangle Calculate the size of angle in the C triangle C. 5 cm Solution 8 cm Firstly identify the sides C is the hypotenuse C Opposite angle is djacent to. Hence we require the Tan ratio i.e. tan 5 8.0 Opposite tan djacent On our calculators we work out 5 8. Then press inverse tan i.e. buttons. tan. Which is usually shift and tan Calculate the size of angle in the triangle C below C 4 cm Solution Labelling the sides we are given the Hypotenuse and the Opposite. 6 cm So we use the Sine ratio 4 Thomas Whitham Sixth Form
sin 4 6 4.8 On our calculators we work out 4 6. Then press inverse sin i.e. sin. Which is usually shift and sin buttons. 5. To calculate the length of a given side in a right angled triangle Calculate the length of PQ in the triangle PQR Q R Solution PR is the Hypotenuse (= 5) QR is the opposite (Not required) P 5 cm PQ is the djacent (to be found) cos djacent Hypotenuse PQ cos 5 5 cos PQ PQ 44.cm Here use our algebra skills and rearrange the equation above by multiplying both sides by 5. Calculate the length of XY in the triangle XYZ. X Solution s we require XY (the hypotenuse) and are given 4.6cm (the adjacent) we must use the cosine ratio. cos djacent Hypotenuse Z 4 Y 4.6 cm 5 Thomas Whitham Sixth Form
4.6 cos 4 XY XY cos 4 4.6 4.6 XY cos 4 XY 6.cm Here use our algebra skills and rearrange the equation above by multiplying both sides by XY. nd then divide both sides by cos 4. 6. The isosceles triangle divides into two identical right angled triangles P P 8cm 8cm 8cm 8cm Q R Q 6 cm cm cm R 70 8 cm 70 4 cm 70 70 4 cm 48 4 4 6 Thomas Whitham Sixth Form
To calculate the size of the angle PQR in the a foregoing triangle PQR P Solution Concentrate on one right angled triangle Here we know the hypotenuse and adjacent cos Q 8 Q 68.0 Q 8cm 8cm cm cm R Find the length of in the triangle below 5 cm C 48 5 cm Solution Start by finding the length x in the Right angled triangle formed C 4 5 cm x Here we are given the hypotenuse and want 5 cm to find the opposite, therefore x sin 4 5 5 sin 4 x x.068 C =.068 4. cm to dp 7 Thomas Whitham Sixth Form
7. ngles of Elevation and depression are always measured from the horizontal level. L ngle of elevation of the top of a lighthouse L from a point P P Horizontal Ground (imaginary) Horizontal line T ngle of depression of a uoy P at sea from the top T of a cliff 8. The general Triangle Notice Side a is opposite angle Side b is opposite angle Side c is opposite angle C 9. The Sine rule Learn the following rule: a sin b sin c sin C 8 Thomas Whitham Sixth Form
0. To calculate the length of a given side in a general triangle Find c in the triangle below c sin 50 = 7 sin 60 c = 7 sin 50 sin 60 c = 6.9cm Find a in the triangle below a sin 7 = 8 sin 0 a = 8 sin 7 sin 0 a =.87cm 9 Thomas Whitham Sixth Form
Find x in the triangle below First we need to work out the angle opposite the 0.6cm. Using angles in a triangle. ngle = 86 x sin 40 = 0.6 sin 86 x = 0.6 sin 40 sin 86 x = 6.8cm. To calculate the size of a given angle in a general triangle Find the size of angle in the triangle below Here we use the sine rule writing it upside down! n additional twist we must first find angle C, as we don t know the length opposite angle. 0 Thomas Whitham Sixth Form sin C 6 sin C = = sin 0 5 sin 0 6 5 sin C = 0.46.. C = 0.
In triangle C, C = cm, angle = 40 and angle = 65. Find the length C. Firstly draw a sketch! C 65 40 C sin 65 = sin 40 C = sin 65 sin 40 C = 4.cm. The Cosine rule a b c bccos b c a a c b ac cos abcos C. To calculate the length of a given side in a general triangle Find the length b in the triangle below Thomas Whitham Sixth Form
Using the cosine rule: b = 5 + 8 5 8 cos b = 8.968 which can be typed into the calculator in one go! b = 0.9cm Find the length x in the triangle below x = 0 + 9 0 9 cos 64 x = 0.09 x = 0.cm Find the length y in the triangle below y =. + 7.5. 7.5 cos 48 y = 84.085 y = 9.7cm Thomas Whitham Sixth Form
4. To calculate the size of an angle when the given the lengths of all three sides of a triangle Find angle C Here we could use the cosine rule and rearrange to find ngle C. Or use the rearranged cosine rule which is cos = b +c a 6 = 7 + 5 7 5 cos 6 = 74 70 cos 70 cos = 8 cos = 8 70 = 57. bc Find angle P Using the rearranged formula cos P = + 8 6 8 cos P = 49 76 P =. Thomas Whitham Sixth Form
5. The area of a triangle rea absin C Find the area of triangle C rea = 8 6 sin 4 =.9cm (i) (ii) Find the length Hence find the area of the triangle. 4 Thomas Whitham Sixth Form
(i) Using the Sine rule we can find sin 7 = 5 sin 6 = 5 sin 7 sin 6 = 5.4cm (ii) rea = 5 5.4 sin 45 Find the area of triangle PQR = 9.4cm Using the cosine rule find an angle inside the triangle 9 = 6 + 4 6 4 cos P 8 = 5 48 cos P 48 cos P = 9 cos P = 9 48 P = 7. rea = 6 4 sin 7. = 9.56cm 5 Thomas Whitham Sixth Form
6. D and D applications From a point at the top of a cliff 50m high the angle of elevation to a balloon, is 4. Given that its horizontal distance from the cliffs edge is 75m, find the height of the balloon above sea level, giving your answer correct to decimal places. Using the right angled triangle given 4 50m 75m x 4 x tan 4 75 75 tan 4 x x 69.94 75m height = 69.94 + 50 = 9.94 m 6 Thomas Whitham Sixth Form
The diagram below shows part of a sea barrier whose constant cross-section is a right-angled triangle. The top DC represents a road and the beach starts at the line. The line joining to D represents one of two paths to the beach, the other being a steeper climb D. ngles DE and CF are 90 each. C D m F E 4m 0m Find a) The length of each path b) The angle of inclination of each path to the horizontal FE Solution (i) Firstly taking from the D diagram the right angled triangle DE we can find the length D D m Using Pythagoras D 5 4 E 4m D 5m D Using the previous result and the right angled triangle taken from the rectangle CD we can find the length of the nd path. 5m D 5 0 0m 5 D.8m 7 Thomas Whitham Sixth Form
(ii) Using the triangles on the previous page we can find the angle of inclination using right angled trigonometry. tan 4 6.87 tan 5 0 6.57 Hence the angle of inclination for path D = 6.87, and for path D = 6.57 On a radar two planes and can be seen on bearings 075 and 54, respectively and at distances 5km and 7km from the control tower. Find the distance between the two planes. N 75 5km 79 7km 8 Thomas Whitham Sixth Form
Using the cosine rule 5 7.9 5.9km 5 7 cos 79 7. rea & Perimeter (i) The Circle C D or C r rc length Perimeter of a circle is called the circumference. Work out the circumference of the circle below l D 60 C D 0 0.4cm What is the length of fencing required to enclose a semi-circular field with diameter 80m. Give your answer correct to decimal place. 9 Thomas Whitham Sixth Form
Circumference = D 80 5.6 Perimeter = 6.6 + 80 = 05.6 m What is the length of an arc of a circle of radius 4cm with an angle of 68:? rc length = D 66 60 66 84 60 48.8cm r r 60 Find the area of a circle which has radius 5cm. 5 5 78.54cm 0 Thomas Whitham Sixth Form
Find the area of the shaded sector below, giving your answer correct to decimal place. 7 cm 5: 7 cm Solution rea 7 57.7cm 5 60 The area of a circle is given as 05 m. What is its diameter? r 05 r 05 r 05 r 8.077 D 6.6m Thomas Whitham Sixth Form
(ii) Irregular shapes Find the area and perimeter of the shape drawn below Here we have a quarter of a circle on top of a rectangle. The difference between 4 and 0 gives the radius as 6cm, and therefore the width is 6cm. rea of rectangle = 6 x 4 = 84 rea of quarter circle = 6 8. 7 4 rea of shape = 84 + 8.7 =.7 cm Thomas Whitham Sixth Form
Calculate the shaded region in the shape below rea of rectangle = 5 x 5 = 65 rea of Circle = 6.5. 7 rea shaded = 65.7 = 49.7 m 8. Volume (i) Prisms Cube Cuboid Cylinder Triangular-based Volume = ase rea x length Thomas Whitham Sixth Form
Which cylinder has the greatest volume? Cylinder ase rea =.5 9. 6 Volume = 9.6 9 = 76.7cm Cylinder ase rea = 4.5 6. 6 Volume = 6.6.5 = 59.04cm Cylinder has the greater volume. Find the volume of the solid below, giving your answers in (a) cm (b) litres ase rea = 9 5 6 4 000 cm = litre Volume = 4 7 = 94m Volume = 94 000 000 cm = 94 000 litres 4 Thomas Whitham Sixth Form
(ii) Pyramids Cone Square-based Tetrahedron Volume = x ase rea x height Find the volume for each of the following pyramids (i) ase rea =.5 9. 6 Volume = 9.68 5.6cm 5 Thomas Whitham Sixth Form
(ii) ase rea = 9 6cm Volume = 9 6 8cm cm cm teal plug is machined into the shape of a truncated cone, with end radii of.5cm and.4cm. Find the volume of the plug Small Cone ase rea =.5 7. 068 Volume = 7.068 4.5 0. 608 Larger Cone ase rea =.4 8. 095 Volume = 8.09558 48. 54 Volume of plug = 48.54 0.608 = 7.65cm 6 Thomas Whitham Sixth Form
(iii) Spheres Volume = 4 r Find the volume of the basket ball whose radius is.9 cm Volume = 4.9 7058.78cm The diagram below is of a bowl which consists of a hemi-sphere of diameter 0cm with a smaller hemi-sphere, of radius 9cm, cut out. Find the volume of the bowl. Volume of smaller hemi-sphere = 9 56. 8 Volume of smaller hemi-sphere = 0 094. 4 Volume of bowl = 094.4 56.8 = 567.58cm 7 Thomas Whitham Sixth Form
9. Surface rea Surface rea 4r s for the other solids the surface area is found by finding individual face areas and adding them together. Find the Surface area for each of the following solids (i) 7cm Curved section unwrapped would represent a rectangle with height 7cm and width equal to the circumference of the circle. 0cm Hence cylinder consists of three regular shapes a rectangle and two circles. ase rea = 0 4. 59 Circumference = 0 6. 88 rea of rectangle = 6.88 7 = 49.8 Surface area = 49.8 + 4.59 = 068.4cm (ii) 5cm 5cm 6 cm 4cm 8 Thomas Whitham Sixth Form
Here we have an * question as we need to use our trigonometry first. We have two faces of these dimensions First find this angle. 5cm 5cm To find this angle we could use either the cosine rule or right angled trig. 4cm Using right angled trig cos 5 66.4 Hence its area = 45 sin 66.4 9. 65 nd we have two faces of these dimensions Likewise with this. 5cm 5cm To find this angle we could use either the cosine rule or right angled trig. 6cm Using right angled trig cos 5 5. Hence its area = 65 sin 5. rea of rectangular base = 4 6 = 4 Surface rea = 4 + + 9.65 = 66.cm 9 Thomas Whitham Sixth Form
0. Ratios: Length, rea & Volume In General Given Length scale factor = a : b Then rea scale factor = a : b Volume scale factor = a : b Two similar cylindrical tins have base radii of 6 cm and 8 cm respectively. If the capacity of the large tin is 5 cm, find the capacity of the small tin. Solution Here length ratio = 6 : 8 = : 4 Hence volume ratio = : 4 = 7 : 64 7 Capacity of the smaller tin = 5 6 cm 64 Two solid spheres have surface areas of 5cm and 45 cm respectively and the mass of the smaller sphere is ½ kg. Find the mass of the larger sphere. Here rea ratio = 5 : 45 = : 9 Length ratio = 9 = Volume ratio = : = : 7 7 Mass of the larger sphere = 40. 5 kg 0 Thomas Whitham Sixth Form
. ngles work out the missing angles for each of the following diagrams (i) a 7 Solution 90 + 7 = 7 a = 80 7 a = 6 (ii) 8 b 64 87 Solution 8 + 64 + 87 = a = 60 a = 7 (iii) c e d Solution d = c = 80 = 48 a = 48 Thomas Whitham Sixth Form
(iv) 07 e g f Solution e = 07 (opposite angles) g = 07 (corresponding angles) f = 80 07 = 7 (straight line) (v) 9 i k m Solution i = 80 9 = 4 (straight line) k = 4 (lternate angles) m = 80 4 = 9 (straight line) ngles subtended from the same arc are equal. The angle subtended at the centre from an arc is double the angle at the circumference. b = a The largest angle in a semi-circle will always be 90 and is positioned on the circumference. Thomas Whitham Sixth Form
Definition: tangent is a line that will touch the circle at one point only. (i.e. it does not cut the circle) The angle between a tangent an its radius will always be 90 Definition: cyclic quadrilateral is any four-sided polygon whose four corners touch the circumference of the circle. Opposite angles in a cyclic quadrilateral add up to 80 a + c = 80 and b + d = 80 Definition: chord is any straight line which touches the circumference at two points. The largest chord possible is called the diameter. The angle between a tangent a chord is equal to the angle in the alternate segment. Work out the lettered angles for each of the following diagrams (i) This angle same as a. (same arc) lso added to b = 90 (semi circle) ngles in a triangle add up to 80 Solution this angle = 80 6 6 = 9 b = 6 (tangent & Chord alternate segment) a = 90 6 = 7 (angles from same arc & angles in semi circle) Thomas Whitham Sixth Form
c = 80 9 = 89 (straight line) (ii) ngles in a triangle add up to 80 this angle = 80 4 05 = 5 Solution = 05 (angles subtended from same arc) i = 5 (opposite angles) (iii) This angle = 60 0 = 0 Solution y = 65 (angle subtended from same arc and at centre) (iv) This angle = 80 5 = 45 (angles on a straight line) 4 Thomas Whitham Sixth Form
Solution a = 90 (angles in a semi circle) b = 80 90 45 = 45 (angles in a triangle) (v) Solution x = 80 4 = 66 (angles on a straight line) g = 80 66 = 4 (angles in a cyclic quadrilateral) = 80 = 59 (angles on a straight line) f = 59 (angles in a cyclic quadrilateral). Transformations ny object can undergo a transformation when it is subject to a translation, a reflection in a given line, rotated by a given angle about a given point or enlargement by a given scale factor about a given point. The transformed object is known as the image of the original object. i. Reflections If the point P is reflected in a given mirror line then its image is usually denoted by P The point P lies on the other side of the mirror line such that the mirror line becomes the perpendicular bisector of PP Here P is the image of P reflected at 90 to the mirror line. 5 Thomas Whitham Sixth Form
Given triangle C with (, 5), (4, 5) and C(, ), obtain the image of C under the following transformations a) Reflection in the y-axis b) Reflection in the line x = c) Reflection in the line y = x a) y 5 4 Reflect each point, & C individually across the mirror line to form the image C C Plot the points & C to form triangle C -5-4 - - - 0 4 5 x - - - Use the y axis as a mirror line so reflect across -4 b) -5 y 5 4 Reflect each point, & C individually across the mirror line to form the image C -5-4 - - - 0 4 5 x - C Plot the points & C to form triangle C - - -4 Draw in the mirror line x = -5 6 Thomas Whitham Sixth Form
c) y 5 4 C -5-4 - - - 0 4 5 x - Reflect each point, & C individually diagonally across the mirror line to form the image C - - -4-5 Draw in the mirror line y = x Describe the transformation that has taken place below. y 0 9 8 7 6 5 4-0 4 5 6 7 8 9 x - 7 Thomas Whitham Sixth Form
Here we can see a reflection in a horizontal line drawn at y = 4. ii. Rotations rotation is specified by the angle of rotation, its direction and its centre. Triangle C has points (, 4) (6, 4) and C(6, 7). Obtain the image of C after a rotation of a) 90 clockwise about the origin b) 80 about the point (, ) ny rotation is easily done using tracing paper. {make sure the object and centre are traced and with a pencil placed on the centre rotate the tracing paper the desired angle. a) y C Plot points, & C and join up 6 4-6 -4-0 4 6 x - Place tracing paper over the grid. Trace the triangle C. Place the pencil on the origin and rotate the tracing paper 90 clockwise. Finally mark the new points of, & C -4-6 C 8 Thomas Whitham Sixth Form
b) y 6 C Place on the centre of rotation 4-6 -4-0 4 6 x C - -4 Place tracing paper over the grid. Trace the triangle C. Place the pencil on the coordinate (, ) and rotate the tracing paper 80. Finally mark the new points of, & C -6 iii. Enlargements ny enlargement is specified by a given scale factor and centre of enlargement. Marks are given for projection lines shown (otherwise known as rays) 9 Thomas Whitham Sixth Form
Enlarge rectangle CD by a scale factor centre (, ) y 0 9 8 7 Draw in the ray lines and from the centre mark times the original distance up! Place on the centre of enlargement 6 5 4-0 - That is distance to ray line is units here. Multiplied by means distance now must be 6 units away from centre. 4 5 6 7 8 9 0 x 40 Thomas Whitham Sixth Form
In the diagram below state the scale factor and centre for the enlargement of onto. y 6 4-6 -4-0 4 6 x - -4-6 Here we can clearly see the enlargement is scale factor. s for the centre we must draw on the ray lines which should all meet at one point! Here you can see the centre as (5, 7) where all the ray lines meet. Hence Enlargement centre (5, 7) with scale factor. 4 Thomas Whitham Sixth Form
iv. Translations If every point on an Object is moved the same distance in the same direction, the transformation is called a translation. Each translation is described as a movement in the x direction followed by a movement in the y direction. Movements and are classed as positive movements in the x and y directions respectively. Whilst movements and are classed as negative movements in the x and y directions respectively. x vector notation for the translation is given as a column vector y Describe the transformation which moves object onto a) y b) C 5 4-5 -4 - - - 0 4 5 x - - - -4-5 C 5 a) onto represents a translation of 6 b) onto C represents a translation of 4 Thomas Whitham Sixth Form
Given triangle C with (-, ) (, ) and C(, 7), draw the image of C under the 4 translation 5 y C 6 Plot points and draw the triangle C 4 C -6-4 - 0 4 6 x Move each of the points 4 places back (-4) and 5 places down (-5) - -4-6 4 Thomas Whitham Sixth Form
. Vectors Definition: vector quantity is one which has both magnitude and direction One of the simplest vectors is displacement.. It has an associated distance (magnitude) and direction. For example if we have two points and then displacement is a vector quantity. It can be conveniently represented by a straight line joining and, and the direction indicated by an arrow. a) Find an expression for in the diagram below. b) Hence find the midpoint of the line For the vector we write either or or in text books a) The vector can be found as it is equivalent to the path to 0 followed by 0 to That is 0 0 Furthermore if 0 a then 0 a 0 0 a b Important result b a 44 Thomas Whitham Sixth Form
45 Thomas Whitham Sixth Form b) For the midpoint we vandalise the diagram by placing M (the midpoint) on and looking for an alternative route. a) a b b a b a a b a a b a a M OM 0 0M is the position of M from the origin
b) OQ b c) P a d) PO a e) PQ b a f) PN b a b a g) ON a b a a b h) N P PN a b a b CDEF is a regular hexagon with centre 0 where 0 a and 0 b a) Find expressions for EC CF b) The positions of points K and L are given by the vectors 0K b a 0L a b Find expressions for K KL L c) What name is given to the quadrilateral KL? a) b a 46 Thomas Whitham Sixth Form
EC CF ED DC b a b b a a b Here we are using the fact that in a regular hexagon is the same as ED, DC is the same as 0 and so on. b) To answer this next part first place K and L roughly onto your sketch! The improved sketch shows which quadrilateral we have but we must prove that it is a trapezium. K 0 0K b b a a KL K0 0L b a a b a b a b L 0 0L a a b b a b KL and KL are parallel. The same is not true for the other two sides in the quadrilateral. Hence KL is a trapezium. 47 Thomas Whitham Sixth Form