raw/sketch/interpret a iagram using earings SS6 To save paper this section has been moved to p79 after SS8. alculate the Surface reas of Prisms, & linders SS7 Surface area is the total area of all the outer surfaces of the object. If ou were painting it, it s all the sides ou would paint. Strateg. onsider the outer surface or draw the net if this helps. Look for faces of the same size and shape. ind the area of each different face. Total together the areas to find the surface area of the solid surface area of the shape can often be the area of the net net is the surfaces of a solid shape folded out flat The nets of the following shapes have been drawn, and the individual shapes that make up the nets listed: cuboid clinder triangular prism circumference of the circle = πr small rectangles large rectangles triangles rectangles rectangle circles What is the surface area of the clinder above? Solution Starting at stage of the strateg:. rea of circle = πr = π = π rea of rectangle = πr h = π = 0π. π + 0π = 8π units or 5 units to s.f. a) ind the surface area of the cuboid above. b) ind the surface area of a cuboid with dimensions cm b cm b cm. c) ind the surface area of a closed clinder 5cm long and with a radius of cm. Give our answer exactl in terms of π. RPI I TST lank out the page above before answering these!. ind the Surface rea of the two solid shapes: i) ii) Give our answer to ii) to s.f. and exactl in terms of π. 6 ZigZag ducation, 00
alculate the Volume of Prisms, & linders SS8 The volume of a solid object is the amount of space it occupies. ) The Volume of Prisms - lobs, uboids and linders prism is a solid () object which has a constant area of cross-section. This means it has the same shape and size from one end to the other. Strateg for cuboids and clinders. Identif the cross-sectional shape. Recall the appropriate area formula.. Recall/work out the appropriate volume formula. Volume of prism = rea of cross-section length Use the V = rea of cross-section length formula to remind ou of the formula for clinders and cuboids. This is Strateg. Watch clinder is a prism with a uniform circular crosssection length, l constant area of cross-section r Your formulae sheet gives ou this formula cross section Remember rea of a circle = πr length Therefore Volume of linder = rea of cross-section length = πr l = πr l cuboid is a prism with a uniform rectangular cross-section length, c constant area of cross-section a b Remember rea of a Rectangle = ab Therefore Volume of uboid = ab c = abc xam stle Question blob of ink marks an area of 0.cm Strateg for blobs and other prisms on some paper.. Identif the shape of the cross-section The blob penetrates the paper uniforml b 0.cm.. alculate the area of this cross-section. alculate the volume of the blob of ink.. Use Volume of prism = rea of cross-section length Solution (Strateg ). &. The cross section is a blob with an area of 0.cm.. V = rea of cross-section length = 0. 0. = 0.0cm. a) Use the above formula to work out the volume of a cuboid with dimensions cm b cm b cm. b) Use the above formula to work out the volume of a clinder with radius cm and height cm. Note: the word height in this question represents the length in the formula above. c) alculate the volume of the following triangular prism using Strateg. 6cm cm length, cm constant area of cross-section RPI I TST lank out the page above before answering these!. ind the volume of a cuboid with dimensions cm b cm b cm.. ind the volume of a clinder with radius 5cm and height 6cm.. ind the volume of an ink blob that marks an area of 0.cm on some paper and penetrates uniforml 0.cm. ZigZag ducation, 00
Know basic ircle Properties, Know & Use the ircle Theorems SS9 n arc is a small part of the circumference of the circle chord is the line joining two points on the circumference of a circle segment is the area bounded b an arc and a chord Segment asic ircle acts Tangents drawn from the same point are equal in length. The perpendicular from the centre to a chord bisects (divides equall) the chord Tangents and radii at the same point on the circumference are perpendicular. hord O The circumference is the distance around the circle The diameter is the distance across the circle through the centre of the circle, O, and is twice the length of the radius. a) Read each of the theorems below three times and carefull stud each diagram. b) Write down each theorem. c) Now draw a diagram to represent each theorem. i) The angle at the centre is twice the angle at the circumference This is sometimes written as the angle subtended (made) b an arc (or chord) at the centre of the circle is twice the angle subtended at an point on the circumference (in the same segment). x O x To help ou spot these look for two points on the circumference (in this case and ) which are both connected to the centre (O) and to a third point on the circumference (). ii) The angle from a diameter is a right angle (90 ) iii) ngles in the same segment are equal 90 x O To help ou spot these look for the x common chord, in this case. major segment minor segment iv) Opposite angles of a cclic quadrilateral sum to 80 In the diagram shown this means + = x + = 80 o d) State two other angles from the diagram that must sum to 80 o. x O cclic quadrilateral is a -sided shape with ever corner touching the circle. e) Giving a reason, state whether GH is a cclic quadrilateral. H G RPI I TST lank out the page above before answering these! In the diagram, is the diameter of the circle and O is the centre of the circle.. If is o, giving a reason, what is angle?. xplain wh is 90 o. Giving a reason, which angle is twice?. i) Giving a reason, which angle is equal to? ii) Is this angle acute (less than 90 ) or obtuse (>90, <80 )? Hint: Giving a reason in this question means stating which theorem ou are using. O ZigZag ducation, 00
raw asic onstructions SS Perpendicular isector, ngle isector and quilateral Triangle Learn the following three ke constructions using a pair of compasses, a straight edge and a pencil. a) op the ke constructions as shown. Perpendicular bisector of a line With compasses centred at, draw arcs above and below the line. With the same radius, and with compasses centred at, draw crossing arcs above and below the line. Join to. This is the perpendicular bisector of. ngle bisector of With the compasses centred on, draw arcs to cut the lines at and. With the same radius, draw arcs centred on and to cross at. Join the points and. This is the bisector of the angle quilateral Triangle Set the compasses to the same length as. raw arcs, centred on and to cross at. Join the points to and to. onstructing ngles of 90, 60, 5 or 0 Notice how the perpendicular construction can be used to construct an angle of 90. The equilateral triangle construction can be used to make an angle of 60. bisection of 90 /60 ou can construct angles of 5 /0. b) Use our right-angle ou constructed from a), and bisect this to make an angle of 5. If ou want ou could bisect the 5 to make an angle of.5. RPI I TST lank out the page above before answering these!. onstruct an angle of 0. ZigZag ducation, 00
onstruct asic Loci SS ircle, Race Track, ngles isector & Perpendicular isector locus is a point, line or shading that shows all the points that satisf a given rule. You need to know the general shape of the solution and how to draw the constructions. The most common loci rules are: The solution shape is a: onstruction Method:. The fixed distance from a single point ircle ompasses centred on the point. The equidistance from two points Perpendicular isector See onstructions. The equidistance from two lines ngle isector See onstructions. The fixed distance from a line Race Track Shape This construction is shown below s. goat is attached to a rope of length 5m, which is attached to a post. raw the maximum area that the goat can move.. ship must sail a route equidistant from two buos and. raw its route. 5m. raw the area of the shape cm awa from the line cm ships route raw a rough ¾ circle centre and. With the compasses near the centre of the line, draw an arc above and below the line. raw a line touching the top of the two ¾ circles and the top arc. raw a line touching the bottom of the two ¾ circles and the bottom arc.. shot-putter throws a shot to a point equidistant between the two lines and. raw the line showing all the points where the shot ma land. a) raw a line 5cm long. raw all the points that are 5cm from. b) raw a line 8cm long. Shade the locus of points P, where P is less than 5cm from and less than cm from. RPI I TST lank out the page above before answering these! raw all the points that are 5cm from the triangle, where is an equilateral triangle of side 5cm. 5cm 5cm Start b constructing (see constructions) or tracing, or roughl drawing the equilateral triangle. 5cm ZigZag ducation, 00
escribe Transformations SS The ke transformations that ou are asked to describe are Reflections, Rotations and Translations. You will also need to describe nlargements but these are easier to spot because the shape gets bigger, what a surprise, or smaller see nlargements on the next page. Reflection is described b a line of reflection (or mirror line). The shape is reflected in the line = x. Rotation is described b properties:. ngle. irection. entre. The shape is rotated 90, clockwise, about (, 0). Translation means move and is described b a vector. The shape is translated means units left, units down written like this the bottom indicates the change in the direction The top number indicates the change in the x direction Meaning the shape has moved units right, units up. inding a Mirror Line - If ou can find a mirror line so that each corner of the object triangle is the same distance Hint as the corresponding corner from the Mirror lines are usuall = x, = x, image triangle, then ou have a line the -axis, the x-axis, vertical lines of reflection. (eg. x = ) or horizontal lines (eg. =.) a) escribe the reflection of to. inding a Translation Vector - Translations are eas to spot as their object and image have the same orientation and size! to is a translation of 5. b) escribe the translation of i) to ii) to. = x 0 x +5 0 x inding the entre of Rotation - Join two corresponding corners and construct, or roughl draw, the perpendicular bisectors see constructions. The centre of rotation is where these two x perpendicular bisectors cross. Use the rd corner and corresponding corner to check its perpendicular bisector should go through the SM point. In this case through (, 0). c) escribe the rotation that takes i) to ii) to. d) op out the grid and draw the two triangles and. Now without looking at the diagram repeat the construction to find the centre of rotation. The angle of rotation can be found b joining a corresponding corner on the object and image to the centre of rotation. In this case the angle of rotation is 90 anticlockwise. raw Rotations/Reflections b using the facts above or follow the following tracing paper method Reflections - op the shape and mirror line onto tracing paper. Turn the paper over so the image appears on the opposite side of the mirror line and align the mirror lines. Mark the corners of the image and join these together. Rotations - op the shape and mark the centre of rotation. Place a compass or pencil point on the centre of rotation. Rotate the tracing paper the required amount. Mark the corners of the image and join these together. x RPI I TST lank out the page above before answering these!. escribe the single transformation that takes: a) to b) to c) to - - - - - - x - ZigZag ducation, 00
escribe nlargements SS5 nlargement is described b a centre and a scale factor of enlargement. or example, the image in fig is an enlargement of object from (, ) b a scale factor of. inding the entre of nlargement Join corresponding corners together and extend them until the cross; this will be the centre of enlargement. In fig the corresponding corners meet at (, ). eciding the Scale actor of nlargement Measure the length from the centre of enlargement to a corner and a corresponding corner. ivide image length b object length, image length to centre of enlargement. lternativel simpl divide two corresponding object length to centre of enlargement lengths of image and object,. This is the scale factor of object length enlargement. In fig the height of the object is and the corresponding image length image height is, = =. object length image length - centre (, ) scale factor - - - - x - - - He Look The enlargement backwards, so from to is an enlargement about (-,-) b a scale factor of ⅓. rawing nlargements raw a line from the centre of enlargement to a corner of the object. Multipl this length b the scale factor and extend our original line to this length. Repeat for the other corners and join these together to make the enlarged shape. In fig one of the corners of the object is left down from the origin and so the corresponding corner of the image will be left and up from the origin. Your turn!! a) In fig another one of the corners of the object is left and down. escribe the position of the corresponding image corner. b) escribe the enlargement from to. c) escribe the enlargement from to. - - - - - - - - x RPI I TST lank out the page above before answering these!. escribe the enlargement from a) to b) to - - - - x ZigZag ducation, 00
alculate Lengths in Similar Triangles SS6 If one triangle is an enlargement of another, then the two are mathematicall similar. Similar triangles have all corresponding angles equal to each other the size of the triangle is different! The sides of the two triangles are in the same ratio. Similar means same shape, different size is parallel to. alculate the length Strateg. stablish that triangles are similar if necessar.. raw the two similar triangles with the same orientation, and label the separate diagrams. 5 cm. ind the ratio of two corresponding sides and so calculate the scale factor of enlargement.. ppl this factor to the side required. 5. heck this is sensible in relation to the original diagram. cm 6 cm Solution. = common to both = -angle in parallel lines = -angle in parallel line. So is similar to... Side corresponds to side. Ratio of corresponding sides to = : 6 Scale factor of enlargement is 6 =.5.. The side which needs to be found is, with a corresponding side on the smaller triangle. =.5 = 5.5 = 7.5 cm. 5. hecking: 7.5cm > 5cm, therefore answer is sensible. I H J a) GH is similar to HIJ because; HG = IHJ (opposite angles), GH = IJH (Z-angle), GH = HIJ (Z-angle). Side G corresponds to IJ. Which side corresponds to side GH? G cm 5cm b) Given that is parallel to, calculate the length of sides and. Hint: Use the ratio of : 6 cm 9 cm RPI I TST lank out the page above before answering these!. These triangles are similar. alculate the length XY. and XZY are equal. o not calculate x. cm 5.6 cm 5 cm x o Z cm x o 6 cm X Y ZigZag ducation, 00
ecide if a ormula is a Perimeter, rea or Volume b considering imensions - SS8 onstants are dimensionless and can be ignored e.g. and π are dimensionless constants. In the following section a, b, c, l, h, w are all measures of length. The formula for lengths involve lengths or added lengths..g. πl or π ( l + h ) or l + h + w The formula for areas involve the product of two lengths..g. π l or πl( l + h ) or l + h + w The formula for volumes involve the product of three lengths..g. πl or πl ( l + h ) or l + h + w Strateg. Write without constants like, and π.. Replace lengths b cm, (areas b cm ).. cm length. cm area. cm volume. (a + πb)c. (a + b)c. (cm + cm) cm = (cm) cm cm + cm are more cm s! = cm. Volume formula. a + b a + b. cm + cm = cm = cm cm + cm are more cm s!. Length formula What tpes of formula are: a) πbc b) π (bc +ac) c) (πc + ½ab) d) xtra note θ is a measure of length (angular length), but sinθ is a constant and can be ignored. xtra What tpes of formulae are e) aθ f) a sinθ πc RPI I TST lank out the page above before answering these! Identif these formulae as perimeter (length), area or volume:. ( ac abc) πc ab. h h ab. abc rom p7 raw/sketch/interpret a iagram using earings SS6 Pthagoras and Trigonometr questions are sometimes asked through a bearings stle question. In such questions ou might need to draw or interpret a bearings diagram. In a bearing of from ; a Reminders person would be standing at!. earings are alwas measured clockwise from North.. earings are alwas given as three figures, so 005 not 5.. Look carefull at the words to and from in questions.. ngles can be calculated using the fact that the Norths are parallel. or example N is 80 5 = 5 see iss for a reminder of Z-angles etc. N 90 5 N raw a diagram where a ship travels from to on a bearing of 05 a distance of 5km, then to from on a bearing of 090 a distance of 5km. It then travels back to from. Solution See iagram N 5 5 In the, = 60 90 5 = 5 (ngles at a point sum to 60 ). a) alculate the other angles in the. b) State the bearings of i) from ii) from iii) from. RPI I TST lank out the page above before answering these!. ship travels to from on a bearing of 70 a distance of 5km and then from to a distance of 5km on a bearing of 060. a) Sketch a diagram to represent this journe. b) alculate the bearing of i) from ii) from. ZigZag ducation, 00
alculate ompound Measures like Speed & ensit SS9 Memorise the formulae: istance verage Speed = Time hange in Speed ccelerati on = Time ensit = Mass Volume alculate the average speed of a car that travels km in hours. Solution istance Sometimes ou need to rearrange the Speed = = km / hour formula, making one of the other Time parts the subject of the formula. a) giant potato has a mass of 0. kg and has a volume of 90 cm. alculate the densit of the potato. Note: Make sure ou can read the calculator displa, see N7 parts e) & f). car travels at 0km/hour for hours. How far does the car travel? Solution istance Speed =. Time So, Speed Time = istance 0 = km b) small orange has a mass of 0.05 kg and a densit of 0.00 kg/cm. alculate the volume of the orange. Hint: Start b making Volume the subject of the equation ensit = Mass. Volume RPI I TST lank out the page above before answering these!. solid stone has a mass of 6 kg and has a volume of 00 cm. alculate the densit of the stone. istance. Make Time the subject of the equation Speed = and hence calculate how long it takes an object Time travelling at 0m/s to cover metres. onvert between Volume Measures including cm and m SS0 onverting asic Units onvert cm to m. Solution 0 = 0.0m onversion actor Reminder cm mm m cm 0 km m 00 onverting Volume Measures onvert cm to m. The cm m conversion factor Volume Measures so cube i.e. the cm m conversion factor Solution (0) = 0.00000 or as appropriate This can be written 6 in Standard orm heck that our answer is reasonable at the end; this will ensure ou have used the correct sign. It is not uncommon for students to use the incorrect smbol. onvert a) 0,000cm to m b) cm to mm. RPI I TST lank out the page above before answering these! onvert. 0.005m to cm. mm to cm. ZigZag ducation, 00