1Unit 3 Revision Notes - Higher Angles: Parallel LinesabCorresponding angles‘a’ and ‘b’ are equalParallel lines are shown by having arrows drawn on themcdAlternate angles‘c’ and ‘d’ are equalNotes: 1) Use the correct words: Alternate and Corresponding in the exam2) Useful for Bearings questions. The Bearing of A from B is 120 degrees what is theBearing of B from A?NBearing of B from A = 300 degreesN120120NBBAA180
2why each angle is the same! 3) Used when showing that two triangles are similar/congruent – remember to explainwhy each angle is the same!ABCDEShow that ABC is similar to CEDAngle A = Angle E (alternate)Angle B = Angle D (alternate)Angles at C are the same (vertically opposite)Angles: Circle ThoeremsChord my not always be visible!The angle subtended by a chord at the circumference is half that at the centre.The angles subtended (made) by a common chord at the circumference are equal.The angle made by a diameter at the circumference is 90 degrees.
3Opposite angles in a cyclic quadrilateral add to 180. The angle between a tangent and chord is equal to the angle subtended (made) by a chord in the alternate segment.Note when doing circle theorems questions watch out for isosceles triangles – in particular triangles with two sides as a radius of the circle.Also remember that a tangent to a circle will always make a 90 degree angle with the radius at the point where they meet.
4THESE FORMULA ONLY WORK FOR RIGHT ANGLED TRIGONOMETRY!! Eg.What are the missing angles? Give reasons.A = 120 (angle at centre = twice that at circumference)B = 60 (angles at circumference are constant)C = 120 (opposite angles in a cyclic quad add to 180)Right Angled TrigonometryTHESE FORMULA ONLY WORK FOR RIGHT ANGLED TRIGONOMETRY!!IF YOU SEE A RIGHT ANGLED TRIANGLE (INDEED A RIGHT ANGLE!) ANYWHERE IN THE EXAM THINK PYTHAGORAS AND (SOHCAHTOA)Eg1 Find the angle XWe know 8 = adjacent12 = hypotenuseCos X = 8/12X = Cos-1 (8/12)X = 48.2 degrees (1 d.p.)12 cmX8 cmRemember to use shift button on calculator to get the angle using cos inverse)
5Both of these are Useful for finding an angle Eg2 Find the side xWe know 12 = opposite70 = angleTan 70 = 12/xx = 12/tan(70)x = 4.4 cm12 cm70oThis is what I calla ‘hard side’ question. You need to get the rearrangement correct from the first to the second line.xEg3 Find the side xWe know 6 = opposite70 = angleSin 70 = x/66 × Sin 70x = 5.6 cm6 cmx cm70oThis is what I call an ‘easy side’ question. You need to get the rearrangement correct from the first to the second line – but it is easier to do than in eg 2 above.Trigonometry for general TrianglesThese rules work for any triangle. If you see a non right angled triangle you have to use them.Useful rearrangements but you need to learn them if you want to use themOn the formula sheet in the exam:Both of these are Useful for finding an angle
6Use the Cosine rule when: You want to find a third side and you have 2 sides and an angle in betweenYou want to find an angle and you have 3 sides50Oa1012A81012Use the Sine rule when:The Cosine rule doesn’t work!This will be when you have a side and an opposite angle as a pair.50O1012B
7Eg 3. A yacht sets off from A and sails 5km on a bearing of 067 degrees to a point B so that it can clear the headland before it turns onto a bearing of 146 degrees. It then stays on that course for 8km until it reaches point C.Find the distance AC.The trick here is to find the angle at B.Clearly this is a cosine rule question. Use a diagram and corresponding angles to do this.NNN2 sides but need angle in between so redraw and use angle facts with parallel lines to find it!146 – 67 = 796714667B67B5AA8CC101b2 = c2 + a2 - 2×c×a×Cos Bb2 = ×5×8×Cos 101b2 = ×Cos 101b2 =b2 = So b = 10.2 kmBe careful with the double negative!!!!! (Cos of a value above 90 is always negative)
8Internal and External Angles of Regular Polygons Internal angles are those on the inside of the polygonFor a triangle the internal angles sum to 180oFor a quadrilateral they sum to 360oFor a pentagon 540o.(Each extra side adds on 180o)Regular shapes are shapes where the internal angles are all the same and where the side lengths are all the same. This is a Regular Pentagon.External angles always add up to 360 degrees and are the amount you turn at each corner.The external angle of a regular polygon with ‘n’ sides is 360 ÷ nThe internal angle of a regular polygon= 180 – external angleAnswer: External angle in a regular octagon = 360 ÷ 8 = 45oInternal angle = 1800 – 45 = 135oAt Angle here= 360 – 135 – 135 = 90o
9Graph Transformations Vertical stretch: Timeswhole expressiony = 2f(x)Move up: Add on the endy = f(x) + 1Original Graphy = f(x)This is drawn as dotted line in each exampleMove sideways: Add to x (but works ‘wrong’ way)y = f(x + 2)Stretch sideways:– Times the x (but works ‘wrong’ way)y = f(3x)
10Look at the graph of y = f(x) opposite. On the same axes plot y = f(x)-3sTranslate down 3b) y= 2f(x)Stretch by factor 2vertically – all thepoints are twice asFar away from theX-axis as theyWere to begin withy = f(x)-3
11½ horizontally – all the points are half as far away from the Look at the graph ofy = f(x) opposite.On the same axes plotthe graph of:y = f(2x)Stretch by factor½ horizontally – all thepoints are half asfar away from they-axis as theywere to begin withb) y = f(x - 4)Translate +4 horizontallyy = f(2x)y = f(x - 4)Note: The graphs y = f(x) and y = -f(x) are related by reflecting the shape in the x axisSimilarly y = -2f(x) is a reflection and a stretch of scale factor 2 vertically
12Using Graphs to solve equations An important principle here is that if you want to solve any equation:Eg 2x + 1 = 3x – 5Then you can do this by plotting the graphs of what is on each side and then seeing where they meet. In this case plot the graphs y = 2x + 1 (y = what is on the left)and y = 3x – 5 (y = what is on the right)Where the graphs meet the x values will give the same y vale ie the equation will be sovled.There may be several values for x.Eg.Use the graph of y = cos Ө shown below to find all the solutions to cos Ө = in the range 0o ≤ Ө ≤ 720oDraw the line y = -0.7Find one solution using cos inverse:cos Ө = - 0.7Then Ө = cos-1 (- 0.7)= 134.4Then use the fact that the graph is symmetrical about 180 to find 3 other solutions:or or 585.6134225.6134225.6and(Remember the graph repeats every 360 degrees, so adding or taking 360 degrees to any solution will get another)
13This is the graph of y = x2 – 5x Use it to solve x2 – 5x = 0 Easy just see where y = x2 – 5x meets the graph y = 0 (ie the x axis)Clearly 0 and 5b) Again just see where y = x2 – 5x meets the graph y = -4Clearly 1and 4 by drawing the line y = -4 on the graphThis is the hard one!Start with the equation you want to solve x2 – 6x + 5 = 0Add and subtract things to both sides to rearrange the equation to make one side look like the equation of the graph you have been given (in this case x2 – 5x).x2 – 6x + 5 = Add x and take 5 from both sides give us:x2 – 5x = x - 5Now plot the graph of y = x – 5 and see where the lines meet. In this case x = 1 & 5.
14Graphs you need to know: General quadratic y = ax2 +bx + c (usually have two solutions whichare the x values where the graphcrosses the x axis)General cubic y = ax3 +bx2 + cx + d (usually have three solutions whichInverse Proportion graph y = k/xStraight line graph y = mx + c straight line graphThe direct proportion graph y = kx looks like this but alwaysGo through the originy = sin(x) y = cos(x) and y = tan(x) you should be familiar with but you don’t need to be able to sketch them for GCSE maths
15Exponential graphs like y = 2x, y = 3x , y = 4x etc..These all look like the shape on the right, which is theGraph of y = 2xwhen x = 1 y = 21 = 2when x = 2 y = 22 = 4when x = 3 y = 23 = 8 etcTransformationsTransformations are ways to change a shape. There are 4 main transformations:Enlargements (shape’s size is changed so that is still in proportion)Translation (shape’s position is changed)Rotation (shape’s orientation is changed)Reflection (shape’s orientation is changed)You can get other transformations by combining these together. You can also define new transformations (like a stretch which enlarges a shape in 1 direction only – these are not on the syllabus apart from as part of graph transformations)
16A reflection is obtained by ‘folding’ the space over a mirror line. ReflectionsA reflection is obtained by ‘folding’ the space over a mirror line.The shortest distance of any point on the original object to the mirrorline should be equal to the shortest distance of its image to the mirror line.To describe a reflection you must have:An ObjectA line of reflection described as an equation(Remember vertical lines are x = lines, horizontal lines are y = lines, the line y = x goes through (0,0), (1,1), (2,2) etcWhen reflecting in a diagonal line it is easier to turn the line of reflection so that it is horizontal or vertical)OBJECTIMAGE
17RotationsTo describe a rotation you must have:An ObjectAn Centre of Rotation (an axis) described as a coordinateAn Angle of RotationA Direction to rotate in (usually clockwise/anticlockwise)(Note: Tracing the object onto scrap paper and rotating the paper using a pencil nib as an axis is a good way to avoidmaking mistakes. Ask for tracing paper if you need it!)A rotation is obtained by turning a space around a fixed point (we call this the centre of rotation). The distance of every point from the centre of rotation will remain constant as the space is turned.
18To describe a translation you must have: An Object A Vector TranslationsA translation is obtained by moving a shape in space. Each corner of the object is moved by an amount described in a vector.The image should be congruent to the object and of the same orientation.To describe a translation you must have:An ObjectA VectorThe top line of the vector describes the horizontal movement (+ve right, -ve left)The bottom line describes the vertical movement (+ve up, -ve down)In this case the Object has been translated by 4 to the right and 2 upAs described in the vector shown.IMAGEOBJECT
193) Multiply this distance by the scale factor Method:Draw lines from the centre of enlargement through all the corners of the object2) Find the distance of the first corner from the centre of enlargement (counting squares is best)3) Multiply this distance by the scale factor4) Measure this new distance from the centre of enlargement and mark it as the first corner in the enlarged imageRepeat for every corner until the enlarged shape is ready to be drawn in.NOTE:If the Scale Factor is –ve, measure in the opposite direction when plotting the image. The image will be upside down and on the other side of the centre (essentially it has been rotated by 180 degrees at the same time as being enlarged).If the Scale Factor is between -1 and 1 the shape will actually get smaller!Eg Enlargement Scale Factor 3 with (2,1) as the centre of enlargement
20Enlarge the shape shown by scale factor -2 using the point B as the centre of enlargement OBJECTBIMAGE
21Enlarge the shape shown by scale factor ½ using the point C as the centre of enlargement OBJECTIMAGEC
22Area and perimeter of 2d shapes Area of a trapezium = ½ (a + b) × hRemember: This is on the formula page!Sector = a pizza slice from a circleArc = part of the circumference of a circleSegment = pieces sliced across a circleArea of a circle = ½ π × radius2Area of a sector = ½ π × radius2 ×ah – perpendicular distance between the two parallel sidesbCircles: Radius = distance from the middle to a circle to the circumferenceDiameter = width through the middle of a circleDiameter = 2 × radiusradiusθ
23Circumference of circle = π × diameter Arc length = π × diameter ×diameterθEg 1 Find the Perimeter of thesector below which is madefrom a circle of perimeter 5 cmEg 2 Find the Area of the segment shown ifthe radius of the circle is 8 cm100o300oArea of segment = Area of sector– Area of triangleArc of sector = π × 82 ×=Area of Triangle = ½ × 8 × 8 × Sin 100=Area of Segment = – = 24.3 cm2Perimeter consists of 2 radii (each 5 cm) and an arcArc length = π × diameter ×=Total perimeter = = cm
24Triangles:Area of a triangle = ½ base × height OR Area of a triangle= ½ a × b × Sin CRemember to use the vertical height(ie from the base to the tip of the triangle when using this formula)Use this formula when you have two sides of the triangle and the angle in between – it is on the formula page!Eg Find the areaEg 13cm50O10 cm12 cm10 cmArea = ½ 10 × 3 = 15 cm2Area = ½ 10 × 12 × Sin 50= 91.9 cm2Eg 24cm8 cmArea = ½ 8 × 3 = 16 cm2
25EgAnswer:The key to this question is that the areas are equal so start by writing down an expression for the area of each shape:(3x +2)(x +1) = Area of rectangle (3x + 5)x + 5x = Area of L shapeAreas are equal: (3x +2)(x +1) = (3x + 5)x + 5x3x2 + 5x + 2 = 3x2 + 5x + 5x5x + 2 = 10x x = 2/5
26Volume and Surface Area of 3d shapes A prism is a 3d shape where the cross section is uniform (unchanged)Examples of prisms – cuboids, cylindersExamples of non-prisms – pyramids, cones, spheresTo find the volume of any prism calculate the area of the cross-section and multiply it by the length of the prism:Eg 1 Find the Volume of the prism belowArea of cross-section (blue)= 6 x x 4 = 64Volume = 64 x 10 = 640 cm3.Eg 2 The prism below has a volume of 200 cm3.If it is 12 cm long what is the area of the cross – section?Volume = Area of cross-section x length= Area x so Area =200 ÷ 12 = cm28 cm4 cm6 cm10cm12 cm
27These are in the formula booklet Volume of a sphere = 4/3 π × radius3Surface area of a sphere = 4 π × radius2Volume of a pyramid = 1/3 Area of Base x heightVolume of a cone =1/3 π × radius2 x height (because it is a pyramid with a circular base, so the area of the base is π × r2 )Surface area of a cone = Area of base Area of curved surfaceπ × radius π × radius x slope lengthThese are in the formula bookletSlope lengthNote: The formula for the curved surface of a cone is the only formula that uses the actual slope length rather than the height.The curved surface part of the formula (π × radius x slope length)is on the formula page.radiusSurface area of a cylinder = Area of top and bottom + Area of curved surface2 x π × radius π × diameter x height(Two circles) (Rectangle that has equal to thecircumference of the top/bottom)
28AnswerRadius of sphere = 3so height of cylinder = 9cmCylinder = π x 32 x 9= 81πHemisphere = 4/3 π x 33 ÷ 2= 18 πVolume = 99 πVolume of liquid is 49.5 π18π in the hemisphereSo 31.5π = Volume of cylinder31.5π = π x 32 x dd= 3.5 cm
29Join these edges together Eg A cone is made by cutting a 60 degree sector from a circle of radius 8 cm as show in in the picture.This is a classic A* grade question.Method:Find the length of the arc from A to B.This length is the same as the circumference of the base of the cone. Use this fact to find the radius of the base of the cone.Arc length = 16 × π × =Then π × diameter =diameter =radius of cone base =Use the fact that the slant height of the cone OA will be equal to the radius of the original circle. Then use Pythagoras with this length and the radius of the base of the cone to find the cone’s height.300Join these edges togetherAOBOWhat is the volume of the cone?8A & BhVolume =1/3 × π × ×= cm32
30Eg The diagram shows a solid made from a cone and a hemi-sphere. The radius of both shapes is rThe slant height of the cone is l.The perpendicular height of the cone is hThe curved surface area of the cone and the curved surface area ofthe hemisphere are equal(a) Show that l = 2r (2)(b) Find the perpendicular height, h, of the cone in terms of r. (2)(c) Find the ratio of the volumes of the cone and the hemisphere. (2)Be prepared to use algebra to solve problems involving volume and surface areaIf the curved surfaces of the two shapes are equal then we can write an equation relatingthem. Curved surface of hemisphere = 4 π × radius2 ÷ 2Curved surface of cone = π × radius× lHence 2 π × radius2 = π × radius× l divide both side s by π and by the radius2 × radius = l - result we need(b) Using pythagoras r2 + h2 = l2 soVolume of cone = π × radius2 × h Volume of hemisphere = 4/3 π × radius3 ÷ 2= π × radius2 × = 2/3 π × radius3= π × radius2 ×= π × radius3 ×Ratio of Cone to hemisphere= (π × radius3 × ) ÷ (2/3 π × radius3)=
31Enlargement Scale Factors If two objects are similar it means that their side lengths are related by a multiplying factor called the enlargement scale factor.To find the enlargement scale factor divide a length in the image by a corresponding length in the object.Note: If the enlargement scale factor for length is xthen the enlargement scale factor for corresponding areas is x2and the enlargement scale factor for corresponding volumes is x3If triangles ABC and CED are similar, find:The length BCThe area of triangle CED given that the area of triangle ABC is 8 cm2ABCDE4 cm6 cm12 cmAnswer:Enlargement scale factor = 6÷4 = 1.5So BC = 12 ÷ 1.5 = 8 cmArea scale factor from ABC to CED = 1.52So area of CED = 8 x = 18cm2
32AnswerThis is a volume scale factor question.Enlargement scale factor forlength = 30÷20 = 1.5So volume scale factor = (1.5)3Small bottle holds 480 ml so bigger bottle holds 480 x (1.5)3 = 1620 mlEg Two solids are similar.The area of one of the faces is 20 cm2 on the larger shape and 5 cm2 on the smaller shape. Ifthe smaller shape has a volume of 8 cm3 what is the volume of the larger shape?Area scale factor = 20 ÷ 5 = 4So length scale factor will be the square root of the area scale factor, so = 2Volume scale factor is the length scale factor cubed, so is 23 = 8 in this case.Hence volume of larger solid is 8 x 8 = 64 cm3
33Answer:You can use enlargement scale factors to make this question a lot quicker.The volume of the cone missing from the top is = 1/3 × π × 22 × 3 = 4πThe larger cone is similar to the smaller one. The length scale factor is 3 by comparing the radii. This makes h = 9 cm.So volume of the whole cone including the missing top which shouldn’t really be there is 1/3 × π × 62 × 3 = 36πThe volume of the frustum = 36π - 4π= 32 π5 marks
34f a If you have a situation like this The vector between the Vector Geometry - Important principles:If you have a situation like thisThe vector between theend points of the vectorsf and a:FA = a - f2) The vector between BA is the negative of the vector from ABBA =- ABThey often ask you to find easier paths at the start of the question and you need to use these paths to help you with the harder paths at the end of the question.4) If you asked to prove that two vectors are parallel you need to show that one of them is a multiple of the otherFAfaO
35AnswerA) OS = OP + ½ of PQWe have OP = pWe need PQ in terms of p and q.Looking carefully PQ = -p + q or q – pSo OS = p + ½(q – p) = p + ½q - ½p = ½q + ½pB) RS = RP + PSRP = ½p and PS = ½PQ = ½(q – p) from part aSo RS = ½p + ½(q – p) = ½p + ½q – ½p = ½qSince OQ = q clearly RS is a multiple of OQ ( ½OQ to be exact), hence parallel
36Solving quadratic equations You need to be able to solve quadratic equations by either factorising, using a graph or by using the quadratic formula. The formula is on the formula page.Expect to see quadratics popping up in strange places and don’t be concerned if you see one as part of a question on area or volume – just rearrange to make the equation equal to zero and solve.If you can’t get the quadratic equation to work then check you r arithmetic – especially the negative numbers
37Solving simultaneous equations If, as part of a question you write down an equation with two variables in it you will not be able to solve the equation uniquely unless you find another equation. You need two equations to solve uniquely for a system containing two variables.IF BOTH EQUATIONS ARE LINEARRearrange the equations and then add and subtract multiples of them together to eliminate one of the variables. You can then solve.Eg Solve 3x = y2x + 3y = 11Rearrange to get letters on the same side in both equations3x – 2y =2x + 3y =Multiply equation 1 by 2 and equation 2 by 3 to get the number of xs in both equations to match:6x – 4y =6x + 9y =Since the signs are the same on the x’s subtract equation 2 from equation 1-13y = now solve to get y = 1 and from that we can see that x = 4Note:Be careful when subtracting/adding the equations.
38IF ONE QUATIONS IS LINEAR BUT THE OTHER IS QUADRATIC Rearrange one equation – the linear one! To get an explicit equation for one variable. Then replace this variable in the second equation using the equation.Eg Solve y2 + 16= 8x3 + y = 2xAnswerRearrange the second equation to get y = 2x – 3 now replace y in the ‘harder’ equation with2x – 3 - be careful to use brackets if they are necessary.(2x – 3) = 8x(2x – 3)(2x – 3) + 16 = 8x4x2 – 12x = 8xIts a quadratic equation so rearrange to make it equal to zero and then solve4x2 – 20x + 25 = 0(2x – 5)(2x - 5) = 0 x = 2.5 or when x = 2.5 , y = 2Note:This question could be phrased by asking you where the curves y2 + 16= 8x and 3 + y = 2x meet. Note also that you expect two values for x and 2 corresponding y values, but the quadratic in this case only gives one solution (because it is repeated)
39Answera) Parallel lines so2x + y = 180 because:This will be y as these are alternate anglesb) 6x + 2y – 100 = 360Since angles in a quadSum to 360 degreesAdding 1006x + 2y = 460Dividing by 23x + y = 230c) Solve simultaneously2x + y = 180SubtractingX = 50 so y = 80(c) Find the values of ‘x’ and ‘y’
40Solving Equations by Trial and Improvement Exam Question:Eg Use Trial and Improvement to Solve x3 + x2 = 200Find your answer to 2 decimal places.You must find the number to 2 decimal place that is nearest to 200.Working:Guess Answer7392625251505.75.65.45.55.555.545.53Show your results clearly and show how you are using intelligence to home in on the solution rather than just random guesssingMethod:Guess a number that might workTry it out – if the answer is too low make a second guess that is higher- if the answer is too high make a second guess that is lower.Keep guessing, each time going half way between your best two resultsNote: Your final answer is the guess with the correct number ofdecimal places that is the closest fit. You must demonstratefor sure that your answer is the best.
41Method:Guess a number that might workTry it out – if the answer is too low make a second guess that is higher- if the answer is too high make a second guess that is lower.Keep guessing, each time going half way between your best two resultsNote: Your final answer is the guess with the correct number ofdecimal places that is the closest fit. You must demonstratefor sure that your answer is the best.Solve (to 2 decimal places)x2 + 2x = ) x2 + x = 80 3) x2 – 3x = 100x3 – x = 75 5) x3 + x2 = 1001) x2 – x = 60 2) x2 – 3x = 100
42straight line going through the origin) ProportionTwo things are in direct proportion if the value of the second can be found from the value of the first by a multiplication.In other words:(This means that the graph of y plotted against x would be astraight line going through the origin)(A quick check is to see if doubling one quantity would lead to an expected doubling in the other quantity.Eg If you travel for twice as long will the distance you travel double?)The other types of direct proportion to watch out for are:A is directly proportional to BA is directly proportional to B 2A is directly proportional tothe square root of B
43m per second when it has been falling for 2 seconds. Eg The speed at which ball falls to the ground when dropped from a tall tower is directlyproportional to the square of the time it has been falling for. The ball reaches a speed of 12m per second when it has been falling for 2 seconds.a) How long will it take for the ball to reach a speed of 100 m per second?b) When the ball has been falling for 20seconds how fast will it be travelling?Answerv = speed t = timev is directly proportional to the time squared so v = k × t2When t = 2 v = so = k × 2212 = k × 4k = 3v = 3 × t2a) Now use the formula to answer the questions by solving the relevant equationsv = what is t? = 3 × t2 so t =b) When t = 20s what is v v = 3 × v = 1200 m per secondMETHODWrite down the proportionality equation with the constant kUse the info in the question to find a value for kWrite down the formula with k now known
44If y and x are inversely proportional then we say: But think of it like this:y × x = kA good way to think about inverse proportion is that as one thing doubles the other halves.Eg If I run twice as fast will I get there in half the time? If so then it is likely to be inverse proportionalNote that:y is inversely proportionalto the square of xBut think of it as y × x2 = kto the square root of xBut think of it as y × = k
45So p = number of people h = number of hours Eg If it take six people 8 hours to complete a job how many hours will it take 4 people?AnswerOften proportionality questions mention the words inverse or direct proportion in the question, but with this example you just need to recognise that this is likely to be an inverse proportion situation. Twice as many people mean the job completing in half the time would seem reasonable!So p = number of people h = number of hoursp is inversely proportional to h so p × h = kUse info from the question to find the value of the constant ‘k’6 × 8 = k so k =48Formula is p × h = k = 48So when p = 4 we need to solve for h × h = 48 implies that h = 12 hoursMETHODWrite down the proportionality equation with the constant kUse the info in the question to find a value for kWrite down the formula with k now known
46PythagorasIf you see a right angled triangle think Trigonometry and Pythagoras!If you need to find side lengths and can’t see a way forwards sometime drawing in a perpendicular to make a right angled triangle is helpful as you can then use pythagoras.Eg Find the area of this isosceles triangleDropping a perpendicular from the top of the triangle makes finding the height obvious using pythagoras132 = h2 + 52169 = h2 + 52169 = h2 + 25144 = h212 = h can ignore -12!13 cm13 cm13 cm13 cmh10 cm10 cmHence area =12 x 5 = 60 cm2
47Eg Find the value of x. Answer If a right angled triangle then (x + 5)2 + (x – 2)2 = 102 (pythagoras theorem)x2 + 10x x2 - 4x +4 = 1002x2 + 6x + 29 = 100This is a quadratic so 2x2 + 6x – 71 = 0make it equal to zeroand solve The positive solution is 4.64 from the formulaExpand out each bracket properly: Remember (x + 5)2(x + 5)(x + 5) = x2 + 10x + 25Note: You can also use pythagoras to prove that a triangle is right angled.It must be if the three sides fit the rule a2 + b2 = c2