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–74–46y32153i.e. x 8 and y 10xyThen, 3(–74)–5(–46)(–46)(3)–(–74)(2)1 2(5)–3(3)y1x –138 14810–9–222 230xy11yx1 and 81011108153xSolution: By cross-multiplication methodSolve: 2x 3y–46 0 –(i)3x 5y–74 0 –(ii)Solution: From eq. (i)x 2–eq.(ii)x1, wehave(4x–4x) (6y–6y) 16–70 9, which is a false statementThe pair of equation has no solutionSolve: 2x 3y 8 – (i)4x 6y 7 – (ii)Solution: From equation (ii), x 3–2y-(iii)substitute value of x in eq. (i)7(3–2y)–15y 219–29y –19 y 291949 In eq. (iii) x 3 – 2 2929BySubBy Eliminationio nionuttistByCross-MultiplicatSolve: 7x–15y 2 –(i)x 2y 3 –(ii)3.x 2y–4 02x 4y–12 02x 3y–9 04x 6y–18 0x–2y 03x 4y–20 01.2.Pair of 20c1c2a1bc 1 1a2b2 c2a1bc 1 1b2 c2a2a1b 1a2b2Coincident LinesIntersecting LinesNo solution –InconsistentInfinitelymany solutions– DependentExactly onesolution –consistent(unique)AlgebraicInterpretationEach solution (x, y), correspondsto a point on the line representingthe equation and vice-versaa1,b1,c1,a2,b2,c2, – Real numbersa1x b1y c1 0a2x b2y c2 0Chapter-1GraphicalRepresentationSolutioGraph nicallyCompare theRatiosGraphicalRepresentationPair of LinearEquations inTwo VariablesS. No.MethodsebraicAlgrml FoeranGeMind map : learning made simpleOswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-X [1

QuadraticCPolyn10Case2- Graph cutsx-axis at exactlyone pointCase3- Graph doesnot cut x-axisNumber of ZeroesDegree of2GraphZerpoly oes ofGra nomialphicallyPolynomialsionentatserpl Remialicaynohlopaic PGrratduaQCase1- Graph cutsx-axis at 2 pointsCaseeroes andonship-ZPolynomialsient ofciffeCoRelatiDivisionAlgoritialomSum of products of thezeroes taken two at atimecαβ βϒ ϒα aProduct of zeroesdαβϒ – aα, β and ϒ are zeroesof Cubic Polynomialax3 bx2 cx dSum of zeroes,bα β ϒ –aα and β are zeroes ofQuadratic Polynomialax2 bx cThen,Sum of zeroes,bα β –aProduct of zeroescαβ aIf p(x) and g(x) are twopolynomials withg(x) 0, then –p(x) g(x) q(x) r(x)where, r(x) 0 ordegree of r(x) degreeof g(x)231ax3 bx2 cx da 0ax2 bx ca 0ax bDegree General FormHighest power of x inPolynomial, p(x)QuadraticCubicChapter-2y x2–3x–4LinearPolynomialPa r a b o laMind map : learning made simpleichmub2] Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-XType s

–74–46y32153i.e. x 8 and y 10xyThen, 3(–74)–5(–46)(–46)(3)–(–74)(2)1 2(5)–3(3)y1x –138 14810–9–222 230xy11yx1 and 81011108153xSolution: By cross-multiplication methodSolve: 2x 3y–46 0 –(i)3x 5y–74 0 –(ii)Solution: From eq. (i)x 2–eq.(ii)x1, wehave(4x–4x) (6y–6y) 16–70 9, which is a false statementThe pair of equation has no solutionSolve: 2x 3y 8 – (i)4x 6y 7 – (ii)Solution: From equation (ii), x 3–2y-(iii)substitute value of x in eq. (i)7(3–2y)–15y 219–29y –19 y 291949 In eq. (iii) x 3 – 2 2929BySubBy Eliminationio nionuttistByCross-MultiplicatSolve: 7x–15y 2 –(i)x 2y 3 –(ii)3.x 2y–4 02x 4y–12 02x 3y–9 04x 6y–18 0x–2y 03x 4y–20 01.2.Pair of 20c1c2a1bc 1 1a2b2 c2a1bc 1 1b2 c2a2a1b 1a2b2Coincident LinesIntersecting LinesNo solution –InconsistentInfinitelymany solutions– DependentExactly onesolution –consistent(unique)AlgebraicInterpretationEach solution (x, y), correspondsto a point on the line representingthe equation and vice-versaa1,b1,c1,a2,b2,c2, – Real numbersGraphicalRepresentationSolutioGraph nicallyCompare theRatiosGraphicalRepresentationPair of LinearEquations inTwo VariablesS. No.MethodsebraicAlgrml FoeranGeChapter-3a1x b1y c1 0a2x b2y c2 0Mind map : learning made simpleOswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-X [3

2x2–5x 3 035x2– x 02255 2 3 0 x–442x 5 15 – 1or x 44443x or x 12x – 5 1 x– 5 14164422x– 5 –4Solution:2–1 016theSquare2–1or x 32Roots are 2 , –13 2(3x–2) 0 or (2x 1) 0x QuadraticEquationsDiscriminantD 0D 0D 01.2.3.No real roots(imaginary)Two equalreal rootsTwo distinctreal rootsRootsFor quadratic equationax2 bx c 0,b2–4ac is Discriminant (D)Quadratic FormulaGeneral FormEquation of degree 2,in one variableS. No.tion of aSoluEquationaticadruQ6x2 3x–4x–2 03x(2x 1)–2(2x 1) 0(3x–2)(2x 1) 0The roots of 6x2–x–2 0Solution:Find roots of 6x2–x–2 0By CompletingBy FactorizationMeaningNature of RootsSolve: 2x2–5x 3 0Chapter-4–b b2–4ac2aRoots of ax2 bx c 0 are given byax2 bx c 0a, b, c – real numbersa 0Mind map : learning made simple4] Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-X

i.e.,a c2s n(2a (n–1)d)2a – first termd – common differencen – total termsWhen first term ofcommon differnce isgiven :b is arithmetic meanb If a, b, c, are in AP,SGeneral formFrom the endan l – (n–1)dFrom beginningan a (n–1)dHerea – first termd – common differencean – nth termHereChapter-5Fixed number in arithmeticprogression which providesthe to and fro terms by adding/subtracting from the presentnumber.Can be positive or negative.a, a d, a 2d, a 3d, .a (n –1) dList of numbers in which each termis obtained by adding a fixed numberto the preceding term except the first term.Fixed number is called common difference.CommonDifferenc eonitiArithmeticProgressionsWhen first & lastterms are given :ns (a an)2orns (a l)2a – first termn – total termsantic meemthArisuman a (n–1)d99 12 (n–1)387n–1 293n 30Let sn 1 2 3 . na 1, last term l nn(a l) n(1 n) sn 22n(n l)n(a l)or sn 2ple(s )Defin2-digit numbers divisible by 312, 15, 18, . 99a 12, d 3, an 99Sum of first n positive integersExa mnth TermHow many 2-digit numbers aredivisible by 3?Mind map : learning made simpleOswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-X [5

5. If one angle of a triangle is equal to one angle ofthe other triangle and the sides including theseangles are proportional, then the two triangles aresimilar.(SAS criterion)4. If in two triangles, sides of one triangle areproportional to (i.e., in the same ratio of ) thesides of the other triangle, then theircorresponding angles are equal and hence thetwo triangles are similiar.(SSS criterion)3. If in two triangles, corresponding angles areequal, then their corresponding sides are in thesame ratio (or proportion) and hence the twotriangles are similar.(AAA criterion)2. If a line divides any two sides of a triangle in thesame ratio, then the line is parallel to the thirdside.1. If a line is drawn parallel to one side of a triangleto intersect the other two sides in distinct points,the other two sides are divided in the same ratio.StatementAEAD ECDBAEAD ECDBABAC & A DDEDFthen, ABC DEFIfBC CAAB EFFDDEthen, A D ; B E , C F ABC DEFIfIf A D, B E C Fthen, AB BC ACDEEFDF ABC DEFthen, DE BCIfthenIf, DE BCFigureArea of SimilarTrianglesMCQNPRar(ABC) ar(PQR)BC 2 QRABPQCARP22Here ABC PQRBACQ ABC PQRPStatementRight angledtrianglPythetagoherasoremBARii) Corresponding sides are in thesame ratioi) Corresponding angles are equalFigureChapter-63. In a triangle, if square of one side is equal to thesum of the squares of other two sides, then theangle opposite the first side is a right angle.2. In a right triangle, the square of the hypotenuse isequal to the sum of the squares of the other twosides.BAACBCIf AC2 AB2 BC2then, B 90 In right ABC,BC2 AB2 AC2B1. If a perpendicular is drawn from the vertex of theright angle of a right triangle to the hypotenusethen triangles on both sides of the perpendicular A DCare similar to the whole triangle and to each other. In right ABC, BD AC,then, ADB ABC BDC ABC ADB BDCTrianglesThe ratio of the areasof two similar trianglesis equal to the squareof the ratio of theircorresponding sidesTheoremsityarSimilMind map : learning made simple6] Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-X

pleamEx1(–7) 2(2) 1(4) 2(–2),1 21 2i.e., (–4, 2)2(–7) 1(2) 2(4) 1(–2),2 12 1Coordinate of Qi.e., (–1, 0) Coordinate of PFind point of Trisection ofline segment AB, A(2, –2) and B(–7, 4)R x1 x2 y1 y2,22idMtoin-ptmenSegeLinArea 1 [x1(y2 – y3) x2(y3 – y1) x3(y1 – y2)] 2Y’IVQuadrantIIIQuadrantof TriangleAreaX’IQuadrantngniXSectionformulam1y2 m2y1m1 m2m1x2 m2x1,m1 m2DistanceformulaCoordinate axisStudy of algebraicequations on graphsA (1, 7); B (4, 2); C (–1, –1); D (–4, 4)Since, AB BC CD DA and AC BD.All four sides and diagonals are equalHence, ABCD is a squareBD (4 4)2 (2 – 4)2 68AC (1 1)2 (7 1)2 68DA (1 4)2 (7 – 4)2 34CD (–1 4)2 (–1 –4)2 34BC (4 1)2 (2 1)2 34AB (1 – 4)2 (7 – 2)2 34y– a xisOrdin a t eX’isaxx–a)cissbsPQ (x2 – x1)2 (y2 – y1)2icalalntVertHorizoAre the following points vertices of asquare : (1, 7), (4, 2), (–1, –1), (–4, 4)?CoordinateGeometryMeaIIQuadrantExampleYInter nall y )( E xte r n a ll y(–)(AMind map : learning made simpleordinateXabscissaY’YChapter-7Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-X [7

0 010Not ( )defined1Not ( )defined Asin Acos Atan Acosec Asec Acot Atan (90 – A) cot Acot (90 – A) tan Asec (90 – A) cosec Acosec (90 – A) sec AmCorytaenemplesglAnTrigonometric Identitiescos Asin A 1 – sin2A 11 32 22 3 32 3 3 3212 2145 1 21 2212 321 330 60 trymeoonIntroduction toTrigonometryTrigsin Acos (90 – A) sin Asin (90 – A) cos Acot2 A 1 cosec2 A0 A 90 1 tan2 A sec2 A0 A 90 cos2 A sin2 A 1tan A Solution : We know that, cos2 A sin2 A 1cos2 A 1 – sin2 A ie. cos A 1 – sin2AExpress tan A, cos A in terms of sin ASide oppositeto C0Not ( )defined1Not ( )defined0190 AHseutenoypTrigonometry RatioSideoppositeto ABCCotangent of ASecant of AABBCACABACBCBCABABACBCACChapter-8Cosecant of ATangent of ACosine of ASine of AStudy of relationships betweenthe sides & angles of a right triangleMind map : learning made simplempleExaValues8] Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-X

PDi.e., BD 3mBDDistplesamExAB 3 15i.e., AB 15 3mIn ABC B 90 , C 60 ABHere, tan 60 BCDetermine heightof object ABt AB (3 3 3)m 3( 3 1)mtan 45 From figure, AB AD DBIn right APD A 30 , D 90 PDtan 30 i.e., AD 3 3mADIn right BPD B 45 , D 90 ObApplicaTrigonoRa mtioSome Applicationof Trigonometryn–tio etricsAMeasuring AnglesAngleofD e prerepDof200 mnsioes(ii)DCFlagChapter-9ChDhα βABxFind flag length60 Find x and hs sio nAnevationf ElolegleABCD30 x B90 (i) BD is a treeAC DCdetermine)(Todistancewb et w eenoobjectsthlengt/hgei n objectaofAngDetermine width ABceaneig htje ctHhMind map : learning made simpleOswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-X [9

2. The lengths oftangents drawnfrom an externalpoint to a circle areequalFigurePORPQ PRQPOPQ 90 YtsOQXsremoeThacntioniQAPOrOnly one commonpoint between circleand PQ line.Tangent andtangent pointCircleslineTwo commonpoints betweenline PQ andcircle.BQOA PrNo common pointbetween line PQand circle.Non-intersectingThe locus of a point equidistant froma fixed point. Fixed Point is a centre& separation of points in the radiusof circle.an1. The tangent at anypoint of a circle isperpendicular tothe radius throughthe point of contactP3. There are exactly two tangentsto a circle through a point lyingoutside the circle.FS ecStatementT2P2. There is one and only onetangent to a circle passingthrough a point lying on thecircle.1. There is no tangent to a circlepassing through a point lyinginside the circle.DefiT1PMind map : learning made simpleQPpOrPOcentrerradiusChapter-1010 ] Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-Xt

DefinitTriangle similarto giventriangleConstructionsin’stiora3. Draw line parallel to B4 C from B3intersecting BC at C’.Draw line to AC from C’intersecting AB at A’A’BC’ is the required triangle2. Locate 4 points (greater of 3and 4 in 3 ) on BX at equal4distance from each other(BB1 B1 B2 B2 B3 B3 B4Join B4 C1. Draw any ray BX making anacute angle with BCConstruct a triangle similar to a given ΔABC with sides 34of the corresponding sides of ΔABCto circlegentTanLineSegmDivisioneoflineTo draw geometricalshapes usingcompasses, ruler etcnt3. Join PQ and PR,required tangents tothe circle2. M as centre andradius MOdraw a circle,intersecting givencircle at Q and R1. Join PO and bisect it atmid-point MionMethod1Given: Circle with centre O and point P outside it.Chapter-11AC : CB 3 : 23. Locate A1, A2, A3 (m 3)on AX and B1, B2 (n 2)on BYJoin A3 B2, intersectingAB at CAC : CB 3 : 22. Draw ray BY AX1. Draw any ray AX makingan acute angle with linesegment ABGiven: Line segment, ratio (3 : 2)3. Through A3 (m 3),draw line parallelto BA5 cutting AB at C2. Locate 5 pointsA1, A2, A3, A4, A5 at equaldistances(A1 A2 A3 A4 A5).Join BA51. Draw any ray AX,Making acute anglewith line segment ABGiven: Line segment, ratio (3 : 2)Mind map : learning made simpleOswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-X [ 112odMeth

eanin14 cm( )PQArea Area of thecorresponding sector– Area of thecorresponding triangleθ πr2 – area of OAB360 ulamorAreas Relatedto CirclesSectorentSegmmulaA Chapter-12θ area of circle360 LeθA πr2ng360 thofarcθL circumference360 θL 2πr360 aAreningMeaPortion of thecircular region enclosedby two radius and thecorresponding arcPortion of the circular regionenclosed between a chordand the corresponding arcgnineaMArea of square ABCD 14 14 cm2 196 cm214Diameter of each circle, D 2 7 cm7For each circle, radius (r) 2 cm2Area of 1 circle πr227 2Area of 4 circles 4 7 2 cm2154 4 4 cm2 154 cm2Area of shaded region Area of ABCD – Area of 4 circles (196 – 154) 42 cm2gination- CombAreaof figuresCircumference π diameter 2πrArea πr2rFoFind Area of shadedregionMTArea of T Area of P Area of QQFPlempExaCircleMind map : learning made simple12 ] Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-X

ArearfaceSulaotπ(r12 r22)TSA πl(r1 r2) TCSA πl(r1 r2)r1l1V 3 πh(r12 r22 r1r2)r2CurvedSurface Arear2where l h2 (r1 – r2)2Volumer1r2r1lConversionof SolidsSurface Areasand Volumes π r 2 1800 2π1r 2 9001r 30 cmThickness Diameter of the cross-section1 15 cm 0.07 cmSolution : Volume of the rod π1 232 8cm3 2π cmLet, r is the radius of cross-section of thewire, volume π r2 1800 cm3A copper rod – Diameter 1cm, length 8cmconverted into a wire of length 18mFind the thickness of the wire.of ConeFrustumhreaeAacSurf5 cmActual capacity Apparent capacity– Volume of hemisphere 98.125 – 32.71 65.42 cm32Volume of hemisphere 3 πr3,if r 2.5cm 2 3.14 (2.5)3 cm3 32.71 cm33Solution :– Apparent capacityof the glass πr2h 3.14 2.5 2.5 5 cm3 98.125 cm35 cmGiven – Inner diameter of theCylindrical glass 5 cmHeight 5 cmFind – Actual capacity of Cylindrical glassSolidsQuantity of 3-D spaceenclosed by a hollow/closed solid.CombinationofVolumeChapter-13Sum of all of the surface areasof the faces of solid.Mind map : learning made simpleOswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-X [ 13

EmpericalOgiven – cfl 2 hfianedMModeRelationshipf1 – f0l h2f1 – f0 – f23 Median Mode 2 MeanRepresentation ofcumulative frequencieswith respect to givenclass intervalsA CumulativeFrequency GraphMeaningGroupedDataassMHere, u x–AharkedLower classlimitHere, d (x – A) fidi fi fixi fi2 x a x Upper classlimitcu M eat)nmsuo(ShrtChapter-14Frequency obtained by addingthe frequencies of all the classespreceding the giving classDirect Method(Long cut)CumulativeFrequencyA collection, analysis,interpretation ofquantitative dataAs fiuix a h fintioMeantionif nieStatisticsDMind map : learning made simpleClStepDevia14 ] Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-X

Event havingprobability tooccur as 1Solution : Number of favorableoutcomes 4Number of possible outcomes 5241P(E) 5213What is the probability of gettingan ace from a pack of 52 cards?Solution : Total outcomes 2Favorable outcomes 11Required Prob. P(E) 2rdExample sytarenmletenSure or Certain EventCaCoineEvent havingonly one outcomeof the experimentytarenntFor event E,complement event,–P(E) 1 – P(E)ElemEveluSum of probabilities of all elementaryevents is 1.For events A, B, C; P(A) P(B) P(C) 1ProbabilityChapter-15P(E) Number of outcomesfavorable to ENumber of all possibleoutcomes of theexperimentNumber of trials inwhich the event happenedP(E) Total number of trialsWhat actually happensin an experimentWhat we expect tohappen in an ityDefinitionsWhen a coin is tossed, whatwould be the probability ofappearing head ?DiceCompVaEvPTSolution : Number of possibleoutcome 62 36Number of favorable outcomes 0As addition of no two number on thedice will give sum 13 is not possible0P(E) 36 0Two dice are rolled, what is probabilityof getting 13 as a sum?Mind map : learning made simpleOswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-X [ 15

Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class-X [ 5 Mind map : learning made simple Chapter-5 s n n(a l) 2 A D rithmetic Progressions e f i n i t i o n A r i t h m e t i c b is arithmetic mean m e a n S u m a (s) List of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. Fixed .

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