Mind map : learning made simple Chapter-1
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1 swl E hptewise Mind Mps, MTHEMTI, lss-i [ Mind mp : lening mde simple hpte- Tem oduct lw m n = m+n uotient lw owe lw Recipocl lw m n = m n ( m ) n = mn Numbe stem,b el numbes n +ve intege 0 uccessive Numbe line mgnifiction Eve point on the line epesents el numbes Rtionliztion s + s Rel numbe Integes (Z) Rtionl numbe () nth oot of el numbe,,... 0,,,... Itionl numbe () = (b) /n nnot be witten in p/q fom Fom Eponents with Integl owes s + s Rtionlising fcto + s s + s s Tnsfom denominto into tionl numbe...,,,0,,,... Whole numbes (W) q 0, (p,q) z Ntul numbes (Z) Emples ube oot oot que b b m m =, = / = () / Emple 5, 7 Emple 7, p q
2 ] swl E hptewise Mind Mps, MTHEMTI, lss-i Mind mp : lening mde simple hpte- (+) + + ( ) + (+) (+b) ( + + z) ( ) ( + ) + ( + b) + b + +z ++z+z = + +z + (+z+z) ( + ) + + ( + ) ( ) ( ) olnomil in one vible: b c + d olnomil olnomils olnomil onstnt (o independent) Zeo Monomil inomil Tinomil, 7/5 egee not defined (constnt polnomil 0) + Emple / is the zeos of the polnomil n lgebic epession of the fom: f() = n n + n n lgebic Identities + +z z (++z) ( + +z z z) (i) + +z = z If + + z = 0 (ii) + + z = z z + (+)( + ) (-)( ++ ) Fcto Theoem Tpes Theoems Tems (i) if ( ) is fcto of p(), then p() = 0 egee olnomil Reminde Theoem ivident = (iviso uotient) + Reminde if p() is polnomil of degee n>, : n el numbe Zeoes of polnomil (ii) if p() = 0, then ( ) is fcto of p(). if p() polnomil of degee n>, is divided b, p() is the eminde Numbe tht stisfies the eqution Line udtic ubic Emple egee Emple () = + find zeoes of the polnomil () = 0 + = 0 = /
3 swl E hptewise Mind Mps, MTHEMTI, lss-i [ Mind mp : lening mde simple hpte- (II) udnt (,+) (III) udnt (, ) (I) udnt (+,+) (IV) udnt (+, ) oodinte Geomet To descibe the position of point in plne tep : elect tlest vlues R(,) of, such tht, I (,0) oodinte sstem tep : w tble fo the 0 5 odeed pi (,) dinte (0, ) tep : lot these odeed points bsciss on the gph ppe Mening bsciss -is Two pependicul tesin sstem lines -is dinte Fied point oint whee & (0,0) is intesect lne divided into pts b co-odinte es (0,0) udnts plne (5,) lotting point in udnt is Emple (I) udnt (+,+) (II) udnt (,+) (III) udnt (, ) (IV) udnt (+, ) > 0, > 0 < 0, > 0 < 0, < 0 > 0, < 0 (, ), (, ) (, ), ( 8, 9) (, ), ( 5, 6), (, 6) (, ) tep : w co-odinte is nd select units tep : tting fom oign, count units on nd is tep : Mk the coesponding points Gph-Line Eqution = tep : onvet given eqution in the fom = m + c tep 5 : w stight line pssing though plotted points. (5,) 5 0 oint, (5,)
4 ] swl E hptewise Mind Mps, MTHEMTI, lss-i Mind mp : lening mde simple hpte- Eqution Intepettion Gphicl epesenttion (,b) (0,0), b, c - constnts Whee, - vibles = 0 = 0 Eqution of -is Eqution of -is teps to find solution Line Eqution in Two Vibles = 0 Emple Line eqution Eqution of the fom + b + c = 0 tep : Wite the eqution in two vibles, if not pesent tep : Reduce it to one vible b putting n bit vlue of n vible, to find pi of solution. tep : Repet step fo nothe bit vlue of vible to find nothe pi of solution. Gphicl Repesenttion = 0 = 0 = K = K tight line pllel to -is tight line pllel to -is =.5 Line eqution: + + tep : Fom eqution, we get = - tep : ut bit vlue of, tep : lot (0,) nd (6,0) on the gph nd join them. It cn hve one, no o infinitel mn solutions. Emple- Gph of Line Eqution =.5 = m Line pssing though oigin = (0,) (6,0)
5 swl E hptewise Mind Mps, MTHEMTI, lss-i [ 5 Mind mp : lening mde simple hpte-5 oint Things which e equl to the sme thing e equl to one nothe. If equls e dded to equls, wholes e equl. If equls e subtcted fom equls,the emindes e equl. iomtic sstem, in which ll theoems e deived fom smll numbe of ioms Euclid s Geomet Euclid s ostultes Hs no width, no length nd no depth ollection of points, cn be etended in both diections. Two - dimensionl collection of points (hs length & bedth onl) oint efinitions Line ufce Line ufce Euclid s ioms 5 stight line cn be dwn fom n one point to n othe point. teminted line cn be poduced infinitel. cicle cn be dwn with n cente nd of n dius. ll ight ngles e equl to one nothe. If stight line flling on two stight lines mkes the inteio ngles on the sme side of it, tken togethe mkes less thn two ight ngles, then the two stight lines, if poduced indefinitel, meet on tht side on which the sum of the ngles is less thn two ight ngles. ente icle Rdius E F = EF = 90 + < 80 Intoduction to Euclid's Geomet Things which coincide with one nothe e equl to one nothe. 5 The whole is gete thn the pt. 6 Things which e double of the sme things e equl to one nothe. 7 Things which e hlves of the sme things e equl to one nothe.
6 6 ] swl E hptewise Mind Mps, MTHEMTI, lss-i Mind mp : lening mde simple hpte-6 R pt of line with one end point Line segment Rs mking Lines n ngle,, - colline points R colline in sme line oints non-colline not in sme line thee o moe points ling efinitions,, - non colline points Tpes of ngles Theoems ngles ngle Vete ms Vlue The inclintion between two stight lines Line collection of points (cn be etended in both diections) Line segment line with two end points m ngles End points Vete cute. If two lines intesect ech othe, then the veticll opposite ngles e equl.. If tnsvesl intesects two pllel lines, then ech pi of ltente inteio ngles is equl. If tnsvesl intesects two lines such tht pi of ltente inteio ngles is equl, then the two lines e pllel. If tnsvesl intesects two pllel lines, then ech pi of inteio ngles on the sme side of the tnsvesl is supplement. 5. If tnsvesl intesects two lines such tht pi of inteio ngles on the sme side of the tnsvesl is supplement, then the two lines e pllel. 6. Lines which e pllel to the sme line e pllel to ech othe. 7. The sum of ll inteio ngles of tingle is 80 R R R R E F = = ioms Right btuse tight line Refle 0 < < 90 = < <80 = < <60. If stnds on line, then the sum of two djcent ngles so fomed is 80. If the sum of two djcent ngles is 80, then the noncommon ms of the ngles fom line.. If tnsvesl intesects two pllel lines, then ech pi of coesponding ngles is equl.. If tnsvesl intesects two lines such tht pi of coesponding ngles is equl, then the two lines e pllel to ech othe. is line then, + = 80 If + = 80 then is stight line R Hee R then = R If R = R nd = R then, then R = R if R = R then then, R + R = 80 if R + R = 80 then nd EF then EF + + = 80
7 swl E hptewise Mind Mps, MTHEMTI, lss-i [ 7 Mind mp : lening mde simple hpte-7 Rule ttement Figue. Tingles closed figue fomed b thee stight lines It hs thee - sides, ngles nd vetices ech Tingle If n thee pmetes of given tingles e sme, the tingles will be conguent. onguence ule Two tingles e conguent if two sides nd the included ngle of one tingle e equl to the two sides nd the included ngle of the othe tingle. onguent. Two tingles e conguent if two ngles nd the included side of one tingle e equl to two ngles nd the included side of othe tingle. E F = ~ EF Inequlities opeties. Two tingles e conguent if n two pis of ngles nd one pi of coesponding sides e equl.. If thee sides of one tingle e equl to the thee sides of nothe tingle, then two tingles e conguent. 5. RH If in two ight tingles the hpotenuse nd one side of one tingle e equl to the hpotenuse nd one side of the othe tingle, then the two tingles e conguent. = ~ EF E F = ~ = E = EF = F E F In nd In = = = In nd EF Given nd = = = = ~ In nd EF = F = E = FE = ~ EF 5 5 In nd EF = F = 5cm = FE = cm F E = - = 5 - = E = F - EF = 5 - = = E Hence = ~ EF ttement Figue ngles opposite to equl side of n isosceles tingle e equl The sides opposite to equl ngles of tingle e equl = = = = ( ule) Hence, = ttement Figue In n tingle, the ngle opposite to the longe side is lge. In n tingle, the side opposite to the lge (gete) ngle is longe. The sum of n two sides of tingle is gete thn the thid side. iffeence of n two sides of tingle is less thn the thid side. is the longest side is lgest if is the lgest is longest In + > + > + > In - < - < - <
8 8 ] swl E hptewise Mind Mps, MTHEMTI, lss-i Mind mp : lening mde simple hpte-8 ttement udiltel udiltels Figue fomed b joining fou points in n ode It hs fou - vetices, ngles nd sides ech is udiltel Mid-point theoem The line-segment joining the mid-points of two sides of tingle is pllel to the thid side. The line dwn though the mid-point of one side of tingle, pllel to nothe side bisects the thid side Tpes opet llelogm Rectngle Rhombus que Tpezium ll sides e conguent pposite sides e pllel nd conguent ll ngles e conguent pposite ngles e conguent igonls e conguent igonls e pependicul igonls bisect ech othe djcent ngles e supplement E E F F Figue l l If E nd F e mid-point of nd, then EF If E is the mid-point of EF, then F = F, F is the mid-point of llel but not conguent ttement Figue. digonl of pllelogm divides it into two conguent tingles.. In pllelogm, opposite sides e equl nd pllel.. If ech pi of opposite sides of qudiltel e equl nd pllel, then it is pllelogm.. In pllelogm, opposite ngles e equl. 5. If in qudiltel, ech pi of opposite ngle is equl, then it is pllelogm. 6. The digonls of pllelogm bisect ech othe. 7. If the digonls of qudiltel bisect ech othe, then it is pllelogm. is pllelogm digonl then opeties In pllelogm,,, nd =, = If,, nd =, = then is pllelogm In pllelogm, =, = If =, = then is pllelogm In pllelogm, = then = If = = then is pllelogm
9 swl E hptewise Mind Mps, MTHEMTI, lss-i [ 9 Mind mp : lening mde simple hpte-9 ttement. llelogms on the sme bse nd between sme pllels e equl in e. () = () = Theoems. Two tingles on the sme bse (o equl bses) nd between the sme pllels e equl in e. Two tingles hving the sme bse (o equl bses) nd equl es lie between the sme pllels Figue if = then (i) If nd e conguent figues, () = () is pllelogm h h opeties = se Height e of pllelogm Numbe (in some unit) ssocited with pt of the plne enclosed b the figue e of tingle = se Height e Medin of tingle divides it into two equl es e of tingle - medin = R E F () () if R EF then () = () (ii) If pln egion fomed b figue T is mde up of two nonovelpping pln egions fomed b figues nd (T) = () + () T e of figue T = e of figue + e of figue
10 clic qudiltel 0 ] swl E hptewise Mind Mps, MTHEMTI, lss-i Mind mp : lening mde simple hpte-0 ttement Figue. Equl chods of cicle subtend equl ngles t the cente.. If the ngles subtended b the chods of cicle t the cente e equl, then the chods e equl.. The pependicul fom the cente of cicle to chod bisects the chod. cente Fied point. The line dwn though the cente of cicle to bisect chod is pependicul to the chod. dius () Fied distnce 5. Thee is one nd onl one cicle pssing though thee given non-colline points. 6. Equl chods of cicle e equidistnt fom the cente. = then = = then = M L M M R M then M = M If M = M then M = then L = M ll the fou vetices of qudiltel lie on cicle icles icle - cclic qudiltel cicle is the locus of cuve equidistnt fom point ente Rdius Relted tems imete () 7. hods equidistnt fom the cente of cicle e equl in length. chod which psses Inteio of the cicle though the cente 8. The ngle subtended b n c t the cente is double the ngle subtended b it t n point on the emining pt of the cicle. 9. ngles in the sme segment of cicle e equl. 0. If line segment joining two points subtends equl ngles t two othe points ling on the sme side of the line contining the line segment, the fou points lie on cicle.. The sum of eithe pi of opposite ngles of cclic qudiltel is 80.. If the sum of pi of opposite ngles of qudiltel is 80, then qudiltel is cclic. R Theoems R If L = M then = Refle = R = R = then,,, lie on the cicle = of c cicle egment = imete of cicle Mjo Inteio Eteio of the cicle Mjo Eteio icle Mino Mino segment Mino c Mjo c Mino L M R + R = 80 + = 80 If + R = 80 + = 80 then R is cclic qudiltel Mjo segment
11 swl E hptewise Mind Mps, MTHEMTI, lss-i [ Mind mp : lening mde simple hpte- tep : w = bse nd mke given ngle. tep : w line equlto sum of sides = Y Y tep : w given two ngles t & Y tep : ut = diffeence of sides join tep : w pependicul bisecto nd let it intesect. Nme it s. tep : Tke s cente dw c intesecting tep : With s cente nd sme dius, dw c intesecting sme c. E tep : w pssing though E, = 60 E 60 onstuctions intesecting cs on both sides. tep : Join tep : With & s cente & dius >, dw M tep : intesect t point M to fom pependicul bisecto ocess of dwing geometicl figue bse ngles constuction Geometicl Given : eimete nd two Given: se, ngles, diffeence of two sides tep : Join, is the equied tingle. Tingle Geomet bo Requiements Given : se, ngles tep : w = bse nd mke given ngle. 60 ngle sum of two sides tep : ut equl to sum of sides, join L tep : isect LY nd MY, let the bisectos meet t. L M tep : w pependicul bisectos of nd Y. Etend them to intesect Y. is the equied tingle. L M M Y Y Y ngle bisecto tep : Tke s cente, dw cs intesecting & (of n dius) tep : & E s centes nd dius > E, dw two c nd join to it tep : F is the equied ngle bisecto. cle. i of setsques. i of divides. ompss. otcto E E F ependicul bisecto tep : w pependicul bisecto of nd let it intesect. Nme it s. tep : Join, is the equied tingle.
12 ] swl E hptewise Mind Mps, MTHEMTI, lss-i Mind mp : lening mde simple hpte- Hee, s = ++0 = cm e = (-) (-) (-0) cm = ( (0) (0) ())cm = 0 cm Heon s Fomul e of tingle = s (s ) (s b) (s c)s whee,,b,c :- sides of tingle s :- semi peimete = +b+c R Heon s fomul e of tingle = s (s ) (s b) (s c) whee, s = +b+c 9 m 8 m 90 m 5 m Find e of tingle of sides cm, cm, 0cm e of qudiltel with given dimensions :- e of = e of + e of Hee, e of = e of = s (s-) (s-b) (s-c) whee s = +b+c = (9+8+) m = 5 m c ltitude b se ltitude se R ; = side of tingle Right-ngled tingle e pplictions Equiltel tingle Heon s fomul Hee, = + e = 5 (5-9) (5-8) (5-) m = 5 (6) (7) () m = 5.96 m e of = (0+5.96) m = m = 5 = 0 m = +5 = m
13 swl E hptewise Mind Mps, MTHEMTI, lss-i [ Mind mp : lening mde simple hpte- e π h cu units Volume = e of bse Veticl height of uboid = (lb+bh+hl) ufce e Volume of cubiod = lbh Volume = π cu units b h l Volume - figue with eve point on its sufce equidistnt fom its cente phee Volume = π h cu units Hemisphee Right cicul cone - object which tpes fom cicul bse to point um of the es of ll fces (o sufces) on - shpe e sic concept ube b h ube e l b h cu units - figue with 6 fces nd ll equl sides phee cut in hlf Volume = π cu units l of n object ufce e Totl sufce e =6 sq units uboid uboid - figue with 6 fces Right cicul clinde h Right cicul clinde clinde with cicul bses nd is joining the two centes of the bses pependicul to the plnes of the two bses Mesuement of evething within the lines of the shpe Volume of n object Ltel sufce e = sq units Volume = cu units Totl sufce e = (lb+bh+hl) sq units Ltel sufce e = (l + b) h sq units untit of thee-dimensionl spce enclosed b closed sufce Volume = Volume = ufce e = π sq units uved sufce e = πh sq units Totl sufce e = π (+h) sq units l e e uved sufce e = π sq units e = π sq units uved sufce Totl sufce e = π (l+) sq unit e = πl sq unit uved sufce h
14 ] swl E hptewise Mind Mps, MTHEMTI, lss-i Mind mp : lening mde simple hpte- Items untities Medin = vlue of th n+ et of vlues of qulittive o quntittive infomtion ttistics gph t obsevtion _ n Men, = i i= n lculted b dding ll the vlues nd dividing it b totl numbe of obsevtions. ttistics bsevtion entl Tendenc Men odd numbe Medin -Ungouped dt Vlue of the middle most obsevtion. 0 bsevtion even numbe Most fequentl occued obsevtion istnce (in km) of 0 students fom thei esidence to school is Mode Tll mks given s onstuct gouped fequenc distibution tble Gphicl epesenttion gouped dt Medin = vlue of th th n n obsevtion+ + obsevtion istnce Tll Fequenc gphs Fequenc polgons Fequenc distibution tble ll ll llll llll llll ll l 5 Histogms gph Fequenc polgon e of stud deling with the pesenttion, nlsis nd intepettion of dt hows vluble dt set Histogm
15 swl E hptewise Mind Mps, MTHEMTI, lss-i [ 5 Mind mp : lening mde simple hpte-5 Totl numbe of tils utcomes fo getting = (5,6) (6,5) Numbe of getting (E) = = Totl numbe of tils 6 = 8 = 6 n, n is the numbe of times dice is thown = 6 = 6 The etent to which something is likel to hppen Impossible Even chnce etin Less likel Moe likel efinition ice dice is thown times. Find the pobbilit of getting utcome Emples Event Relted tems obbilit Til Numbe of tils in which event hppened Totl numbe of tils ction which esults in one o sevel outcomes obbilit vlue Empeicl pobbilit oin coin is tossed times. Find pobbilit of heds? t gete thn Totl numbe of tils = n, n is numbe of times coin is tossed [0,] = = 8 i.e. HHH, HHT, HTT, HTH, TTH, THH, TTT, THT (E) = omething tht follows s esult o consequence ollection of some outcomes of n epeiment (E) = Numbe of times heds occued Totl numbe of tils = 8
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