Maths Concept King PDF 2024 Edition

 Symbol Table of Symbols (izrhdksa dh rkfydk) Appendix   implication Symbol Refer...

 Symbol Table of Symbols (izrhdksa dh rkfydk) Appendix   implication Symbol Reference  negation, equivalence, relation = equal to ,  quantifier  not equal to {} set  identity  empty set, void set, null set  approximately equal to ! Factorial  congruent to i Imaginary unit r approaches, ray  Union Si  proportional to  Subset of < less than  Superset not less than  Intersection  greater than  Subset or equal to ap not greater than  Mean (average)  less than or equal to Conversion of Units  greater than or equal to Conversion of Length > much greater than 10 centimetres = 1 decimetre (dm) io ra  infinity 10 decimetres = 1 metre (m) at or  sigma (Summation) 10 metres = 1 decametre (dam) ic % percentage 10 decametres = 1 hectometre (hm) bl 10 hectometres = 1 kilometre (km) P + plus, positive Pu Conversion of Area – minus, negative  plus or minus 100 square millimetres = 1 square centimetre on an 100 square centimetres = 1 square decimetre a  b 100 square decimetres = 1 square metre pi  multiplication a  b 100 square metres = 1 square decametre am 100 square decametres = 1 square hectometre a  b g  division 100 square hectometres = 1 square kilometre Ch a / b 1 hectare = 10000 square metres Ga  therefore Conversion of Volume  since 1000 cubic millimetres = 1 cubic centimetre — line segment 1000 cubic centimetres = 1 cubic decimetre  acute angle 1000 cubic decimetres = 1 cubic metre  perpendicular 1000 cubic metres = 1 cubic decametre  parallel 1000 cubic decametres = 1 cubic hectometre  triangle 1000 cubic hectometres = 1 cubic kilometre rectangle Conversion of Capacity 10 millilitres = 1 centilitre square 10 centilitres = 1 decilitre logba logarithm (to base b) 10 decilitres = 1 litre log10a common logarithm 10 litres = 1 decalitre loge a or ln a natural logarithm 10 decalitres = 1 hectolitre  conjunction (and) 10 hectolitres = 1 kilolitre  disjunction (or) 1  Symbol Conversion of Weight Equivalents of Units 10 milligrams = 1 centigram Units of Lengths 10 centigrams = 1 decigram 12 inches = 1 feet (ft) = 0.3048 metres 10 decigrams = 1 gram (g) 3 feet = 1 yard (d) 10 grams = 1 decagram 1 yard = 0.9144 metres 10 decagrams = 1 hectogram 22 yards = 1 chain 10 hectograms = 1 kilogram (kg) 1 kilometre = 0.621 mile or 103 metres 100 kilograms = 1 quintal 1 mile = 1.6093 kilometres or 1760 yards 10 quintals or 1000 kg = 1 metric tonne 1 inch = 2.54 centimetres Conversion of Time 1 hectare = 2.471 acres 60 seconds = 1 minute 1 mile = 5280 feet 60 minutes = 1 hour Units of Area r 24 hours = 1 day 1 square feet = 144 square inches 7 days = 1 week = 0.0929 square metres Si 15 days = 1 fortnight 1 square metre = 1.196 square yards 28, 29, 30 or 31 days = 1 month 1 square yard = 0.836 square metres 12 months = 1 year 1 square kilometre = 0.3861 square miles 365 days = 1 year = 1000 hectares 366 days 10 years 25 years = = = 1 leap year decade silver jubilee ap 1 square mile 1 acre = = = 2.59 square kilometres 640 acres 4840 square yards n 50 years = golden jubilee = 4046.86 square metres t 60 years = diamond jubilee 1 hectare io = 10000 square meters ra at 75 years = radium jubilee or platinum jubilee ic 100 years = century bl P 1000 years = 10 centuries or Pu 1 millennium on an pi am g Ch Ga 2  Geometry Geometry (Line & Angle) (js k vkSj dks.k) Line and Angle (js[kk vkSj dks.k) L  Point () : Zero dimension figure or a circle with d d M zero radius. AB CD and EF is transversal line fcanq % 'kwU; vk;keh vkÑfr ;k 'kwU; f=kT;k okyk ,d o`ÙkA AB CD vkSj EF ,d fr;Zd js k gS  Types of point/fcUnq ds izdkj E (i) Collinear point/ljsa[k fcUnq 2 1 A B r 3 4 If 3 or more than 3 points lie on a line close to or Si far from each other, then they are said to be collinear. 6 5 C D 7 8 ;fn 3 ;k 3 ls vf/d fcanq ,d js[kk ij ,d nwljs ds fudV ;k nwj F fLFkr gks] rks os ljsa[k dgykrs gSA Corresponding angles @ laxr dks.k 1 = 5, 4 = 8 ap Ex. Point P, Q, R, S are collinear/P, Q, R, S ljsa[k fcanq gSA 2 = 6, 3 = 7 P Q R S Alternate Angles @ ,dkarj dks.k 3 = 5, 4 = 6 (ii) Non-collinear point/vljsa[k fcanq 4 + 5 = 180° n t 3 + 6 = 180° In 3 or more points are not situated on a straight io ra  Concurrent line/leorhZ js[kk at line, these all point are called non-collinears point. Three or more than three lines, which pass from ic ;fn 3 ;k vf/d fcUnq ,d lh/h js[kk ij fLFkr ugha gSa] rks ;s lHkh a single point is called concurrent lines. bl P fcanq vljsa[k fcanq dgykrs gSaA rhu ;k rhu ls vf/d js[kk,a] tks ,d fcanq ls gksdj xqtjrh gSa] leorhZ Pu Ex. js[kk,a dgykrh gSaA on  Line: (One dimension figure) line is a set of points an C having only length with no ends.   E F pi js k % (,d vk;keh vkÑfr) js k fcanqvksa dk ,d lewg gS ftlesa am A B dsoy yackbZ gksrh gS ftldk dksbZ var ugha gksrk gSA g  Line segment: A line with a fixed length. Ch G H D js k aM % ,d fuf'pr yackbZ okyh js kA Ga P Q  Linear pair angle/jSf[kd ;qXe dks.k  Ray: A line with uni-direction length. A B A linear pair is a pair of adjacent angle whose non-common sides are opposite rays. fdj.k % ,d fn'kk yackbZ okyh js kA ,d js[kh; ;qXe vklUu dks.kkas dk ;qXe gksrk gS] ftldh xSj mHk;fu"B  Parallel lines: two or more line that never intersects L  M Hkqtk,a foijhr js[kk gksrh gSaA lekukarj js k,¡ % nks ;k vf/d js k,¡ tks ,d nwljs dks dHkh ugha c° dkVrh gSaA L M  Transversal Line : A line which intersects y° x° (touches) two or more lines at distinct point is called transversal lines of the given lines. x + y = 180° fr;Zd js k % og js k tks nks ;k nks ls vf/d js kvksa dks vyx&vyx fcanq ij dkVrh (Li'kZ) gS] nh xbZ js k dh fr;Zd js k dgykrh gSA  Linear pair angle are supplementry.  Angle: inclination between two lines is called angle. 3  Geometry dks.k% nks js[kkvksa ds chp ds >qdko dks dks.k dgrs gSaA Types of Angles (dks.kksa ds çdkj) ABC =  A A  Acute Angle @ U;wu dks.k 0° <  < 90° B C B C A  Adjacent angles/vklUu dks.k  Right Angle @ ledks.k 90° AB  BC =90° Two angles are said to be adjacent if B C  Obtuse Angle @ vf/d dks.k A D Common A 1 90° 15 c2 = a2 + b2 r OR 7+15 > 12 OR 12 + 15 > 7 A III. Obtuse Angle Triangle @ vf/d dks.k f=kHkqt Si c b A c b B a C 1. Sum of any two sides is always greater than 3rd C a B side. ap fdUgha Hkh nks Hkqtkvksa dk ;ksx ges'kk rhljh Hkqtk ls cM+k gksrk gSA a+b > c C = largest @ C = lcls cM+k dks.k side c = largest @ Hkqtk c ¾ lcls cM+k c2 > a2 + b2 n t b+c > a io  sides of triangle : 11.7, 16.9, 23.4. which type of  c+a > b ra it is? at 2. Difference of any two sides is always less than f=kHkqt dh Hkqtk,¡ % 11.7, 16.9, 23.4. ;g fdl çdkj dk  gS\ ic 3rd side. Take ratio of sides 11.7 : 16.9 : 23.4 bl fdUgha nks Hkqtkvksa dk varj lnSo rhljh Hkqtk ls de gksrk gSA P 9 : 13 : 18 Pu  b > c–a 182 > 92 + 132  is obtuse angle triangle. b > a–c Pythagoras Triplets (ikbFkkxksjl f=kXkq.k@f=kd) on an   |c–a| < b < c+a A  pi  If 10, 17, x are sides of a , x integer c b am Then 7 < x < 27  x {8, 9, 10,.... 26} B a C g Ch 2 2 2 xmin = 8, xmax = 26 b =c +a (3,4,5), (5,12,13), (7,24,25), Ga xtotal = 19 values possible @ 19 eku laHko (8,15,17), (9,40,41), (11,60,61),  19 's possible @ 19 f=kHkqt laHko gSA (12,35,37), (16,63,65), (13,84,85), Possible values of x = 2×small side –1 (20,21,29), (28,45,53), (33,56,65),  2×10–1 = 19 (39,80,89), (36,77,85), (65, 72, 97), x ds laHkkfor eku ¾ 2×NksVh Hkqtk –1  2×10–1 = 19 (20, 99, 101) Relation between 3 sides of Triangle multiplication and division on these triplets will (f=kHkqt dh 3 Hkqtkvksa ds chp laca/) also result in triplets. I. Acute Angle Triangle @ U;wu dks.k f=kHkqt bu f=kd ij xq.kk vkSj Hkkx dk ifj.kke Hkh f=kd gksxkA (5,12, 13)  2 (10, 24, 26) A (3,4,5) (6,8,10), (9,12,15), (12,16,20), (15,20,25) b c Ex: 1 ? 3.5 3 C a B 12 3 7  Geometry 7 24 25  Vertically Opposite Angle @ ('kh"kkZfHkeq[k dks.k) ÷2 ÷2 ÷2 D A 3.5 12 12.5  3.5 3 , 12 3 , 12.5 3 Ex: 2 180– 180– 14 ? B C 3  2 = 13 2 2  Some other properties @ dqN vU; xq.k 21 7 ×2 7 ×3 7 × 13 = 7 13 Ex: 3 42 3 6  7  7  3  6 7  21 ++ = 3×360° – 180° = 900° ? r  A Si 18 7 6 7 3 3rd side = 6 7 × 21  9 = 6 7 × 12 = 12 21 B C 1 2 Ex: 4 1 + 2 = 180° + A ap 9.6 ? 9.6 : 18 ×1.2  If angles of a  are in A.P., middle angle is always 8 : 15 1.2 20.4 : 17  60°/ (;fn  ds dks.k lekarj Js.kh esa gS]a rks eè; dks.k ges'kk 60° 18  Exterior angle is equal to sum of opposite interior gksrk gS) n t angles. (a–d), a, (a+d) io  a–d+a+a+d = 180° ra ckgjh dks.k foijhr vkarfjd dks.kksa ds ;ksx ds cjkcj gSA at 3a = 180° A+B+C = 180° ic a = 60° A+B = 180° – C bl  A + C = 120° & B = 60° P A Pu A B C exterior angle at vertex C    60°– 60° 60°+ on an B C  D C pi sum of all exterior angles = 360° y° z° am lHkh ckgjh dks.kksa dk ;ksx = 360° x° A g B Ch Angle Bisector (dks.k f}Hkktd) B (internal @ vkarfjd) = 360° – (x+y+z) Ga  A E B (external @ ckgjh) = x°+y°+z° D  A C 36° B B 21° BE exterior angle bisector of ABC C x 19 BE ABC dk cká dks.k lef}Hkktd ° 2 + 2 = 180° D + = 90° = EBD  Angle between internal angle bisector and x° = 36° + 21° + 19° = 76° external angle bisector of an angle is 90°.  D C fdlh dks.k ds vkarfjd dks.k lef}Hkktd vkSj cká dks.k lef}Hkktd x°x2 b° x1 ds chp dk dks.k 90° gksrk gSA y2 BD is interior angle bisector of ABC a° y1 y° BD, ABC dk vkarfjd dks.k lef}Hkktd gS A B a+b = x+y 8  Geometry x1 + y1 = a° A x2 + y2 = b° x1 + x2 +y1+y2 = a+b h1 o x+y = a+b h2 h3  Altitude / Height / Perpendicular B C 'kh"kZ&yac @ ÅapkbZ @ yacor O = circumcentre @ ifjdsUæ The perpendicular drawn from the vertex of the Cevian (dsfo;u) triangle to the opposite side.  Cevian Any line which joins vertex to opposite f=kHkqt ds 'kh"kZ ls foijhr fn'kk esa hapk x;k yacA side. A dsfo;u dksbZ Hkh js k tks 'kh"kZ dks foijhr Hkqtk ls tksMr+ h gS A A Position of cevian r h h Si B C C D B D B C D E  A line that splits an angle into two equal angles. ap ,d js k tks ,d dks.k dks nks cjkcj dks.kksa esa foHkkftr djrh gSA AD, AE are cevians @ AD, AE dsfo;u gSa  ABC is scalene @ ABC fo"keckgq  gS A A n t io ra at B D E F C B C D AC > AB ic AD is the angle bisector of BAC, BD and DC need   B > C bl P not be equal  AD will be near to largest among B and C i.e Pu AD] BAC dk dks.k lef}Hkktd gS vkSj BD, DC dk cjkcj gksuk angle B and far from small angle C. vko';d ugha gS  AD, B vkSj C dks.k esa ls lcls cM+s B ds fudV gksxk vkSj on an Median (ekfè;dk) NksVs dks.k C ls nwj gksxkA pi  Line drawn from a vertex to opposite side which AE Angle bisector of A am divides the opposite side into equal parts. AE A dk dks.k f}Hkktd g fdlh 'kh"kZ ls foijhr fn'kk esa haph xbZ js k tks foijhr Hkqtk dks AF median i.e. BF = FC Ch leku Hkkxksa esa foHkkftr djrh gSA AF ekfè;dk ;kuh BF = FC Ga A  AD Altitude @ AD ÅapkbZ AE Angle bisector of A AE A dk dks.k f}Hkktd B C A D AD is the median of side BC. @ AD Hkqtk BC dh ekfè;dk gS BD = DC Perpendicular bisector (yac f}Hkktd) B D E C  A Perpendicular bisector is a line that bisects a line segment in two equal parts and makes an B – C DAE = angle of 90° at the point of intersection. 2 yac lef}Hkktd ,d js k gS tks ,d js k aM dks nks cjkcj Hkkxksa esa A DAE = – 90° + B foHkkftr djrh gS vkSj çfrPNsnu fcanq ij 90° dk dks.k cukrh gSA 2 9  Geometry A A B C  A = – – – +B 2 2 2 2 A B C B C  + + =90° 2 2 2 O B C B –C = – = A 2 2 2 BOC = 90° – 2  A+B+C = 180° A = 2(90°–BOC) A  In any quadrilateral bisector of A & B meet at P. fdlh prqHkqZt esa A o B ds lef}Hkktd P ij feyrs gSA C D r I P Si 2 C/ Q B/ 2 S B/2 C/2 R B C A B +C 180° – A B A = 180° – (B + C)  = C+ D ap 2 2 APB = 2  B  C  BIC = 180° –    A B  2  APB = 180° –     2 2 n t  180°  A  A C D = 180° –   = 90° + io APB =  ra  2  2 2 2 at A + B + C + D = 360° ic A  BIC = 90° + A B C D 360 2 bl    = =180° 2 2 2 2 2 P  Pu A  A C D B  180° –  2  2  =    2 2 on an Bisector of C and D meet at R pi I A + B am  DRC = 2 g B C P + R = 180° Ch S + Q = 180° Ga  PQRS will be a cyclic quadrilateral. PQRS ,d pØh; prqHkZqt gksxkA O  A E A BIC = 90° + 2 2  A D BOC = 90° – 2 AIO will be a straight line and bisect angle A. AIO ,d lh/h js k vkSj dks.k A f}Hkkftr gksxkA B C F BIC + BOC = 180° BICO will be a cyclic quadrilateral. A BEC = BICO ,d pØh; prqHkqZt gksxk 2 10  Geometry Area side properties (f=kHkqt dk {ks=kiQy) Area of triangle (f=kHkqt dk {ks=kiQy) h sinC =  h = bsinC 1 b  Area of  = × base × height 2 c b CsinB = bsinC  = 1 sinC sinB  dk {ks=kiQy = × vkèkkj × Å¡pkbZ 2 1 1 1 = acsinB = absinC = bcsinA A 2 2 2 r Side-Angle ratio of some triangles c b a+b+c s= (dqN f=kHkqtksa dk Hkqtk&dks.k vuqikr) Si 2 B a C  B Area of  = s s  a  s  b  s  c  45° 2 ap  In any ABC, AD  BC 1 A 45° C A n O t 1 io ra at 45° 45° 90° B C    ic D AB + OC2 = OB2 + AC2 2 bl sides  1: 1: 2 P Pu O is any point on altitude @ O ÅapkbZ ij dksbZ fcanq gS Sine Rule (T;k fu;e) B on an A 60°  pi a c am c b h g 30° Ch C b A Ga B C D a a:b:c = sin30° : sin60° : sin90° a b c = = = K (constant) 1 3 sinA sinB sinC : :1 2 2 a:b:c = KsinA : KsinB : KsinC a:b:c=1: 3 :2 a:b:c = sinA : sinB : sinC 1 B Area of ABC = × base × height 2 75° 1 Area of  = ×a×h a c 2 1 = ×a×csinB 15° 2 A C b h sinB =  h = CsinB c a:b:c= 3 –1 : 3 +1 : 2 2 11  Geometry C Length of Angle bisector (dks.k f}Hkktd dh yackbZ)  CD is angle bisector of BCA b 120° a CD, BCA dk dks.k lef}Hkktd gS C 30° 30° A c B a:b:c=1:1: 3 a b x Cosine Rule (dksT;k fu;e) B A  A m D n C x2 = ab – mn c b Exterior Angle bisector theorem r (ckã dks.k lef}Hkktd çes;) Si B a C A b2 + c 2 – a 2  cosA =  a2 = b2+c2 –2bc.cosA  AB DB 2bc  = ap AC DC c 2 + a 2 – b2 B cosB =  b2 = a2+c2 –2accosB C D 2ca  A a 2 + b2 – c 2 n cosC =  c2 = a2+b2 –2abcosC t io 2ab ra at Stewarts Theorem (LVhoVZ çes;)  C ic  180- B C bl D P Pu a b 1 x ArADB  AD  BD  sin  BD C=m+n 2 = = ArADC 1 DC on  AD  DC  sin 180    an B m n A 2 pi C  ftl ratio esa cevian base dks divide djsxh] Area Hkh a2n+b2m=x2c+mnc am mlh ratio esa divide gksxkA In isosceles triangle a = b If AD is median BD = DC g a2n+a2m = x2c + mnc Ch  Ar ADB = Ar ADC a2(m+n) = c(x2+mn) Ga  C D a2 = x2+mn x2=a2–mn Interior Angle bisector theorem (vkarfjd dks.k lef}Hkktd çes;) A AB BD  = AC DC A B   Ar ABC = Ar ABD AB  CD If AB  CD, same parallel line ds chp same base ij B C cus  dk Area cjkcj gksrk gSA D 12  Geometry Similarity of triangles (f=kHkqt dh le:irk) (~)  Similarity of triangles : Two triangles are similar  If two angle is same in a triangle then third angle if they have the same ratio of corresponding sides will be similar. and equal pair of corresponding angles. ;fn ,d f=kHkqt esa nks dks.k leku gSa rks rhljk dks.k Hkh leku gksxkA f=kHkqtksa dh le:irk % nks f=kHkqt le:i gksrs gSa ;fn mudh laxr A = D corresponding Hkqtkvksa dk vuqikr leku gks vkSj laxr dks.kksa dk ;qXe leku gksA B = E Angles C = F  Similarity of triangles : size may be different but shape should be same. A r f=kHkqtksa dh le:irk % vkdkj fHkUu gks ldrs gSa ysfdu vkÑfr D Si leku gksuh pkfg,A 50° 50° ~ 70° 60° 70° 60° E F ~ B C ap sides opposite to corresponding angles is called corresponding sides. laxr dks.kksa dh lEeq Hkqtk,¡ laxr Hkqtk,¡ dgykrh gSaA n t 4 4 ~ 2 2 io ABC  DEF ra at BC AC AB 2    (Property) ic 4 EF DF DE bl  Conditions of Similarity (le:irk dh 'krsZ)%&  In similar triangle ratio of each corresponding P length is equal. Pu 1. A-A (Angle-Angle) (dks.k&dks.k) le:i f=kHkqt esa çR;sd laxr yackbZ dk vuqikr cjkcj gksrk gSA ABC ~ PQR on an BC AC AB h1 Angle bisector1 median1    A P EF DF DE h2 = Angle bisector2 = median2 pi am r1 R1 perimeter of ABC = r = R = perimeter of DEF g B CQ R 2 2 Ch 2. S-S-S(Side-Side-Side) (Hkqtk&Hkqtk&Hkqtk) Ga 1 ×BC×h1 2 2 2 AB = PQ, BC = QR, AC = PR Area of ABC 2  BC   AC   AB  = = = = Area of DEF 1 ×EF×h2  EF   DF   DE  A P 2 = Ratio of square of corresponding length. B C q R laxr yackbZ ds oxZ dk vuqikrA Thales Theorem (FksYl çes;) 3. S-A-S(Side-Angle-Side) (Hkqtk&dks.k&Hkqtk)  If a line (DE) is drawn parallel to one side (BC) of B = Q AB = PQ, QR = BC triangle then it will divide other two sides in the A P same ratio. Hence AD : DB = AE : EC ;fn f=kHkqt dh ,d Hkqtk (BC) ds lekarj ,d js k (DE) haph tk, rks og vU; nks Hkqtkvksa dks leku vuqikr esa foHkkftr djsxhA vr% B C q R AD : DB = AE : EC 13  Geometry DE  BC ;fn D, AB dk eè; fcanq gS vkSj DE BC gS rks E, AC dk eè; fcanq gksxkA A Similar figures (le:i vkÑfr;k¡)   D E 90- 90-  90- B C   90-  ADE  ABC AD AE DE A = = r AB AC BC If AD : DB = 8:5 E  Si D area ADE 82 64 then  2  area ABC 13 169  ArADE 64 64 ap = = B C Ar DECB 169–64 105 A = common Convergence of thales theorem ABC = ADE =  (given) FksYl çes; dk vfHklj.k  3rd angle will also be equal @ rhljk dks.k Hkh cjkcj gksxk n t If D & E two points on AB and AC such that io  ABC  ADE ra AD AE at = then DE  BC  If we make a right angle triangle in an right angle DB EC triangle then big and small right triangle  are ic Mid point theorem (eè; fcanq çes;) always similar. bl P  The line segment in a triangle joining the mid ;fn ge ,d ledks.k f=kHkqt esa ,d ledks.k f=kHkqt cukrs gSa rks cM+k Pu points of two sides of triangle will be parallel to its vkSj NksVk ledks.k f=kHkqt ges'kk le:i gksrs gSaA third side and is also half of the length of third on an side. A pi f=kHkqt dh nks Hkqtkvksa ds eè; fcanqvksa dks feykus okyk js k aM mldh rhljh Hkqtk ds lekarj gksxk vkSj rhljh Hkqtk dh yackbZ dk vk/k Hkh gksxkA 90- am A D g 1 Ch ABC  EDC Ga 2 E 90-  D C B E 1  A B C  D, E mid points, AD = DB & AE = EC DE  BC BC ADE ABC, DE = 2 Ar ADE : Ar ABC = 1 : 4  B C Ar ADE : Ar DECB = 1 : 3 D  Convergence of mid point theorem C = common eè; fcanq çes; dk vfHklj.k A = ADC =  (given) If D is mid point of AB and DE BC then E will be  3rd angle will be equal ABC = DAC mid point of AC.  ABC DAC 14  Geometry  A B AO A1 A 4     OC A3 A2 O A1×A2 = A3×A4 A1 A4 =   A3 A2 D C A1, A 2, A 3 and A 4 are the areas of respective AB  CD triangles. AOB  COD A1, A2, A3 and A4 Øe'k% f=kHkqtkas ds {ks=kiQy gSA Proof:-  D C r y x 14 O 10 Si h1 h2 A B x and y is the heights of triangles. h1 10  h2 14 h1 : h2  5 : 7 ap Area of  = 1 2 × Base × height n 1 t Area of triangle A1 = × AO × x...(1) B io 2 ra at  1 Area of triangle A2 = × OC × y...(2) ic 2 D bl x Q P 1 Pu y Area of triangle A3 = × CO × x...(3) 2 Z A C P on 1 an Area of triangle A4 = × AO × y...(4) 2 pi AB  PQ  CD Multiply eq (1) and eq (2) am xy Z = x y 1  1  A1 × A2 =   AO  x    CO  y  g 2  2 Ch  z CP z AP By commutative property Ga  & y  AC x CA 1  1  z z CP + AP z z AC A1 × A2 =   CO  x    AO  y       =1 2  2  x y AC x y AC A1 × A2 = A3 × A4 Hence proved 1 1 1 Alternatively:- + = (Result) x y z D C In any quadrilateral (fdlh prqHkZqt esa) A3 – b 80° c 1 C A1  A2  D O A3 A4 a d A1 A2 O A4 A B [sin (180° – ) = sin] A B 15  Geometry 1 (In a trapezium the triangle formed on non-parallel A1 =  a  b sin...(1) sides have equal area) 2 (,d leyac prqHkwZt eas] vlekarj Hkqtkvkas ij cus  dk {ks=kiQy 1 cjkcj gksrk gSA) A2 = × c × d sin...(2) 2 K×K=a×b 1 K= A3 = × b × c sin...(3) ab 2 Medial Triangle (eè; dk f=kHkqt) 1 A4 = × a × d sin...(4)  A 2 [Multiply eq. (1) and (2) or (3) and (4)] P Q 1 1 × abcd sin2 = × abcd sin2 r 2 2 A1 × A2 = A3 × A4 Hence proved B C Si R In a trapezium (,d leyEc prqHkqZt esa) P, Q, R are mid points @ P, Q, R eè; fcanq gSa  D C a 1 Perimeter of  PQR = × perimeter of ABC 2 ap k k O 1 b Area of  PQR = × Area of ABC 4 n t A B io ra Ar  ADB = Ar  ACB at ic Common Area = AOB bl AOB ~ COD P Pu  Ar  AOD = Ar BOC on an pi am g Ch Ga 16  Geometry Congruency of triangle (f=kHkqt dh lok±xlerk) Congruency of triangle (f=kHkqt dh lok±xlerk) (iv) RHS (Right angle-hypotenuse-side)  Two triangles are called congruent if all three RHS (ledks.k&d.kZ Hkqtk) corresponding sides are equal and all the three A D corresponding angles are equal.  nks f=kHkqt lok±xle dgykrs gSa ;fn rhuksa laxr Hkqtk,¡ cjkcj gksa vkSj rhuksa laxr dks.k cjkcj gksaA B C E F  Congruency of triangle size and shape is same ADB  ADC r ABC  DEF AD  angle bisector of A @ AD dks.k dk  f}Hkktd Si f=kHkqt dh lok±xlerk vkdkj o vkÑfr leku gksrh gSA AD Median (ekfè;dk) ABC  DEF AD  bisector of BC A D AD All 4 centres lie on AD. ap A  90- - B C E F 90 n t Condition of congruency (lok±xlerk dh 'krZ) I io H G ra (i) SSS (side-side-side) Corresponding sides are at O equal. B C ic D SSS (Hkqtk&Hkqtk&Hkqtk) laxr Hkqtk,a cjkcj gksA bl  P P A D Pu  on an B C E F pi A B (ii) SAS (side-angle-side) Two side and angle C am between them is equal. g SAS (Hkqtk-dks.k-Hkqtk) Ch nks Hkqtk,a vkSj muds chp dk dks.k PCA  PCB cjkcj gksrk gSA PA = PB Ga P is any point A D Mass point geometry (æO;eku fcanq T;kfefr)   l1 l2 B C E F m1 m2 (iii) ASA (Angle-side-angle) Two angle and side between them is equal. Center of mass(nzO;eku dsUnz) ASA (dks.k&Hkqtk&dks.k) nks dks.k vkSj muds chp dh Hkqtk m1l1 = m2l2 cjkcj gksrh gSA m1 l 2 = m2 l1 A D  5m 7.2     B C E F 48 kg x kg Find x = ? 17  Geometry 48 7.2 100 A  x= kg = 33.3 kg x 5 3 7.8 m F E l1 l2 O 45 kg 72 kg B C D Find l1 = ? m 45 : 72 AF BD CE × × =1 5 : 8 FB DC EA l 8 : 5 OD OE OF   1 r ×0.6 AD BE CF 4.8 m Si AO BO CO   2  13  7.8 0.6 OD OE OF Ceva's Theorem (lsok dh çes;)  In ABC, AD, BE and CF are the cevians i.e. any ap line from vertex to opposite side. f=kHkqt ABC eas, AD, BE vkSj CF dsfo;u gSa ;kuh 'kh"kZ ls foijhr Hkqtk ij dksbZ js kA n t io ra at ic bl P Pu on an pi am g Ch Ga 18  Geometry Centre of Triangle (f=kHkqt ds dsUæ) Centres of Triangle P, AOB ds dks.k f}Hkktd ij dksbZ fcanq gS  PR = PQ  ID = IE = IF = r, ADI  AFI Incentre Orthocentre (H) A (I)  BIC = 90° + Centroid 2 Circumcentre (C) (G) B AIC = 90° + 2 r Incentre (I) (vUr%dsUæ) C AIB = 90° + Si  Incentre is the intersection point of all three 2 internal angle bisectors of ABC. abc  In ABC = Semi-perimeter (s) vUr%dsaæ ABC ds lHkh rhu vkarfjd dks.k lef}Hkktdksa dk 2 çfrPNsnu fcUnq gSA  Area ABC =  = r × s ap Δ r= A s Area Inradius = Semiperimeter n t D io ABC = BIC + AIC + AIB (Area) ra at c r F r a  b  c 1 1 1 ic I b = ar + br + cr = r   =r×s 2 2 2  2  bl r Incircle P  If altitudes h1, h2, h3 are given then Pu B E ;fn Å¡pkbZ h1, h2, h3 nh xbZ gks rks C a on an 1 1 1 1 = + + (Result) r h1 h 2 h 3 pi Centre of incircle is called incentre and its radius is called inradius (r)  A am var%o`Ùk ds dsaæ dks var%dsaæ vkSj bldh f=kT;k dks var%f=kT;k (r) dgk g Ch tkrk gS c b F E Ga  Incentre always lies inside the triangle. I var%dsaæ ges'kk f=kHkqt ds vanj fLFkr gksrk gSA B C  Incentre is equidistant from all three sides of D triangle. a var%dsUæ f=kHkqt ds rhuksa vksj ls lenwjLFk gSA AI b  c  ID a A BI c  a  IE b R CI a  b  IF c P Proof:- AI AC b In ACD,  ...(i) B ID CD CD O Q AB BD P is any point on angle bisector of AOB In ABC,  AC CD 19  Geometry c BD P+B – H 