Maths Concept King Book PDF

 Symbol Table of Symbols Appendix implication Symbol Reference...

 Symbol Table of Symbols Appendix implication Symbol Reference negation, equivalence, relation = equal to , quantifier not equal to {} set identity empty set, void set, null set approximately equal to Conversion of Units congruent to Conversion of Length r approaches, ray 10 millimetres (mms) = 1 centimetre (cm) proportional to 10 centimetres = 1 decimetre (dm) Si < less than 10 decimetres = 1 metre (m) not less than 10 metres = 1 decametre (dam) greater than 10 decametres = 1 hectometre (hm) not greater than 10 hectometres = 1 kilometre (km) p less than or equal to Conversion of Area 100 square millimetres = 1 square centimetre greater than or equal to ta 100 square centimetres = 1 square decimetre > much greater than 100 square decimetres = 1 square metre io 100 square metres = 1 square decametre a infinity at 100 square decametres = 1 square hectometre or sigma ic 100 square hectometres = 1 square kilometre Pr % percentage bl 1 hectare = 10000 square metres + plus, positive Conversion of Volume Pu – minus, negative 1000 cubic millimetres = 1 cubic centimetre plus or minus 1000 cubic centimetres = 1 cubic decimetre on n a b 1000 cubic decimetres = 1 cubic metre pi multiplication a b 1000 cubic metres = 1 cubic decametre ga am 1000 cubic decametres = 1 cubic hectometre a b 1000 cubic hectometres = 1 cubic kilometre division Ch a / b Conversion of Capacity Ga therefore 10 millilitres = 1 centilitre since 10 centilitres = 1 decilitre — line segment 10 decilitres = 1 litre acute angle 10 litres = 1 decalitre perpendicular 10 decalitres = 1 hectolitre parallel 10 hectolitres = 1 kilolitre Conversion of Weight triangle 10 milligrams = 1 centigram rectangle 10 centigrams = 1 decigram square 10 decigrams = 1 gram (g) logba logarithm (to base b) 10 grams = 1 decagram log10a common logarithm 10 decagrams = 1 hectogram loge a or ln a natural logarithm 10 hectograms = 1 kilogram (kg) conjunction (and) 100 kilograms = 1 quintal disjunction (or) 10 quintals or 1000 kg = 1 metric tonne 1  Symbol Conversion of Time Equivalents of Units 60 seconds = 1 minute Units of Lengths 60 minutes = 1 hour 12 inches = 1 feet (ft) = 0.348 metres 24 hours = 1 day 3 feet = 1 yard (d) 7 days = 1 week 1 yard = 0.9144 metres 15 days = 1 fortnight 22 yards = 1 chain 28, 29, 30 or 31 days = 1 month 1 kilometre = 0.621 mile or 103 metres 12 months = 1 year 1 mile = 1.6093 kilometres or 1760 yards 365 days = 1 year 1 inch = 2.54 centimetres 366 days = 1 leap year 1 hectare = 2.471 acres 10 years = decade Units of Area 25 years = silver jubilee 1 square feet = 144 square inches 50 years = golden jubilee = 0.0929 square metres r 60 years = diamond jubilee 1 square metre = 1.196 square yards Si 75 years = radium jubilee or 1 square yard = 0.836 square metres platinum jubilee 1 square kilometre = 0.3861 square miles 100 years = century = 1000 hectares 1000 years = 10 centuries or 1 millennium 1 square mile = 2.59 square kilometres p = 640 acres ta 1 acre = 4840 square yards = 4046.86 square metres 100 square metres = 1 acre n io 100 acres = 1 hectare a at = 10,000 square metres ic Pr bl Pu on n pi ga am Ch Ga 2  Geometry Geometry (Line & Angle) Line and Angle Supplementary angle of an angle is 90° more than complementary angle. Point: Zero dimension figure or a circle with zero 90° radius. Transversal Line : A line which inte rests (touches) two or more lines at distinct point is Line: One dimension figure line is a set of points called transversal lines of the given lines. having only length with no ends. _________ r _________ Parallel lines: two or more line that never Si Line segment: A line with a fixed length. intersects L M P Q L M L p Ray: A line with uni-direction length. A B d d M Angle: inclination between two sides is called ta AB CD and EF is transversal line n angle. AB CD EF io a E at ABC = 2 1 A B ic 3 4 Pr A bl B 6 5 Pu C C D Complementary Angle : If sum of two angles is 7 8 90º then they are Complementary to each other. F on n 90º Corresponding angles 1 = 5, 4 = 8 pi 2 = 6, 3 = 7 ga am Alternate Angles 3 = 5, 4 = 6 Complementary Angle/ + = 90° 4 + 5 = 180° Ch 3 + 6 = 180° Ga Supplementary Angle : If sum of two angles is If AB CD then find the value of + + ? 180º then they are supplementary to each other. AB CD + + 180º A B Supplementary Angles/ + = 180° O C D Angle Complementary Supplementary + + = 360° A a B 43° +90 c 47° 137° C b 12° 78° +90 168° x 90° – +90 180° – D b=a+c+x 3  Geometry A B LHS RHS A a D m C D B E Sum of angle on RHS = LHS RHS = LHS n C F + = + b A E AB DE m a p = = BC EF n B F an+bm b q BE = C G m+n r c r Si D H a:b:c=p:q:r p a p = p+q+r a+b+c ta n io a at ic Pr bl Pu on n pi ga am Ch Ga 4  Geometry Types of Triangles Triangle Scalene Triangle A triangle is a 3-sides polygon that consists of three A edges and three vertices. c b 3 B a C three unequal sides 3 sides, 3 vertices, 3 altitudes, 3 angles three unequal angles 3 3 3 3 A B C & a b c r A By Angle Si Acute Angle Triangle h1 c b h2 h3 B a C p A+ B+ C=180° ta All three angles < 90° < 90° Right Angle Triangle 1 1 1 1 Area ×a×h 1 = bh 2 = ch 3 = × Base × A 2 2 2 2 n Corresponding height. io a at ah1= bh2= ch3 = constant B C ic 1 1 1 Pr h1 : h2 : h3 = : : One angle is 90°. B = 90° and A + C = 90° a b c bl Obtuse Angle Triangle Pu Type of Triangle One angle > 90° > 90° on n By side By Angle pi Let B = largest angle B ga am By side side AC = b = largest side Equilateral Triangle AC = b = Ch A Ga c b B a C Equilateral triangle has 3 equal sides, each angle Let C = smallest angle C = 60° side AB = c = smallest side 60° 3 AB = c = Isosceles Triangle Inequality of triangle The triangle inequality states that for any triangle the sum of the lengths of any two sides must be greater than the length of the remaining side. two equal sides two angle same 5  Geometry |b–c| < a < b+c II. Right Angle Triangle A A c b b c B a C C a B |a–c| < b < a+c C = largest C = |a–b| < c < a+b side c = largest c Inequality of Triangle c =a +b 2 2 2 eg 4, 9, 15 not possible 4 + 9 < 15 III. Obtuse Angle Triangle 5, 10, 15 not possible 5 + 10 = 15 A 7, 12, 15 is possible 7+12> 15 c b r OR 7+15 > 12 OR 12 + 15 > 7 A C a B Si c b C = largest C = B C side c = largest c a Sum of any two sides is always greater than 3rd c2 > a2 + b2 p side. sides of triangle : 11.7, 16.9, 23.4. which type of ta it is? a+b > c b > c–a 11.7, 16.9, 23.4. n b+c > a b > a–c Take ratio of sides 11.7 : 16.9 : 23.4 c+a > b |c–a| < b < c+a io 9 : 13 : 18 a at Difference of any two sides is always less than 18 > 9 + 13 is obtuse angle triangle. 2 2 2 ic 3rd side. Triplets Pr bl A Pu If 10, 17, x are sides of a , x integer c b Then 7 < x < 27 B a C on n x {8, 9, 10,.... 26} b =c +a 2 22 xmin = 8, xmax = 26 pi (3,4,5), (5,12,13), (7,24,25), ga xtotal = 19 values possible 19 am (8,15,17), (9,40,41), (11,60,61), 19 's possible 19 (12,35,37), (16,63,65), (13,84,85), (20,21,29), (28,45,53), (33,56,65), Ch Possible values of x = 2×small side –1 (39,80,89), (36,77,85), (65, 72, 97), Ga 2×10–1 = 19 (20, 99, 101) x 2× –1 2×10–1 = 19 multiplication and division on these triplets will Relation between 3 sides of Triangle also result in triplets. 3 I. Acute Angle Triangle (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 C = largest C = 7 24 25 side c = largest c ÷2 ÷2 ÷2 c < a +b 2 2 2 3.5 12 12.5 3.5 3 , 12 3 , 12.5 3 6  Geometry Ex: 2 Angle Bisector A 14 ? E D 32 22 = 13 21 C B 7 ×2 7 ×3 7 × 13 = 7 13 BE exterior angle bisector of ABC Ex: 3 BE ABC 2 + 2 = 180° 42 3 6 7 7 3 6 7 21 ? + = 90° Angle between internal angle bisector and 18 7 6 7 3 external angle bisector of an angle is 90°. 3rd side = 6 7 × 21 9 = 6 7 × 12 = 12 21 90° r Ex: 4 BD is interior angle bisector of ABC Si 9.6 ? 9.6 : 18 BD, ABC ×1.2 1.2 20.4 8 : 15 : 17 Vertically Opposite Angle 18 Exterior angle is equal to sum of opposite interior D A p angles. ta 180– 180– A+B+C = 180° B C A+B = 180° – C n Some other properties A io a at exterior angle at vertex C ic Pr B C bl sum of all exterior angles = 360° Pu + + = 3×360° – 180° = 900° = 360° A Types of Angles on n A pi B C ga Acute Angle 0° < < 90° am 1 2 B C 1 + 2 = 180° + A A Ch Right Angle AB BC =90° If angles of a are in A.P., middle angle is always 90° Ga B C 60° A.P. 60° Obtuse Angle (a–d), a, (a+d) a–d+a+a+d = 180° A 3a = 180° 90°< C AD is the angle bisector of BAC, BD and DC need not be equal AD will be near to largest among B and C i.e angle B and far from small angle C. AD BAC BD, DC AD, B C B C Median AE Angle bisector of A Line drawn from a vertex to opposite side which divides the opposite side into equal parts. AE A AF median i.e. BF = FC AF BF = FC 8  Geometry AD Altitude AD In any quadrilateral bisector of A & B meet at P. AE Angle bisector of A A B P AE A D C A P S Q R A B B D E C C+ D APB = 2 B – C DAE = A B 2 APB = 180° – 2 2 A r C D DAE = – 90° + B APB = 2 2 2 Si A + B + C + D = 360° A A B C = – – – + B A B C D 360 2 2 2 2 = =180° 2 2 2 2 2 A B C + + =90° p 2 2 2 A C D B 180° – 2 2 = 2 2 B C B – C = 2 – 2 = 2 ta Bisector of C and D meet at R n A+B+C = 180° A + B io DRC = 2 a at A P + R = 180° ic Pr S + Q = 180° bl PQRS will be a cyclic quadrilateral. Pu I PQRS 2 C/ B/ 2 A on n B/2 C/2 B C pi ga B + C 180° – A am A = 180° – ( B + C) = 2 2 I Ch B C BIC = 180° – 2 B C Ga 180° A A = 180° – 2 = 90° + 2 A BIC = 90° + O 2 A A BIC = 90° + 2 A BOC = 90° – B C 2 AIO will be a straight line and bisect angle A. O AIO A A BIC + BOC = 180° BOC = 90° – 2 BICO will be a cyclic quadrilateral. A = 2(90°– BOC) BICO 9  Geometry Area side properties Area of triangle h sinC = h = bsinC 1 b Area of = × base × height 2 c b CsinB = bsinC = 1 sinC sinB × × 2 1 1 1 = acsinB = absinC = bcsinA A 2 2 2 Side-Angle ratio of some triangles r c b a+b+c s= 2 Si B a C B Area of = s s a s b s c 45° 2 In any ABC, AD BC p 1 A ta C 45° A O n 1 io 45° 45° 90° a at B C D ic Pr AB + OC2 = OB2 + AC2 2 sides 1: 1: 2 bl O is any point on altitude O Pu Sine Rule B 60° on A n a c pi ga c b am h 30° A Ch C b B C D a Ga 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 C b A h sinB = h = CsinB c a:b:c= 3 –1 : 3 +1 : 2 2 10  Geometry C Length of Angle bisector 120° CD is angle bisector of BCA a CD, BCA C 30° 30° A c B a:b:c=1:1: 3 a b x Cosine Rule B A A m D n C x2 = ab – mn c b Exterior Angle bisector theorem r Si B a C A b2 + c2 – a 2 cosA = a2 = b2+c2 –2bc.cosA 2bc p c2 + a 2 – b2 B cosB = b2 = a2+c2 –2accosB C D 2ca ta AB DB a 2 + b2 – c 2 = cosC = c2 = a2+b2 –2abcosC AC DC n 2ab Stewarts Theorem io 1 a Ar ADB AD BD sin at 2 BD = = Ar ADC 1 DC AD DC sin 180 ic C Pr 2 bl A Pu a b x C=m+n on n B m n A pi C ga 180- am a n+b m=x c+mnc 2 2 2 B C D In isosceles triangle a = b ratio cevian base divide Area Ch a2n+a2m = x2c + mnc ratio divide Ga a2(m+n) = c(x2+mn) If AD is median BD = DC a2 = x2+mn Ar ADB = Ar ADC x2=a2–mn Interior Angle bisector theorem C D A A B B C D Ar ABC = Ar ABD AB BD If AB CD, same parallel line same base = AC DC Area 11  Geometry Similarity of triangles (~) Similarity of triangles : Two triangles are similar if they have the same ratio of corresponding sides BC AC AB h1 Angle bisector1 median1 and equal pair of corresponding angles. EF DF DE h2 = Angle bisector2 = median2 r1 R1 perimeter of ABC = r = R = perimeter of DEF Similarity of triangles : size may be different 2 2 but shape should be same. 1 r ×BC×h1 2 2 2 Area of ABC 2 BC AC AB = 1 = = = Si Area of DEF ×EF×h2 EF DF DE 2 = Ratio of square of corresponding length. ~ p Thales Theorem If a line (DE) is drawn parallel to one side of ta triangle (BC) then it will divide other two sides in n the same ratio. Hence AD : DB = AE : EC 4 4 ~ 2 2 io (DE) (BC) a at 2 4 ic AD : DB = AE : EC Pr If two angle is same in a triangle then third angle bl DE BC will be similar. Pu A on A = D n corresponding B = E Angles E pi C = F D ga am A D Ch B C 50° 50° Ga ~ ADE ABC 70° 60° AD AE DE 70° 60° E F = = B C AB AC BC sides opposite to corresponding angles is called If AD : DB = 8:5 corresponding sides. Ar ADE 64 64 = = Ar DECB 169–64 105 ABC DEF Convergence of thales theorem BC AC AB (Property) EF DF DE If D & E two points on AB and AC such that In similar triangle ratio of each corresponding AD AE = then DE BC length is (equal) DB EC 12  Geometry Mid point theorem If we make a right angle triangle in an right angle triangle then big and small right triangle are The line segment in a triangle joining the mid always similar. points of two sides of triangle will be parallel to its third side and is also half of the length of third side. A 90- A 1 D ABC EDC 2 E D 90- C r B E 1 Si B C A D, E mid points, AD = DB & AE = EC DE BC BC p ADE ABC, DE = 2 Ar ADE : Ar ABC = 1 : 4 Ar ADE : Ar DECB = 1 : 3 ta B C D n Convergence of mid point theorem io C = common a at If D is mid point of AB and DE BC then E will be A = ADC = (given) ic 3rd angle will be equal ABC = DAC Pr mid point of AC. bl D, AB DE BC E, AC ABC DAC Pu A B Similar figures on n O pi ga 90- 90- am D C Ch 90- AB CD Ga 90- AOB COD A E D 10 14 h1 h2 B C A = common ABC = ADE = (given) h1 10 h2 14 3rd angle will also be equal ABC ADE h1 : h2 5 : 7 13  Geometry In a trapezium B D C a D k k x O Q y b Z A C P A B xy Ar ADB = Ar ACB Z = x y Common Area = AOB z CP z AP & y AC Ar AOD = Ar BOC x CA (In a trapezium triangle formed on non-parallel r z z CP + AP z z AC sides have equal area) Si =1 x y AC x y AC 1 1 1 + = (Result) K×K=a×b x y z K= p ab In any quadrilateral ta Medial Triangle D C A A3 n A1 A2 io a O P Q at A4 ic Pr B C bl R A B Pu P, Q, R are mid points P, Q, R A1×A2 = A3×A4 1 on Perimeter of PQR = × perimeter of ABC n A1 A4 2 = A3 A2 pi ga 1 Area of PQR = × Area of ABC am 4 Ch Ga 14  Geometry Congruency of triangle Congruency of triangle (iv) RHS (Right angle-hypotenuse-side) Two triangles are said to be congruent if all three RHS corresponding sides are equal and all the three corresponding angles are equal. Congruency of triangle size and shape is same ADB ADC r ABC DEF AD angle bisector of A AD Si AD Median ABC DEF AD bisector of BC A D AD All 4 centres lie on AD. A p 90- B C E F ta 0- n9 Condition of congruency I (i) SSS (side-side-side) Corresponding sides are io H G O a at equal. B D C SSS ic Pr P bl A D Pu on n B C E F pi A B (ii) SAS (side-angel-side) Two side and angle C ga am between them is equal. SAS - - PCA PCB Ch PA = PB Ga P is any point A D Mass point geometry l1 l2 B C E F m1 m2 (iii) ASA (Angle-side-angle) Two angle and side m1l1 = m2l2 between them is equal. ASA m1 l 2 = m 2 l1 5m 7.2 48 kg x kg Find x = ? 15  Geometry 5 x 100 Ceva's Theorem x= kg = 33.3 kg 7.2 48 3 In ABC, AD, BE and CF are the cevians i.e. any line from vertex to opposite side. 7.8 m l1 l2 ABC , AD, BE CF 45 kg 72 kg A Find l1 = ? m 45 : 72 F E 5 : 8 O l 8 : 5 B C ×0.6 D r 4.8 m Si 13 7.8 0.6 AF BD CE × × =1 FB DC EA p ta n io a at ic Pr bl Pu on n pi ga am Ch Ga 16  Geometry Centre of Triangle Centres of Triangle P is any point on angle bisector of AOB P, AOB PR = PQ Incentre Orthocentre ID = IE = IF = r, ADI AFI (I) (H) Circumcentre Centroid A (G) BIC = 90° + (C) 2 Incentre (I) B r AIC = 90° + 2 Incentre is the intersection point of all three Si internal angle bisectors of ABC. C AIB = 90° + ABC 2 a b c In ABC = Semi-perimeter (s) p 2 A ta Area ABC = = r × s Δ r= n D io Area a c r at r F Inradius = Semiperimeter I b ic Pr ABC = BIC + AIC + AIB (Area) r bl B 1 1 1 a b c Pu = ar + br + cr = r =r×s E C 2 2 2 2 a If altitudes h1, h2, h3 are given then on n Centre of incircle is called incentre and its radius h 1, h 2, h 3 pi is called inradius (r) ga 1 1 1 1 am = + + (Result) (r) r h1 h2 h 3 Ch A Incentre always lies inside the triangle. Ga c b Incentre is equidistant from all three sides of F E I triangle. B C D a A AI b c ID a R BI c a IE b P CI a b IF c B O Q 17  Geometry Point = I, O, G, H (lie at same place P) A AP = R, PD = r r 3 h = AD = a 2 P O r 3 a2 a h 4 B r= r= ,r= s 3a 2 3 3 2 PA = PB 2h a AOP BOP R= = 3 3 In right angle triangle R 2 Area of circumcircle 4 r 1 , r P-r Area of incircle 1 Si P-r A H x x P I B-r P R p 2r r r y z O r B B-r ta B y Q z C n Orthocentre io S= x+ y+ z a at P+B – H Area of ( ) ABC = (x y z )· xyz r= ic 2 Pr xyz (x + y + z) bl P–r+B–r=H r= = s x+y+z Pu P + B – 2r = H P B H xyz =r on r= n 2 x+ y+z Distance between orthocentre and incentre = OI pi In any quadrilateral ga = am 2r D b C = OI = 2r P Ch Sides (triplet) Radius (r) d c Ga (r) Q (3, 4, 5) r=1 (3k, 4k, 5k) r=k A a B (39, 52, 65) r = 13 (a + b) – (c + d) PQ = 2 For equilateral Triangle A A r2 r1 R a a P O B r r1 1 sin B C D r2 = 1 sin a 18  Geometry In an Equilateral Triangle A 30° 30° r3 r R 3 r B C r1 r2 1 1 r 1 sin 30 2 1 = = r R 1 sin 30 1 3 1 r= r1r2 r2 r3 r3 r1 2 Si r 1 A = R 3 Area small circle 1 p = Area large circle 9 E I A circile is inscribed in ABC. Three tangents PQ, ta a RS and TU are drawn of this circle PQ AB, RS n AC and TU BC. Three others incircle are also drawn as shown in figure find correct relation? B io a b D C at ABC ic Pr PQ, RS, TU PQ AB, BI = ab = 2r bl RS AC TU BC r = radius of ABC Pu BI ab r= = 2 2 on n A In Any Triangle pi r3 ga T U am A S Q Ch r Ga r2 r3 I D E C B R P C r = r1 + r2 + r3 B C Find relation between r, r1, r2, r3? DE BC r, r1, r2, r3 I = Incentre (I) DE = BD + EC 19  Geometry Circumcentre and Orthocentre Circumcentre Circumradius (R) (R) Intersection of all 3 perpendicular bisectors. A 3 Circumcircle C R b A o R A A R r R B a a a C Si 2 2 O a a R sin A = R = 2 sin A R 2R B C p ta x Centre of circumcircle is called circumcentre (O) n x (O) io R = 2sin a Radius of circumcircle is called circumradius (R) at (R) a b c ic Pr R = 2 sin A = 2 sin B = 2 sin C O is equidistant from all 3 vertex of ABC bl O, ABC 3 Pu a b c OA = OB = OC = R = sin B = sin C = 2R sin A Circumcentre may lie inside, outside or on the. on n 1 2 Area of triangle ( ) = bc sin A sin A = pi 2 bc ga O is the intersection of all 3 bisectors of sides am O 3