Government ofTamilnadu A publication under Free Textbook Programme of Government of Tamilnadu Department of School Education © Government of Tamilnadu First Edition - 2012 Revised Edition - 2013 (PublishedunderUniformSystemof SchoolEducationschemein TrimesterPattern) Textbookiareparatian andcomspilation College Road, Chennai - 600 006. TextbookPrinting 2 :3 College Road, Chennai - 600 006 This book has been printed on 80 GSM Maplitho Paper Price: Rs. Printedby Offsetat: Textbook available at wvvw.textbookson|ine.tn.nic.in co~Te~ws MATHEMATICS - (1-129) % , X scIENcE-(130-240) Vi} § T L VV TSOCIAL SCIENCE - (241-316) I VL L VHVi } V AM V _ _ * VL H W 3 mm .z.;'ET'f~§E d;zH;'2?E§"£ wx;s.V&V~ w "§/;§::1A~:$AW;wJ:"_ wzv z. mu .z.;'ET'f~§E d;zH;'2?E§"£ wx;s.V&V~ w "§/;§::1A~:$1«*F;wJ:"_ W 3% ;._ r. , . Lius MATH EMATICS STANDARD EIGHT 3% %.,\t.,. aw? 3% ;._ r. , . xm Elma 3.2 _ . e/%_mw»,«\_ Real Number System 1.1 Introduction 1.2 Revision : Representationof Rational Numbers 1.3 on the Number Line Four Properties of Rational Numbers 1.4 Simplication of Expressions Involving Three PaulW05 2°Se*°°"W 1.5 [26March,1913 Brackets Powers: Expressing the Numbers in Exponential Form with Integers as Exponent 1.6 Laws ofExponents withIntegral Powers 1.7 Squares, Square roots,Cubes, Cube roots 1.8 mathematician Approximations of Numbers in istory, working with hundreds r collaborators 1.1 Introduction n many elds ncluding number Number theory as a fundamental body of knowledge has played a . _ . pivotal role in the developmentof Mathematics. The Greek Mathematician 1 His fascination S heory. withmathematicsPythagorasand his disciplesbelievedthat everything is number and { eveloped early at he age of three. Hecouldcalculate ow many seconds person had that the central explanation of the universe lay in numbers. . The sY stem of writin g numerals was develo Ped some 10,000 Years ,1 ago. India was the main centre for the development of the number ived. HislifeWas system whichweusetoday.It tookabout5000yearsforthecomplete 1:: t d is; Number: A Portrait f Paul Erdos, whilehewasSun alive. Erdossaid,I knownumbers development ofthenumber system. TheWhole numbers arefountain head ofallMathematics. Thepresent __ ealltiflllaI10thing _ _ Inthissystem, weusethe_numbers 0,l, 2,3,4,5,6,. 7,8,9.It isalso _ called the decimal system with base 10. The word decimalcomesfrom Latin S. _ system of writing numerals 1Sknown as Hindu-Arabic numeral system. word Decem which means Ten. Mathematics is the Queen of Science and s 0 Number theory IS the Queen or l\/lathematics. Real Number class VII.wehave learntabout Natural numbers N R2, numbers W = {0, 1, 2, }, Integers Z = {--- , 2, -1, 0, 1, 2, S stem },Whole I } and Rational numbers Qand also the four fundamental operations onthem. 9I . .O o State Whether the following statementsare True Or False a) All Integers are Rational Numbers. b) All Natural Numbers are Integers. c) All Integers are Natural Numbers. d) All Whole Numbers are Natural Numbers. e) All Natural Numbers are Whole Numbers. f) All Rational Numbers are Whole Numbers. 1.2 Revision : Representation of Rational Numbers on the Number Line Rational numbers The numbers oftheform 7;Where pand qare integers and q75 0are known asrational numbers. Thecollection of numbers of theform3, whereq > 0 is 4;. 35 Q denotedby Q. Rationalnumbersinclude natural numbers, whole numbers, integers and all negative and positive fractions. Herewecanvisualizehowthegirl collectedall EnCirC1¬i%}¬COrr¬C{ typegmumbsr therational numbers inabag. Sysm 2» I»F El 3 TE «~<-~-+~-~»~~4:»-~+~-t-~»~-:»--+~~»-+~--~>» -3 V 2%- §0§§ 11 2 3 Rational numberscan also be representedon thenumber lineandhere wecansee apicture ofagirl walking onthenumber line. To expressrational numbers appropriately on the number line, divide each unit length intoasmany number ofequal partsasthedenominator oftherational number andthen mark thegiven number onthenumber line. (i)Express A onthenumber line. 7 571lies between 0and 1. Chapter 1 is a rational Is the converse number. true? 1.3 Four Properties of Rational Numbers 1.3.1 (:1) Addition (i) Closureproperty 1 The sumof any two rationalnumbersis alwaysa rationalnumber.This is calledi Closure propertyof addition of rationalnumbers. Thus,Q is closed underaddition. (i) 9 +A = 9 9 = 2 3 is arationalnumber. V (11) 5+%==f+% =15;1=-136: 5%isarationalnumber. (ii) Commutative property 1 Addition of two rational numbers is commutative. 12,%-wehave LHS=;+§ RHS.-=%+% ___5+4=__9_ 10 =__4+5=_9__ 10 10 LHS = RHS Commutative property is true for addition. (iii) Associative property Addition of rational numbers is associative. 10 Real Number S stem AForthreerationalnumbers A L and2, wehave 32 2 3 LHS=%+(%-+2) RHS=(%+%)+2 =%+(%+i> =(%+%>+2 =%+e+%>=%+% =4+15___%=3% ___7+12=%=3_é_ LHS = RHS Associative property is true for addition. (iv) Additive identity The sum of any rational number and zero is the rational number itself. (V) Additive inverse (La) ' thenegative oradditive inverse of%. b (i) Additive inverse ofg3--3isT (ii) Additive inverse of~'753 isQ; (iii) Additive inverse of 0 is 0 itself. Chapter 1 1.3.1 (1)) subtraction (1) Closure Property The difference between any two rational numbers is always a rational number. Hence Q is closed under subtraction. (i) %-~-~%=-2isarational number. 7 (ii) 1--%« =2-3§-1=-% isarational number. (ii) Commutative Property Subtraction of two rational numbers is not commutative. atFor tworational numbers 6 and %,wehave 4 .4. .. .2, 2. .. .4. 9 5# 5 9 __4 .. A LHS.. 9 5 :l = 20 - 18 RHS= 25 _.A9 f 45 18 ~ 20 L =L 45 Q 45 :2. T 45 LHS ¢ RHS Commutative property is not true for subtraction. 5 (iii) Associativeproperty Subtraction V of rational numbers is not associative. Forthreerational numbers 2, 1 1 1 1 1 1 2 (3 4>75 (2 3) 4 LHS L(L..i) RHS: Real Number S stem 1.3.1 (C)Multiplication (i) Closure property _ The product of two rational numbersis alwaysa rational number.HenceQ is closedundermultiplication. (i) %>< 7=-7=21isarational number. 3 3 (ii) %><% =% isarational number. (ii) Commutative property Multiplication of rational numbers is commutative. For two rational numbers3- and ?-§we have 5 ll %x<%8> <%>x% LHs=%><(113) RHs=1'1§x(%) ._.:_2_4_ ___ :24 55 -. LHS 55 = RHS Commutative property is true for multiplication. (iii) Associative property Multiplication of rational numbers is associative. Cha ter 1 Associative property is true for multiplication. (iv) Multiplicative identity The product of any rational number and l is the rational number itself. One is the multiplicative identity for rational numbers. ls l themultiplicative {V} Multiplication identityfor integers? by 0 Every rational number multiplied with 0 gives 0. (1) .5>+% S stem Chapter l LHS 75 RHS Associative property is not true for division. 1.3.1 (9) Distributive Property (i) Distributivepropertyof multiplication overaddition Multiplication of rational numbers is distributive over addition. For three rational numbers 3-,5-and -3-, wehave 2 2 4 3 _ 2 A 4 2 1 §X(§+5) . 3X9""3 5 LHS=%><<%+%> RHs=%><%+%><% ___2_ 3X(20+27 45 > lX4_7:& 3 5 _ _8_ 27+_2_ 5 i 135 i LHS = __. 135 135 RHS Multiplication is distributive over addition. (ii) Distributive property of multiplication over subtraction Multiplication of rational numbers is distributive over subtraction. Real Number System LHS 4 RHS Multiplication is distributive over subtraction. EXERIS 1.1 1. Choose i) the correct answer: The additive identity of rational numbersis (A) 0 (B) 1 . (C) - 1 (D) 2 ii) The additive inverse of3753is :1 1 (m5 (m3 i :i Cg (m3 iii) The reciprocal of"1'35 is . _5_ "- 13 mM3 m>5 iv) The multiplicative inverse of (M7 v) +i=i+<;3> <9 4+<7+1>=<4+7>+: 7 9 9 (iii)8+%=%+8 9 7 9 8 2 9 8 (iv)('i57')+o='1&# 3. Name the property under multiplication used in eachof the following: 9 :>x+;*e=f:-X9» 9 <:;5>x1=e5~=1x<:;»s~> Cha ter 1 <*T§7>x<1%8>=1 ;>< %x<.%+§>=%x.%+%x§ 4. Verify whether commutative property is satised for addition, subtraction, multiplication and division of the following pairs of rational numbers. mm m%m%- 5. Verify whether associativeproperty is satised for addition, subtraction, multiplication and division of the following pairs of rational numbers. @§%m% @%%m% 6. Use distributive property of multiplication of rational numbersand simplify: - :5. §_ _5_ -- 2 (1) 4 X(9+7> 1.3.2 To nd rational 1 __1.. ()7X<4 2) numbers between two rational numbers Can you tell the natural numbers between 2 and 5? 1 2 3 4 5 6 7 8 9 10 They are 3 and 4. Can you tell the integers between 2 and 4? -3 -2 -1 0 1 2 3 4 5 They are 1, 0, 1, 2, 3. Now, Can you nd any integer between 1 and 2? No. But, between any two integers, we have rational numbers.For example between 0and 1,wecan ndrational numbers ---which can bewritten as; 0.1, 0.2, 0.3, 0 0.1 0.2 0.3 0.4 0.5 .1- .2. .3. .4. 5 6 __7_...8- .2. 10 10 10 10 10 1 10 10 1 0.6 0.7 0.8 10 0.9 1 Real Number System Now,consider 25 and3. Canyound anyrational number between 15 and1? 5 5 Yes.Thereis a rationalnumber-3-. 5 W% i 0 % 2 L 2 2 A 5 5 5 5 1 Inthe same manner, weknow that the numbers and garelying between 0 and 1. Canyound morerational numbers between A and-3-? 5 5 Yes. Wewrite35 asQ and15 asQ then wecanndmanyrational numbers E 50 50 between them. 2o;01;l22_3a;s2_62_i7;82_i9a:; 50505050505050505050505 We can nd nine rational numbers 21 22 23 24 25 26 27 28 and 29 5050505050505050 50' i If wewantto nd somemorerationalnumbersbetween Q andQ wewrite Q 50 50 50 as~220 andgi as~230.Thenwe getnine rational numbers 221 222223224225 500 50 500 500 500 500 500 500 226 227 228and229 500 500 500 I 500' Letusunderstand -03-52- thisbetter withthehelpof the number line shown I in the adjaeentgure. Observe the number linebetween 0and 1using iamagnifying lens. .%.=.25% %.l 205 216 207 208 209 30 3.63.6E6.57). .5.6=% Chapter 1 Similarily,wecanobserve manyrationalnumbers in theintervals1 to 2, 2to 3 and so on. If we proceed like this, we will continue to nd more and more rational numbers between any two rational numbers. This shows that there is high density of rational numbers between any two rational numbers. S0, unlike natural numbers and integers, To nd rational numbers between two :~ rational : numbers We can nd rational numbersbetween any two rational numbers in two methods. 1. Formula 3 method Letaandbbeanytwogiven rational numbers. Wecanndmanynumber of rationalnumbers ql, q2,q3, in between a andb asfollows: q1=-12~(a+b) 0 q2=%(a+q1) l 5 L 61, 1? b I3 12 I1 q3=~é(a+qz)and soon. Thenumbers q2, q3lietotheleftofql.Similarly, q4,qsare therational numbers between aandblietotherightofqlasfollows: C 94=%(q1+ 17) a q5=é-(q4 +b)and soon. a I1 14 b ql q4q51, u-«----------4»----+-----+----1 k.;ae 2. Aliter Let a and b be two rational numbers. g (i) Convert thedenominator ofboththefractions intothesame denominator bytakingLCM.Now,if there isanumber between numerators there isa rational number between them. V (ii) If thereis no number between theirnumerators, thenmultiplytheir numerators anddenominators by10togetrational numbers between them. Togetmorerationalnumbers, multiplyby 100,1000andsoon. 14 L Real Number S stem Example 1.1 Find arational number between -4% and Formula method: Given: a=%,b=% Letqlbetherational number between -3and g1q1= § 2< 20 > L L 3_1 2X(20> 40 The rational number is Miter: - . _1 _A Gwen. a 4,b 5 Wecanwriteaandbas ix; = andSIX:= Tondarational number between % and £6,wehave tomultiply the 1 numerator anddenominator by10. ' 5 15X10=150, 16X10=160 20 20 10 200 10 200 The rational numbers betweenand are 151 152 153 154 155 156 157 158 and 159 200 200 200 200 200 200200 200 200' Example 1.2 Find tworational numbers between "#3and Let ql andq2be two rational numbers. _1 q1_?(a+b) l -3 l__l --6+5 _l --~1_----l 41 2X(5 +2)"2X( 10 >"2X(10>20 _1 __1 -1 42 §<+%> ?X(T+(20>> EXERCISE 1.2 1. Find one rational number between the following pairs of rational numbers. (i)%and%(ii)T2and% (iii)%and % 2. Find two rational numbers between (i)%and % (ii)g and % 3. Find three rational numbers (iii)%and % (iv)"'?1 and % between (i)%and%(ii)%and % (iii)T1and% 1.4 Simplication (iv)%and § of Expressions Involving (iv)%and Three Brackets Let us seesome examples: (i) 2+3=5 (ii) (iii) %><%=% L 51o=5 (iv)42><%-.=? Inexamples, (i),(ii)and(iii),there isonlyoneoperation. Butinexample (iv)we have twooperations. I Doyouknowwhichoperation hasto bedonerst in problem (iv)? L In example (iv),if wedonotfollowsome conventions, wewillgetdifferent solutions. I Forexample (i) (42)><%=2><%=l (ii) 4 (2x =4 1=3,weget different values. So,to avoidconfusion, certainconventions regarding theorderof operations arefollowed. Theoperations areperformed sequentially fromlefttorightintheorder of BODMAS. Newwe winstudy more about brackets and epeiatieii . ei. Brackets Some grouping symbols areemployed toindicate apreference intheorder of operations. Most commonly used grouping symbols are given below. . . .. Parenthesis or commonbrackets t Bracesor Curly brackets. I [ll I BracketsiorSquarebrackets= ' Real Number System Operation Of We sometimes come across expressionslike twice of 3, one fourth of 20, half of 10 etc. In these expressions,of means multiplication with. For example, (i)twice of3iswritten as 2X3, (ii) one-fourth of20iswritten as%:x 20, (iii)halfof10iswritten as%x10. If more than one grouping symbols are used, we rst perform the operations within the innermost symbol and remove it. Next we proceed to the operations within thenextinnermost symbols andsoon. Example 1.3 Simplify: (1% + >< -1 2 8 _ 4 2 8 (13+3>"15 (3"'"3)>15 l3>"15 _ £_&=L 2X15 15 115 Example1.4 ' '-.1. .3. § Simpl1fy.52 +4 of 9. 1 3 52+40f9 8 _ 11 3 2+4X9 8 __l_1, .2.4..=_1..1.. .2. 2+36 2+3 = 33+4=g=6r 6 Example 1.5 Simplify: (""31 X5)+[3:(1 1 6 6' Example 1.6 Choose the correct answer: 2x1: 3 10 (ii) (A)-5; 2 (iii) 2: 5X7 (A)% .2. (iv) (B)2-2. (B)% .4. 5+91S (A)g 1.1--7 5 --1s_ 2 (B)g Real Number S Istem .5 Powers: Expressing theNumbers in Exponentlal FormwithIntegersI as Exponent In this section,we are going to study how to expressthe numbers in exponential form. We can express 2 X 2X 2 X 2: 24 , where 2 is the base and 4 is the index or In general, a" is the product of a with itself n times, where a is any real number and n is any positive integer fa is called the baseand n is called the index Aor power. Denition If n is a positive integer, then x meansx.x.x.....x LW :1factors i.e, x = x X x x x x Xx ( where n is greater than 1) ""'T/"*" ntimes Note : x = x. HOW 3:Exponent; i 73 is read as a Or Power ._4 Here 7 is called the base, 3 is known as To illustrate this more clearly, let us look at the following table <%>x<%>X<%>> Chapter 1 (iii) 32=2><2><2><2><2=25 (iv) 128=2><2><2><2><2><2><2=27 (v) 256=2><2><2><2><2><2><2><2=28 1.6 Laws of Exponents with Integral Powers With the above denition of positive integral power of a real number, we now establish the following properties called laws of indices or laws of exponents. (i) Product Rule _a" xa" = a*",wherea5is arealnumber andm,ntarepositiveintegers 2:=6 =62 (Using thelaw if =a"",where a26,m=4,11:2) (111)Power Rule <32)4=32X 32X 32X 32= 32+2+2+2 = 38 we can get the sameresult by multiplying the two powers i.e, (32)4= 32 = 38. (iv) Number with zero exponent 5 Using these two methods, m3+ m3= m° = . From the above example, we come to the fourth law of exponent Real Number System v) Law 01Reciprocal The value of a number with negative exponent is calculated by converting into multiplicative inverse of the samenumber with positive exponent. (D 4.4: 414= 4><4><4><4 1 1 = 256 (iii)W2= iir = 1o>1<1o =% Reciprocal of3isequal to%=3:)=30= '1. Similarly, reciprocal of62=~61; =~g~:~ =6° =6 Further, reciprocal of isequal to 1 =(~§)'3 (%3 From the above examples,we come to the fth law of exponent. EIf fa? yisgalreahnimberleand isipaniipnteger,rthepnTa*?il ;; , (vi) Multiplying ii numbers with same exponents Considerthe simplications, (i) (ii) 43><73= (4><4><4)><(7><7><7)=(4><7) = (4><7)3 E 5~3><43 = §><$=(%)3><(%)3 = éxéxéxlxlxl - (§Xi>X(§Xi>X(%Xi>=<§o> = 2o3=(5x4)-3 erxe>r = <§x§>x<:x:>=<§x:>x<§x:> In general, for any two integers a and 19we have a2X b2 = (a X b)2= (ab)2 We arrive at the power of a product rule as follows: (a X a X a X times) X (b X b X b X times) = ab ><......m times =(ab)"' (i.e.,) a><4* =(3><4)*=12x (ii) 72><22 =(7><2)2 =142=196 (vii) Power of a quotient rule Consider the simplications, <» (av an en (;2=<;)=:=<:>i (wage) 5 52 I _ §_ _5___5><5__5__2__ 2 __1___ 2 2_~l><3_2 3X3'3><3325X 2'5 52 Z 3w2 5-2 a2 a2 HCIICC canbeWI1t{CI1 asF gm _ 1 2 2 ~ <19) _ <....mtimes ~' (9) = Q31 b 19'" m i(= %,(where baés O,aand bare real) numbers, isaninteger ex = amer = = an<%>3= = Example 1.8 Simplify: (i) 25><23 (ii) 1o9+1o6 (iii) (x°)4 (v) (vii)(2x3)4 (vi)(2) (viii) If 21= 32, nd the Value of p. (i) 25><23=25+3 = 28 (ii) 1o9+106 =10 (iii) (x°)4= (1)4 = 1 (iv) (23)°= 8° = 1 = 103 [-.- a0: 1] ['.' a0= 1] (iv) (23) Real Number System m;=%% (Vi) (25)2= 25 = 21°= 1024 (vii) (2><3)4= 64 = 1296 (or) (2><3)4=24><34= 16><81 =1296 (viii) Given : 21= 32 2p=25 2 3 2 4 22 Therefore p = 5 (Herethebaseonbothsidesareequal.) 1 Example 1.9 Find the Value of the following: (1)3><3"3(ii)31,4 (iii) (iv)10-3 (V) (Vi)(%>0><3 (vii)(gff (viii)%)5><(%)+(%) 34 am (iv) 10* % ("71>5==%1 woX(8>(8>= 8_3__9 = 39:1 (8) wwea$=1 (8) Example 1.1] Simplify - 3-2 22 -- (22)3 (1)(2)><(3) (11) (32)2 (23)2 X(32)22(3><~2)X 3(2><2) 4 2"6><34 =%><34 =%=E711 23 (11) X (222 = 32:: =.2; =.6§4__ (3) X 3 81 Example 1.12 Solve (i) <%>x<%>=<%>6 Given 12 12 = 144 = 122 .. x = 2 ,, (11) 2 2x 2 x _ ('3 The base on both sides are equal) 2 6 (§) ><(§)- (§> (%>2x+x = (%>6 ('3 The base onboth sides are equal) 2x+x = 6 3x = 6 x = g: 2. Example 1.13 . . . <33>*x<22>*3 Simplify. (24)_2X3_4X4_2 Real Number System EXERCISE1.4 1. Choose the correct answer for the following: (i) a"' X a is equal to (A) a' + a" (B) a (C) a"' (D) am" (ii) p0 is equal to (A) 0 (iii) (B) 1 1 (D) p In 102,the exponentis (A) 2 (iv) (C) (B) 1 (C) 10 (D) 100 (B) 1 (C)mg- (D)g. (C) 24 (D)4 (C) 5 (D) 6 6 is equal to (A)6 (v) The multiplicative inverse of 2* is (A) 2 (vi) (B) 4 (- 2)"5 X (- 2) is equal to (A) 2 (vii) ( 2) is equal to (A)§~ (viii) (B) 2 (B)~31» (C)'21- (D) 1 (2° + 4) X 22is equal to (A) 2 (B) 5 (C) 4 (D) 3 (B) 34 (C) l (D) 3-4 (B) 50 (C) 1 -4 (ix) (32 isequal to (A) 3 (X) ( 1) is equal to (A) -1 50 (D) 1 2. Simplify: (i) (- 4) + (- 4)) (ii) (iii)(.3)4><(-§~)4 (v) (3-7:31°) x 3-5 (Vi) 26x 32X23><37 (%>5X(%)2X(%>2 23><36 yabX ybc><(2p)2><5°-(811 (x) (L) 3><82/3><40 +(~9~ )l/2 4 3. 16 Find the value of: (i) (3° + 4")><22 (iv) (3-1+ 4-1+ 5-1)° (ii) (24 x 4-1)+ 2* (in <~:)+< (V)[<"2>T (vi) 740- 7-21. >1/2 Chapter 1 Find the value of m for which (1) 5'2 5* = 55 (iv) (a3)l" = a9 (ii) 4'" = 64 (iii) 8"'3 = 1 (v) (5m)2><(25)3><1252 = 1 (vi)2m=(8)'13" + <23)? 5. (a) If 2 =16, nd (1) x (ii) 2% (iii) 2 (iv) 2 (v) /F (b) If 3x = 81, nd (1) x (ii) 3M (iii) 3*/2 (iv) 32* (V) 3 6.Prove that (i)3m ><(i>m =1,(ii) <%)m+n.n+l.(Ll =1 3x(x+1) 3 x xl xm 1.7 Squares, Square roots, Cubes and Cube roots 1.7.1Squares When anumber ismultiplied byitself wesay that thenumber issquared. Itis denoted byanumber raised tothepower 2. For example: (i) 3><3 = 32= 9 (ii) 5><5 = 52: 25. Inexample (ii)52isread as5tothepower of2(or) 5raised tothepower 2(or) 5 squared. 25isknown asthesquare of5. Similarly, 49 and 81 are the squaresof 7 and 9 respectively. In this section, we are going to learn a few methods of squaring numbers. Perfect Square The numbers 1,4,9,16, 25,---are called perfect squares orsquare numbers as 1=12,4=22,9=32,l6=42andsoon. I Anumber iscalled aperfect square if it isexpressed asthesquare ofanumber. Properties of Square Numbers We observethe following propertiesthrough the patternsof squarenumbers. 1. Insquare numbers, thedigits attheunitsplace arealways 0,l, 4,5,6 9.Thenumbers having 2,3,7 or8 atitsunitsplacearenotperfect square numbers. Real Number System 144' If a number has 1 or 9 in the unit's If a number place then its squareends in 1. ill) 3- If a number place then its squareends in 4. 1 . lV) -49 169 H130 ;g1 has 3 or 7 in the unit's If a number ..5 . . 0 15 - t I - 1 ;36 ~ ' 196 has 4 or 6 in the unit's place then its squareends in 6. . . 0 25 t . If a number has 5 in the unit's place 05225 . t thenits squareendsin 5. .025. 3. 1* e 14 place then its squareends in 9. V) has 2 or 8 in the unit's 0.60250 - Consider the following squarenumbers: We have one zero 0 102= 100 But we have 202 = 400 0two zeros (i) Whenanumber ends with O , its square ends with double zeros. 302= 900 (ii) If a number ends with 1002 : 10000 2002 : 40000 We have two zeros BUEWS have oddnumber ofzeros then it four Zeros ' 0 isnotaperfect square. 7002 = 490000 4. Consider the following: (i) 100 = 102 (Even number of zeros) (ii) 81,000 = T 100 is a perfect square. 81 X 100 X 10 = 92X l02><10 81,000is not aperfect square. (Odd number of zeros) 27 1 Chapter 1 5. Observe the following tables Square of even numbers Square of odd numbers From the above table we infer that, (i) Squares of evennumbers areevené (ii)Squares ofoddnumbers areodd. Example 1.14 Find the perfect squarenumbers between (i) 10 and 20 (ii) 50 and 60 (iii) 80 and 90. (i) The perfect squarenumber between 10 and 20 is 16. (ii) (iii) There is no perfect squarenumber between 50 and 60. The perfect squarenumber between 80 and 90 is 81. Example 1.15 Byobserving theunitsdigits, which ofthenumbers 3136, 867 and 4413 can not be perfectsquares? T Since 6 isin unitsplace of3136, there isachance thatit isaperfect square. 867and4413aresurelynotperfectsquares as7 and3 aretheunitdigitof these: numbers. 1 Example 1.16 Write down the unit digits of the squaresof the following numbers: (1) 24 (i) (ii) 78 The square of 24 Therefore, we have 4x4 (iii) 35 24 X 24. Here 4 is in the unit place. 16. 6 is in the unit digit of square of 24. 28 Real Number System (ii) 78 X 78. Here, 8 is in the unit place. The square of 78 Therefore, we have 8 x 8 = 64. (iii) 4 is in the unit digit of squareof 78 The square of 35 = 35 X 35. Here, 5 is in the unit place. Therefore, we have 5 X 5 = 25. 5 is in the unit digit of square of 35. Some interesting patterns of square numbers Addition of consecutive odd numbers: 1 =1=12 1+3 1+3+5 1+3+5+7 1+3+5+7+9 = 4:22 = 9:32 = 16:42 = 25:52 1 + 3 + 5 + 7 + ---+ n = n2 (sum of the rst n natura odd numbers) The above gure illustrates this result. Tondthesquare ofarational number -5. *- EXERCISE 1.5 1. Just observethe unit digits and statewhich of the following are not perfect squares. (i) 3136 (ii) 3722 (iii) 9348 (iv) 2304 (v) 8343 2. Write down the unit digits of the following: (i) 782 (ii) 272 (iii) 412 (iv)352 (V) 422 3. Find the sum of the following numberswithout actually adding the numbers. (i)1+3+5+7+9+11+13+15 (ii) 1+3+5+7 Chapter 1 4. Expressthe following as a sum of consecutiveodd numbersstarting with 1 (i) 72 (ii) 92 (iii) 52 (iv) 112 5. Find the squaresof the following numbers mg m% m§ mg m% 6. Find the values of the following: (near anew? <2><2><2><2><2=22><22><22 28 «/64=«/22><22><22=2><2><2=8 22 Chapter 1 Example 1.18 Find the squareroot of 169 169 =13><13 =132 1313 1 x/169 = 132 =13 Example 1.19 Find the squareroot of 12.25 § INS:/mwmm ' im = «/1225 = t/52><72= 5><7 ¢1o0 ¢1o2 10 ¢m%=«%=w Example 1.20 Find the squareroot of 5929 5929 = 7><7><11><11=72><112 \.-v_/ «/5929 = .'.«/5929= \V/ «/72><1l2=7><11 77 Example 1.2] Find theleast number bywhich 200must bemultiplied to make it a perfect square. 200= 2><2><2><5><5 \Y/\Y~/ 2 remains without a pair. Hence, 200 must be multiplied by 2 to make it a perfect square. Example 1.22 Find the least number by which 384 must be divided to make it a perfect square. Real Number System (ii) Long division method In case of large numbers, factors can not be found easily. Hence we may use another method, known as Long division methed. Using this method, we can also nd square roots of decimal numbers. This method is explained in the following worked examples. Example 1.23 Find the squareroot of 529 using long division method. Step 1 : We write 529 as 5 29 by grouping the numbersin pairs, starting from the right end. (i.e. from the units place ). Step 2 : Find the number whose squareis less than (or equal to) 5. Here it 1S 2. Step 3 : Put 2 on the top, and also write 2 as a divisor as shown. Step 4 : Multiply 2 on the top with the divisor 2 and write 4 under 5 and subtract. The remainder is 1. Step 5 : Bring down the pair 29 by the side of the remainder 1, yielding 129. 29 Step6 : Double2 and take the resultingnumber4. Find that 51 number n such that 4n X n is just less than or equal to 2 Forexample:42><2=84;and43><3=129and ;4 4. 431 SO11= 29 11 29 Step 7 : Write 43 as the next divisor and put 3 on the top along with 2. Write the product 43 X 3 = 129 under 129 and subtract. Since the remainder is O,the division is complete. Hence x/529 = 23. Exampie 1.24 Find V 3969 by the long division method. Step 1 : We write 3969 as 3959 by groupingthe digits into pairs, starting from right end Chapter l Find the number whose square is less than or equal to 39. It is 6. Step 3 : Put6onthetopand also Write 6asadivisor. 6 6EE 596 Step4 : Multiply6 with 6 andwritetheresult36under39and6 -356-9- subtract. The remainder is3. 36 3 Step5 : Bringdownthepair69bythesideof thisremainder 3, 6 35@yielding 369. 36 \l/ 3 69 Step 6 : Double 6, take the result 12 and nd the number n. Such that 12nX n is just less than or equal to 369. Since l22><2 = 244; l23><3 =: 369, rt = 3 Step 7 : Write 123 asthe next divisor and put 3 on the top along with 6. Write the product 123X 3 = 369 under 369 and subtract. Since the remainder is O,the division is complete. Hence v 396 = 63. 1.7.2 (a) Square roots of Decimal Numbers Toapply thelongdivision method, wewritethegiven number bypairing off thedigitsasusualin theintegral part,andpairingoff thedigitsin thedecimal part fromleftto rightafterthedecimal part. i For example, we write the number 322.48 as }{}a:~:t;:§.ma§p{3i§2§T Weshould know howtomark thedecimal pointinthesquare root.Forthiswe note thatforanumber withl or2digits, thesquare roothas1digitandsoon.( Refer Table l). The following worked examples illustrate this method: Real Number System Example 1.25 Find the squareroot of 6.0516 We write the numberas 6 . E E. Sincethe numberof digits in the integralpart is 1,the squareroot will have 1 digit in its integral part. We follow the sameprocedure that we usually use to nd the squareroot of 60516 4. .6 2g6.051'6 4 31;". . 4462 05 486} 29 16 0 From the above working, we get V 6.0516 = 2.46. Example 1.26 Find the least number, which must besubtracted from 3250 tomake itaperfect square 2; 107 50 i7 5 1 Thisshows that572 isless than 3250 by1.Ifwesubtract theremainder from the number, wegetaperfect square. Sotherequired leastnumber is 1. Example 1.27 Find the least number, which must be addedto 1825 to make it a perfect square. This shows that 422 < 1825. Chapter 1 Next perfect square is 432 = 1849. Hence, the number to be added is 432 - 1825 1849 1825 24. Example 1.28 Evaluate «/0.182329 0' 29 asO.KE 5. Since thenumber to82 238to hasnointegralpart,thesquare root 1 54 847 also will have no integral part. We 815191129 1 Hence We write the number 0.182329 59 29 x/0.18232 thenproceed asusualforndingthe squareroot of 182329. = 0.427 Note: Since the integral part of the radicand is O,the squareroot also has 0 in its integral part. Example 1.29 Find the squareroot of 121.4404 x/121.4404 = 11.02 Example 1.30 Findthesquare rootof 0.005184 Note:Sincetheintegralpartof theradicand is 0, a zerois 2 decimal point in the quotient. A O is written in the quotient after the decimal point since the rst left period following the decimal point is 00 in the radicand. 1.7.2 (b) Square root of an Imperfect Square An imperfect square is a number which is not a perfect square.For example 2, , 5, 7, 13,...are all imperfect squares.To nd the squareroot of such numbers we use he Long division method. If the required square root is to be found correct up to n decimal places, the squareroot is calculatedup to n+1 decimalplacesandroundedto n decimalplaces. Accordingly, zerosareincluded in thedecimal partof theradicand. Example 1.3] Find the squareroot of 3 correct to two places of decimal. Since we need the answer correct to two places of decimal, we shall rst nd the squareroot up to three places of decimal. For this purpose we must add 6 ( that is three pairs of ) zeros to the right of the decimal point. «/3 1.732up to three places of decimal. /§ 1.73correct to two places of decimal. Example 1.32 Findthesquare rootof 10;3 correct totwoplaces ofdecimal. 102-:-3-2=10.666666 ........ 3 3 In order to nd the square root correct to two places of »'» 3: 1 . édecimal, wehave tondthesquare rootuptothree places. Therefore wehave toconvert 3 asadecimal correct tosix 646 3 §P1aC¬S- if . \@ = 3.265 (approximately) 6525 .. Chapter 1 EXERCISE 1.6 1. Find the square root of each expression given below : m3x3x4x4 (m2x2x5x5 (iii)3><3><3><3><3><3 (iv)5><5><11><11><7><7 2. Find the square root of the following 2 (1)614 (ii)% (111) 49 (iv)16 3. Find the square root of each of the following by Long division method : (i) 2304 (ii) 4489 (iii) 3481 (iv) 529 (v) 3249 (vi) 1369 (vii) 5776 (viii) 7921 (ix) 576 (x) 3136 4. Find the squareroot of the following numbersby the factorization method : (i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100 5. Find the square root of the following decimal numbers : (i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (vi) 0.2916 (vii) 11.56 (viii) 0.001849 (V) 31.36 6. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square : (1) 402 (11)1989 (111)3250 (iv) 825 (v) 4000 7. Findtheleast number which must beadded toeach ofthefollowing numbers so as to get a perfect square: (i) 525 (ii) 1750 1 (iii) 252 (iv) 1825 (v) 6412 8. Find the square root of the following correct to two places of decimals : (1) 2 (11) 5 (111) 0.016(iv)-;- (v)1-11-2- 9. Find the length of the side of a squarewhere areais 441 m2. 10. Find the square root of the following : - 225 .- 2116 529 . 7921 (1)3136 (11)3481 <1)1764 (1V)5776 Real Number System Introduction This is an incident about one of the greatestmathematical geniuses S. Ramanujan. Once mathematician Prof. G.H. Hardy came to Visit him in a taxi whose taxi number was 1729. While alking to Ramanujan, Hardy described that the number 1729 was a dull number. Ramanujan quickly pointed out that 1729 was indeedan interestingnumber.He said,it is the smallest Si"5a Ram3" _ number that can beexpressed asasum oftwocubes 111f ( 1887..-1920 wo different ways. ie., 1729 = 1728 +1 = 123+ 13 and 1729 = 1000 + 729 = 103 + 93 1729is known as the Ramanujan number. There are many other interesting patterns o icubes, cuberootsandthefactsrelated tothem. Cubes We know that the word Cube isusedin geometry. 3 A cube isasolid gurewhich has allitssides are equal. If theside ofacube intheadjoining gureisa then its volume is given by a x a x a = a3cubic units. Herea3is calleda cubed" or "a raisedto the powerthree or "a to the power 3". Now, consider the number 1, 8, 27, 64, 125, These are called perfect cubes or cube numbers. Each of them is obtained when a number is multiplied by itself three times. Examples: 1><1><1=-.13, 2><2><2 = 23, 3><3><3= 33,5><5><5 =53 Example 1.33 Find the Value of the following : (1)153 (ii)<4>3(iii)<1.2>3(iv)(%)3 (1) 153 15X 15X 15 = 3375 Chapter 1 1.2 1.2>< 1.2= 1.728 <3>><<3)><<3>___ -27 4><4><4 64 Observe the question (ii) Here ( 4)3 = negative nurnber 64. itselfantieveniinlirnberr the pprodijéii ispositive; Brut whenit istiirnnltipliedrbytitselftieaiiiioddgnumber ortrmes, the produetijs alsotrnegative, ie,e(-:1) -7-.A;:{""*1tiitf niseven753 + 3xooo\1o Table : 2 Properties of cubes From the above table we observethe following properties of cubes: Example 1.34 Is 64 a perfect cube? 64 = 2><2><2><2><2><2 &:...,...j.Jmaria 23><23=(2><2)3= 43 64 is a perfect cube. Example [.35 Is500aperfect cube? intheproduct but 1 only two 2s. 500 = 2 X 2X 5X 5 X 5 ,.«s/»~;?'WW~M/i é.v.:.J 2 So 500 is not a perfect cube. O0 2 250 Is243aperfect cube? If notnd thesmallest number by which 243 must be multiplied to get a perfect cube. 243 = 3><3><3><3><3 L.___..\,_.__.J T l\Jl\J 5 Example 1.36 125 2 5 In the above Factorization, 3 X 3 remains after grouping the3's in triplets. 243is notaperfectcube. To make it a perfect cube Wemultiply it by 3. 243><3= 3><3><3><3><3><3 £.v2:l _,.I 729 = 33><33= (3><3)3 729 = 93which is a perfect cube. 1 >-*l Chapter 1 1.7.4 Cube roots If the Volume of a cube is 125 cm3,what would be the length of its side?To get the length of the side of the cube, we need to know a number whose cube is 125. To nd the cube root, we apply inverse operation in nding cube. Forexample: 5Y"l Weknowthat23= 8,thecube rootof8is2. 3x/denotes cube- root WeWrite it mathematically as ........7 .......................... 3/§ = (8)1/3= (23)1/3 = 23/3 : 2 Some more examples: (1) 37125 = V? = (53)/3= 53/3 =5 (ii) =5 a/671 = 3/E = (43)3 = 43/3= 41: 4 (iii) 3./1000 = 3./103= (1o3)1/3 =1o3/3=1o1 =10 Cube root through prime factorization rnethod Method of nding the cube root of a number Step 1 2 Resolve the given number into prime factors. Step 2 : Write these factors intriplets such that allthree factors ineach triplet areequal. S Step 3 : From theproduct ofallfactors, take one from each triplet that gives the cube root of a number. E Example1.37 Find the cube root of512. a/512 = (512)% = ((2><2><2)><(2><2><2)><(2><2><2))% = (23><23><23)% = <29>%=23 3/512 = 8. Example 1.38 Find the cube root of 27 X 64 3 27 . . . Resolving27 and64 into primefactors,weget 3./27= (3><3><3)%=(33)? 3 9 33 1 Real Number System V27 3 3/64 =(2><2><2><2><2><2)% _ =(2$§=22=4 264 2 32 3«/64 :4 216 V27><64= 3/27><%/64 28 24 =3><4 V 27 x 64 2% l2 Example 1.39 Is 250 a perfect cube?If not, then by which smallest natural number should 250 be divided so that the quotient is a perfect cube? 250 = 2><5><5><5 &:"..J The prime factor 2 does not appear in triplet. Therefore 250is nota perfectcube. 2 250 5 12; 5 25 Since inthe Factorization, 2appears only one time. IfWe 55 1 divide the number 250 by 2, then the quotient will not contain 2. Restcanbeexpressed in cubes. 250+ 2 = 125 = 5x 5x 5 = 53. i Thesmallest number bywhich 250should bedivided tomake it aperfect L cube is 2. Cuberootof a fraction C ub 6 mot - ("6) Cube root of its numerator . denominator . 0f a f mot' Ion = Cuberootof its a_3«/3=al=(a)% 3Vb W (by (wt 5 210 2x2x2x3x3x3 mm II .23 216 _a3 125 || o\|u1 0\ Example 1.4] Find the cube root of §~ 5><5><5><2><2><2 10 O o 3 512 1000 8 10 -4 [0 G 3 5 II 2» 1 U1 >* N I 22m Real Number System 3x/51 83 2 V% =%/?= 7 . 3/729V27 ___ 9-3 "a/512 +V343 8+7 :L=; 2 155 ERCI 1.7 1. Choose the correct answer for the following 2 (i) Which of the following numbers is a perfect cube? (A) 125 (ii) C) 75 (D) 100 Which of the following numbers is not a perfect cube? (A) 1331 (iii) (B) 36 (B) 512 (C) 343 (D) 100 The cube of an odd natural number is (A) Even (C) May be even,May be odd (B) Odd (D) Prime number (iv) The number of zeros of the cube root of 1000 is (A) 1 (B) 2 (C) 3 (D) 4 (V) The unit digit of the cube of the number 50 is (A) 1 (B) 0 (C) 5 (D) 4 (vi) The number of zerosat the end of the cube of 100 is (A) 1 (B) 2 (C) 4 (D) 6 (vii) Find the smallestnumber by which the number 108 must be multiplied to obtain a perfect cube (A) 2 (B) 3 (C) 4 (D) 5 (viii) Find the smallestnumberby which the number 88 must be divided to obtain a perfect cube (A) 11 (B) 5 (C) 7 (D) 9 (ix) Thevolumeof a cubeis 64 cm3. The sideof the cubeis (A) 4 cm (B) 8 cm (C) 16 cm (X) Which of the following is false? (A) Cube of any odd number is odd. (B) A perfect cube does not end with two zeros. (D) 6 cm Chapter 1 (C) The cube of a single digit number may be a single digit number. (D) There is no perfect cube which endswith 8. 2. Check whether the following are perfect cubes? (i) 400 (ii) 216 (V) 1000 (Vi) 900 (iii) 729 (iv) 250 3. Which of the following numbersare not perfect cubes? (i) 128 (ii) 100 (V) 72 (Vi) 625 (iii) 64 (iv) 125 4. Find the smallest number by which each of the following number must be divided to obtain a perfect cube. (i) 81 (ii) 128 (V) 704 (Vi) 625 (iii) 135 (iv) 192 5. Find the smallestnumber by which eachof the following number must be multiplied to obtain a perfect cube. (i) 243 (ii) 256 (iii) 72 (iv) 675 (V) 100 M 6. Findthecuberootof eachof thefollowingnumbersby primeFactorization method: (i) 729 (ii) 343 (iii) 512 (iV) 0.064 (V)0.216 (Vi) 52731 (Vii) 1.331 (Viii)27000 7. The Volume ofacubical box is19.683 cu. cm. Find the length ofeach side ofthe box. 1.8 Approximation of Numbers In our daily life we need to know approximate Vall.1¬S 01' measurements . Benjamin bought a Lap Top for ? 59,876. When he wants to convey this amount to others, he simply saysthat he has bought it for ? 60,000. This is the approximate Value which is given in thousands only. : Vasanth buysapairof chappals for?599.95. Thisamount maybeconsidered approximately as?600for convenience. A photo frame has the dimensions of 35.23 cm long and 25.91 cm wide. If we want to check the measurements with our ordinary scale, we cannot measureaccurately becauseour ordinary scaleis marked in tenths of centimetre only 46 1 8 Real Number S stem Insuch cases, wecan check thelength ofthephoto frame 35.2 cmtothenearest tenth or 35 cm to the nearestinteger value. In the abovesituationswe havetaken the approximatevalues for our convenience. Thistypeof considering thenearest valueis calledRoundingoff thedigits.Thusthe approximate value corrected totherequired number ofdigits isknown asRounding off thedigits. Sometimes it is possible only to give approximate value, because (a) If we want to say the population of a city, we will be expressing only in § approximate value say 30 lakhs or 25 lakhs and so on. (b)When wesay thedistance between twocities, weexpress inround number 350 km not 352.15 kilometres. While rounding off the numbers we adopt the following principles. (i) If thenumber next tothedesired place ofcorrection isless than 5,givethe answer up to the desired place as it is. (ii) If thenumber next tothedesired place ofcorrection is5and greater than 5add1tothenumber inthedesired place ofcorrection andgivetheanswer. iiThe symbol forapproximation isusually denoted by Takean th andbreadth. Howdo you expressit in cms approximately. Letusconsider some examples tondtheapproximate values ofagiven number. Take the number 521. Approximation nearest to TEN Consider multiples of 10 before and after 521.(i.e. 520 and 530 ) We nd 519 that 521 is nearer 520 521 522 to 520 than to 530. 523 524 525 526 527 528 529 530 The approximate value of 521 is 520 in this case. Approximation nearest to HUNBRED (i) Consider multiples of 100 before and after 521.( i.e. 500 and 600 ) 47 300 400 We nd that 521 is nearer to 500 than to 600. So, in this case,the approximate alue of 521 is 500. (ii) Consider the number 625 Supposewe take the number line, unit by unit. 623 p 624 625 626 627 628 629 J Inthiscase, wecannot saywhether 625isnearer to624or626because it is exactly midway between 624and626.However, byconvention wesaythatit isnearer to 626andhence itsapproximate Value istakentobe626. 2 Suppose weconsider multiples of100, then 625 willbeapproximated to600 and not 700. Some more examples For the number 47,618 (a) Approximate Value correct to the nearesttens = 47,620 (b) Approximate Valuecorrect to the nearesthundred = 47,600 (c) Approximate Valuecorrect to the nearestthousand = 48,000 ((1)Approximate Value correct to the nearestten thousand = 50,000 Becimal Approximation Consider the decimal number 36.729 (21) Itis36.73 correct totwodecimal places. ( Since thelast digit 9>5, weadd to 2 andmakeit 3 ). i 36.729 2 36.73 ( Correct to two decimal places ) (b)Look atthesecond decimal in36.729, Here it is2which isless than 5,sowe eave7 asit is. 36.7292 36.7( Correctto onedecimalplace) A Consider the decimal number Ravi has the following numbered cards 36.745 (a) lts approximation is 36.75 correct to wodecimal places. Since thelastdigitis5,WeHelp him tondheapriate dd 1 to 4 and make it 5_ valuecorrectto thenearest 20,000. (b) lts approximation is 36.7correct to one decimal lace. Since the second decimal is 4, which is less than 5, e leave 7 as it is. 36.745 _~_36.7 Find the greatest number using the method of approximation a. 20ll20ll20ll + T7g~ ? Considerthe decimalnumber2.14829 (i) Approximate value correct to one decimal placeis 2.1 (ii) Approximate value correct to two b. 201120112011 c. 201120112011 X (1. 201120112011 + decimal placeis2.15 ahapah (iii) Approximate value correct to three decimal place is 2.148 (iv) Approximate value correct to four decimal place is 2.1483 Example 1.44 Round off the following numbers to the nearestinteger: (a) 288.29 (b) 3998.37 (a) 288.29 2 288 (c) 4856.795 (d) 4999.96 (b) 3998.37 2 3998 6 (Here,thetenthplacein theabove numbers arelessthan5.Therefore all the integersareleft astheyare.) (c)4856.795 2 4857 (d)4999.962 5000 [Here,thetenthplacein theabove numbers aregreater than5.Therefore the integer values are increasedby 1 in each case.] EXERCISE 1.8 A 1. Express the following correct to two decimal places: (i) 12.568 (ii) 25.416kg (iii) 39.927in (iv) 56.596m (v) 41.056m (vi) 729.943 km 2. Express the following correct to three decimal places: (i) 0.0518m (ii) 3.5327 km (iii) 58.29361 (iv) 0.1327 gm (V) 365.3006 (vi) 100.1234 l by Chapter 1 3. Write the approximatevalue of the following numbersto the accuracystated: (i) 247 to the nearest ten. (ii) 152 to the nearest ten. (iii) 6848 to the nearest hundred. (iv) 14276 to the nearest ten thousand. (V) 3576274 to the nearest Lakhs. (vi) 104, 3567809 to the nearest crore 4. Round off the following numbers to the nearest integer: (i) 22.266 (ii) 777.43 (iii) 402.06 (iv) 305.85 (V) 299.77 (vi) 9999.9567 1.9. Playing with Numbers Mathematics isasubject withfull offun,magic andwonders. Inthisunit, aregoingto enjoywith someof this fun andwonder. A (a) Nunibers in General form Letustakethenumber 42andwriteit as 42 = 40+2=10><4+2 Similarly, the number 27 can be written as 27 = 20+7= l0><2+7 In general,anytwo digit numberab madeof digits a and b canbe written as ab: l0><3+l0><5+1><1 In general, a 3-digit number abc made up of digit (1,E9and Cis written as abet = l00> 2 K%\*--...,........w"x ,,,.«M"'"" , /,»-°"°*-»- Now dividE$3, (\\ Um OK\§ answer by (igot 13) £1 / a.,__WM_,,,_.,. \*-..W__M__MW_,.wvV/ fTherewontbe . anyremainder! \ } , '; V \ gmBut hm? d'dr __~ (xyou know It? ~k,~«______,_,.,.» ~~~..,M_,_~W,,,»~ Now letussee if wecan explain Venus trick.Suppose, Manoj chooses the number ab,whichisa shortformforthe2 -digitnumber 10a+ b. Onreversing the digits, hegetsthenumber ba= 1019 + a.When headds thetwonumbers hegets: : (10a+b)+(10b+a) 11a+11b 11(a+ 1)) Sothesum isalways amultiple of11, justasVenu hadclaimed. Dividing the answerby 11,we get (a + b) (i.e.) Simply adding the two digit number. (0)Identify thepattern and findthenext three terms .0 Studythepattern in thesequence. (i) 3, 9, 15, 21, (Each term is 6 more than the term before it) If this pattern continues, then the next terms are ___ , ___ and ___ (ii) 100, 96, 92, 88, ___ , ___ , ___ . (Each term is 4 lessthan the previous term ) (iii) 7, 14, 21, 28, ___ , ___ , ___ . (Multiples of 7) (iv) 1000, 500, 250, ___ , _____,___. (Each term is half of the previous term) (v) 1, 4, 9, 16 _ . (Squaresof the Natural numbers) Chapter 1 ((1)Number patterns in Pasca1sTriangle The triangular shaped, pattern of numbers given below is called Pascais Triangle. Identify the number pattern in Pascalstriangle and completethe 6'1 row. 3 x 3 Magic Square T Look at the abovetable of numbers.This is called a 3 X 3 magicsquare. In amagicsquare, thesumof thenumbers in each row,each column, andalong each diagonal isthesame. Inthismagic square, themagic sum is27.Look atthemiddle number. Themagic sumis3times themiddle number. Once 9islledinthecentre,§ there areeight boxes tobelled.Four ofthem willbebelow9andfourofthem above; it.Theycould be, 5 (a)5,6,7,8 and10,11,12,13 withadifference of 1between each number. 5 (b)1,3,5,7 and11,13,15,17 withadifference of2 between themorit can anysetofnumbers withequal differences such as 11,6, 1,4 and14,19,24,29 withadifference of5. Once wehave decided onthesetofnumbers, say1,3,5,7and11,13,15,17drawi fourprojections outside thesquare, asshown inbelow gureandenterthenumbers; inorder, asshown inadiagonal pattern.The number fromeach oftheprojected box transferred to the empty box on the opposite side. Murugan has 9 pearls each of worth 1 to 9 gold coins. Could you help him to distribute them among his three daughters equally. MAGIC STAR In the adjacent gure, use the numbers from 1 to 12 to ll up the circles within the star such that the sum of each line is 26. A number can be used twice SU DO atmost. KU Use all the digits 1, 2, ..., 9 to ll up each rows, columns and squaresof different colours inside without repetition. A three digit register number of a car is a squarenumber. The reverse of this number is the register number of another car which is also a square number. Can you give the possible register numbers of both cars? The Revolving Number if ; 14 2 8 5 7 Firstsetoutthedigitsin acircle.Nowmultiply142857 bythe number from 1to6. 142857 1//>4\A2 142857 nal;;:gg&c X 4 3 142857 142857 \./ x5 x 6 We observe that the number starts revolving the same digits in different combinations.Thesenumbersare arrived at starting from a different point on the circle. Chapter 1 EXERCISE 1.9 1. Complete the following patterns: (i) 40, 35, 30, (ii) 0, 2, 4, , , , , Choose Add a number 9 to it Double the answer Add 3 with (iii) 84, 77, 70, (iv) 4.4, 5.5, 6.6, _ (V) 193969109 (vi) , Multiply the result by 3 , Subtract $$%$ have chosen rst from the answer. 5 9 (This sequenceis called FIBONACCI SEQUENCE) (vii) it Subtract the number that you 9 1, l, 2, 3, 5, 8, 13, 21, 3 from Divide it by 6 , __, 9 the result $ What isyour answer? 1, 8, 27, 64, .2-a?\if}}a.=»/"l,i : Ten 2. A water tankhassteps inside it.A monkey issittingonthetopmoststep. ( ie,thej rst step) Thewaterlevel is at theninth step. (a) He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level ? (b) After drinking water, he wants to go back. For this, he jumps 4 stepsup and thenjumps back 2 stepsdown in every move. In how many jumps will he reachback the top step ? 3. A vendor arrangedhis applesas in the following pattern : (a) If there are ten rows of apples,can you nd the total number of appleswithout actually counting? (b) If there are twenty rows, how many appleswill be there in all? Canyou recognizea patternfor the total numberof apples?Fill this chartandtry! EII3 Eli i Real Number System Rational numbers are closedunder the operationsof addition, subtraction and multiplication. The collection of nonzero rational numbers is closed under division. The operationsaddition andmultiplication arecommutativeand associative for rational numbers. :3? 0 is theadditiveidentityfor rationalnumbers. *9 1is themultplicative identityfor rationalnumbers. ii?Multiplication of rational numbers is distributive overaddition and subtraction. 1 5* Theadditive inverse of%is_Taandviceversa. % The reciprocal ormultiplicative inverse of%is5. Q? Betweentwo rationalnumbers,therearecountlessrationalnumbers. Q39Thesevenlawsof exponents are: If cl and b are real numbers and /72,n are whole numbers then amxan : am+n (ii) a"+ a a'"", where a aé0 (iii) a0 = 1 , where a 750 (iv)a I %,where aaé 0 (V) (am)? : amn (vi) a" X b' '-= (ab)" (v11) bm (b) where 19 7E 0 :39 Estimated valueof anumberequidistant fromtheothernumbers is always greaterthan the given number and nearerto it. 2.1 Introduction 3 3 2.2 Semi Circles and Quadrants 2.3 Combined Figures § Measurements 3 i § 2.1 Introduction Measuring isaskill. Itisrequired forevery individual inhis/ her life.Everyone ofushasto measure something ortheotherin ourdaily life. For instance, we measure Measurements Recall Let us recall the following denitions which we have learnt in class VII. (i) Area Area is the portion inside the closed gure in uTheeiword.peri.in Greek _ a plane surface. g;II1ean?s. pIar0unAdilAandlt meter if . *meansf.mea(sure . VA (ii) Perimeter The perimeter of a closed gure is the total measureof the boundary. Thus, the perimeter means measuring around a gure or measuring along a icurve. Can you identify the shapeof the following objects? Fig. 2.2 Theshapeof eachof theseobjectsis a circle. (iii) Circle Let O be the centre of a circle with radius r units (OA). Area of a circle, A = 7rr2sq.units. Perimeter or circumference of a circle, P = 27rr units, 2» . where 7:2 722 or3.14. Q,.Cmwe° Fag. 2.4 Take a cardboard and draw different circles circles radii. and nd of Cut the their areasand perimeters 57 Chapter 2 2.2 Semi circles and Quadrants 2.2.1 Semicircle Have you ever noticed the sky during night time after 7 days of new moon da or full moon day? What will be the shapeof the moon? It looks like the shapeof Fig. 2.6. How do you call this? Fig. 2.6 This is called a semicircle. [Half part of a circle] The two equal parts of a circle divided by its diameter are called semicircles. $emiCirc 15I How willyou get asemicircle from acircle? L A B Take acardboard ofcircular shape and cut it G through its diameter E g..m.m\.\x W Fig.2.7 U) (a) Perimeter of a semicircle Perimeter, P L2 X(circumference ofacircle) +2Xr unitsAr o = -é-><27rr+2r P = 7rr+2r=(7r+2)runits Fig-2-9 (b) Area of a semicircie Area,A = ix (Areaof acircle) 2 A B y A = 77;sq. units. D Fig. 2.10 4.2.2 Quadrant of a circle c Cut the circle through two ofits perpendicular diameters. We Q get four equal parts of the circle. Each part is called a quadrant of A the circle. We get four quadrants OCA, OAD, ODB and OBC while cutting the circle as shown inthe Fig. 2.11. Measurements (1 Perimeter, P = ix (circumference ofacircle) +2runits =i X27rr +2r , _7_c_r_ 2 D P = 525 +2r= + 2)runits Fig,213 Area, A = A11><(Area ofacircle) A = ixnrzsq.units D Fig. 2.14 Example 2.1 Find the perimeter and area ofasemicircle whose radius is14 cm. 14 cm Given: Radius ofasemicircle, r=14 cm (7 Perimeter ofasemicircle, P = (7:+2)r units Fig.2.15 P = (%+2)><14=3-6-><1%1 =308 cmz. Example 2.2 The radius of a circle is 21 cm. Find the perimeter and area of aquadrant ofthecircle. Given: Radius of a circle, r = 21 cm Perimeter ofaquadrant, P= _ _ F ig. 2.16 + 2)r units 22 __ 22 (<2 +2>><21._(14 +2)><21 P _( 22+-28 _ 14 )><2114><21 = 75 cm. Example 2.3 Thediameterof a semicirculargrassplot is 14m. Find A 14m 33 the cost of fencing the plot at ? 10 per metre . . _ Given: Fig. 2.17 Diameter, d = 14 m. Radius oftheplot,r = l 2 = 7m. To fence the semicircular plot, we have to nd the perimeter of it. Perimeterof a semicircle,P = (7r+ 2) X r units (%+2)><7 <22:e>>>r=7m Area of the park = Area of the semicircle A = 732 sq. units = %X7%7= 77m2 Area of the park = 77 m2. A rod is bent in the shapeof a triangle as shown E inthegure. Find thelength oftheside if it is bent in the shapeof a square? g~ 3 3 cm Wlsllimtwlnlllllinntwsinlonmiliiliinm 60 Measurements EXERCISE 1. Choose (i) the correct (ii) (iii) times the area of the circle. (B) four (C) one-half Perimeter of a semicircle is (D) one-quarter _ (A)CT:2)runits (B)(7z +2)runits (C) 2r units (D) (7r+ 4) r units If the radius of a circle is 7 m, then the area of the semicircle is (A) 77 m2 (iv) answer: Area of a semicircle is (A) two 2.1 (B) 44 m2 (C) 88 m2 (D) 154 m2 If the areaof a circle is 144 crnz,then the areaof its quadrantis (A) 144 crnz (B) 12 cm2 (C) 72 cm2 (D) 36 cm2 (V) The perimeter of the quadrantof a circle of diameter 84 cm is (A) 150 cm (vi) (D) 42 cm (C) 3 (D) 4 of the circle. (B) one-fourth (C) one-third (D) two-thirds (C) 180° (D) 360° (C) 270° (D) 0° The central angle of a semicircle is (A) 90° (ix) (B) 2 Quadrant of a circle is ___ (A) one-half (viii) (C) 21 cm The number of quadrantsin a circle is (A) 1 (vii) (B) 120 cm (B) 270° The central angle of a quadrant is (A) 90° (B) 180° (X) If the areaof a semicircle is 84 cm2,then the areaof the circle is (A) 144 crnz (B) 42 cm2 (C) 168 cm2 (D) 288 cm2 2. Find the perimeter and areaof semicircleswhoseradii are, (i) 35 cm (ii) 10.5 cm (iii) 6.3 m (iv) 4.9 m 3. Find the perimeter and areaof semicircleswhosediametersare, (i) 2.8 cm (ii) 56 cm (iii) 84 cm (iv) 112 In 4. Calculate the perimeter and areaof a quadrantof the circles whoseradii are, (i) 98 cm (ii) 70 cm (iii) 42 m (iv) 28 m 5. Find the areaof the semicircleACB and the quadrantBOC in the given gure. 6. A park is in the shapeof a semicircle with radius 21 In. Find the cost of fencing it at the cost of ? 5 per metre. 2.3 Combined Figures (0) (d) (6) Fig. 2.19 What do you observefrom thesegures? , Some eciombinationsepofe plsailc In Fig. 2.19 (a), triangle is placed overa T 2 A :s . : . . . . . _ i gureszplacedgadjacently, with semicircle. In Fig. 2.19 (b), trapezium 1Splaced A A V A= over asquare etc. 1 isfgonesidefequalein*length»toa1 T Two or three plane gures placed adjacently to forma newgure.These are of S'idVek:okf:4tl1kékkk0, 1~r EJuxtapgsioingofgufes; A H A AT T combined gures.Theabovecombined guresareJuxtaposition of someknowngures;triangle,rectangle,semi-circle, etc. Can we seesome examples? Two scalene triangles ~ lQuadrilateral. A Two righttriangles and arectangle Trapezium iA Sixequilateral triangles eHexlagjon A (a) Polygon A polygon is a closed plane gure formed by :1 line segments. Aplane gure bounded bystraight linesegments is arectilinear gure. A rectilinear gure of three sides is called a triangle and four sides is called a Quadrilateral. F. 220 lgf ' ~ Theiword Polygon? a rectilinear. gure with .;threeormore sidesg,s Measurements (b) Regular polygon If all the sidesand angles of a polygon are equal, it is called a regular polygon. For example, (i) An equilateral triangle is a regular polygon with three sides. D (ii) C Square is a regular polygon with four sides. A Fig. 2.22 (c) Irregular polygon Polygons not having regular geometric shapesare called irregular polygons. (d) Concave polygon A polygon in which atleastone angle is more than 180°,is called aconcave polygon. Fig. 2.23 (e)Convex polygon A polygon in which each interior angle is less than 180°, is called a convex polygon. Polygons areclassiedasfollows. F1'8» 2-24 Triangle. . Quadrilateral { Pentagon L L L i . illexatgon Ileptagon L Vijay has fencedhis land with 44m barbed wire. Which of the following shape will occupy the maximum area of the land? . pi7(i)ctag0ni a) Circle b) Square C) Rectangle 2m X20111 L l i 2L.Decagon , _ d) Rectangle 7mX 15m Most of the combinedgures are irregular polygons.We divide them into known plane gures.Thus, wecanndtheirareasandperimeters byapplying the formulae ofplane gures which wehave already learnt inclass VII.These arelisted inthefollowing table. Chapter 2 l Triangle AB + BC + CA (base + height + hypotenuse) Right triangle AB+BC+CA Equilateral = 3a ; Altitude, h=/2;a triangle units Isosceles triangle Scalene triangle 2a +2 a2_ h2 S(S__a)(S__b)(S_c) AB+BC+CA wheres: a+b+c = (a + [7.1.C) Quadrilateral Parallelo gram Rectangle Trapezium Measurements Divide the given shapesinto plane gures as you like and discuss among yourselves. Example 2.5 i) Find the perimeter and area of the following combined gures. 1t(M(.. (i) It is a combined gure made up of a square ABCD and asemicircle DEA. Here, are DEA ishalf the circumference of a circle whose diameter is AD. A Given:Side ofasquare = 7m Diameter of a semicircle = 7m 7 m Radius ofasemicircle, r = %m Perimeter of the combined gure D = /I 7m C B + BC + CD + DEA P = 7+7+7+-£>< (circumference ofacircle) = 21+%><27rr =21+ %><% P = '. Perimeter of the combined gure 32 m. Area of a semicircle + Area of a square Area of the combined gure A 21 + 11 = 32 m _ _ 7Z'I2 2 2 +a 22 7X7 2= 7><2X2><2+7 4 +49 Area of the given combinedgure = 19.25+49 = 68.25m2. Chapter 2 (11) The given combined gure 1smade up of a squareABCD and an equilateral triangle DEA. Given: Side of a square 4 cm Perimeter of the combined gure AB + BC + CD + DE + EA 4+4+4+4+4=20cm Perimeter of the combined gure 20 cm. Area of the given combined gure Area of a square + Area of an equilateral triangle 4X4+ T X4X4 16 + 1.732 x 4 Area of the given combined gure = Area of the given gure 16 + 6.928 = 22.928 2 22. 93 cm2. Example 2.6 Find the perimeter and area of the shadedportion Fig. 2.29 E (i) Thegivengure is acombination of arectangle ABCDandtwosemicircles AEB and DFC of equal area. Given: Length of the rectangle, l = 4 cm Breadth of the rectangle, b = 2 cm Diameter 2cm of a semicircle Radiusof a semicircle, r = 3 = 1cm 2 AD+BC+ Perimeter of the given gure I'D IT AEB + DFC 4+4+2X%>< (circumference ofacircle) 8+2>< %><27zr 8+2 xxi 8+2><3.l4 8+6.28 66 7 Measurements . . Perimeter of the given gure 14.28 cm. Area of the given gure = Area of a rectangle ABCD + 2 x Area of a semicircle 7272 2 4x2+2><22><1><1 lxb+2x 7><2 Total (ii) area = 8 + 3. 14 = 11. 14 cm2. Let ADB, BBC and CFA be the three semicircles I, II and III respectively. Given: > Radius of asemicircle I, r1 = 7G = 5 cm Radius ofasemicircle II,r2= § :4 cm Radius ofasemicircle III,r3= g =3cm Perimeter ofthe shaded portion Perimeter of a semicircle I + Perimeter ofasemicircle II+ Perimeter of a semicircle III = (7r+ 2)><5+(7r+ 2)><4+(72'+ 2)><3 = (72-+~ 2)(5 +4+3) = (gym = (7r+2)><12 =%><12 =61.714 Perimeter of the shadedportion 2 61.7lcm. Area of the shadedportion, A = Area of a semicircle I + Area of a semicircle II + Area of a semicircle III A ___7rr2 21+ 7rrZ 22+ rrrz 23 =»¥Qx5x5+¥Q~x4x4+§Lx3x3 7X2 7X2 7X2 Example 2.7 A horse is tethered to one corner of a rectangular eld of dimensions 70 m by 52 In by a rope 28 m long for grazing. How much area can the horse graze inside? How much area is left ungrazed? F Fig. 2.30 Length of the rectangle, l = 70 In Breadth of the rectangle, b = 52 in Length of the rope = 28 In Shaded portionAEFindicates theareain whichthehorsecangraze.Clearly,it is the area of a quadrant of a circle of radius, r = 28 m Areaofthequadrant AEF = Zl Xrrrzsq.units = -411-><-2-7-2->< 28x28=616 m2 Grazing Area = 616 ml. Area left ungrazed = Area of the rectangle ABCD Area of the quadrant AEF Area of the rectangle ABCD = l X 19sq. units 70 X 52 = 3640 m2 3640 Area left ungrazed 616 = 3024 m2. D Example 2.8 p * In the given gure, ABCD is a square of side 14 cm. Find the * area of the shaded portion. Side of a square,a = 14 cm Radius ofeach circle, r = -3cm Area of a square 4 X Area of a circle Area of the shadedportion I an I #/'\ «>1 N 2 2 C? ' A? Measurements Example 2.9 A copper wire is in the form of a circle with radius 35 cm. It is bent into a square.Determine the side of the square. Given: Radius of a circle, r = 35 cm. Since the samewire is bent into the form of a square, Perimeter of the circle = Perimeter of the square . Perimeter . of the circle = i . Fig. 2.33 27rr units = 2x-3% x35cm P = 220 cm. 0 Let a be the side of a square. Perimeter of a square = 4a units 4a = 220 a = 55 cm Side of the square 61 0 61 Fig. 2.34 55 cm. Example 2.10 L Fourequalcirclesare described aboutfour cornersof ' asquare sothateach touches twooftheothers asshown in theIt_ 2.35. Findtheareaof theshaded portion, eachsideof the square measuring 28cm. 3 Let ABCD be the given square of side a. '. a = 28cm 2.8. Radius of each circle, r 2 14 cm Area of the shadedportion Area of a square 4 x Area of a quadrant a24X-111-><7rr2 28x28-4 ><%><272><14><14 784 - Area of the shadedportion 616 168 cm2 Example 2.11 A 14In wide athletic track consistsof two straight sections each 120 m long joined by semicircular ends with inner radius is 35 m. Calculate the area of the track. Given:Radius of the inner semi circle, r = 35 m Width of the track = 14 m Radius of the outer semi circle, R R = 35 + 14 = 49 m = 49 m Area ofthetrack isthesum oftheareas ofthesemicircular tracks andtheareas of the rectangular tracks. Area of the rectangular tracks ABCD and EFGH = 2 X (l X b) = 2 x 14 X 120 = 3360 m2. Area of the semicircular tracks = 2 x (Area of the outer semicircle Area of the inner semicircle) L 2-1 2 2><<27rR 27rr> 2><-5-><7r(R2~-r2) %x(492 352) (-.atbi =(a+b)(a b)) %(49 +35) (49 35) -Z7-2-><84><14 =3696 m2 Area of the track = 3360 + 3696 = 7056 ml. Example 212 Inthegiven Fig.4.37, PQSR represents aflower bed. If OP=21mand; OR= 14m, nd theareaof the shaded portion. Given : OP 3 21 m and OR = 14 in Area of the ower bed = Area of the quadrantOQP Area of the quadrant OSR L 2-1 2 47r><7z><212L><2z><142 4 ><7r><(212 .-142) ><%2><(21+14)><(21-.14) ><% X35X7=192.5 m2. Area ofthe owerbed H Example 2.13 Find the area of the shadedportions in the Fig. 2.38, where ABCD is a squareof side 7 cm. Let us markthe unshaded portionsby I, II, III andIV as B shownin theFig.2.39. Let P,Q,R and S be the mid points of AB, BC,CD and DA respectively. Side of the square,a = 7 cm Radiusof thesemicircle, r = 1 cm 2 Area of I + Area of III = Area of a squareABCD Area with of two semicircles centres P and R a22><%-><7rr2 L Q 1 1 7x7. 2><2><7><2><2 Area of I + Area <4-%>cm2=221cm2. of III Similarly, we have Area ofII +Area ofIV (4 - cmz =221cm2. Area of the shadedportions = Area of the square ABCD (Area of I + Area of II + Area of III + Area of IV) _ 49_ (A2 +A2 ) = 49 21: 28 cm2 Area of the shadedportions = 28 cm2. Example 4.14 A surveyor has sketchedthe measurementsof a land as below. Findthearea oftheland. Let J, K, L, M be the surVeyorsmarks from A to D. KB=6rn, LE=9m,MC=l0m, AK= 10 In, AL: AM: l5rnand 12 m, AD=20m. The given land is the combination of the trapezium KBCM, LEFJ and right angled triangles ABK, MCD, DEL and JFA LX (KB+ MC)XKM ('3 parallel sides are f MC and height isKM = ~2><(6+10)><5 I<;B=6m,Mc=:10m, 1 2 A1 = §Xl6><5=40lI1. i KM:Al\/IMAK _-::15__1g_-;5m) ('.' parallel sides are A2= %X(H:+LE)XJL IFand height isJL J? = 7m,LE=9m, =.%.x(7 +9)x7 .¥L:AL~A} 2l2~5m7mi) A2 = L><16><7=56m2. 2 A5= %>
<(ADAL)>< 9 Measurements 1 A6 = §><5><7 2 l7.5m.2 Areaoftheland = A1+A2+A3+A4+A5+A6 40+56+30+25+36+ Area of the land = 204.5 m2. EXERCISE 2.2 1. Find the perimeter of the following gures 2cm 8cm 2cm 10 cm (V) 2. Find the area of the following gures 17.5 Chapter 2 3. Find the areaof the coloured regions m 2cm q 3.5cm 3.5c 3.5cm 4. In the given gure, nd the areaof the shadedportion if AC = 54 cm, BC = 10 cm, and O is the centre of bigger circle. 5. Acow istiedupforgrazing inside arectangular eldofdimensions 40mX36m in one comer of the eld by a rope of length 14 m. Find the areaof the eld left ungrazedby the cow. 6. A squarepark has eachside of 100 m. At eachcomer of 100m the park there is a ower bed in the form of a quadrant of radius 14 m as shown in the gure. Find the area of the remaining portion of the park. 1cm 7. Find the area of the shaded region shown in the gure. The . four cornersare quadrants. At the centre,there is a circle of diameter 2 cm. 8. A paper is in the form of a rectangle ABCD in which AB = 20 cm and BC = 14 cm. A semicircularportion with BC as diameteris cut off. Find the areaof the remaining part. Measurements On a squarehandkerchief,nine circular designseachof radius 7 cm are made. Find the area of the remaining portion of the handkerchief. 10. From eachof the following notes in the eld book of a surveyor,make a rough plan of the eld and nd its area. Can you help the ant? An ant is moving around a few food piecesof different shapesscatteredon the oor. For which food-piece would the ant have to take a shorter round and longer round? a circle inscribed in it? Chapter 2 Which one of these gures has perimeter? The central angleof a circle is 360°. Perimeterof a semicircle= (7f+ 2) X r units. 7rr2 2 sq . units. Area of a semicircle= 5555 The central angle of a semicircleis 180°. Perimeter ofaquadrant = +2)Xr units. Area of a quadrant = 7172 sq . units. 4 The central angle of a quadrant is 90°. Perimeter of a combined gure is length of its boundary. A polygon is a closedplane gure formed by n line segments. Regular polygons arepolygons inwhich allthesides and angles areequal. Irregular polygons are combination of plane gures. Geometry 3.1 Introduction 3.2 Properties of Triangle 3.3 Congruence of Triangles 3.1 Introduction ii Geometry was developed by Egyptians more than 1000 years l3ucli'd was before Christ, to help them mark out their elds after the oods from the agreatGreek Nlathematiciaii Nile. But it was abstractedby the Greeks into logical system of proofs who gave birth to with necessarybasic postulates or axioms. ogicalthinking ii geometry. Geometry plays a vital role in our life in many ways. In nature, Euciidcollected he various Kwe come across many geometrical shapes like hexagonal bee-hives, spherical balls, rectangular water tanks, cylindrical wells and so on. The . . . . . . nforiiiation on geomify'dr0lmd . 3008.0 c d construction ofPyramids ISaglaring example forpractical application pub, iSh 6($521 em of geometry.Geometry has numerous practical applications in many . . . . . . titheformof . l3 books in a eldssuch asPhysics, Chemistry, Designing, Engineering, ArchitectureWstenmm mmm. and Forensic Science. These boeks are called Euclid V The word Geometry is derived from two Greek words Geo which means earth and metro which means to measure.Geometry _ , , , , Eimmsa EuclidSaid: , , The wholeis is abranchof mathematics whichdealswith theshapes, sizes,positions gmam. Wm, anyOf and other properties of the object. In class VII, we have learnt about the properties of parallel lines, transversal lines,anglesin intersectinglines,adjacentandalternative angles. Moreover, wehave also come across theangle sum property of atriangle. S133? S7 Chapter 3 Let us recall the results through the following exercise. REVISION 1. In Fig.3.l, x° = 128°.Find y°. C EXERCISE 2. Find ABCE and LECD in the Fig.3.2,whereLACD = 90° 13 D Fig. 3.2 3. Two angles of a triangle are 43° and 27°. Find the third angle. 4.Find x°intheFig.3.3, if PQHRS. 5.IntheFig.3.4, twolines ABand CD P i Q ¬'°::?g:":":%§' ;;2x°+l5° intersect at the point 0. Find the value of x° and y°. 6. In the Fig. 3.5 AB [[CD. Fill in the blanks. (i) AEFB and LFGD are .................. .. angles. (ii) AAFG and AFGD are ................. .. angles. (iii) AAFE and LFGC are .................... .. angles. 3.2 Properties of Triangles Geometry 3.2.1. Kinds of Triangles Triangles can be classied into two types based on sides and angles. Based on sides: (a) Equilateral Triangle (b) IsoseelesTriangle (c) ScaleneTriangle (d) Acute Angled (e) Right Angled (f) Obtuse Angled Triangle Triangle Triangle Based on angles: 465° 703 Three acute angles One right angle 3.2.2 Angle Sum Property ofaTriangle One obtuse angle H' km Theorem 1 f ix The sum ofthe three angles ofatriangle is180°. .2g;'/ \X Given : ABCisaTriangle. ToProve : AABC + ABCA + ACAB = 180 Construction: ThroughtheVertexA drawXY parallelto BC. Fig 3-7 Proof v (i) BC ||AXYandAB is a transversal AABC = AXABT ~ uA [T Alternateangles; A(ii)ACis agtransverslal, ABCA= AYAC Alternate angles-A A (iii)AABC +AABCA =i4XAB +AYAC A Byadding (i)and (ii).ij (iv) (AABC A+ABCA) + 4CAB = AA (AXAB+?4YAC)+ AACAB A. A (V) _V ' L V , L ; Byfadding 4BAConbothsides-: .7 AABC + 4BCAs+ACAB 18o° - Tf Theangleofa straightline is 180°.: Chapter 3 «W» p (i) Triangleis a polygonof threesides. (ii) Anypolygon could bedivided intotriangles byjoining thediagonals. (iii) Thesumof theinteriorangles of apolygoncanbegivenby the formula (11 2) 180°,where n is the number of sides. Theorem 2 If a side of a triangle is produced, the exterior angle so formed, is equal to the sum of the two interior oppositeangles. Given : ABC is a triangle. BC is produced to D. To Prove : AACD =-. AABC + ACAB Proof (1) InAABC,4ABC + ABCA + ACAB =180° Anglesumpropertyof a triangle- (ii) ABCA+ 4ACD2 1800 Sumoftheadjacent angles of astraight (iii) AABC + ABCA + 4CAB= line. ABCA + AACD Equating (i) and (ii). UV) 4ABC + 4CAB=4ACD Subtracting LBCA onboth sides of(iii). (v) TheexteriorangleLACD is equalto the Hence proved sum of the interior opposite angles AABC and ACAB. Example 3.1 InAABC, AA =75°, AB =65° ndLC. We know that in AABC, AA + LB + LC = 180° 75° + 65° + LC 180° 140° + AC 180° LC 180° '. AC 40°. 140° Example3.2 InAABC, given thatAA= 70°andAB=AC.Findtheother angles ofAABC. Let AB = x° and AC = y°. Given that AABC is an isoscelestriangle. AC xv In AABC,4A + LB + AC = AB yo 180° 70 + x + y 180° 70 + x + x" 180° 8. . 2 x° 180° 70° 2 x° 110° x° , 1 1 Fig. 3.10 =55°.Hence 413 =55° and Ac=55°. Example 3.3 The measuresof the angles of a triangle are in the ratio 5 : 4 : 3. Find the angles ofthetriangle. Given that in a AABC, 4A:4B:4C = 5 : 4 : 3. Chapter 3 We know that the sum of the angles of a triangle is 180° . 5x° +4x° +3x° = 180° =>12x° = 180° x° = 1§g° =150 So, the angles of the triangle are 75°, 60° and 45°. Example 3.4 Find the angles of the triangle ABC, given in Fig.3.1l. BD is a straight line. Weknowthatanglein thelinesegment is 180°. x°+ 110° =180° 1 11 xo : 1800_110o 1 D Fig. 3.11 x° = 70° We know that theexterior angle isequal tothesum ofthetwointerior opposite angles. x°+y° = 110° 70° +y° == 110° y° = 110° 70° = 40° Hence, x° = 70° and y° = 40°. Example 3.5 Find the Valueof ADEC from the given Fig. 3.12. We know that in any triangle, exterior angle is equal to the sum of the interior angles opposite to it. In AABC, AACD = AABC + ACAB .'.4ACD = Also, AACD = 70°+50° AECD = 120° = 120°. Considering AECD, 4ECD+4CDE+ADEC =180° 120°+22°+4DEC LDEC = 180° = l80°l42° Geometry all thetypesof trianglesT1,T2,T39 T4') T5andT6.Let usnamethetriangles as ABC.Let a, b, c be the sides opposite to the vertices A, B, C respectively. What do Theorem ou observe from this table ? 3 Any two sides of a triangle together is greater than the third side. (This i known as Triangle Inequality) Consider thetriangleABCsuchthatBC» 2 c (1) AB= 8cm, AB+ BC= 20cm .'~"..iTW .. (11)BC= 12Cm9BC+ CA= 21Cm (iii) CA = 9 cm, CA + AB : 17 cm NOW Cleay (i) ormatriangle using straws oflength 3cm,4cmand 5cm. Similarlytry to formtriangles of the following length. a)5cm,7cm,1lcm. AB + BC > CA b) 5 cm,7 cm, 14cm. (ii) BC + CA > AB c)5cm,7cm,12cm. (iii) CA+AB>BC Conclude your ndings. In all the cases,we nd that the sum of any two sides of a triangie is greater than the third side. Example 3.6 Which of the following will form the sides of a triangle? (i) 23cm, 17cm, 8cm (i) (ii) 12 cm, 10 cm, 25 cm (iii) 9cm, 7cm, 16cm 23 cm, 17cm, 8cm are the given lengths. Here23+17>8,17+8>23and23+8>17. 23 cm, 17cm, 8cm will form the sides of a triangle. (ii) 12cm, 10cm, 25cm are the given lengths. Cha ter 3 Here 12 + 10 is not greater than 25. ie, .: it 12cm, 10cm, 25 cm will not form the sides of a triangle. (iii) 9cm, 7cm, 16cm are given lengths. 9 + 7 is not greater than 16. .'.9cm, 7cm and 16cm will not be the sides of a triangle. From theabove results weobserve thatinanytriangle thedifference between 1 he length of any two sidesis lessthan the third side. EXERCISE 1. (i) (ii) Choose the correct 3.1 answer: Which of the following will be the angles of a triangle? (A) 35°, 45°, 90° (B) 26°, 58°, 96° (C) 38°, 56°, 96° (D) 30°, 55°, 90° Which of the following statement is correct? (A) Equilateral triangle is equiangular. (B) Isoscelestriangle is equiangular. (C) Equiangular triangle is not equilateral. (D) Scalenetriangle is equiangular (iii) The three exterior angles of a triangle are 130°, 140°, x° then x° is (A) 90° (iv) (B) 100° (C) 110° (D) 120° Which of the following set of measurements will form a triangle? (A) 11 cm, 4 cm, 6 cm (B) 13 cm, 14 cm, 25 cm (C) 8 cm, 4 cm, 3 cm (D) 5 cm, 16 cm, 5 cm (V) Which of the following will form a right angled triangle, given that the two angles are (A) 24°, 66° (B) 36°, 64° (C) 62°, 48° (D) 68°, 32° 2. The anglesof a triangle are (x Find the three angles. 35)°, (x 20)° and (x + 40)°. 3. In AABC, the measureof AA is greaterthan the measureof LB by 24°. If exterior angle LC is 108°. Find the anglesof the AABC. Geometry 4. The bisectors of AB and AC of a AABC meet at O. Show that 41300 .=90° +TA. 5. Find the value of x° and y° from the following gures: y 50; (i) 6. Find the anglesx°, y° and z° from the given gure. 3.3 Congruence of Triangles We are going to learn the important geometrical idea Congruence. To understand what congruence is, we will do the following activity: Tatwo tenrupee notes. Place them oneover theother. What doyouobserve? note covers the other completely and exactly. V From the above activity we observe that the gures are of the same shapeand thesame size. a Ingeneral, if twogeometrical guresareidentical in shape andsizethenthey aresaidtobecongruent. Check whether the following objects are congruent or not : (a) Postal stamps of same denomination. (b) Biscuits in the samepack. (c) Shaving blades of samebrand. 85 Chapter 3 Now we will consider the following plane gures. F A U E C B D Fig. 3.13 P T R Q S Fig. 3.14 Observe the above two gures. Are they congruent? How to check? We use Step 1 : Take a trace copy of the Fig. 3.13.We can use Carbon sheet. 2? Step2 : Place thetrace copy onFig.3.14 without bending, twisting and stretching. Step 3 : Clearly the gure covers each other completely. Therefore the two gures are congruent. 3.3.1 (a) Congruence among Line Segments Two line segmentsare congruent, if they have the same length B 09¢ ,........._........§..i.."3......._.........., A Here,the lengthof AB = the lengthof CD. Hence E E CD (b) Congruence of Angles Two angles are congruent, if they have the samemeasure. M 0 40° N P 40° R Here the measuresare equal. Hence AMON 2 LPQR. Q (c) Congruence of Squares Two squares having same sides are congruent to each other. Here, sides of the square ABCD = sides of the squarePQRS. A 2cm B P 2cm Q Square ABCD 2 Square PQRS (d) Congruence of Circles Two circles having the same radius are congruent. In the given figure, r a d i u s ofcircleC1= radius ofcircleC2. A Circle C13 Circle C2 Cut this figure into two pieces through the dotted lines What do you understand from these two pieces? Theabove congruences motivated ustolearn about thecongruence oftriangles. Let us consider the two triangles as follows: A 1? 0%u\ 0 gm 60 B g 8 cm C 65$} 60 Q 8 cm R If we superpose AABC on APQRwith A on P,B on Q andC on R suchthat the twotriangles covereachotherexactly withthecorresponding vertices, sides and angles. We can match the corresponding parts as follows: 3.3.2. Congruence of Triangles Two triangles are said to be congruent, if the three sides and the three angles of one triangle are respectively equal to the three sides and three angles of the other. 3.3.3. Conditions for Triangles to be Congruent We know that, iftwotriangles are congruent, then sixpairs oftheir corresponding parts (Three pairs of sides,three pairs of angles) are equal. Buttoensure thattwotriangles arecongruent insome cases, it issufcient to Verify that only three pairs of their corresponding parts are equal,whicharegivenasaxioms. There are four such basic axioms with different combinations ofthethree pairsofcorresponding parts. These axioms helpusto identifythecongruent triangles. ifI If Sdenotes thesides, Adenotes theangles, Rdenotes theright angle and denotes thehypotenuse of a trianglethentheaxiomsareasfollows: (i) SSS axiom i) SSS Axiom (ii) SAS axiom (SideSideSide (iii) ASA axiom (iv) RHS axiom axiom) If three sides ofatriangle arerespectively equal tothethree sides ofanother riangle then the two triangles are congruent. A P Geometry We consider the triangles ABC and PQR such that, AB = PQ, BC = QR and CA: RP. Take a trace copy 0fAABC and superpose on APQR such that AB on PQ , BC on QR and AC on PR Since AB = PQ A => A lies on P, B lies on Q Similarly BC=QR=>CliesonR 0 Now, the two triangles cover each other exactly. AABC 2 APQR *W Whatwill happen when the ratio Exple H is not V Fromthefollowing gures,state whether thegivenpairsof triangles are icongruent bySSS axiom. > A :0 Compare the sides of the APQR and AXYZ PQ =XY = 5cm, QR=YZ =4.5cm and RP =ZX= 3cm. If we superposeA PQR on A XYZ. P lies on X, Q lies on Y, R lies on Z and APQR covers AXYZ exactly. Example 3.8 val; In thegure,PQSR is a parallelogram. §PQ =4.3cmandQR=2.5cm.IsAPQR 2 APSR? Consider APQR and APSR. Here, PQ = SR = 4.3 cm and PR =QS = 2.5cm. PR = PR APQRE ARSP - APQR 32 APSR [ ¢:..;:.] If any two sides and the included angle of a triangle are respectively equal to any two sides and the included angle of another triangle then the two triangles are congruent. A B C Q R We consider two triangles, AABC and APQR such that AB = PQ, AC = PR and includedangleBAC = includedangleQPR. Wesuperpose thetrace copy ofAABC onAPQR withABalong PQand AC alongPR. Now, Alies onPand Blies onQand Clies onR.Since,AB =PQ and AC=PR, BliesonQandCliesonR.BCcovers QRexactly. AABC covers APQR exactly. Hence, AABC E APQR (iii) ASA Axiom (Angle-Side-Angle Axiom) If twoangles andaside ofonetriangle arerespectively equal to anglesand the corresponding side of another triangle then the two triangles are A congruent. Consider the P triangles, AABC and APQR Here, BC = QR,413= 4Q,4C = AR pg Bythemethod ofsuperposition, B t is understood that AABC C Q covers So, B lies on Q and C lies on R. Hence A lies on P..'. AABC covers APQR exactly. Hence, AABC E APQR. As the triangles are congruent, we get remaining corresponding parts are also equal. (' )AB-PQ AC-PR d4A 4P Prove the following axioms using the papercuttings a) SSS Axiom and b) ASA Axiom R Geometry Representation: TheCorresponding Partsof Congruence TrianglesareCongruent is represented in shortformasc.p.c.t.c. Hereafter thisnotation will beusedin the problems.5 A A A A Example 3.9 AB and CD bisect Given : each other at O. Prove that AC = BD. O is mid point of AB and CD. A0 = OB and CO = OD To prove : AC = BD Proof Consider : AAOC () and ABOD A0 = OB C CO = OD LAOC = ABOD AAOC E ABOD ;~.t] Hence we get, AC = BD [:~ 5. Example 3.10 D C In the given gure, ADAB and ACAB are on thesame base AB.Prove thatADAB EACAB Consider ADABandACAB Fig315 ADAB = 35° + 20° = 55° = ACBA [. LDBA = LCAB = 20° AB is common to both the triangles. ADBA 2 ACAB Hypotenuse Do you know what is meant by hypotenuse ? Hypotenuse is a word related with right angled triangle. C hypotenuse Consider the right angled triangle ABC. AB is a right angle. The side opposite to right angle is known as the hypotenuse. H AC ' h t 91 X10111 ( lb ang e ypo enuse - 6) If the hypotenuse and one side of the right angled triangle are respectively B C E F Consider AABC and ADEF where, AB = LE = 90° Hypotenuse AC = Hypotenuse DF [ Side AB = Side DE [s By the method of superposing,we seethat AABC E ADEF. 3.3.4 Conditions which are not sufcient for congruence of triangles (i) AAA (Angle - Angle - Angle) it It isnota sufcient conditionforcongruence oftriangle. Why? Let us nd out the reason. Consider the following triangles. P 70° A 70° In the above gures, AA = AP, AB = 4Q and AC = AR But size of AABC is smaller than the size of APQR. When AABC issuperposed ontheAPQR, they willnotcover each other; exactly. AABC32APQR. (ii) SSA (SideSideAngle) Wecananalysea caseasfollows: ConstructAABC with the measurements AB = 50°, AB = 4.7 cm and AC=4cm.Produce BCtoX. WithAascentre andACasradius draw anarcof4cm. ' t BX t C dD 3 Geometry AD is also 4cm [ Consider AABC and AABD. ABiscommon. ABiscommon andAC= AD = 4cm SideAC,sideAB and413of AABCandsideB D , side AB and 4B of AABD are respectively ongruent to each others. But BC and BD are not equal. A ABC 3: A ABD. Example 3.1] Prove that the angles opposite to equal side of a triangle are equal. ABC is a given triangle with, AB = AC. To prove : Angle opposite to AB = Angle opposite to AC (i.e.) AC = AB. Construction : Draw AD perpendicular to BC. LADB = AADC = 90° Proof Condiser AABD and AACD. AD iscommon AB=AC LADB AADB Hence LABD (or) AABC = LADC E AADC = AACD = LACB. AB = LC. This is known as = 90° Hence the proof. at Example 3.12 Prove that the sides opposite to equal angles of a triangle are equal. Chapter 3 ZADB AD LADC = 90° is common side. AADB E AADC (by AAS axiom) Hence, AB = AC. B So, the sides opposite to equal angles of a triangle are equal. Example 3.13 In the given gure AB = AD and ABAC = ADAC. Is AABC E AADC? If so, state the other pairs of corresponding parts. In AABC and AADC, AC is common. ABAC AB = LDAC = AD [,»e.t..t:r] AABC 2 AADC [; So, the remaining pairs of corresponding parts are BC = DC, LABC = AADC, AACB = AACD. [:.; .1 Example 3.14 APQR isanisosceles triangle withPQ=PR,QPisproduced toSand bisects theextension angle2x°.ProvethatAQ= xandhenceprovethatPT ||QR. Given : APQR is an isoscelestriangle with PQ = PR . Proof : PT bisects exterior angle ASPR and therefore ASPT = ATPR = x°. Also Weknow that in any triangle, exterior angle = sum of the interior opposite angles. In APQR, Exterior angle ASPR = APQR + APRQ 2x° = 4Q + LR : 2x x LQ+ZQ = ZLQ = 4Q X O 94 I Geometry To prove : PT ||QR Lines PT and QR are cut by the transversal SQ. We have ASPT = x°. We already proved that AQ = x". Hence, ASPT and APQR are corresponding angles. PT ||QR. EXERCI3. 1. Choose the correct answer : (i) In the isoscelesAXYZ, given XY = YZ then which of the following anglesare equal? (A) AX and AY (ii) (C)AZ and AX (D) AX, AY and AZ In AABC and ADEF, AB = AE, AB = DE, BC = EF. The two triangles are congruentunder axiom (A) SSS (iii) (B) AY and AZ (B) AAA (C) SAS (D) ASA Two plane figures are said to be congruentif they have (A) the same size (B) the same shape (C) the same size and the same shape (D) the same size but not same shape (iv) In a triangle ABC, AA = 40° andAB = AC, then ABC is (A) a right angled (B) an equilateral (C) an isosceles triangle. (D) a scalene (V) In the triangle ABC, when AA = 90° the hypotenuse is ---- -(A) AB (B) BC (C) CA (D) None of these (Vi) In the APQR the angle included by the sidesPQ and PR is (A) 4P (B) 4Q (C)AR (D)Noneof these (Vii) Inthegure,theValueofx°is--------(A)80° (B)100° (C) 120° (D) 200° 2. In the gure, ABC is a triangle in which AB = AC. Find x° and y°. 3. In the gure, Find x°. 4. In the gure APQR and ASQR 5. In the gure, it is given that BR = PC are isoscelestriangles. Find x°. and AACB = AQRPand AB ||PQ. Prove that AC = QR. P B 6. In the gure, AB = BC = CD , AA = x. 7. Find x°, y°, z° from the gure, Prove that ADCF = 34A. where AB = BD, BC = DC and ADAC A B = 30°. D 8. Inthe gure, ABCD isaparallelogram. 9.Ingure, BObisects AABC of AB isproduced toEsuch thatAB=BE. AABC. PisanypointonB0.Prove AD produced to F suchthatAD = DF. thattheperpendicular drawnfromP Show that%FDC EACBE. A B toBAand BCareequal. A E 10. The Indian Navy ights y in a formation that can be viewed as two triangles with commonside.ProvethatASRT E A QRT, if Tisthemidpoint of soandSR=RQ. 96 L L, _ _ LV I III Geometry The sum of the three anglesof a triangle is 180°. :5 If thesidesof atriangleis produced, theexterioranglesoformed,is equal to the sum of the two interior opposite angles. %?Any two sidesof atriangletogetheris greaterthanthethird side. U? TwoplanefiguresareCongruent if eachwhensuperposed on the other coversit exactly.It is denotedby the symbol E % Twotrianglesaresaidto becongruent, if threesidesandthethreeangles of one triangle are respectivelyequal to three sidesand three anglesof the other. til? SSSAxiom: If threesidesof a trianglearerespectively equalto thethree sidesof another triangle then the two triangles are congruent. at SASAxiom: If anytwo sidesandthe includedangleof a triangleare respectively equal to any two sides and the included angle of another triangle then the two triangles are congruent. W!»ASAAxiom: If two angles andasideof onetrianglearerespectively equal to two anglesand the corresponding sideof another triangle then the two triangles are congruent. Q? RHS Axiom: If thehypotenuse andonesideof therightangledtriangle are respectivelyequal to the hypotenuseand a sideof another right angled triangle, then the two triangles are congruent. Chapter 3 THE IMPORTANCE OF CONGRUENCY In our daily life, we use the concept of congruencein many ways. In our home, we use double doors which is congruent to each other. Mostly our house double gate is congruent to each other. The wings of birds are congruent to each other. The human body parts like hands, legs are congruent to each other. We can saymany exampleslike this. Birds while flying in the sky, they fly in the formation of atriangle. If youdrawamedian through theleading birdyou canseea congruence. If thecongruency collapses thenthe birdsfollowing attheendcouldnotflybecause theylosses their stability. V Now, try to identify the congruence structuresin the nature and in your practical life. QI'D' A . i C'I'D'I'D'I'i . - . . . _ E ii Practical Geometry 4.2 2 Quadrilateral E immmmma l 4.3Trapezium i i Parallelogram i E l %\~a.a...M..m.. Fig. 4.1 is 360°. . (AB,AD),(AB,BC), (BC,CD), (CD,DA) areadjacent sides. A6andi BD arethediagonals. AA,41-3, 4c and 41)(orLDAB, LABC, ABCD, ACDA) are theanglesi of thequadrilateral ABCD. 4A+4B+4C+4D=360° 4.2.2 Area of a Quadriiateral Let ABCD be any quadrilateralwith §5 as one of its diagonals. Let AE andW be theperpendicularsdrawn from the VerticesA and C on diagonalE From the Fig. 4.2 Area of the quadrilateral ABCD = Area of A ABD + Area of A BCD Fig. 4.2 %><(AE+CF) =%xdx(h1+ hz) sq. units. Practical Geometry where BD d, AE h, and CF = hz. Area of a quadrilateral is half of the product of a diagonal and the sum of the altitudes drawn to it from its opposite vertices. That is, t tivit qr C Byusing paper folding technique, verify A=%d(hl+hz) 4.2.3Construction ofa Quadrilateral In this class, let us learn how to construct a quadrilateral. p To constructa quadrilateral rst we constructa triangle from the given data. Then, wendthefourthvertex. Toconstruct a triangle, werequire threeindependent measurements. Also we needtwomoremeasurements to nd thefourthvertex. Hence,weneedfive independent measurements toconstruct aquadrilateral. Wecanconstruct, aquadrilateral, when thefollowing measurements aregiven: if (ii) Foursides andoneangle V Vg Threesides,one diagonal andoneangle Three sides andtwojanglesp C ACDA (V) Twosides andthreeangles 4.2.4 Construction ofaquadrilateral when foursides andonediagonal aregiven Example 4.1 Construct a quadrilateral ABCD with AB = 4 cm, BC = 6 cm, CD = 5.6 cm DA = 5 cm and AC = 8 cm. Find also its area. Given: AB = 4 cm, BC = 6 cm, CD = 5.6 cm Rmlgh Diagram DA=5cmandAC=8cm. Toconstruct aquadrilateral 5-6 " C Steps for construction A Step1 : Drawa roughgureandmarkthegivenV: W II1CE1S1.1I'C1I1¬11tS. Step 2 : Draw a line segmentAB = 4 cm. Step3 : WithA andB ascentres drawarcsofradii 8 cm and 6 cm respectively and let them cut at C 101 Fig4.3 Chapter 4 A Fig. 4.4 Step4 : JoinR andBC. V Step 5 : WithA andC ascentres drawarcsof radii5 cm,and5.6 respectively andlet themcut at D. A Step 6 : Join E and CD. ABCD is the requiredquadrilateral. Step7 : FromB drawE _l_ ACandfromD drawW J. AC,thenmeasure thelengths ofBBandDF.BE= h1=3cmandDF= h2=3.5cm. A AC = d = 8 cm. Calculation of area: In the quadrilateralABCD, d = 8 cm, h1= 3 cm andh2= 3.5 cm. Area ofthequadrilateral ABCD =£-d(hl+k2) =%(8)(3 +3.5) _..1_ - 2><8><6.5 = 26 cm2. 4.2.5 Construction of a quadriiaterai when four sides and one angle are given Example 4.2 Construct aquadrilateral ABCDwithAB= 6 cm,BC= 4 cm,CD= 5 cm, DA = 4.5cm, AABC = 100°andnd its area. Given: AB 6 cm, BC = 4 cm, CD = 5 cm, DA = 4.5 cm LABC = 100°. Practical Geometry To construct a quadrilateral Rough Diagram X A, 0112 Steps for construction Step 1 : Draw a rough diagram and mark the given measurments. Step 2 : Draw a line segmentBC = 4 cm. Step 3 : At B on BC make ACBX whosemeasureis 100°. Step4 : With B ascentreandradius6 cm drawan are.This cutsB5?at A. Join CA Step 5 : With C and A as centres, draw arcs of radii 5 cm and 4.5 cm respectively and let them cut at D. Step 6 on Join CD and AD. ABCD is the required quadrilateral. Step 7 : From B draw B15J. AC andfrom D draw D-B_LAC . Measurethe lengthsof BF and DE. BF = h] = 3 cm, DE = h2= 2.7 cm and AC = d = 7.8 cm. Calculation of area: In the quadrilateral ABCD, d = 7.8 cm, h1= 3 cm and h2= 2.7 cm. Area of the quadrilateral ABCD 1 §d@+@ %(7.8) (3+2.7) =%><7.8><5.7 =22.23 cmz. Chapter 4 4.2.6 Construction of a quadrilateral when three sides, one diagonal and one angle are given Example 4.3 Construct a quadrilateral PQRS with PQ = 4 cm, QR = 6 cm, PR = 7 cm, PS = 5 cm and APQS = 40° and nd its area. Given: PQ = 4 cm, QR = 6 cm, PR: 7 cm, PS: 5 cm and 4 PQS = 40°. Rough Diagram To construct a quadrilateral Fig. 4.7 Steps for construction Step 1 : Draw a rough diagram and mark the given measurements. Step 2 : Draw a line segment PQ = 4 cm. Step 3 : With Pand Qascentres draw arcs ofradii7cmand 6cmrespectively and let them cut at R. : Step4 : JoinK and Step5 : AtQonE make |l_Qlwhosemeasure is40°. Step6 : WithP ascentreandradius5 cmdrawanare.ThiscutsEl: atS. Step7 : JoinRS. PQRS is the required quadrilateral. Step 8 : From Qdraw QX_L E and from Sdraw SY_I_ W.Measure the lengthsQXandSY.QX= h1= 3.1cm,SY= h2= 3.9cm. PR=d=7cm. it 71647 Practical Geometry Calculation of area: In the quadrilateralPQRS,d = 7 cm, h1= 3.1cm andh2= 3.9cm. Area ofthe quadrilateral PQRS =%d(hi+hz) =%wx31+3m =ix7x7 2 = 24.5 cmz. .2.7 Construction of a quadrilateral when three sides and two angles are given Example 4.4 Construct a quadrilateral ABCD with AB = 6.5 cm, AD = 5 cm, CD = 5 cm, ABAC= 40°andAABC= 50°,andalsond itsarea. Rough Diagram Given: AB = 6.5 cm, AD = 5 cm, CD = 5 cm, um '4 ABAC= 40°andLABC= 50°. To construct a quadrilateral Y x M» K? \@@§ 05% 6.5 cm Fig. 4.10 Steps for construction Step 1 : Draw a rough diagram and mark the given measurements. Step 2 : Draw a line segmentAB = 6.5cm. Step 3 : At A on AB make ABAX whosemeasureis 40° and at B on AB make LABY whose measure is 50°. They meet at C. Chapter 4 With A and C as centres draw two arcs of radius 5cm and let them cut at D. Step 5 : Join @ and W ABCD is the required quadrilateral. Step 6 : FromD draw DB J. AC andfrom B draw B-CJ. AC . Thenmeasure the lengthsofBC andDE. BC = hi =4.2 cm, DE = hz :43 cm and AC = d = 5 cm. Calculation of area: In the quadrilateralABCD, d = 5 cm, BC = h1= 4.2 cm andh2= 4.3 cm. Area ofthe quadrilateral ABCD =%d(h,+h2) =-%(5)(4.2 +4.3) =%>< 5x 8.5=21.25 cmz. 4.2.8Construction ofaquadrilateral whentwosides andthreeangles aregiven Example 4.5 L Construct aquadrilateral ABCD withAB=6cm,AD=6cm,AABD = A BDC=40°andA DBC=40°.Findalsoitsarea. Given: AB = 6 cm, AD = 6 cm, LABD = 45°, .4BDC=40° and4DBC=40°. To construct a quadrilateral Rough Diagram Practical Geometry Steps for construction Step 1 : Draw a rough diagram and mark the given measurements. Step 2 : Draw a line segmentAB = 6 cm. Step 3 : At B on E make AABX whosemeasureis 45°. Step4 : WithA ascentreand6 cmasradiusdrawanarc.Letit cutB? atD. Step5 : JoinE Step6 : At B on BD makeA DBY whosemeasure is 40°. Step7 : At D on BD makeA BDZwhosemeasure is 40". Step8 : Let B? andD2 intersect atC. ABCD is the required quadrilateral. Step 9 : FromA draw E _LW andfrom C draw § J. §. Thenmeasure the lengthsof AE andCF. AE = hl : 4.2 cm, CF : hz :3.8 cm and BD : d : 8.5 cm. Calculation of area: In the quadrilateralABCD, d : 8.5 cm, hi: 4.2 cm andhz: 3.8 cm. Area ofthe quadrilateral ABCD : ~12d(h}+h2) =12(8.5) (4.2 +3.8) =i><8.5><8 = 34cm2. 2 EXERCISE Draw quadrilateral 4.1 ABCD with the following measurements. Find also its area. 1. AB: 5cm,BC: 6cm,CD: 4cm,DA:5.5cmandAC: 7cm. 2 AB : 7 cm, BC : 6.5 cm, AC : 8 cm, CD: 6 cm and DA : 4.5 cm. 3 AB : 8 cm, BC : 6.8 cm, CD : 6 cm, AD: 4 AB : 6 cm, BC : 7 cm, AD : 6 cm, CD: 5 cm, and A BAC = 45°. 5. AB : 5.5 cm, BC : 6.5 cm, BD : 7 cm, AD: 6.4 cm and A B = 50°. 5 cm and A BAC= 50°. 6 AB : 7 cm, BC : 5 cm, AC : 6 cm, CD: 4 cm, and A ACD = 45°.. 7 AB : 5.5 cm, BC : 4.5cm, AC : 6.5 cm, 4 CAD = 80° and A ACD = 40°. 8 AB:5cm,BD:7cm,BC:4cm, 9 AB : 4 cm, AC = 8 cm, 4 ABC = 100°, 1. ABD = 50° and 4 CAD A BAD=l00° andé DBC=60. = 40°. 10. AB = 6 cm, BC = 6 cm, A BAC = 50°, 4 ACD = 30° and .4 CAD = 100°. Chapter 4 4.3 Trapezium 4.3.1 Introduction Inthe class VIIwe have learnt special quadrilaterals such astrapezium and isoscelestrapezium. We have also learnt their properties. Now we recall the denition of a trapezium. 4.3.2 Area of a trapezium Let us consider the trapezium EASY Y _ s ig.4.13 H AAA Wecanpartition theabove trapezium intotwotriangles by drawing; adiagonal W. if Onetrianglehasbasea ( EA = a units) TheothertrianglehasbaseVS"( YS = b units) Weknow a || YF = HA = h units Now,theareaof A EAYis 1ah.Theareaof A YASis 1bh. 2 2 Hence, the area of trapezium EASY = Area of A EAY + Area of A YAS _2 ,1,ah + .1. 2 bh = -12h(a+b)sq. units =%>< height >< (Sum ofthe parallel sides) sq. units Area of Trapezium Practical Geometry onstruction o a trapezium In general to construct a trapezium, we take the parallel sides which has greater measurementas base and on that base we construct a triangle with the given measurementssuch that the triangle lies between the parallel sides.Clearly the vertex opposite to the base of the triangle lies on the parallel side opposite to the base. We draw the line through this vertex parallel to the base. Clearly the fourth vertex lies on this line and this fourth vertex is xed with the help of the remaining measurement. Then by joining the appropriate Verticeswe get the required trapezium. To construct a trapezium we need four independent data. We can construct a trapezium with the following given information: an 0n;ai;;g¢n;1 Three sides andoneangle A (iii) Twosides andtwoangles (iv) Four sides 4.3.4 Construction of a trapezium when three sides and one diagonal are given Example 4.6 Constructa trapeziumABCD in which AB is parallelto DC, AB = 10cm, BC = 5 cm, AC = 8 cm and CD = 6 cm. Find its area. Rough Diagram Given: E isparallel toW AB=10cm, BC=5cm, AC=8cmand CD=6cm. To construct a trapezium 109 . . CK \,~ Chapter 4 Steps for construction Step 1 : Draw a rough diagram and mark the given measurements. Step 2 : Draw a line segmentAB = 10 cm. Step 3 : With A and B ascentresdraw arcs of radii 8 cm and 5 cm respectively and let them cut at C. Step4 : JoinE andW. Step5 : Draw65 parallelto Q Step6 : WithC ascentreandradius6 cmdrawanarecutting(Di)atD. Step7 : JoinAD. ABCD is the required trapezium. Step 8 : From C draw @_L E andmeasurethe length of CE. CE = h = 4 cm. AB=a=10cm,DC=b=6cm. Calculation of area: In the trapezium ABCD, a = 10 cm, b = 6 cm and h = 4 cm. Area ofthe trapezium ABCD -12h(a+b) %.(4)(1o +6) L 2><4><5.6><4.8><11 2 2.4><5.6>A quadrilateral with onepairof oppositesidesparallelis calledatrapezium. To constructa trapeziumfourindependent measurements arenecessary. If nonparallel sidesareequalin a trapezium, it is calledanisosceles trapezium. Q? To constructanisosceles trapezium threeindependent measurements arenecessary. 249A quadrilateral with eachpairof oppositesidesparallelis calleda parallelogram. To constructaparallelogram threeindependent measurements arenecessary. ti'>The area ofaquadrilateral, A=%(1(hi+hz) sq. units, where disthe diagonal, hi andhzarethealtitudes drawnto thediagonal fromitsopposite Vertices.E Qi9The area ofatrapezium, A=-12h(a+b)sq. units, where aandbarethelengths of theparallel sides and/9 istheperpendicular distance between thetwo parallel sides. i$The area ofaparallelogram, A-"I bhsq. units, where bisthebaseoftheparallelogram and/9is theperpendicular distance between theparallelsides. The golden rectangle is a rectangle which has appeared in art and architecture through the years. The ratio of the lengths of the sides of a golden rectangle is approximately : 1. This ratio is called the golden ratio. A golden rectangle is pleasing to the eyes. The golden ratio was discovered by the Greeks about the middle of the fth century B.C. The Mathematician Gauss, who died in 1855, wanted a 17-sided polygon drawn on his tombstone, but it too closely resembled a circle for the sculptor to carve. 0 Mystic hexagon: A mystic hexagon is a regular hexagon with all its diagonals drawn. 123 i Answers AANSWERS Chapter 1. Number System 1. i) A ii) C 2. i) Commutative iii) ii) iv) Additive identity 3. i) Commutative B iv) D V) A Associative iii) Commutative v) Additive inverse ii) Multiplicative identity iii) Multiplicative Inverse iv) Associative v) Distributive property of multiplication over addition . 505 6' 1) 252 .. :_1_ 11)14 1.1)% 11) -3 -£51 2' 1) 7o1 0 "@243 11) 110220 - .3: ..§.. .2. -- 3' 1) 8 16 32 L L 111) %-71% 1V) 33:111) 30 20 1V)24 12 .4...1._83 167 11)60 120240 -_5 - 111) 12 8 48 L A Q 1) 48 96 192 Note: In the above problems 1, 2 and 3; the given answers are one of the possibilities. 1. 1) A 11) B 111) C 1V) A V) B 2. 1)2% 11) % 111) %i 1V)1% V)11498 V1) 4% V11) 4 V111) -5% 1. 1) C 11) B Vi) A Vii) B . -1 2. 1)67 V1) 54 .. 111) A viii) B 1 1V) D V) C ix) B X) D . 2 1 11)64 111) 625 iv) 67:5 v) 372 V11) 1 V111) 256[)4 ix)231 124 x) 5-1- Answers (ii), (iii), (V) are not perfect squares. 2. 1) 4 ii) 9 iii) 1 3. i) 64 ii) iii) 81 4. 16 i) 1+3+5+7+9+11+13 iii) 1+3+5+7+9 iv) 5 V) 4 ii)1+3+5+7+9+11+13+15+17 iv)1+3+5+7+9+11+13+15+17+19+21 5. 1)694 11)% 6. 1)9 11) 49 7. a) 42+52+2_(F=212 52+ Q3+ 302= 312 111) 2151v)% 111) 0.091v)3 v) 196% v) % v1)0.36 b) 10000200001 100000020000001 6--+72+Q3=3 1. 1) 12 11) 10 2. 1)% 11) 211- 111) 7 1v)4 3. 1) 48 11) 67 111) 59 1v) 23 v) 57 v1) 37 v11) 76 v111) 89 ix) 24 x) 56 4. 1) 27 11) 20 111) 42 1v) 64 v) 88 v1) 98 v1) 77 v111) 96 ix) 23 x) 90 5. 1) 1.6 11) 2.7 111) 7.2 1v) 6.5 v) 5.6 v1) 0.54 v11) 3.4 111) 27 1v) 385 v111) 0.043 6. 1) 2 11) 53 111) 1 1v) 41 v) 31 7. 1) 4 11) 14 111) 4 1v) 24 v) 8. 1) 1.41 11) 2.24 111) 0.13 1v) 0.94 v) 1.04 9.21m 149 10.1)-§%11)¬51-3111) % 1v)1% 125 Answers 1. i) 12.57 iv) 56.60m 2. i) 0.052 m iv) 0.133 gm 3. i) 250 ii) 25.42 kg iii) V) 41.06In vi) 729.94 km ii) iii) 58.2941 vi) 100.123 3.533 km V) 365.301 ii) 150 iii) 39.93m 6800 iv) 10,000 iii) 402 iv) 306 ii) 6, 8, 10 iii) 63, 56,49 v) 15,21,28 vi) 34, 55, 89 v) 36 lakhs vi) 104 crores 4. i) 22 ii) 777 1. 1) 25, 20, 15 iv) 7.7, 8.8, 9.9 v) 300 vi) vii) 125,216, 343 2. a) 11jumps b) 5 jumps 3. a) 10 rows of apples = 55 apples b) 210 apples AI26 10,000 Answers Chapter 2. Measurements LDC vi) D mB m)A vii) B viii) 2. i) 180 cm, 1925 cm2 C ii) iwn wA ix) A X) C 54 Cm, 173.25 cm2 iii) 32.4 m, 62.37 m2 iV)25.2 m, 37.73 m2 3. i) 7.2 cm, 3.08 crnz ii) iii) 216 Cm, 2772 C1112 144 cm, 1232 cm2 iv) 288m, 4928 m2 4. i)350 cm, 7546 cm2 ii) 250 cm, 3850 cm2 iii)150 In, 13861112 iv) 100m, 616 H12 5. 77 cm2, 38.5 cm2 6. Rs.540 1. i) 32 cm ii) 40 cm iii) 32.6 cm iv) 2. i) 124 cm2 ii) 25 m2 iii) 273 cm2 iv) 49.14 cm? v) 10.401112 3. i) 24 m2 ii) 284 cm2 iii) 308 cm2 iv) 10.5 cm2 V) 135.625 cmz 1286 m2 40 cm v) 98 cm Vi) 6.125cm2 4. 770 cm2 5. 6. 9384 m2 7. 9.71 cm? 8. 203 cm2 9. 378 cm2 10. i) 15,100 m2, ii) 550000 m2 Chapter 3. Geometry 1.y°=52° 2. x°=40° 5. x° = 105° 6.i) Corresponding angle, ii) Alternate angle, iii) Corresponding angle 1. i) B ii) A 2. 3. x° = 65° 3. 4A=110° iii) A 4. x°=40° iv) B V) A x° =42° 5. i) x° = 58°, y° = 108° ii) x° = 30°, y° = 30° iii) x° = 42°, y° = 40° 6. x° = 153°, y° = 132°, z° = 53°. mcm imc 2. x° = 66°, y° = 132° mc wB mA 3. x° = 70° 7 x°=30° 127 y =60° vmB Answers Play wifhNumiaers Sequential. Inputs ofnumbers with8 1X8+1 = 9 12X8+2 = 98 123X8+3 = 987 1234X8+4 = 9876 12345 X 8 + 5 = 98765 123456 X 8 + 6 = 987654 1234567 X 8 + 7 = 9876543 12345678 X 8 + 8 98765432 123456789 X 8 + 9 987654321 Sequential 8swith9 9X9+7 98 X 9 + 6 8 wgthom 8 = 88 12345679X 9 = 111111111 = 888 12345679 X 18 = 222222222 987 X 9 + 5 = 8888 12345679 X 27 = 333333333 9876 X 9 + 4 = 88888 12345679 X 36 = 444444444 98765 X 9 + 3 = 888888 12345679 X 45 = 555555555 987654 X 9 + 2 = 8888888 12345679 X 54 = 666666666 9876543 X 9 + 1 = 88888888 12345679 X 63 = 777777777 98765432 X 9 + 0 = 888888888 12345679 X 72 = 888888888 12345679 X 81 - 999999999 Numeric Palindrome with 1 X 1 = 11 X 11 1 = 121 111X111 = 12321 1111 X 1111 = 1234321 X 11111 = 123454321 X 111111 = 12345654321 X 1111111 = 1234567654321 11111 111111 1111111 1s 11111111 X 11111111 111111111 X 111111111 123456787654321 1234567898765432 8112988 '3 can, 5did Student's Activity Recard . 129