TOTAL 90 DETAILS OF THE CONCEPTS TO BE MASTERED BY EVERY CHILD OF CLASS X WITH EXERCISE AND EXAMPLES OF NCERT TEXT BOOK SA-I SYMBOLS USED *:-Important Questions, **:- Very important Questions, ***:- Very very important Questions S.No TOPIC CONCEPTS DEGREE OF IMPORTANCE References(NCERT BOOK) 01 Real Number Euclid’s division Lemma & Algorithm *** Example -1,2,3,4 Ex:1.1 Q:1,2,4 Fundamental Theorem of Arithmetic *** Example -5,7,8 Ex:1.2 Q:4,5 Revisiting Irrational Numbers *** Example -9,10,11 Ex: 1.3 Q:1.2 Th:1.4 Revisiting Rational Number and their decimal Expansion ** Ex -1.4 Q:1 02 Polynomials Meaning of the zero of Polynomial * Ex -2.1 Q:1 Relationship between zeroes and coefficients of a polynomial ** Example -2,3 Ex-2.2 Q:1 Forming a quadratic polynomial ** Ex -2.2 Q:2 Division algorithm for a polynomial * Ex -2.3 Q:1,2 Finding the zeroes of a polynomial *** Example: 9 Ex -2.3 Q:1,2,3,4,5 Ex-2.4,3,4,5 03 Pair of Linear Equations in two variables Graphical algebraic representation * Example:2,3 Ex -3.4 Q:1,3 Consistency of pair of liner equations ** Ex -3.2 Q:2,4 Graphical method of solution *** Example: 4,5 Ex -3.2 Q:7 Algebraic methods of solution Substitution method Elimination method Cross multiplication method Equation ** Ex -3.3 Q:1,3 Example-13 Ex:3.4 Q:1,2 Example-15,16 Ex:3.5 Q:1,2,4 Example-19 Ex-3.6
1. 2. 3. 4. 5. Circle Semi- circle Ring (Shaded region) Sector of a circle Segment of a circle or + 2r 2 + R) l+2r= +2r Sin 2 ½ 2 (R2-r2) or - sin 2 A B l Area of major segment= Area of a circle – Area of minor segment Distance moved by a wheel in 1 rotation=circumference of the wheel Number of rotation in 1 minute =Distance moved in 1 minute / circumference LEVEL-I If the perimeter of a circle is equal to that of square, then the ratio of their areas is 22/7 14/11 7/22 11/14 [Ans-ii]
28. Required fig. 1 Since, tangents from an external point to a circle are equal in length BP = BQ -----------(i) CP = CR -----------(ii) And, AQ =AR -----------(iii) 1 or, AB + BQ = AC + CR or, AB + BP = AC + CP 1 Now, Perimeter of = AB + ( BP +PC) + AC = (AB + BP) + (AC +PC) = 2 (AB+BP) = 2(AB+BQ) = 2AQ Proved. 1 Or Required fig. 1 We know that the tangents to a circle from an external point are equal in length. --------------(i) BP = BQ -------------(ii) CR = CQ ------------(iii) DR =DS ------------(iv) 1 Adding (i), (ii), (iii) & (iv), we get (AP+BP) + (CR+DR)= (AS+DS) + (BQ+CQ) or, AB + CD = AD +BC 1 or, 2AB = 2BC or, AB = BC so, AB = BC = CD = AD ABCD is a rhombus. Proved. 1 29. For correct fig. Let AB = Building, CD = Tower In, tan300= or, 1 1 In, tan600= or, = -------------(ii) 1 By (i) & (ii) 1
155-160 160-165 6 5 157.5 162.5 2 3 12 13 Mean=a+h x =147.5+5 x 23/51 =147.5+115/51 =(147.5+2.255)cm=149.755cm Let us find median of the data:- Class of height (in cm) Frequency number of girls Cumulative frequency 135-140 140-145 145-150 150-155 155-160 160-165 4 7 18=f 11 6 5 4 11=cf 29 40 46 51 Total n we have median class (145-150) it gives l=145,h=5,f=18,cf=11 median=l+ x h=145 + x5 =145+14.5 x5 18 =145+4.028 =149.028cm Let us find mode of the data:- Class of heights (in cm) FREQUENCY (No of Girls) 135-140 140-145 145-150 150-155 155-160 160-165 4 7=f1 18=fm 11=f2 6 5 Total 51 (Modal class) Modal class is 145-150 Thus l=145, h=5, fm=18, f1=7, f2=11 Mode=H xh=145 + x 5 =145+55/18 =145+3.055 =148.055 cm CONCLUSION:- Mean=149.755, median=149.028 and mode=148.055 3x median=3x149.028=447.084 Mode + 2 x mean=148.055+2x149.755 =148.055+299.510=447.565 Thus we have verified that 3x median =mode + 2 x mean(Approx) ACTIVITY – 5