CSS Middle Standard “Mathematics” 101 –x _2x 3 –3x –3 10xyz 7x4 – 10xy + 10xyz _3x4 15xy 13xyz 4x4 – 25xy + 23xyz (vii) 18x2yz – 13xy + 10xy from 10x2yz + 10xy 18x2yz – 3xy from 10x2yz + 10xy 10x2yz + 10xy _18x2 yz 3xy –8x2yz + 13xy Q4. Simplify the following expressions. (i) (a + b) + (2a – b) = a + 2a + b – b = 3a + 0 = 3a (ii) (x2y + 4) – (13x2y + 14y2 – 6) = x 2y + 4 – 13x2y – 14xy2 + 6 = x 2y – 13x2y – 14y2 + 4 + 6 = –12x2y – 14y2 + 10 (iii) (14a2 – 3a) – {(–3a + 6a2 ) – (29 – 8a2 )} = (14a2 – 3a) – {–3a + 6a2 – 29 + 8a2} = 14a2 – 3a + 6a – ba2 + 29 – 8a2 = 14a2 – 6a2 – 8a2 – 3a + 3a + 29 2 2 = 14a – 14a – 3a + 3a + 29 = 29 (iv) {(16m2 – 3n) – (2m2 + 5n)} 16m2 – 3n – 2m2 – 5n = 16m2 – 2m2 – 3n – 5n = 14m2 – 8n (v) (10x2yz + 2yz) – [3yz + {2x2y + z + 5y2 – (2yz + 6)}] = (10x2yz + 2yz) – [3yz + {2x2y + z + 5y2 – 2yz – 6}] = 10x2yz + 2yz – 3yz – 2x2y – z – 5y2 + 2yz + 6 = 10x2yz + 2yz – 3yz + 2yz – 2x2y – 5y2 – z + 6 = 10x2yz + yz – 2x2y – 5y2 – z + 6 Exercise (8d) Q1. If x = 2, evaluate the following. (i) x + 6 Put x = 2, x + 6 = 2 + 6 = 8 (ii) 16x2 Put x = 2, 16x2= 16(2)2 = 16(4) = 64 (iii) 2x + 1 4x = 2(2) + 1 4(2) = 4 + 1 8 = 33 8 (iv) 14x2 – 13x = 14(2)2 – 13(2) = 14(4) – 26= 30 (v) 16 x + 18 x = 16 2 + 18 2 = 8 + 9 = 17 Q2. Evaluate the following for x = 1 y = 2. (i) 16x2 + 4y2 . Put x = 1, y = 2= 16(1)2+ 4(2)2 = 16(1) + 4(4) = 16 + 16 = 32 (ii) x + y + 14xy. Put x = 1, y = 2 = 1 + 2 +14(1)(2) = 1 + 2 + 28 = 31 (iii) 3x – 4xy + 15x2 . = 3(1) – 4(1)(2) + 15(1)2 = 3 – 8 + 15= 18 – 8 = 10 (iv) 2x – 1 5 y. = 2(1) – 1 5 (2) = 2 – 2 5 = 10 2 5 = 8 5 = 1.6
CSS Middle Standard “Mathematics” 102 (v) 10x2 – xy + y2 .= 10(1)2 – (1)(2) + 22= 10 – 2 + 4 = 12 Q3. Evaluate the following for x = 2, y = 3, z = –1. (i) 2x3 + 4xy – 3z. = 2(2)3 + 4(2)(3) – 3(–1)= 2(8) + 4(6) + 3 = 16 + 24 + 3 = 43 (ii) 4x3 – 2y2 – 2z. = 4(2)3 – 2(3)2 – 2(–1) = 4(8) – 2(9) + 2 = 32 – 18 + 2 = 16 (iii) 15x2 + 10z – 4xz + 2y. = 15(2)2 + 10(–1) – 4(2)(–1) + 2 (3) = 15(4)–10 + 8 + 6 = 60 – 10 + 8 + 6 = 64 (iv) 41x + 10xy – 17xyz. = 41(2) + 10(2)(3) – 17(2)(3)(–1) = 82 + 10(6) + 17(6)= 82 + 60 + 102= 244 (v) 14xyz – 4z + 10xyz. = 14(2)(3)(–1) – 4(–1) + 10(2)(3)(–1)= –14(6) + 4 – 10(6) = –84 + 4 – 60= –140 (vi) 105x + 14xy – 100z. = 105(2) + 14(2)(3) – 100(–1) = 210 + 14(6) + 100 = 210 + 84 + 100 = 394 (vii) 10x2 y 2 + 10z – 18z2 . = 10(2)2 (3)2– 10(–1) – 18(–1)2 = 10(4)(9) – 10 – 18 = 10(36) – 10 – 18 = 360 – 10 – 18 = 332 Q4. If a = 1, b = 2, c = –1. (i) 4abc. = 4(1)(2)(–1) = –8 (ii) 2a + 2b2 – 3c. = 2(1) + 2(2)2 – 3(–1) = 2 + 2(4) + 3 = 2 + 8 + 3 = 13 (iii) a 2 – 2b2 + 2c. = (1)2 –2(2)2 + 2(–1) = 1 – 2(4) –2 = 1 – 8 – 2 = –9 (iv) (a + b)2 – (b – c)2 . = (1 + 2)2 – (2 + 1)2 = 9 – 9 = 0 (v) 3a2 – 14c2 . = 3(1)2 – 14(–1)2 = 3(1) – 14(1) = 3 – 14 = –11 (vi) 19b2 + 16ab – 3c2 . = 19(2)2 + 16(1)(2) – 3(–1)2 = 19(4) + 16(2) – 3 = 76 + 32 – 3 = 105 Review Exercise 8 Q1. Choose the correct answer and fill the circle. i. What is added to 2a3 – 2a2 + 3a – 1 to get a3 + 3a2 – 3a + 1? –a3 + 6a2 – 6a + 2 –2a3 + 3a2 – 3a + 2 –a3 – 6a2 + 6a + 2 4a3 + 3a2 – 3a + 2 ii. 4x + 3y is an algebraic: expression sentence equation term iii. x 2yz and xy2 z 2 are: constants like terms unlike terms sentence iv. z is called: constant variable coefficient expression v. The coefficient in 4x2 is: 14x2 x 2 4 2 vi. 4x in this expression how many terms are used: 13 3 6 1 vii. If x = 6, y = 5, the value of x + 2y is: 12 14 16 18 viii. The sum of 3a2 + 5a2 + 7a2 is: 10a2 12a2 15a2 20a2
CSS Middle Standard “Mathematics” 103 ix. 4x is which type of algebraic expression: polynomial monomial binomial trinomial Q2. Write T or F if statement is true or false respectively. (i). 13<6 is true statement. T (ii). 3x2 and 3x are like terms. F (iii). The exponent in 3x4 + 1 is 1. F (iv). 16x + 14y = 20 is false statement. F (v). In open statement, the sentence must be true. T Q3. Evaluate each of the following algebraic expression. (i). 13x – 4y + 2z when x = 2, y = –3, z = –5 = 13(2) – 4(–3) + 2(–5)= 26 + 12 – 10 = 38 – 10 = 28 (ii). 18(x – 2y) when x = 5, y = 3 = 18(5 – 2(3)) = 18(5 – 6) = 18(–1) = –18 (iii). 4 5 a – 5 8 b when a = 10, b = 16 4 1(10 5 2 5 ) 16 8 – ( 2 ) = 8 – 10= –2 (iv). 3 9 b + 18 3 c 2 when b = 27, c = 6 3 18 2 27 6 9 3 ( ) ( ) 3 12 3 27 9 ( 18 3 ) ( 36 ) = 9 + 216 = 225 (v). x + 14xy – z2 when x = 2, y = 1, z = 2 = 2 + 14(2)(1) – (2)2 = 2 + 28 – 4 = 26 Q4. Translate each of the following algebraic expression into words. (i). 3a Ans: 3 times of a. (ii). 14a +b Ans: Sum of 14 times of a and b or 14 times a is increased by b. (iii). 3x + 4y Ans:Sum of 3 times a and 4 times y. (iv). (4x + 6) 5 Ans: Four times x increased by 6, divided by 6. (v). 1(a + 4) Ans: a is increased by 4 times. (vi). 13x + 14y + 13 Ans: 13 times x is increased by 14 times y and thirteen. Q5. Find the sum of following expression. (i). 4x2 + 4x – 3x, 12x + 4x, 13x3 – 10x 4x2 + x, 16x, 3x2– 10x 4x2 + x 16x 13x2 – 10x 17x2 + 7x (ii). 18z2 – 14z, 10z3 + 15z – 2z2 , 4z3 – 10z 18z2 – 14z 10z3 + 15z – 15z 4z3 – 10z 14z3 + 16z2 – 9z (iii). 100p2 – 14p +13, 50p2 + 28p, 14p3 – 10p2 100p2 – 14p +13 50p2 + 28p 14p3 – 10p2 14p3 + 140p2 + 14p + 13 (iv). 18a3 – 14a2 , 13a4 – 11a3 + 100a2 , a3 + 10a2 18a3 – 14a2 13a4 – 11a3 + 100a2 a3 + 10a2 13a4 + 8a3 + 96a2 (v). 117q3 + 14q2 – 10q, 100q4 – 3q2 , 10q2 – 14q 117q3+ 14q2 – 10q 100q4 – 3q2 10q2 – 14q (vi). 71s2 – 14s – 14, 18s2 – 10s + 11s, 118s2 – 14s + 14 71s2 – 14s – 14 18s2 + s
CSS Middle Standard “Mathematics” 104 100q4 + 117q3 + 21q2 – 24q 118s2 – 14s + 14 207s2 – 27s (vii). 118x3 – 174x2 , 148x4 – 114x3 + 100x2 , 114x + 10x2 118x3 – 174x2 148x4 – 114x3 + 100x2 –110 x2 + 114x 148x4 + 4x3 – 184x2 + 114x Q6. Simplify the following. (i). (3 + x) + (2z – x) = 3 + x + 2z – x = 3 + x – x + 2z = 3 + 2z (ii). (p2q + 4) – (13p2q + 14q2 – 6) = p2q + 4 – 13p2q – 14q2 + 6 = p2q – 13p2q – 14q2 + 4 + 6 = –12p2q – 14q2 + 10 (iii). –2[3c –{d – 2(c + d)}] = –2[3c – {d – 2c – 2d}] = –2[3c – d + 2c + 2d] = –2[3c + 2c – d + 2d] = –2[5c + d] = –10c – 2d (iv). (14x2 – 3x) – (–3x + 6x2 ) = 14x2 – 3x + 3x – 6x2 = 14x2 – 6x2 – 3x + 3x = 8x2 (v). (16a2 – 3b) – (–3a2 + 5b) = 16a2 – 3b + 3a2 – 5b = 16a2 + 3a2 – 3b – 5b = 19a2 – 8b (vi). (10p2qr – 2pq) – (2p2qr + 5q2 – 2qr + 6) = 10p2qr – 2pq – 2p2qr – 5q2 + 2qr – 6 = 10p2qr –2p2qr – 2pq – 5q2 + 2qr – 6 = 8p2qr – 2pq –5q2 + 2qr – 6 (vii). 10{q2 + 3r(p2 + r) + 3(r2 + rp2 )} = 10{q2 + 3rp2 + 3r2 + 3r2 + 3rp2} = 10{q2 + 3rp2 + 3rp2 + 3r2 + 3r2} = 10{q2 + 6rp2 + 6r2} = 10q2 + 60rp2 + 60r2 (viii). (17x2 – 12x2 – 3xy2 ) – (2x2 + 12x2 – xy 2 ) = (5x2 – 3xy2 ) – (14x2+ xy2 ) = 5x2 – 3xy2 – 14x2 – xy 2 = 5x2 – 14x2 – 3xy2 + xy2 = –9x2 – 2xy2 Unit No. 9 Linear Equation Teacher Objective: To introduce an algebraic equation. To explain the difference between algebraic expression and algebraic equation. To explain the construction of linear expression and linear equation with one variable. To explain the solution of simple linear equation involving fractional and decimal coefficients. To explain the solution of real life problems involving linear equation. Learning Outcomes: At the end of this Unit, the students will be able to: Define an algebraic equation. Differentiate between algebraic equation and algebraic equation. Define linear equation in one variable. Construct linear expression and linear equation in one variable. Solve simple linear equations involving fractional and decimal coefficient like 1 2 x + 5 = x – 1 3 Solve real life problems involving linear equations. Teaching Material: ● CSS Middle Standard Mathematics Book 6 ● Writing Board ●Marker ● Eraser Procedure: Greet the students and ask them about open statements and algebraic expressions with variables. Ask them about the relationship of expression through the symbols. Invite them for reading books.
CSS Middle Standard “Mathematics” 105 Note for the teacher: Focus on all examples given in the book. Always solve few questions on the board by yourself. Then encourage students to solve other questions there. For notebooks work, begin by pair work finally ending with individual work. Suppose you have an exercise consisting of 5 questions. Each question further consists of 5 parts then your strategy should be Question No. Done by the teacher Board practice by the students Pair work Individual work Home work 1 Part I Part ii Part iii Part iv Part v Focus on the definitions and ask students to learn the definitions with understanding. Once in a week, homework can be assigned based on the definition given in the chapter / lesson / unit going on in the classroom. Exercise (9a) Q1. Write each sentence as an algebraic equation. (i) Twice a number, decreased by twenty nine, is seven. 2x – 29 = 7 (ii) Thirty two is twice a number increased by 8. 32 = 2x + 8 Or 2x + 8 = 32 (iii) Twelve is sixteen less than four times a number. 12 = 4x – 16 Or 4x – 16 = 12 Q2. Write each an algebraic equation as sentence. (i) x + 13 = 24. Ans: A number increased by thirteen is twenty four. (ii) 25b = 200 Ans: Twenty five times a number is two hundred. (iii) y + 43 = 82 Ans: A number increased by forty three is eighty two. (iv) z – 18 = 43 Ans: A number decreased by eighteen is forty three. (v) 350 – d = 280 Ans: Three hundred fifty decreased by a number is two hundred eighty. (vi) 15 p = 3 Ans: Fifteen divided by a number is three. (vii) p + 10 = 40 Ans: A number increased by ten is forty. (viii) 8–1 p = 4 Ans: Eight decreased by one divided by a number is four. Q3. Which of the following equations are true or false for given value of variables? (i) 3y –7 = 0 where y = 7 3 7 3 3 –7 = 0 7 – 7 = 0 0 = 0 True (ii) 4 + m = 1 where m = 9 4 + 9 = 1 13 1 False (iii) 4 5 h – 1= –1 5 where h = 1 4 5 (1) – 1 = –1 5 4 5 – 1 = –1 5 4–5 5 = –1 5 True (iv) 3 2 z – 5 = 2z + 4 where z = 8 4 3 (8) 2 – 5 = 2(8) + 4 12 – 5 = 16 + 4
CSS Middle Standard “Mathematics” 106 7 = 20 False (v) f + 8 = 5f – 9 where f = 6 6 + 8 = 5(6) – 9 14 21 False Exercise (9b) Q1. Solve the following equations to satisfy the equation. (i) 13x – 10 = 29 Step-1 13x – 10 = 29 13x – 10 + 10 = 29 + 10 (Add 10 on both sides) 13x = 39 Step-2 13x 13 = 39 13 (Divide by 13 on both sides) x = 3 (ii) 29x – 50 = 95 Step-1 29x – 50 = 95 29x – 50 + 50 = 95 + 50 (Add 50 on both sides) 29x = 145 Step-2 29x 29 = 145 29 (Divide by 29 on both sides) x = 5 (iii) 4 + y = 3y Step-1 4 + y–y = 3y – y (Subtracting y from both sides) 4 = 2y Step-2 2y 2 = 4 2 (Divide by 2 on both sides) y = 2 (iv) 17 – y = 10 Step-1 17 – y = 10 17 – y – 17 = 10 – 17 (Subtracting 17 from both sides) – y = –7 Step-2 –y –7 = –7 –7 (Divide by –7 on both sides) y = 7 (v) 14 + z = 12 Step-1 14 + z – 14= 12 – 14 z = –2 (vi) 100 + z = 2z Step-1 100 + z – z = 2z – z (Subtracting z from both sides) 100 = z z = 100 Q2. Which of the following equations are true or false for the given value of the variables? (i) 4x + 1 = 9 for x = 2 4x + 1 = 9 4(2) + 1 = 9 8 + 1 = 9 9 = 9 True (ii) 13x – 1 = 14 for x = 10 13(10) – 1 = 14 130 – 1 = 14 129 14 False (iii) 13x + 4 = 13 for x = 3
CSS Middle Standard “Mathematics” 107 13(3) + 4 = 13 39 + 4 = 13 43 13 False (iv) 18x + 10 = 46 for x = 11 18(11) + 10 = 46 198 + 10 = 46 208 46 False (v) 20x – 10 = 30 for x = 10 20(10) – 10 = 30 200 – 10 = 30 190 30 False (vi) 180x + 10 = 70 for x = 10 180(10) + 10 = 70 1800 + 10 = 70 1810 70 False Exercise (9c) Q1. Write any equivalent equation of the following equations. (i) x + 4 = 0 2x + 8 = 0 (Multiplying both sides by 2) (ii) 2x – 10 = 0 4x – 20 = 0 (Multiplying both sides by 2) (iii) x – 4 = – 9 3x – 12 = –27 (Multiplying both sides by 3) (iv) 4x – 6 = x + 4 8x – 12 = 2x + 8 (Multiplying both sides by 2) (v) 8x + 2 = 4x 16x + 4 = 8x (Multiplying both sides by 2) Q2. Solve the following equations and find the values of variables. (i) 0.8x + 0.6 = 1.6x – 1.4 0.8x + 0.6 + 1.4 = 1.6x – 1.4 + 1.4 (Add 1.4 on both sides) 0.8x + 2 = 1.6x 0.8x + 2 – 1.6x = 1.6x – 1.6x (Subtract by 1.6x on both sides) (0.8 – 1.6)x + 2 = 0 – 0.8x + 2 = 0 0.8x 2 0.8 0.8 (Divide by 0.8 on both sides) x = 2.5 (ii) 2x – 10 = 4 2x – 10 + 10 = 4 + 10 (Add by 10 on both sides) 2x = 14 2x 14 2 2 (Divide by 2 on both sides) x = 7 (iii) 3x + 27 = 33 3x + 27 – 27 = 33 – 27 (Subtract by 27 on both sides) 3x = 6 3x 6 3 3 (Divide by 3 on both sides) x = 2 (iv) 4x – 16 = –36 4x – 16 + 16 = –36 + 16 (Add by 16 on both sides) 4x =– 20
CSS Middle Standard “Mathematics” 108 4x 20 4 4 (Divide by 4 on both sides) x = –5 (v) 4x – 6 = 2x + 2 4x – 6 + 6 = 2x + 2 + 6 (Add 6 on both sides) 4x = 2x + 8 4x – 2x = 2x + 8 – 2x (Subtract by 2x on both sides) 2x = 8 2x 8 2 2 (Divide by 2 on both sides) x = 4 (vi) 1.4x + 2.4 = 3.6x– 4.2 1.4x + 2.4 + 4.2 = 3.6x – 4.2 + 4.2 (Add 4.2 on both sides) 1.4x + 6.6 = 3.6x 1.4x + 6.6 – 1.4x = 3.6x – 1.4x (Subtract by 1.4x on both sides) 6.6 = 2.2x 6.6 2.2x 2.2 2.2 (Divide by 2.2 on both sides) 3 = x x = 3 (vii) 4x 2 – 5 3 = 2x + 4 12x–10 15 = 2x + 4 12x – 10 = 15(2x + 4) (Multiply by 15 on both sides) 12x – 10 = 30x + 60 12x – 10 – 60 = 30x + 60 – 60 (Subtract by 60 on both sides) 12x – 70 = 30x 12x – 70 – 12x = 30x – 12x (Subtract by 12x on both sides) –70 = 18x –70 18x 18 18 (Divide by 18 on both sides) –35 9 = x x = –35 9 (viii) 4 4 x – 5 3 = 2x + 4 12x–20 15 = 2x + 4 12x – 20 = 15(2x + 4) 12x – 20 = 30x + 60 12x – 20 – 60 = 30x + 60 – 60 (Subtract 60 from both sides) 12x – 80 = 30x 12x – 80 – 12x = 30x – 12x (Subtract 12x from both sides) –80 = 18x –80 18x 18 18 (Divide by 18 on both sides)
CSS Middle Standard “Mathematics” 109 –40 9 = x x = –40 9 (ix) 4 3 1 1 x – x 5 5 5 5 4x – 3 x 1 5 5 4x – 3 = x + 1 (Multiplying both sides by 5) 4x – 3 + 3 = x + 1 + 3 (Add 3 on both sides) 4x = x + 4 4x – x = x + 4 – x (Subtract x from both sides) 3x = 4 3x 4 4 x 3 3 3 (Divide by 3 on both sides) (x) 6 3 1 1 1 x 10 6 x 5 5 5 5 11 3 51 31 x x 5 5 5 5 11x 3 51 31x 5 5 11x + 3 = 51 + 31x (Multiplying by 5 on both sides) 11x + 3 – 51 = 51 + 31x – 51 (subtract by 51 on both sides) 11x – 48 = 31x 11x – 48 – 11x = 31x – 11x (Subtract 11x from both sides) – 48 = 20x –48 20x 20 20 (Divide by 20 on both sides) –12 5 = xx = –12 5 (xi) 1.6 – 2x = 1.4 1.6 – 2x – 1.6 = 1.4 – 1.6 (Subtract 1.6 from both sides) –2x = –0.2 –2x –0.2 –2 –2 (dividing by –2 on both sides) x = 0.1 Q3. A woman is now 3 times as old as his son. In 10 years’ the sum of their ages will be 76. How old was the women when her son was born? Solution: Let son’s age = x Then women’s age = 3x After 10 years Son’s age = x + 10 Women’s age = 3x + 10 According to given condition (x + 10) + (3x + 10) = 76 x + 3x + 10 + 10 = 76 4x + 20 = 76 4x = 76 – 20
CSS Middle Standard “Mathematics” 110 4x = 56 x = 56 4 x = 14 Hence women’s age = 3x = 3(14) = 42 Q4. Bilal buys a novel and note book and pays Rs.500 for both. The novel costs Rs. 80 more than twice the cost of note book. Find the cost of novel? Solution: Let cost of note book = x Then cost of novel = 2x + 80 According to given condition x + 2x + 80 = 500 3x + 80 = 500 3x = 500 – 80 3x = 420 x = 420 3 = 140 Now the cost of novel = 2x + 80 = 2(140) + 80 = 280 + 80 = Rs. 360 Q5. The length of a rectangular pool is 6cm more than its width and the perimeter of rectangle is 36cm. What is its length and width? Solution: Let width = x Then length = x + 6 Perimeter = 36 2(2x + 6) = 36 2x + 6 = 182x = 18 – 6 2x = 12x = 6 Hence width = x = 6cm Length = x + 6 = 6 + 6 = 12cm Q6. Two friends have Rs. 8000. If one of them had four times as much as the other, how much money did each of them have? Solution: Let share of one friend = x Then share of other friend = 4x According to given condition x + 4x = 8000 5x = 8000x = 8000 5 = 1600 Hence share of one friend= x = Rs. 1600 share of other friend= 4x = 4(1600) = Rs. 6400 Q7. Age of a father is 3 times the age of his daughter. If father is 18 years old than his daughter, then find the age of father and daughter? Solution: Let daughter’s age = x Then father’s age = 3x At time of birth, the difference between their ages are is i.e 3x – x = 18 2x = 18x = 18 2 = 9 Hence daughter’s age = x = 9 years father’s age = 3x = 3(9) = 27 years Review Exercise 9 Q1. Choose the correct answer and fill the circle. i. A mathematical sentence with an equality sign is called: algebra expression equation equality
CSS Middle Standard “Mathematics” 111 ii. A/an ______ is a phrase: sentence variable expression symbol iii. When left side is equal to right side,______is completed: equation sentence linear equation none iv. Seven times a number increased by 2: 2x + 7 7x + 2 6 – 2 7 none v. Three forth of a number subtracted from 6: 6 – 3 4 x 4 – 6 3 3 4 3 4 – 6x all vi. A number divided by 6 decreased by 5: x 5 – 6 6 x – 5 x 5 + 6 all vii. Five less than three times a number is 46: 5 – 3x = 46 3x – 5 = 46 5x = 46 3 none of these viii. Twice number is 28: x + 2 = 28 28 = x 2x = 28 2 2x = 28 ix. Equation 2m – 9 = 3 for m = ? 1 3 5 6 Q2. What should be the value, which balance the following equations. i. 9 5 a – a = 1 9a – 5a 5 = 1 4a 5 = 1 4a = 5 × 1 (Multiplying by 5 on both sides) 4a = 5 a = 5 4 (Divide by 4 on both sides) ii. 6a – 3 = 15a 6a – 3 – 6a = 15a – 6a (Subtract 6a from both sides) –3 = 9a 3 9 9a 9 3 9 =a 3 9 1 3 a iii. 0.4a + 9.5 = 9.99 0.4a + 9.5 – 9.5 = 9.99 – 9.5 (Subtract 9.5 from both sides) 0.4a = 0.49 a = 0.49 0.4 = 1.225 iv. 1.65a + 385 = –a 1.65a + 3.85 + a = –a + a (Add a on both sides) (1.65 + 1)a + 3.85 = 0 2.65a + 3.85 = 0 2.65a + 3.85 – 3.85 = 0 – 3.85 2.65a = –3.85 a = – 3.85 2.65 = –1.436 Q3. Solve each of the following equations. i. 8x + 13 = 0 8x + 13 – 13 = 0 – 13 ii. 3 4 5 x – 5 = 0
CSS Middle Standard “Mathematics” 112 8x = –13 x = –13 8 23 5 x – 5 = 0 23 5 x – 5 + 5 = 0 + 5 (Multiply by 5 on both sides) 23 5 x = 5 23 5 x × 5 = 5 × 5 23x = 25 x = 25 23 iii. 5(x – 4) = 3x + 2 5x – 20 = 3x + 2 5x – 20 + 20 = 3x + 2 + 20 5x = 3x + 22 5x – 3x = 3x + 22 – 3x 2x = 22 2x 22 2 2 x = 11 iv. 2x + 0.05 = 0.8x + 1.25 2x + 0.05 – 0.05 = 0.8x + 1.25 – 0.05 2x = 0.8x + 1.2 2x – 0.8x = 0.8x + 1.2 – 0.8x (2 – 0.8)x = 1.2 1.2x = 1.2 1.2x 1.2 1.2 1.2 x = 1 v. 5 2 n = 4 3 + 2n 5 2 n – 2n = 4 3 + 2n – 2n 5 – 2 2 n = 4 3 5–4 2 n = 4 3 1 2 n = 4 3 n = 4 3 × 2 n = 8 3 vi. 12n + 32 = – 4 12n + 32 – 32 = –4 –32 12n = –36 12n 12 = –36 12 n = –3 Q4. Area of right angled triangle is b c 1 2 square meter. If the area is equal to 15 square meter b = 5 then c = ? Solution: As Area = b × c 1 2 15 = 1 2 5 × c 15 × 2 = 5 × c 2 1 2 30 = 5c 30 5 = 5c 5 c=6 Q5. Find the number whose five times when added to 12 give the result 17. Solution: Let the required number be x According to given condition 5x + 12 = 17 5x + 12 – 12 = 17 – 12 5x = 5
CSS Middle Standard “Mathematics” 113 5x 5 = 5 5 x = 1 Q6. Find three consecutive numbers in which x is middle number and their sum is 99. Find the number. Solution: If the middle number = x Then first number = x – 1 3 rd number = x + 1 Now according to given condition x – 1 + x + x + 1 = 99 3x = 99 3x 3 = 99 3 x = 33 Now first number = x – 1 = 33 – 1 = 32 Middle number = x = 33 Third number = x + 1 = 33 + 1 = 34 Q7. Find the three consecutive even numbers whose sum is 54. Solution: Let first even number = 2x 2 nd even number 2x + 2 3 rd even number = 2x + 4 According to given condition Sum of consecutive even number = 54 2x + 2x + 2 + 2x = 4 = 54 6x + 6 = 54 6x = 54 – 6 6x = 48 x = 48 6 = 8 x = 8 Now first even number = 2x= 2(8)= 16 2 nd even number = 2x + 2 = 2(8) + 2 = 16 + 2= 18 3 rd even number = 2x + 4 = 2(8) + 4= 16 + 4 = 20 Unit No. 10 Geometry Teaching Objectives To explain the addition of measure of two or more line segments. To explain the subtraction of a line segment from a longer one. To explain the method of drawing right bisector of gives line segment using compass. To explain the drawing of perpendicular to given line, form a point on it by using compass. To explain the method of drawing perpendicular, from point outside the line, using compass. To explain use of compass to construct a an angle equal and twice of a given angle, bisect the given angle, divide given angle in four equal parts. To explain the construction of triangle when three sides or two sides and one angle or two angle and one sides is given. To explain the construction of a right angle triangle when hypothesis and one sides are given. Learning Outcomes: At the end of this unit, students will be able to: Add measure of two or more line segments.
CSS Middle Standard “Mathematics” 114 Subtract measure of a line segment from longer one. Draw the right bisector of a given line segment by using compasses. Draw the perpendicular to a given line from point on it using compasses. Draw the perpendicular to line from a point outside the line, using compass. Use of compasses to: Construct an angle equal in measure of a given angle. Construct an angle twice in measure of a given angle. Bisect a given angle. Divide a given angle into four equal angles. Construct the following angles: 60°, 30°, 15°, 90°, 45°, 1 22 2 , 75°, 1 67 2 , 120°, 156°, 165°, 135°, 105° Construct a triangle when three sides (SSS) are given. Caution: Sum of two sides should be greater than third side. Construct a triangle when two sides and their included angle (SAS) are given. Construct a triangle when two angles and their included side (ASA) are given. Construct a triangle when hypotenuse and one sides (RHS) for a right angle triangle are given. Teaching Material: ● CSS Middle Standard Mathematics Book 6 ● Writing board ●Marker ● Eraser Procedure: Greet the class. Ask them about geometry and construction of line segment, angle and triangle. Introduce the geometry instruments and guide the students about the uses of these instruments. Invite them for book reading. Note for the teacher: Focus on all examples given in the book. Always solve few questions on the board by yourself. Then encourage students to solve other questions there. For notebooks work, begin by pair work finally ending with individual work. Suppose you have an exercise consisting of 5 questions. Each question further consists of 5 parts then your strategy should be Question No. Done by the teacher Board practice by the students Pair work Individual work Home work 1 Part i Part ii Part iii Part iv Part v Focus on the definitions and ask students to learn the definitions with understanding. Once in a week, homework can be assigned based on the definition given in the chapter / lesson / unit going on in the classroom. Exercise (10a) Q.1 Draw a line segment whose length is equal to (a) the sum of the length given below (b) The difference of lengths given below. (i) 4.8cm, 2.3cm (a) Construction: (i) Draw a line OA. (ii) With center P on OA , draw an arc of radius 4.8cm, which interest OA at point Q. (iii) With center Q on OA , draw an arc of radius 2.3cm, which intersect OA at point R. (iv) PR is required line segment Hence mPR = mPQ + mQR (v) With help of ruler, measure the required line segment, which is 7.1cm. (b) (i) Draw a line OA
CSS Middle Standard “Mathematics” 115 (ii) With center P on OA , draw an arc of radius 4.8cm. Which intersect OA at point Q with center Q on OA draw an arc of radius 2.3cm across the left which intersect OA at point R. PR is required line segment hence PR PQ QR and its length is 2.5cm? (ii) 5cm, 3cm (a) Construction: (i) Draw a line OA. (ii) With center P on OA , draw a line segment of radius 5 cm, which intersect OA at point Q. (iii) With center Q on OA , draw an arc of radius 3cm, right which intersect OA at point R. (iv) PRis required line segment. Hence PR PQ QR , the length of is 8cm. (b) (i) Draw a line OA. (ii) With center P on OA , draw an arc of radius 5cm which intersect OA at point Q. (iii) With center Q, draw an arc of radius 3cm across left, which intersect OA at point R. (iv) PR is required line segment. Hence PR PQ QR . The length of PRis 2cm. (iii) 8.2cm, 1.3cm (a) Construction: (i) Draw a line OA. (ii) With center P on OA , draw an line arc of radius 8.2cm, which intersect OA at point Q. (iii) With center Q on OA , draw an arc of radius 1.3 cm across right, which intersect OA at point R. (iv) PRis required line segment hence PR PQ QR , the length ofPR is 9.5cm. (b) (i) Draw a line OA. (ii) With center P on OA , draw an arc of radius 8.2cm, which intersect OA at point Q. (iii) With center Q on OA , draw an arc of radius 1.3cm across left, which intersect OA at R. (iv) PRis required line segment. Hence PR PQ QR , the length of PRis 6.9cm Q.2 Draw the following line segment and bisect them using a compass. (i) 7.5cm
CSS Middle Standard “Mathematics” 116 Construction: (i) Draw a line segment PQ = 7.5cm. (ii) Adjust the compass to slightly longer than half of line segment and draw arcs, with center P, below and above the line segments. (iii) Keeping the same compass width, draw arcs from Q above and below. These arcs intersect each other at points R and S. Join R with S. RS bisect PQ at point T. (ii) 5.4cm Construction: (i) Draw a line segment PQ = 5.4cm. (ii) Adjust the compass to slightly longer than half of line segment and draw arcs from P above and below the line segment. (iii) Keeping the same compass width draw arcs from Q below and above. The arcs intersect the previous arcs at point R and S. Join R to S. RS bisect PQ at point T. (iii) 8.9cm Construction: (i) Draw a line segment PQ = 3.8cm (ii) Adjust the compass slightly longer than half of the line segment length. And draw the arcs from P below and above the line segment. (iii) Keep the same compass width, draw arcs fromQ below and above line segment. These arcs intersect the previous arcs at point R and S. (iv) Join R with S. (v) RS bisect PQ at point T. Q.3 Draw a line segment PQ = 6cm. Take any point R outside it. Draw the perpendicular from R on PQ. Construction: (i) Draw PQ = 6cm. (ii) Take a point R outside PQ .With centers Q and P, draw arcs of same radius which intersect each other at point S. Join S with R.RS is required perpendicular on PQ . Q.4 Draw a line segment AB = 5cm. Take any point D on the line segment. Draw a perpendicular on AB at point D.
CSS Middle Standard “Mathematics” 117 Construction: (i) Draw a line segment AB = 5cm. (ii) Selected a point D somewhere in the middle. (iii) With center D, draw arcs of radius 1cm on both sides of D, which intersect ABat P and Q. (iv) Adjust the compass and draw an arc with center P on either above or below side of AB (v) With center Q draw an arc of same radius. (vi) Both arcs intersect each other at point R. (vii) Joins R with D. (viii) DR is perpendicular to ABat D. Q.5 IfmKL = 4.5cmand m PQ = 3.5cm .Draw ABsuch that m AB = m KL + m PQ and m EF = m KL m PQ Solution: Same as question 1. Q.6 mAB = 7cmdraw a perpendicular to a point C on ABsuch that AC = 3cm. Construction: (i) Draw mAB 7cm (ii)Take a point C on ABsuch that AC 3cm (iii) With center C draw two arcs of radius 1cm on both sides of C which intersect ABat P and Q. With center P, draw arc of some suitable radius on either side above or below of AB. (iv)With same radius draw an arc from Q. (v)Both arcs intersects each other at R. (vi)Join C to R. (vii)CDis required perpendicular of AB at C. Q.7 A line segment ABis 8cm long. I am interested to draw a bisector PQ which is perpendicular to AB. Guide me what steps I will follow. Construction: (i) Draw AB8cm. (ii)Adjust the compass slightly longer than half of AB and with center A draw the arcs on both sides of AB. (iii) With center B, draw arcs of same radius on both sides of ABwhich intersect previous arcs at P and Q. (iv) Join P with Q. (v) PQ is required perpendicular bisector of AB. Q.8 Take a point on each segment and draw perpendicular to the given line segment (use ruler and compass only)
CSS Middle Standard “Mathematics” 118 (i) 5.3cm Same as question 4 (ii) 7.2cm Same as question 4 (iii) 2.9cm Same as question 4 (iv) 8.9cm Same as question 4 Q.9 Draw CE whose length is equal to difference of length of CD and EF given is figure. Construction: (i) Draw CD= 8cm. (ii) With center D, draw an arc of radius 3cm = EFacross lift, which intersect CDat point E. hence CEis required line segment whose length is 5cm. Exercise (10b) Q.1 Construct the following angles with help of a protractor. (i) 56° Construction: (i) Draw a ray AB. (ii) With center A, draw an angle of 56° with help of protractor. (iii) CAB = 56° is required angle. Take protractor, put at point A in such way that base line meets AB and draw a point C on 56°. Remove protector and draw a line from A to C. (ii) 110° Construction: (i) Draw a ray AB. (ii) With center A, draw an angle of 110° with help of protractor. (iii) CAB = 110° is required angle. (iii) 85° Construction: (i) Draw a ray AB. (ii) With center A, draw a angle of 85°. (iii) CAB is required angle. (iv) 120°
CSS Middle Standard “Mathematics” 119 Construction: (i) Draw a ray AB. (ii) With center A, draw an angle of 120° with help of protractor. (iii) CAB = 120° is required angle. Q.2 Draw the following angles using protractor and bisect them using compass. (i) 135° Construction: (i) Draw a ray AB. (ii) With center A, draw an angle of CAB 135° with help of protractor. (iii) With A as enter draw an arc of suitable radius with compass which cut CAB at E and F. With center E draw an arc of radius slightly larger than half of EF. (iv) With center F draw an arc of same radius. (v) Both arcs intersect each other at D. (vi) Join A to D. (vii) AD is the required bisector of CAB. (ii) 90° Construction: (i) Draw an angle CAB = 90° with help of protractor. (ii) With center A, draw an arc which intersect CAB at E and F. (iii) With center E, draw an arc of radius slightly larger than half of EF. (iv) With center F, draw an arc of same radius. (v) Both arcs intersects each other at D. (vi) Join A to D. (vii) AD is required bisector of CAB. (iii) 30° Same as part (i) and (ii) (iv) 105° Same as part (i) and (ii) (v) 165° Same as part (i) and (ii) (vi) 1 67 ° 2 Same as part (i) and (ii) Q.3 Divide the following angles into four equal parts by compass but draw them with protractor. (i) 110°
CSS Middle Standard “Mathematics” 120 Construction: (i) Draw CAB = 110° with help of protractor. (ii) Bisect the given angle CAB into two equal angles CAD, DAB by using compass. (iii) AD is bisector of CAB. (iv) With center O and F, draw arcs of suitable radius, which intersect each other at T. Join A to T. (v) With center O and G. draw arcs of suitable radius which cut each other at S.Join A to S. Now CAB = 110° has been divided into four equal angles. i.e m BAT = m TAD = mDAS = mSAC (ii) 150° Construction: (i) Draw an angle CAB = 150° with the help of protractor. (ii) Bisect CABinto two equal angles CAD and DAB using compass. (iii) AD is bisector of CAB. (iv) With center O and F, draw arcs of suitable radii which intersect each other at T. Join A to T. (v) With center O and G, draw arcs of suitable radii which intersect each other at S. Join A to S. (vi) Now given angle CAB= 150° has been divided into four equal angles m BAT = m TAD DAS SAC (iii) 140° Construction: (i) Draw an angle CAB = 140° with the help of protractor. (ii) Bisect CABinto two equal angles CAD and DAB using compass. (iii) AD is bisector of CAB. (iv) With center O and F, draw arcs of suitable radii which intersect each other at T. Join A to T (v) With center O and G, draw arcs of suitable radii which intersect each other at S. Join A to S. (vi) Now given angle CAB= 140° has been divided into four equal angles m BAT = m TAD DAS SAC (iv) 85° Construction: (i) Draw an angle CAB = 85° with the help of protractor. (ii) Bisect CABinto two equal angles CAD and DAB using compass. (iii) AD is bisector of CAB. (iv) With center O and F, draw arcs of suitable radii which intersect each other at T. Join A to T.
CSS Middle Standard “Mathematics” 121 (v) With center O and G, draw arcs of suitable radii which intersect each other at S. Join A to S. (vi) Now given angle CAB= 85° has been divided into four equal angles m BAT = m TAD m DAS m SAC Q.4 Draw PQ = 5cm. Construct angles of 60° at P then construct angles given below on point Q using protractor and bisect them using compass. (i) 150° Construction: (i) Draw PQ = 5cm (ii) With center P, draw an angle of 60° with help of protractor. (iii) With center Q, draw an angle of 150° with help of protractor. (iv) Both angles intersect each other at R. (v) Bisect the angles P and Qwith help of compass. (vi) QS is required bisector of Q= 150° and PT is required bisector of P = 60° (ii) 1 67 ° 2 Construction: (i) Draw PQ = 5cm (ii) With center P, draw an angle of 60° with help of protractor. (iii) With center Q, draw an angle of 67.5° with help of protractor. (iv) Both angles intersect each other at R. (v) Bisect the angles P and Qwith help of compass. (iii) 120° Construction: (i) Draw PQ = 5cm (ii) With center P, draw an angle of 60° with help of protractor. (iii) With center Q, draw an angle of 120° with help of protractor. (iv) Both angles intersect each other at R. (v) Bisect the angles P and Qwith help of compass.
CSS Middle Standard “Mathematics” 122 (iv) 150° Construction: (i) Draw PQ = 5cm (ii) With center P, draw an angle of 60° with help of protractor. (iii) With center Q, draw an angle of 150° with help of protractor. (iv) Both angles intersect each other at R. (v) Bisect the angles P and Qwith help of compass. Q.5 Draw the following angles. (i) 135° Construction: (i) Draw a point A and make a ray AB. (ii) With center A, draw an arc, of suitable radius, which intersect AB at point E. With center E and without changing the radius, draw an arc to intersect the first arc at F. (iii) With center F and without changing the radius, draw an arc to intersect the first arc at G. (iv) With center G and without changing the radius, draw an arc to intersect the first arc at H. (v) Bisect GH through I. With center I and G draw arcs of suitable radius which intersect each other at C. (vi) Join A to C. (vii) Hence m CAB = 135° is required angle. (ii) 165° Construction: (i) Draw a point A and make a ray AB. (ii) With center A draw an arc of some suitable radius which interest AB at E. (iii) With center draw an arc of same radius which intersects the first arc at F. (iv) With center F, draw an arc of same radius which interests the first arc at G. (v) With center G, draw an arc of same radius which intersects the first arc of same radius which intersects the first arc at H. (vi) Bisect G H and get point I. With center I and H, draw arcs of some suitable radii which intersect each other at C. (vii) Join A to C. (viii) mCAB = 165°is the required angle. (iii) 75° Construction: (i) Draw a point A and make a ray AB . (ii) With center A, draw an arc of suitable radius which intersect AB at E. With canter at E draw an arc of same radius, which intersect the first arc at F. (iii) With center F, draw an arc of same radius which intersect the first arc at G. R
CSS Middle Standard “Mathematics” 123 (iv) Bisect FG and get point I. With center I and F, draw arcs ofsome suitable radii, which intersect each other at C. (v) Joint A to C. (vi) Hence CAB = 75° is required angle. (iv) 105° Construction: (i) Draw a point A and make a ray AB . (ii) With center A, draw an arc of some suitable radius, which intersect AB at E with compass. (iii) With center E, draw an arc of some radius which intersect first arc at F. (iv) With center at F, draw an arc of same radius which intersects the first arc at G. Bisect FG and get point I. (iv) With centers I and G draw arcs of some suitable radii which intersect each other at C. (v) Join C to A (vi) CAB = 105°is required angle. (v) 15° Construction: (i) Draw a point A and make a ray AB . (ii) With center A, draw an arc of suitable radius which intersect AB at E (iii) With center at E, draw an arc of same radius, which intersect the first arc at F. (iv) Bisect EF and get point G. Taking centers G and E draw arcs of suitable radius which intersect each other at C. (v) Join C with A. (vi) CAB = 15°is required angle. (vi) 1 67 ° 2 Construction: (i) Make DAB = 135°with help of compass. (ii) Bisect DAB with help of compass. (iii) AC is the bisector of DAB . (Hence 1 CAB = 67 2 is required angle. (vii) 120° Construction: (i) Draw a point A and make ray AB . (ii) With center A, draw an arc of suitable radius, which intersect AB at E. (iii) With center E, draw an arc of same radius which cut the first arc at F. (iv) With center F, draw an arc of same radius which cut first arc at G. (v) Through G draw CA .
CSS Middle Standard “Mathematics” 124 (vi) CAB = 120°is required angle. (viii) 150° Construction: (i) Draw a point A, and make AB . (ii) With center A draw an arc of suitable radius, which cut AB at E. (iii) With center E, draw an arc of same radius, which intersect the first arc at point F. (iv) With center F, draw an arc of same radius which intersect the first arc at G. (v) With center G, draw an arc of same radius which intersect the first arc at H. (vi) With center G and H, draw the arcs of same radii, which intersect each other at C. CAB = 150°is required angle. (x) 45° Construction: (i) Draw a point A and make ray AB . (ii) With canter A, draw an arc of suitable radius, which intersect at E. (iii) With center E, draw an arc of same radius which intersect the first arc at F. (iv) With center F, draw an arc of same radius which intersect the first arc at G. (v) With center F and G, draw the arcs of same radii, which intersect each other at I. Join I to A. (vi) AI passes through J. (vii) With center E and J, draw arcs of same radii which intersect each other at C. Join A to C. (viii) CAB = 45°is required angle. (ix) 1 22 ° 2 Construction: (i) Draw angle of DAB of 45° with help of compass. (ii) Bisect DAB with help of compass. (iii) AC is bisector of DAB . (iv) 1 CAB = 22 2 is required angle. Q.6 Construct angles of the following with help of compass and divide these angles into four equal angles. (i) 120° Construction: Construction is same as Q# (vii), and divide it into four equal parts by using the procedure explained in question #3. (ii) 60° Construction: (i) Draw a point A and make AB . (ii) With center A, draw an arc of suitable radius which cut AB at E. (iii) With center E, draw an arc of same radius, which intersect first arc at F. (iv) CAB = 60°is required angle divided it into four equal parts by using procedure explained in Q # 3.
CSS Middle Standard “Mathematics” 125 (iii) 150° Construction: Construction is same as Q#5 (viii) and divide it into four equal parts by using procedure explained in Q#3. (iv) 135° Construction is same as Q#5 (i) and divide it into four equal parts by using procedure explained in Q#3. Exercise (10c) Q.1 Draw AB = 4cmConstructs angles of 60° and 45° at point A and B and draws a triangle. Construction: (i) Draw AB 4cm . (ii) With center A, draw an angle of 60° and with center B draw an angle of 45°. Both angles intersect each other at C. (iii) Hence ABC is required triangle. Q.2 Draw AB = 6.4cm. Construct angles of 30o and 90o at A and B respectively and draw triangle. Construction: (ii) Draw AB = 6.4cm. (ii) With center A, draw angle of 30° with the help of compass and an with center B, draw an angle of 90°. Both angles intersect each other at C. ABCis required triangle. Q.3 Construct a triangle, when: (i) mPQ = 7cm,mQR = 3cm and mRP = 6cm Construction: (i) Draw PQ = 7cm. (ii) With center P, draw an arc of 6cm and with center Q draw an arc of 3cm. Both arc intersect each other at point R. Join R with P and Q. Hence PQRis required triangle (ii) m AB = 4.6cm, mBC = 3.4cm and mCA = 2.5cm Construction: (i) Draw AB= 4.6cm (ii) With center A, draw an arc of radius 2.5cm. (iii) With center B, draw an arc of radius 3.4cm. (iv) Both arcs intersect each other at C. Join C with A and B. (v) Hence ABCis required triangle. Q.4 Draw XY=7.2cm. Construct angles of 1 22 ° 2 and 75° at X and Y.Complete the triangle.
CSS Middle Standard “Mathematics” 126 Construction: (i) Draw XY= 7.2cm. (ii) With center X, draw angle 1 22 ° 2 with help of compass. (iii) With center Y, draw an angle of 75° with help of compass. (iv) Both angles intersect each other at Z. (v) Hence XYZ is required angle. Q.5 Construct ABC when: (i) m AB = 5cm,m B = 45° and mBC = 4cm Construction: (i) Draw AB 5cm . (ii) Draw an angle of 45° with help of compass at point B. (iii) With center B, draw an arc of 4cm, which intersect B at C. (iv) Join C with A. (v) ABCis required triangle. (ii) m AB = 5.3cm,m A = 60° and m AC = 4.2cm Construction: (i) Draw AB=5.3cm. (ii) Draw an angle of 60o at A with the help of compass. (iii) With center A, draw an arc of radius 4.2cm which intersect A at C. (iv) Join C to B. (v) ABCis required triangle. Review Exercise 10 Q.1 Choose the correct answer and full the circle. i. A closed three sided figure is called an a ____________: angle triangle quadrilateral any of them ii. When two rays meet together they formed a/an ______: angle bisecting angle triangle all iii. A ______is a part of line which is made by joining two points together. It has start and end point: line line segment triangle none iv. ____________ is the longest side of a right angled triangle. The side opposite of right angle: hypotenuse multi tense bisecting angle all v. When we bisect90°. It gives an angle of ____________? 75° 45° 60° 115° vi. When we double the angle we bisects the angle? No yes why none
CSS Middle Standard “Mathematics” 127 Q.2 Draw PQ and RS such m PQ =3.6cm and mRS = 1.9cm. Draw KL where KL= m PQ + m RS . Measure the length RS of KL. Construction: Take a line OA . With center K on OA , draw an arc of 3.6cm = PQ which intersect OA at point C. With center C, draw an arc of radius 1.9cm = RS which intersect. OA at point L. Join K to L. Hence KL PQ RS is required line segment. Its length is 5.5cm. Q.3 Draw AB , CD and EF , Such that m AB =2cm, mCD=2.8cm and m EF =1cm. Draw XY intersect XY =m AB +mCD-m EF . Measure XY Construction: Take a line OL. Draw a point X on OL . With center X, draw an arc of 2cm = mABwhich intersect OL at point K across right. With center K, draw an arc of 2.8cm = mCDacross right which intersect OL at point M with center M draw an arc of radius 1cm = EF across left, which intersect OL at Y. Join X to Y. XY is required line segment such that mXY = mAB + mCD mEF . The length of is 3.8cm. Q.4 Draw LM = 5cm long. Cut off LN where mLN = 3cm. Measure length of NM . Construction: Draw LM = 5cm. With center L, draw an arc of radius 3cm which intersect LM at point N. NM LM LN is required line segment and the length NM = 2cm. Q.5 Using ruler and compass only draw a triangle PQR, where: (i) mPQ = 7.5cm, mQR = 4cm and PR = 3.5cm Construction: (i) Draw PQ 7.5cm . (ii) With center Q, draw an arc of 4cm radius. (iii) With center P, draw an arc of 3.5cm radius. (iv) Both arcs intersect each other at point R. (v) Join R to P and Q. (vi) XPQR is required triangle . (ii)m PQ = 4.5cm, mQR = 3.6cm and PR = 2.8cm Construction: Same as Q#5(i) (iii)mPQ = 3.8cm, m P = 30°and PR = 2.3cm Construction: (i) Draw PQ=3.8cm . (ii) With center P, draw and angle of 30° with help of compass. (iii) With center of P draw an arc of 2.3cm radius, 3.5cm
CSS Middle Standard “Mathematics” 128 which intersect P at R. (iv) Join R to Q. (v) Hence PQR is required triangle. (iv) mPQ = 5.2cm, m P = 45° and mPR = 2.5cm Construction: Same as Q #5 (iii) (v)m PQ = 7.2cm, m P = 60°and m Q = 75° Construction: (i) Draw PQ=7.2cm . (ii) With center P, draw an angle of 60° with help of compass. (iii) With center Q, draw an angle of 75° with help of compass. (iv) Both angle intersect each other at R. (v) Hence PQR is required triangle. Q.6 Draw ABC such that m ABC = 60° using ruler and compass only. (i) Draw the bisector of ABC. Construction: (i) Draw ABCwith help of compass. (ii) With center E and F draw arcs of same radii, which intersect each other at D. (iii) Join B to D. (iv) AD is required bisector of ABC (ii) Construct angle in measure of ABC. Construction: Same as Q# 10, 2, i (iii) Construct an angle twice in measure of ABC. Construction: Same as 10, 2, ii (iv) Divid ABCinto four equal angles. Construction: Same as (Use of compass to divide a given angle into four equal angles). Q.7 Draw AB = 5.8 cm long. Take a point C on AB such that mAC =2.6cm. At C, draw perpendicular to AB by using ruler and compass only. Construction: (i) Draw AB = 5.8 cm. (ii) With center A, draw an arc of 2.6cm, which intersect AB at C. Draw perpendicular bibector on point C. (iii) With center C draw two arcs of 1cm radii on both sides of C, which intersect AB at D and E. (iv) With center D and E, draw arcs of same radii on both sides of AB . (v) These intersect each other at points P and Q. (vi) Join P to Q. PQ is required perpendicular of AB passing through C. D E
CSS Middle Standard “Mathematics” 129 Q.8 Draw PQ = 4cm long. Take a point R on PQ such that mPR=1.9cm. At R, draw a perpendicular to PQ by using ruler and compass. Construction:Same as Q# 7. Q.9 Draw AB = 5.9cm long. Take a point C outside AB and draw perpendicular to AB form C (use ruler and compass). Construction:Same as Q# 3 (Exercise 10 a) Q.10 Draw CD= 5.2cm long. Draw its right bisector with help of ruler and compass. Construction:Same as Q#7 (Exercise 10a) Q.11 Construct the given angles using ruler and compass only: (i) 135° Construction:Same as Q#5 (i) Exercise 10b. (ii) 105° Construction:Same as Q#5 (iv) Exercise 10b. (iii) 180° Construction: (i) Draw a point A and make ray AB . (ii) With center A, draw an arc of suitable radius, which intersect AB at E. (iii) With center E, draw an arc of same radius which intersect the first arc at F. (iv) With center F, draw an arc of same radius, which intersect first arc at G. (v) With center G, draw an arc of same radius, which intersect radius, which intersect first arc at H. (vi) Draw AC passing through H. (vii) CAB= 180° is required angle. (iv) 90° Construction:Same as (construction of an angle of 90°) (v) 45° Construction:Same as Q#5 (x) Exercise 10b. (vi) 15° Construction:Same as Q#5 (v) Exercise 10b. Model Paper No. 1 Section A Time Allowed: 50 Minutes (Objective Type) Total Marks: 40 Q.1: Fill the circle of correct answer. (40 Marks) 1. The integer was introduced in the year. 1463 1563 1663 1763 2. Additive inverse of 0 is __________. 1 –0 0 1 0 3. |–5| = __________. –5 5 5 5 4. Sum of two negative integers is __________ : positive negative both None 5. (+35) + (–25) = 45 –45 10 –10
CSS Middle Standard “Mathematics” 130 6. (–13) × (+5) –65 65 45 75 7. 5 × __________ = –30 6 –6 15 10 8. (–169) ÷ (–13): 10 –10 13 –13 9. __________ ÷ (–18) = 25. 450 –450 35 25 10. 12 3 5 7 __________: 89 35 99 35 36 35 17 35 11. 3 7 of 49 is: 63 14 9 21 12. 42.3 – 7.48 35.82 42.82 34.82 36.82 13. 144 100 is __________: 144 36 25 14.4 .144 14. BODMAS stand for: bracket observation, division, multiplication, addition, subtract bracket operation, division, multiplication, addition, subtraction all of them none of them 15. 1.48 + (18.4 – 12.3) = 7.58 5.58 7.52 8.56 16. The simplest form of 75:90 is: 6:5 5:6 5:4 4:6 17. In expression a b a is called __________: denominator antecedent consequent none 18. 1 1 : 4 12 into the whole number ratio is: 1:3 3:1 4:3 3:4 19. The ratio of Ali’s monthly in came to expenses is 3:5. If the expenses is Rs. 7000, then the income end saving are: 11200, 4200 10000, 3000 12000, 5000 12450, 5450 20. If 30:5::x:4, then x = 1 4 4 37.5 3.7 21. If the cost of 13kg mangoes is Rs.1950,, what is price of 18kg of mangoes. Rs. 2700 Rs. 2500 Rs. 2300 Rs. 2100 22. If 7 girst and 9 boys earn Rs. 630 per day. How much would 9 girls earn per day? 630 540 810 750 23. If 12 men can do a piece of work in 8 days. How many days require to do it by 16 men? 4 days 7 days 6 days 5 days
CSS Middle Standard “Mathematics” 131 24. If 13 pencils cost Rs. 312, how many pencils can be bought for Rs. 3367? 12 13 14 15 25. 5 1 4 3 4 3 1 10 12 11 12 26. If 8:x::4:2, then x= 10 4 16 1 4 27. Which of the following is equal to 7 20 ? 3.5 14 to 100 35% None 28. 37.5%is equal to: 375 to 50 375 to 100 .375 4 37.5 100 29. Percentage of 7 to 50 is : 18% 16% 14% 12% 30. 24% =: 7 25 4 8 6 25 9 25 31. Profit % = Profit ×100 C.P Profit ×100 S.P 100 ×S.P 100+profit% None 32. Hina sold a table for Rs. 2400 at the profit of 700. Find the price at which she bought it: Rs.1600 Rs.1700 Rs.1500 Rs.1800 33. If a monitor was sold for Rs.5400 at the loss of Rs.600. Find the cost of the monitor. Rs.4800 Rs.6000 Rs.4200 Rs.6200 34. What is added to 4a2 – 3a + 9 to get a2 + 4a –7? –3a2 + 7a –16 3a2 – 7a + 16 3a2 + 7a – 16 –3a2 – 7a + 16 35. 5x is __________ type of allegoric expression: monomial binomial trinomial None 36. 7 times a number decreased by 14 equals to 2: 7x + 14 = 2 7x – 2 = 14 14 + 7x = 2 7x – 14 = 2 37. If 4 8 y + 4 = 10, then y= 12 14 18 8 38. A line perpendicular to a line segment, which divide it into two equal parts is called. perpendicular bisector right bisector None 39. The side opposite to right angle in a right angle triangle is called: perpendiculer base hypotenoes bisecting angle Section B Time Allowed: 2:10 minutes (Subjective Type) Total Marks: 60 Attempt all questions. Each question carries equal marks. Q.1: Find the sum of the following in three steps method. (i) –85316, –42539 (ii) 35467, 39513 (iii) –93568, 53215
CSS Middle Standard “Mathematics” 132 Q.2: Simplify the following: (i) (–18) × (–5) × (–7) × 9 × 11 (ii) 25 × (–15) × (–16) × (–3) × (–12) Q.3: simplify the following: 130 – [140 – {13 + (27 + 3 –12)}] Q.4: Anoral’s income is Rs. 75,000. She paid 1 10 of her income as house rent, 1 15 of her income on gass bill, 4 7 on other expenditure and remaining she saves calculate what amount she save? Q.5: In a boarding school food for 180 students is sufficient for 24 days. 60 more students join them, for how many days the same food sufficient for the students? Q.6: A certain sum of money is divided among Ali, Amna and Ahmad in the ratio 3:4:5. If Amna’s share is Rs. 24000, find the share of Ali and Ahmad. Q.7: During a summer mega sale an electricity brand reduce the prices of all its goods by 20% calculate the original selling price of the refrigerator which was sold for Rs. 32500 during sale? Q.8: The length of a rectangular pool is 8cm more than its width and perimeter of rectangle is 38 cm. What is its length and width? Q.9: Simplify the following expression: (3x2yz + 7yz) + [4yz + {13x2y + y + 9y2 – (3x2yz + 7)}] Q.10: Draw AB = 4cm. Construct angles of 60° and 45° at point A and B and draw a triangle. Model Paper No. 2 Section A Time Allowed: 50 Minutes (Objective Type) Total Marks: 40 Q.1: Fill the circle of correct answer. (40 Marks) 1. Negative numbers were accepted into number system in __________. 17th 18th 15th 19th 2. (–3) + (–5) = –8 8 –2 2 3. (+20) – (+15) = 35 –35 5 –5 4. (–4) × (–17) : 64 –64 72 84 5. __________ × (–2) = 24 –12 12 6 8 6. 4 × (–3) × 8 = –84 76 88 –96 7. (–300) ÷ 2 = 150 200 –150 100 8. 0 ÷ 400 = __________ 440 0 1 None 9. __________ ÷ 12 = –15. 160 –180 –150 3 2 10. 2:4 ÷ 0.3 0.8 0.08 8 4 11. 1 8 of 48,000:
CSS Middle Standard “Mathematics” 133 5,000 6,000 4,000 7,000 12. 34.4 + 26.8 61.2 62.8 7.6 63.2 13. [ ] is called __________: vinculum square bracket braces parranthese 14. 3.54 – (12.5 + 3.4 – 1.2) = 11.16 11.19 –11.16 11.30 15. 1800 100 180 18 1.8 .18 16. 7.8 × 1.3 = 12.14 11.14 10.14 10.24 17. 3 1 5 5 0.675 0.565 0.575 0.535 18. The simplest form of 40:80 is: 4:8 8:4 1:2 2:1 19. In expression a b , b is called: numerator antecedent consequent none 20. 1 1 : 16 20 into the whole number ratio is __________ : 4:5 5:4 3:4 5:3 21. If in a book shop the ratio of science books to Urdu books, is 4:5. If there to are 2000 books of Urdu what will be the number of science books. 1800 1600 1400 1200 22. If x:20::7:10, then x = 7 14 1 7 45 23. If Irfan earns Rs.1015 per week. In how many days he earns Rs. 2610. 16 14 18 22 24. If the cost of 14 milk packs, each weighing 1kg, is Rs. 1540. Find the cost of 16 milk packs of the same kind and weight. 1460 1560 1660 1760 25. Percentage 12 to 25 is: 38% 48% 52% 40% 26. 48% = 9 25 6 25 12 25 14 25 27. Loss % = Loss 100 C.P Loss×100 S.P 100 C.P 100 Loss% None 28. Ahmad sold a camera worth Rs. 30,000 at a loss of Rs. 12,000. Find the selling price of camera. Rs. 42,000 Rs. 18,000 Rs. 15,000 Rs. 48,000 29. At Eid, Anoral gave gifts to 36 out of 100 friends and Alishaba gave 9 out of 25 friends what percent of each friend received gift. 42% 30% 36% 9%
CSS Middle Standard “Mathematics” 134 30. If 20% discount on a shalwar kameez is Rs. 1500, find the marked price of it. 6000 7500 7200 8000 31. 123xy + 21xy __________ 144xy: > < = None 32. 7×30 __________ 200: > < = None 33. 3a + 7y is an allegoric: expression sentence equation terms 34. 3 items a number increased by 5: 3x + 5 5x + 3 3x – 5 3 – 5x 35. Equation 3x + 5 = 29 for x =? 7 6 8 9 36. A mathematical sentence with an equality sign is called: allegoric expression equation symbol 37. We bisects < 270°. It gives an angle of __________? 175 105 135 95 38. When two rays meet together they form a/an __________: angle triangle bisecting all 39. The distance between two point on a line is called: triangle ray line line segment 40. A closed three sided figure is called __________: rectangle square triangle angle Section B Time Allowed: 2:10 minutes (Subjective Type) Total Marks: 60 Attempt all questions. Each question carries equal marks. Q.1: For the following numbers write ascending and descending order and also arrange the absolute. Value of integer in descending order. (i) 845,739,910 – 1100 – 500 (ii) 4578,4138,4258 – 7853,5678 Q.2: Find the sum of the following in three steps method: (i) 25317, –59321 (ii) 95473, 843712 (iii) –43876, 63791 Q.3: simplify the following: 875.32 + 12.8 × [4.32 + {1.5 (7.8 × 4 –3)}] Q.4: Humaira had ot travel from Islamabad 10 Lahore which is 392km. She travel 7 9 of the distance by bus, 1 13 by train and remaining on local van. Find the distance. She travelled by bus train and van. Q.5: 35 men got a project to construct a house in 24 days. 13 of them went on leave, in how many days the remaining men will construct house? Q.6: Anoral has pencils of Rs. 10, Rs. 25, Rs. 35. The ratio of these pencils is 3:5:2 respectively and total amount of all kind of pencils is Rs. 200. Find the number of pencils of each kind. Q.7: A company finds that 3 5 % 6 of their bike made in the year 2016 to 2018 has some problems in FGS. The company made 15000 bikes in this time duration. How many bikes were defective. Q.8: Find three consecutive numbers in which x is middle number and their sun is 99. Find the numbers. Q.9: Subtract the following:
CSS Middle Standard “Mathematics” 135 (i) 3x2y + 4y2 z – 9zx from –9x2y –7y2 z + 6zx (ii) a4 –9b2 + 6a2b from 7a4 – 3b2 + 3a2b (iii) –5a from a + 5 Q.10: Draw CD= 5.2cm long. Draw its right bisector with help of ruler and compass. Model Paper No. 3 Section A Time Allowed: 50 Minutes (Objective Type) Total Marks: 40 Q.1: Fill the circle of correct answer. (40 Marks) 1. Europe started using the negative number in __________. 1445 1554 1545 1645 2. (+17) + (–19) = 2 –2 –36 36 3. (–135) – (–455)= –590 –320 320 None 4. 24 × 12 = 268 288 320 358 5. –7 ×__________ = –82 –12 14 16 12 6. 3 × (–5) × (18) × (–4) = 1000 –1080 1080 1320 7. (–369) ÷ 3 = 125 –123 129 139 8. –1600 ÷ __________ = 40 40 –40 30 50 9. __________ ÷ 19 = –25. –675 575 –475 375 10. Integers were denoted by N W I Z 11. Absolute value is the distance of a number form__________: total sum 1 zero number line 12. 2 9 of 63 is: 10 21 14 27 13. 32.1 + 4.81: 37.91 36.81 36.91 39.81 14. 7 1 8 9 70 72 52 72 60 70 71 72 15. 3.8 × 9.3 = 34.34 42.36 35.34 24.92 16. 154 100 is 1.54 0.154 15.4 154 17. 2.16 – (3.2 + 1.9) = 0 – 2.94 8.6 0.86 18.6
CSS Middle Standard “Mathematics” 136 18. 1 3 of 9000: 3000 300 4500 4200 19. BODMAS stand for: bracket observation, division, multiplication, addition, subtraction bracket operation, division, multiplication, addition, subtraction all of them none of them 20. 25% of 600 : 120 150 180 300 21. Find the ratio of marks in maths of Anoral and Alishaba. Anoral got 65 marks and Alishaba got 80 marks. 12:16 16:13 14:16 13:16 22. The lowest form of ratio 630 700 is: 10 9 5 7 9 10 8 7 23. Which of the following is equal to 6 25 ? 4.3 24% 12 to 100 none 24. Selling price = 100 ×S.P 100+profit% 100 ×S.P 100 + Loss% 100 Loss%×C.P 100 none 25. Hina sold a table for Rs.2400 at the profit of 700. Find the price at which she bought it. Rs. 1600 Rs. 1700 Rs. 1500 Rs. 1800 26. If a monitor was sold for Rs. 5400 at the loss of Rs. 600. Find the cost of monitor. Rs. 4800 Rs. 6000 Rs. 4200 Rs. 6200 27. 32% = 7 25 9 25 8 25 6 25 28. Percentage of 8 out of 40 is. 15% 30% 20% 25% 29. 28.5% = 2.85 285 to 100 57 to 200 none 30. Ali gets Rs. 25,000 salary per month. If his salary is increased by 10%, how much will be get? 25,750 26,500 27,500 27,000 31. 2P + 5P __________ 7P: > < = None 32. 72 ÷ 8 __________ 7: < > = None 33. What is added to 3a + 5 to get 7a – 9: 4a – 6 3a – 14 4a – 14 10a – 4 34. 3x + 5 = 7 is an algebraic __________. expression equation sentence term 35. A number multiplied by 3 increased by 8: 3x + 8 3 + 8x x 8 3 3(x + 8) 36. Two fifth of a number subtracted from 7:
CSS Middle Standard “Mathematics” 137 2x 7 5 2 7x 5 2x 7 5 2 7 x 2 37. If 5x – 5 = 20, then a = ? 4 15 3 5 38. The word triangle cones from __________: Greek French English Latin 39. An angle 15° is obtain by dividing 60° into __________ equal parts: 2 3 4 5 40. The longest side in a right angle triangle opposite two right angle is called: base perpendicular hypotenuse none Section B Time Allowed: 2:10 minutes (Subjective Type) Total Marks: 60 Attempt all questions. Each question carries equal marks. Q.1: Simplify the following: (i) 25 × (–13) × (–9) × (–7) × 8 × 2 (ii) (–12) × (+20) × (–15) × 3 × (–8) × 17 Q.2: For the following numbers write ascending and descending order and also arrange the absolute value of integers in descending order. (i) 235, 453, 310, –900, –750 (ii) 2315, 2158, 2437, –3415, –1100 Q.3: simplify the following: 1 1 8 12 100 12 13 9 8 3 7 5 Q.4: Alishba has Rs. 15000 she gave 3 7 to Anoral. Anoral gave 1 8 of her won share to her friend Sana. How much anoint Sana got? Q.5: The ratio of numbers of male and female employees in a company is 4:5. If there are 2400 male employees, find the number of female employees. Q.6: If 2 women by a micro for Rs. 6000 each. First woman shell the micro after one year for Rs. 500 gain and second woman sell the micro in Rs. 1000 Loss. Find the selling. Q.7: Age of father is 3 times the age of his daughter. If father is 18 years old then his daughter, then find the age of father and daughter? Q.8: Ibrahim purchased 32 dozens of pencils at rate of Rs. 288 per dozen. He sold each one of them at rate of Rs. 27, what was his percentage profit? Q.9: Simplify the following expression: 3x2y – [7xy + {3y2 + x + 9z – 2 (3x2y + 4xy – 2x)}] Q.10: Construct ABC, when: mAB=4cm, mBC=3.5, m =60° Unit No. 11 Perimeter and Area Teacher Objectives: To introduce the perimeter and area of square and a rectangle. To recognize area of path (inside or outside) of a rectangle and square.
CSS Middle Standard “Mathematics” 138 To introduce the method to solve the real life problem related to perimeter and area of a square and rectangle. To introduce the altitude of a geometric figure as the measure of the shortest distance between base and it top. To introduce the area of parallelogram when altitude and base given formula. To introduce trapezium and to calculate its area when attitude and measure of the parallel sides given. To introduce formula to find the area of a triangle when altitude and base given. Learning Outcomes: Student should be able to: Find the perimeter and area of a square and a rectangle. Find area of path (inside or outside) of a rectangle or square. Solve real life problem related to perimeter and area of a square rectangle. Recognize altitude of a geometric figure as the measure of the shortest distance between the base and it top. Find area of a parallelogram when altitude and base are given. Define trapezium and find its are when altitude and measure of the parallel sides are given. Find area of a triangle when measures of the altitude and base are given. Teacher materials. ● CSS middle Standard Mathematics Book 6 ● Writing Board ●Marker ● Eraser Procedure: Greet students and ask them to draw a rectangle and square on the board. Ask them what they know about area and perimeter. Invite them for book reading and focus on the formulas. Note for the teacher: Focus on all examples given in the book. Always solve few questions on the board by yourself. Then encourage students to solve other questions there. For notebooks work, begin by pair work finally ending with individual work. Suppose you have an exercise consisting of 5 questions. Each question further consists of 5 parts then your strategy should be Question No. Done by the teacher Board practice by the students Pair work Individual work Home work 1 Part i Part ii Part iii Part iv Part v Focus on the definitions and ask students to learn the definitions with understanding. Once in a week, homework can be assigned based on the definition given in the chapter / lesson / unit going on in the classroom. Exercise (11a) Area of path (Inside or outside) of rectangle and square. Area of path = Area of large rectangle – Area of small rectangle Q1. Complete this table. A B C D Length 7cm 3.1cm Width 10.1cm 4.1cm 2.8cm Area 5.3cm2 44.8cm2 8.3cm2 Perimeter 34.2cm2 For A: Area = length × width = 7 × 10.1= 70.1cm2 For B: Area = length × width 5.3 = 3.1 × width 5.3 3.1 = width width = 1.71cm
CSS Middle Standard “Mathematics” 139 Perimeter = 2(length + width) = 2 (3.1 + 1.7) = 9.62 cm 84 = 7 × base Base = 847 Base = 12cm For C: Area = length × width 44.8 2 cm = length × 4.1cm 2 44.8cm 4.1cm = length 10.93 = = 10.93cm Perimeter = 2( + w) = 2(10.93 + 4.1) = 30.06 cm For D: Area = × width 8.4 = × 2.8 2 8.4cm 2.8cm = 3 = Perimeter = 2( = width) = 2(3 + 2.8) = 11.6cm Q2. Find the area and perimeter of each square. (i) x = 3.3mm Area of square = x2=( 3.3)2= 10.89 mm2 Perimeter of square = 4x= 4×3.3 = 13.2 mm (ii) x = 2.8mm Area = x2=( 2.8)2= 7.84 mm2 Perimeter = 4x = 4(2.8)=11.2mm (iii) x = 4.8 m Area = x2= (4.8) 2=23.04 m2 Perimeter = 4x = 4(4.8)=19.2 m (iv) x = 8.8mm Area = x2 = (8.8)2=77.44 mm2 Perimeter = 4x = 4(8.8) = 35.2mm (v) x = 7.3mm Area = x2= (7.3)2= 53.29 mm2 Perimeter = 4x = 4(7.3) = 29.2 mm (vi) x = 4.3m Area = x2=( 4.3)2 = 18.49 m2 Perimeter = 4x = 4(4.3) =17.2 m Q3. Complete the following rectangle. (i) A = , P = , = 5cm, w = 3cm As area = × w = 5 × 3 = 15cm2 P = 2 ( × w) = 2(5 + 3) = 2(8) = 16cm
CSS Middle Standard “Mathematics” 140 (ii) A = , P = 20, w = 4cm, = As P = 2( + w) 20 = 2( +4) 10 20 2 4 10 = + 4 10 – 4 = 6 = = 6 cm A = × w = 6 × 4 = 24cm2 (iii) A = 20cm2 , P = , w = 4cm, = As A = × w 20 = × w 20 4 = 5 = = 5 cm P = 2( + w) = 2(5 + 4)= 2(9)= 18 cm (iv) P = 18cm , A = , = 7cm, w = As P = 2( + w) 18 = 2(7 + w) 9 18 2 7 w 9 – 7 = w w = 2 cm Area = × w = 7 × 2 = 14 cm2 Exercise (11b) Q1. How many square tiles of size 9cm will be needed to fit in a square floor of a bathroom of side 720cm. Find the cost of tiling of Rs. 75 per tile. Solution: Area of tiles = a2 = 81cm2 Area of floor = 7202 = 51,8400cm2 No. of tiles = 518400 81 = 6400 Cost of tilling = Rs. 75 × 6400 = Rs. 480,000 Q2. If it cost Rs. 1600 to fence a rectangular park of length 20m at the rate of Rs. 25 per m2 . Find the width of park and its perimeter. Also find the area of field. Solution: Length of park = 20m Cost of fencing the park = Rs. 25 × perimeter Rs. 1600 = Rs. 25 × perimeter Primeter = 1600 25 m Perimeter = 64 m
CSS Middle Standard “Mathematics” 141 As perimeter = 2(length + width) 64 = 2(length + width) 32 = 20 + width width = 32 – 20 = 12 m NowArea = length × width = 20 × 12 = 240m2 Q3. Danial has a rectangle rose garden that measure 8m by 10m. One bag of fertilizer can cover 16m2 . How many bags will be needed to cover the entire garden? Solution: Area of rectangular garden = 8 × 10= 80m2 One bag of fertilizer cover = 16m2 Then the bags needed to cover the entire garden = 80 16 = 5 bags Q4. The area of rectangle is 117 square meters and the width is 9m, what is its length? Solution: Given area of rectangle = 117m2 width = 9m As area of rectangle = length × width 117 = length × 9 Length = 117 9 = 13m Q5. The perimeter of a square country yard is 144m. Find the cost of cementing it at the rate of Rs. 50 per m2 . Solution: Cost of cementing = 50 × 144 = Rs. 7200 Q6. A square swimming pool has length 20m. Find the area and perimeter of pool. Solution: Area of square swimming pool = length × length = 20 × 20 = 400m2 Perimeter of square swimming pool = 4 × 20 = 80m Q7. A man has square space that measures 37m on each side. His house is 9m by 11m. If the rest of space is a plot of grass. What area of grass does have now? Solution: Area of square space = 37 × 37 = 1369m2 Area of house = 9 × 11 = 99m2 Area of rest of space = 1369 – 99 = 1270m2 Area of grass = 1270m2 Q8. Ali wants to run a wire around the space to keep his sheep in. if the space measures 110m by 165m, how much wire does he need? Solution: Perimeter of space = 2(110 + 165) = 550 The required wire = 550m Exercise (11c) Area of triangle: Area of triangle = 1 b×h 2 Area of parallelogram
CSS Middle Standard “Mathematics” 142 Area of parallelogram = b×h sq unit Q1. Find the Area of each triangle (i) Area of = b × h 1 2 1 4 3 2 = 6cm2 (ii) Area of = b × h 1 2 = 1 2 ×5.5 × 4 = 1 2 × 22.0 = 11cm2 (iii) Area of = b × h 1 2 = 1 2 × 3.5 × 3.5 = 6.125cm2 Q2. A parallelogram has sides 12cm and 9cm. If the shortest distance between its shortest sides is 8cm. Find the shortest distance between the longer sides. Solution: Area of parallelogram ABCD is AD × EB = AB × DF 9 × 8 = 12 × x 72 = 12 × x x= 72 12 = 6cm Shortest distance between longer sides = 6cm. Q3. Copy and complete the following. CD (cm) BF (cm) AD (cm) BE (cm) Area of ABCD (i) 9 6 7 (ii) 16 7 11
CSS Middle Standard “Mathematics” 143 (iii) 13 9 11 (i) Area of ABCD = CD × BF= w × h = 9 × 6 = 54 cm2 Also area of ABCD = AD × BE 54 = 7 × BE BE = 54 7 = 7.714 (ii) Area of ABCD = AD × BE = 7 × 11 = 77 Also area of ABCD = CD × BF 77 = 16 × BF BF = 77 16 = 4.8125 (iii) Area of ABCD = AD × BF = 13 × 9 = 117 Also area of ABCD = AD × BF 117 = AD × 11 AD = 117 11 = 10.64 Q4. Find area of each parallelogram. (i)Area of parallelogram = b × h = 3.5 × 2.5 = 8.75cm2 (ii) Area of parallelogram = b × h = 2.5 × 4 = 10cm2 Exercise (11d) Area of trapezium: Area of trapezium = 1 (a×b)×h 2 Q1. The length of the parallel side of a trapezium are in the ratio 3:2, the distance between them is 10cm. If the area of trapezium is 325cm2 . Find the length of parallel sides.
CSS Middle Standard “Mathematics” 144 Solution: The ratio of length of parallel sides = 3:2 Distance between parallel sides = 10cm Area of trapezium = 325cm2 As area of trapezium = 1 2 height × sum of parallel sides 325 = 1 2 (10) × sum of parallel sides 325 = 5 × sum of parallel sides 325 5 = sum of parallel sides Sum of parallel sides = 65cm As the ratio of parallel sides are 3:2 So the length of first parallel sides = 3 5 × 65 = 39cm Length of 2nd parallel sides = 2 5 × 65 = 26cm Q2. Find the area of trapezium whose parallel sides are CD = 12cm AB = 36cm and non-parallel sides are BC = 15cm and AD = 15cm and the distance between parallel sides is 9cm. Solution: Sum of parallel sides = 12 + 36 = 48cm h = the distance between parallel sides = 9cm Area of trapezium = 1 2 h × sum of parallel sides = 1 2 × 9 × 48 = 216cm2 Q3. Copy and complete the area of each trapezium. b1 cm b2 cm H cm Area of ABCD (i) 6 9 7 ? (ii) 4 10 ? 42 (iii) 7 11 10 ? (i) Area of trapezium = 1 2 × h × (b1 + b2) = 1 2 × 7 × (6 + 9) = 1 2 × 7 × (15) = 52.5cm2 (ii) Area of trapezium = 1 2 ×h × (b1 + b2) 42 = 1 2 h × (4 + 10) 42 = 1 2 h × 14 42 = h × 7 h = 42 7 h = 6cm (iii) Area of trapezium = 1 2 × h × (b1 + b2)= 1 2 × 10 × (7 + 11) = 1 2 × 10 × 18 = 90cm2
CSS Middle Standard “Mathematics” 145 Review Exercise 11 Q1. Choose the correct answer and fill the circle. i. Area can be measured in: centimeter kilometer square meter fit ii. Perimeter of a square is multiplication of: 3 2 4 5 iii. Area of foot path has always _________ area on the outer side as compare to its inner area: grater lesser equal none iv. An altitude of a geometry figure is the _______ distance from its top to its opposite side: longer shorter bigger longest v. A parallelogram is a quadrilateral in which the opposite pairs of sides are parallel and ________: non equal equal bisects none of them vi. A trapezium is a quadrilateral that has _______ of parallel sides: none only one pair both none of them Q2. A square park has sides of 250m. Find the area of a jogging track 5m wide which is constructed on the inner side of its boundary. Solution: Area of park = 250 × 250= 62500m2 Area of park without jogging area = 240 × 240 = 57600m2 Area of jogging track = 62500 – 57600 = 4900m2 Q3. Find the area and perimeter of square for the following. Solution:(i) = 4.3cm Area = × = 4.3 × 4.3 = 18.49cm2 Perimeter = 4 × = 4 × 4.3 = 17.2cm (ii) = 73.3cm Area = 2= (17.3)2= 299.29cm2 Perimeter = 4l = 4(17.3)= 69.2cm (iii) = 3.2 Area = 2= (3.2)2= 10.24cm2 Perimeter = 4= 4(3.2) = 12.8cm (iv) = 18.2cm Area = 2= (18.2)2= 331.24cm2 Perimeter = 4= 4(18.2) = 72.8cm (v) = 5.7cm Area = 2= (5.7)2= 32.49cm2 Perimeter = 4 = 4(5.7) = 22.8cm (vi) = 10.3cm Area = 2= (10.3)2= 106.09cm2 Perimeter = 4= 4(10.3) = 41.2cm Q4. If we have two parallel sides of a figure of length 4cm and 7cm respectively and their height is 5cm. Find the area of figure. Solution: Sum of parallel sides = 4 + 7 = 11cm Height between them = 5cm Area of trapezium = 1 2 h × sum of parallel sides= 1 2 × 5 × 11 = 55 2 = 27.5cm2
CSS Middle Standard “Mathematics” 146 Q5. If we have area of parallelogram 84cm2 and its height is 7cm. Find base of parallelogram. Solution: Area of parallelogram = height × base 84 = 7 × basebase = 84 7 = 12cm Unit No. 12 Three Dimensional Solid Teacher Objectives: To recognize 3D figure (Cube, cuboid, sphere, cylinder and cone) with respect to their faces, edges and vertices. To explain and identify units of surface area and volume. To introduce the formula to find surface area and volume of cube and cuboid. To explain the method to solve real life problems involving volume, and surface area Learning Outcomes: At the end of this unit, students will be able to: Identify 3D figure (cube, cuboid, sphere, cylinder and cone) with respect to their faces, edges and vertices. Define and recognize units of surface area and volume. Find surface area and volume of cube and cuboid. Solve real life problem involving volume and surface area. Teacher materials. ● CSS Middle Standard Mathematics Book 6 ● Writing Board ● Marker ● Eraser Procedure: Greet the students and show them the pictures of 3 dimensional building. Encourage them to describe the shape of the building. Tell them that the shapes which can be calculated in three directions are called three dimensional shapes. Invite the class for book reading. Note for the teacher: Focus on all examples given in the book. Always solve few questions on the board by yourself. Then encourage students to solve other questions there. For notebooks work, begin by pair work finally ending with individual work. Suppose you have an exercise consisting of 5 questions. Each questions further consists of 5 parts then your strategy should be Question No. Done by the teacher Board practice by the students Pair work Individual work Home work 1 Part i Part ii Part iii Part iv Part v Focus on the definitions and ask students to learn the definitions with understanding. Once in a week, homework can be assigned based on the definition given in the chapter / lesson / unit going on in the classroom. Exercise (12a) Q1. Find the volume of each of the following cuboids. Solution: (i) length = 14cm, breadth = 12cm, height = 10cm volume = length × breadth × height = 14 × 12 × 10= 1680cm3 (ii) length = 10dm, breadth = 6dm, height = 80dm volume = length × breadth × height
CSS Middle Standard “Mathematics” 147 = 10 × 6 × 80= 4800dm3 (iii) length = 6m, breadth = 4.5m, height = 2.5m volume = length × breadth × height = 6 × 4.5 × 2.5= 67.5cm3 (iv) length = 4m, breadth = 1.2m, height = 2m volume = length × breadth × height = 4 × 1.2 × 2 = 9.6m3 (v) length = 3m 50cm, breadth = 2.5m, height = 1m 50cm volume = length × breadth × height = 3.5 × 2.5 × 1.5= 13.125cm3 (vi) length = 10cm, breadth = 7cm, height = 9cm volume = length × breadth × height = 10 × 7 × 9= 630cm3 Q2. Find the volume of each of cubes whose edges are given belwo. Solution: (i) = 2.5m volume = 3=( 2.5)3= 15.625m3 (ii) = 80m volume = 3= (80)3= 152000m3 (iii) = 22 3 m volume = 3 = 3 22 3 = 10648 27 m 3 = 394.37m3 (iii) = 150dm volume = 3= (2150)3= 3375000dm3 (iv) = 17dm volume = 3= (17)3= 4913dm3 Q3. A water storage room contains liquid if the dimension of the room are 6m × 5m × 4m, find how many littres of liquid it holds. Solution: Volume of water storage room = 6m × 5m × 4m= 120m3 = 120 × 1m3= 120 × 1000 litre = 120000 litres 1m3 = 1000 litre Q4. Find the cost of painting a refrigerator whose length, width and height are respectively 3m, 2m, 1m at the rate of Rs. 10 per meter. Solution: Surface area of refrigerator = 2 length × width + 2 × length × height + 2 height × width = 2(3 × 2 + 2 × 1 + 1 × 3) = 2(6 + 2 + 3)= 22m2 Cost of painting = 10 × 22= Rs. 220 Q5. A cuboid has dimension 7m × 5m × 4m. is its surface area greater or smaller than a cube has a side of length 6m. Solution:
CSS Middle Standard “Mathematics” 148 Surface area of cuboid = ( w + h + hw) Surface area of cuboid = 2(7 × 5 + 5 × 4 + 4 × 7) = 2(35 + 20 + 28)= 2(83)= 166 Surface area of cube = 6 2= 6(6)2= 216 The surface area of cuboid is smaller than surface area of cube. Q6. Find the surface area of a cuboid power bank with length 12cm, width 8cm and height 6cm. Solution: Surface area of cuboid = 2 ( w + h + hw) = 2(12 × 8 + 8 × 6 + 6 × 12) = 2(96 + 48 + 72) = 2(216) = 432 cm2 Q7. The cost of the paint is Rs. 36per kg .If 1kg of paint cover 16sq feet, how much will it cost to paint outside of a cube having 8 feet each side. Solution: Cost of paint = Rs. 36per kg 1kg paint covers = 16sq feet Surface area of cube = 6 2= 6(8)2= 384 sq feet Paint required for area of 384 sq feet = 386 16 = 24 kg Cost of paint outside the cube = Rs. 36 × 24= Rs. 864 Review Exercise 12 Q1. Choose the correct answer and fill the circle. i. A cuboid is a______ D figure: 2D 3D all none ii. All solid bodies are: 2D 4D 3D none iii. Volume of cuboid is______: × × w 6 2 × B × H 3 iv. Surface area of cube is: × 6 2 2 6 2 none v. Volume is a measure of how much________ a particular object occupies: space fluid liquid all vi. Three cubes each of side 5cm are joined and to end. What is the surface area of resulting cuboid? 300cm2 350cm2 375cm2 400cm2 vii. If a cuboid is 3.2cm high, 8.9cm large and 4.7cm wide then the total surface area is: 170.7cm2 180cm2 205.7cm2 325.8cm2 Q2. Identify the missing elements in the following sets of cuboids. No. Length (cm) Width (cm) Height (cm) Volume (cm)3 Surface area (cm)2 1 14 6 672 2 11 8 616 3 6 5 270 Fro 1
CSS Middle Standard “Mathematics” 149 As volume = L × W × H 672 = 14 × W × 6 672 = 84 W 672 84 = W 8 = WW = 8cm Surface area = 2(LW + WH + HL) = 2(14 × 8 + 8 × 6 + 6 × 14) = 2(112 + 48 + 84) = 488cm2 For 2 volume = L × W × H 616 = 11 × 8 × H 616 = 88 H 616 88 = H 7 = HH = 7cm Surface area = 2(LW + WH + HL) = 2(11 × 8 + 8 × 7 + 7 × 11) = 2(88 + 56 + 77) = 2(221) = 442cm2 For 2 volume = L × W × H 270 = L × 6 × 5 270 = L × 30 270 30 = L L = 9cm Surface area = 2(LW + WH + HL) = 2(9 × 6 + 6 × 5 + 5 × 9) = 2(54 + 30 + 45) = 2(129) = 258cm2 Q3. A cuboidal water tank is 6m long, 5m wide and 4.5m deep. How many litres of water can it hold. (1m3 = 1000l) volume = L × W × D = 6 × 5 × 4.5 = 135m3 = 135 × 1000 = 135000 Q4. A cooking oil contains in a tin can with the dimensions of 30cm × 25cm × 15cm. Find how many litres of oil it holds. 1000cm2 = 1 litre volume = 30cm × 25cm × 15cm = 11250cm3= 11.25 litre Q5. How many cubes of 10cm edge can be put in a cubical box of 1m edge. Surface area of cube = 6(10)2= 600cm2 Surface area of cubical box = 6(100)2 cm2 1m = 100cm = 60000cm2 No. of cube put in cubical box = 60000 600 = 100 Q6. How many bricks each measuring 25cm × 11.25cm × 6cm will be need to build a wall 8m × 6m × 22.5m. volume of bricks = 25cm × 11.25cm × 6cm = 1687.5cm3 volume of wall = 8m × 6m × 22.5m
CSS Middle Standard “Mathematics” 150 = 1080m3= 1080 × 1000000cm3 = 1080000000cm3 No. of bricks = volume of wall volume of brick = 1080000000 1687.5 = 640000 Unit No. 13 Information Handling Teacher Objectives: To introduce data and data collection. To make differentiation between grouped and ungrouped data. To introduce the construction of horizontal and vertical bar graph. To explain how to read pie chart. Learning Outcomes: At the end of this unit, students will be able to: Define data and data collection. Distinguish between grouped and ungrouped data. Draw horizontal and vertical bar graph. Read a pie graph. Teacher materials. ● CSS Middle Standard Mathematics Book 6 ● Writing Board ●Marker ● Eraser Procedure: Greet the class and ask the students what they know about data. Ask them why do we need data. Enlist the responses on the board. Invite the students for book reading. Note for the teacher: Focus on all examples given in the book. Always solve few questions on the board by yourself. Then encourage students to solve other questions there. For notebooks work, begin by pair work finally ending with individual work. Suppose you have an exercise consisting of 5 questions. Each question further consists of 5 parts then your strategy should be Question No. Done by the teacher Board practice by the students Pair work Individual work Home work 1 Part i Part ii Part iii Part iv Part v Focus on the definitions and ask students to learn the definitions with understanding. Once in a week, homework can be assigned based on the definition given in the chapter / lesson / unit going on in the classroom. Exercise (13a) Q1.(i) Define and differentiate between primary and secondary sources with some examples. (ii) What is the difference between group and ungrouped data explain it. Solution: (i) answer on page 176 of CSS middle standard math 6th . (ii) answer on page 176 and 178 of CSS middle standard math 6th . Q2 Given data is from ungroup data arrange it in ascending order. (i) 67,56,89,34,65,87,5,8,11,35,26,47,59,89,94,100,12 Ans: 5,8,11,12,26,34,35,47,56,59,65,67,87,89,89,94,100 (ii) 100,106,804,657,854,987,367,724,679,111,298,176,333,456 Ans: 100, 106, 111, 176, 298, 333, 367, 456, 657, 679, 724, 804, 854, 987 (iii) 1009, 8976, 4876, 5987, 3098, 2987, 6043, 1296 Ans: 1009, 1296, 2987, 3098, 4876, 5987, 6043, 8976 (iv) 234, 56, 873, 123, 341, 766, 555, 324, 567, 778, 234, 543, 678, 543