Acme Mathematics 7

Acme Mathematics 7 151(c) SectorThe shaded part is the sector of the circle. It is the area between the two radii.Classwork1. Study the given figures and fill in the blanks.(a) O is called ....................... of the circle. (b) AB is called ....................... of the circle. (c) OC is called ....................... of the circle. (d) OA and OC are ....................... of the circle. (e) The shaded part is ....................... of the circle.(f) The unshaded part is ....................... of the circle.2. Write True or False. (a) The diameter is twice of its radius. (b) The circumference is the full distance around a circle.(c) The radius divides the circle into two equal segments.(d) Diameter is the longest chord. 3. Name the shaded part.(a) (b)Exercise 3.41. Name the parts of the given circle. BCAOA COBA COBOAC BA COB

152 Acme Mathematics 72. Name the chords. ADCBO3. Find the diameter of the circle whose radius is given below. (a) 2 cm (b) 3.5 cm (c) 4 cm (d) 5 cm 4. Find the radius of the circle whose diameter is given below. (a) 8 cm (b) 6.6 cm (c) 70 cm (d) 7 cm 5. Fill in the blanks. (a) A diameter is the ............. chord of a circle. (b) Perimeter of a circle is called the .......... of the circle. (c) A ............ divides the circle into two semicircles.6. Study the given circles.Now,(a) Draw radius on the circles.(b) Draw diameter on the circles.(c) Measure the radius and diameter of the circles and complete the table given below.Fig. Radius (r) Diameter (d) Relation(i)(ii)Figure (i) Figure (ii)O D

Acme Mathematics 7 153Relation between circumference and diameter of a circle and its usesA circle is a closed figure bounded by curve lines. In the figure, ABC is a circle. O is centre of the circle. OA is the radius of the circle. AOB is the diameter of the circle. Perimeter of a circle is called circumference. The circumference and diameter of a circle holds a constant ratio. We can prove it by taking some circle of different radii. We measure their circumferences by thread. Now, draw three circles with radii 1 cm, 1.5 cm and 2 cm respectively. 1 cm 1.5 cm 2 cmNow, measure the length of each thread separately by using rural, like, 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 156 5 4 3 2 1The length of thread gives the circumference of the circle. Double of the radius gives the diameter. The measurements of circumference and diameter of the circle are given in the table.Figure Diameter (d) Circumference (C) Ratio = Cd 1. 2 cm 6.28 cm 6.282 = 3.14 approximately2. 3 cm 9.43 cm 9.43 3 = 3.14 approximately3. 4 cm 12.57 cm 12.574 = 3.14 approximatelyWe see that circumference (C) divided by diameter (d) is almost the same for every circle. This ratio is called π (The Greek Letter, pi), which is constant. r is the radius of the circle O A BC

154 Acme Mathematics 7Hence, We have, Cd = π or, C = πdor, C = 2πr (d = 2r) Thus, C = 2πr π is not equivalent to any rational number because it is a non-terminating and nonrecurring decimal. Why we use the ratio π =22 7 to find the area or perimeter? Can't we use other concept why?For practical purpose we take π = 22 7 or 3.14. RememberWe take value of π as 22 7 until or otherwise it is stated.Solved ExamplesExample 1 Find the circumference of a circle whose radius is 14 cm. [Take π = 22 7 ]Solution: Here, radius (r) = 14 cm Circumference (C) = ? Using the formula, C = 2πr or, C = 2 × 22 7 × 14 cmor, C = 88 cm Hence, circumference of the circle is 88 cm. Example 2 Find the radius of a circular field whose circumference is 2.2 kilometers. [Take π = 22 7 ] Solution: Here, Circumference (c) = 2.2 km,Radius (r) = ? Using the formula,C = 2πr or, 2.2 km = 2 × 22 7 × r or,22 × 7 10 × 2 × 22 km = r or, r = 22 × 7 10 × 2 × 22 kmor, r = 7 × 1000 10 × 2 m (1 km = 1000 m)or, r = 350 m Hence, radius of the circle is 350 m.r is the radius of the circle

Acme Mathematics 7 155Example 3 The radius of the moving wheel of an engine is 1.75 m. How many revolutions will it make in travelling 22 km ? [Take π = 22 7 ]Solution: Here, Radius of the wheel (r) = 1.75 m Now, Circumference of the wheel (c) = 2πr or, C = 2 × 22 7 × 1.75 mor, C = 44 7 × 175 100 mor, C = 11 m In travelling 11 meters, the wheel makes 1 revolution In short,11 meters = 1 revolution 1 meters = 1 11 revolution 22 km = 22000 × 1 11 revolution [Hints; 1 km = 1000 m] = 2000 revolution Hence, the wheel makes 2000 revolutions.Classwork1. Write true or false for the following statements.(a) Circumference of circle is 2πr.(b) Circumference of semi-circle is πr2.(c) When circumference increases its area also increases.(d) When circumference decreases its area also decreases.(e) Circumference decreases when radius increases.(f) Circumference increases when diameter decreases.2. Draw the circle whose circumference is 22 cm.3. Draw a circle and mark its any four parts.

156 Acme Mathematics 7Exercise 3.51. Calculate the circumference of a circle whose radius is given below. [Take π = 22 7 ](a) 4.9 cm (b) 21 cm (c) 7 cm (d) 154 cm 2. Calculate the circumference of a circle whose diameter is given below. [Take π = 22 7 ](a) 49 cm (b) 63 cm (c) 1.4 km (d) 28 m 3. Complete the table given below. [Take π = 22 7 ]S.N. Circumference (C) Diameter (d) Radius (r)(a) ......................... 4 cm ....................(b) ....................... .................... 1.5 cm(c) 47.1 cm ................... ...................(d) ....................... 2.1 cm ...................(e) ......................... ........................... 7 cm(f) 9.42 cm ............................. .........................4. The radius of a wheel of a bicycle is 0.77 m. How far does it reach after revolving the wheel 200 times. 5. The diameter of a circular garden is 140 meters. On its outside, there is a road 7 meters wide. Find the outer circumference of the road.6. A carriage wheel makes 1500 revolutions in going over a distance of 4.5 km. Find its radius. 7. A circular road runs round a circular garden. If the circumference of the outer circle and inner circle are 100 m and 80 m, find the width of the road.

Acme Mathematics 7 1578. The radius of a circular pond is 42 m. Find the length of fencing wire with 4 rounds.9. If 550 m of wire is required to fence a circular field with 4 rounds, find the radius of the field. 10. The hour hand of a clock is 4.27 cm long. What distance does its tip move in 6 hours ?11. The length of a minute hand on a wall clock is 6 meter. Calculate the distance cover by minute hand in thirty minutes.12. Measure round your head's diameter and give an approximate value for your head's diameter.[Use π = 3.14]13. Draw circles with given circumferences:(a) 25 cm (b) 9.42 cm (c) 47.1 cm14. Aahana tied her pet dog with a 6 m long rope to a pole in the garden. The dog started to take circular rounds around the pole with straight rope. Find the distance covered by the dog in 40 rounds. [use π = 3.14]15. Calculate the perimeter of the given figure. [use π = 22 7 ]A BCO 14 cm

158 Acme Mathematics 7Project Work1. Objective : To find the relationship between the circumference and diameter of a circle2. Materials required :a sheet of paper with at least four circle of different diameter.thread scale, scissors.3. Activities:Circle number length of diameter length of circumference length circumference length of diameter1 .....2 .....3 .....4 .....” Number the circle 1, 2, 3 and 4.” Cut the circles.” Take a circle and round the thread to fit its circumference” Cut the thread and measure its length.” Measure the length of diameter and complete the table.” Repeat the activity with all the circle.” Divide the length of circumference by the length of the diameter each time.” Present your result in your classroom.

Acme Mathematics 7 1591. (a) From the figure write a radius and a diameter.(b) Calculate the total surface area of given cuboid.(c) Find the circumference of a circle whose radius is 14 cm. (Take = 227 )2. (a) Find the perimeter of given triangle.(b) Write the formula to find, the total surface area of a cuboid.(c) Find the total surface of area of cube when l = 3 cm.(d) A box has length 6 cm, breadth 5 cm and height 4 cm then find its total surface area.3. Mr. Sharma had a triangular plot of land having length of edges 40 m, 30.5 m, and 48.5 m.(a) What is the perimeter of his plot?(b) If he constructed a circular pond of radius 7 m inside the plot, find the circumference of the pond.(c) If he built a cuboidal house on his plot with length 20 m, width 18 m and height 25 m, what is the volume of his house?(d) If one of the room of his house is 10 m long, 6 m wide and 4 m high, calculate the total surface area of the room.4. There is a triangular pond with edges 17 m, 23 m and 25 m in the ground. A man wants to round on the edges of the pond 6 times by wire to keedp safe from falling babies on the pond.(a) Write formula to find perimeter of scalene triangle.(b) Find the perimeter of the pond.(c) How long wire does he need?(d) Find the cost of wire at Rs. 50 per m.A O BC10 cm6 cm 8 cm4 cm5 cm6 cmAB CMixed Exercise

160 Acme Mathematics 75. (a) A rectangular ground is 40 m long and 30 m broad. If a girl runs around the ground,(i) Write the formula to calculate the perimeter of ground.(ii) What distance does she cover in 5 round.(b) A chalk box has length 18 cm, breadth 12 cm and height 10 cm,(i) What is the total surface area of chalk box?(ii) When top surface of the box is removed, find the surface area of the box.6. The total surface area of a cuboid with length 11 cm and breadth 8 cm is 537 cm2.(a) Write the formula for finding the total surface area of cuboid.(b) Find the height of the cuboid.(c) Find the perimeter of triangular pond with edges 13 m, 15 m and 20 m.7. Given cuboid tank's l = 8 cm, b = 7 cm, h = 6 cm.(a) Find the total surface area of the given cuboid.(b) Find the volume of given cuboid.(c) How many litre of water is holded of given cuboid tank.(d) Write the difference between cuboid and cube.8. (a) The perimeter of an equilateral triangle is 48 cm.(i) Find the length of each side of that triangle.(ii) Write down the formula to find the total surface area of cube having length each side 'l'.(b) The length, breadth and height of a cuboid are 3 m , 2 m and 1.5 m respectively.(i) Find the volume of cuboid.(ii) How much liters of water does the given cuboid contain, if it is half filled?9. A cuboidal box has its length two times of its breadth. The height of it is 10 cm and volume 2000 cm3.(a) Write the formula for finding the volume of cuboid.(b) Find the length, of the cuboid if its breadth is 10 cm.(c) How many cuboids of 100 cm3 can be kept in this box?8 cm 7 cm6 cm

Acme Mathematics 7 16110. (a) Find the perimeter of triangle from the given figure.B 5 cm CA(b) Two sides of a triangle are 30 cm and 40 cm respectively. If the perimeter of the triangle is 120 cm, what is the length of third side.(c) Find out how much Sita spent if she used metal wire worth Rs. 20 per meter for fencing a triangular field having side length of 8 m, 10 m and 12 m for 5 times.11. The volume of a cube and a cuboid are equal. The length of the cube is 6 cm and length and breadth of the cuboid are 9 cm and 4 cm respectively.(a) Find the volume of the cube.(b) Find the volume of the cuboid.(c) Find the height of the cuboid.(d) How much volume will be increased if the length of the cube will be doubled?12. Draw a circle.(a) Name the centre.(b) Draw radius and measure it.(c) Draw diameter and measure it.(d) Calculate its circumference.(e) If radius is doubled, what will its circumference?13. Draw a circle and level the following parts.(a) Chord(b) Major segment(c) Minor arc(d) Semi-circle(e) Major sector

162 Acme Mathematics 7CA 4cm T5cm 6cmDO 14cm G11cm 16cm21m22mEvaluationTime: 48 minutes Full Marks: 201. A chalk box has length 15cm, breadth 10cm and height 6cm.(a) Write the formula to calculate the total surface area of cuboid. [1](b) Find the volume of chalk box. [2](c) When the box is kept on the table, how much surface of the table is occupied by it ? [1](d) Write the relation between radius and diameter of a circle. [1]2. (a) Find the perimeter of given triangle. [1](b) Write the formula to find, the total surface area of a cuboid. [1](c) Find the total surface of area of cube when L = 3cm. [1](d) A box has length 6cm, breadth 5cm and height 4cm then find its total surface area. [2]3. Study the given figure. (a) Write the formula to find total surface area of a cube. [1](b) Find the total surface area of given figure. [2](c) If total surface area of cube is 96cm2 , find the length of the cube. [1](d) Find the perimeter of given triangle. [1]4. Kamala has a triangular yard whose measurements are 20m, 21m and 22m. She makes a circular pond of radius 0.7m in the yard. Also she has a dog house in the yard. The measure of dog house are 5m, 4m and 3m. (a) Find the length of boundary of the yard. [1] (b) Find the circumference of the pond. [1](c) Find the volume of the dog house. [1](d) Find the total surface area of the dog house. [2] Chalk Box3 cm5 cm4 cm

Acme Mathematics 7 1634UNIT Algebra1. Express the following in expanded form(a) 23 (b) 34(c) x5 (d) a32. Write the following in index form.(a) 4 × 4 × 4 (b) 5 × 5 × 5 × 5(c) p × p × p × p × p × p 3. Name the base in the following.(a) 513 (b) 2p4(c) xy7(d) 704. Name the index(a) 2p4(b) 2p4(c) 2p4(d) 2p45. If x = 2 and y = 1, find the value of the following. (a) 3x (b) y5 (c) (xy)2 (d) x4(e) 10y (f) 4xy (g) x0 (h) y06. Divide: (a) 32x4y4 by 8xy (b) (10x3y2 + 5x3y3) by 5xy 7. Fill in the blanks: (a) x3 ÷ x = ..............(b) a2 × a3 = ..............(c) b3 + b2 = ..............(d) a6 – a4 = ..............(e) 32 = ..............(f) 83 = ..............4.1 IndicesWarm Up Test

164 Acme Mathematics 7A Index FormConsider the number 81. We know that, 81 = 3 × 3 × 3 × 3 Therefore, 81 = 3434 is the index (exponential) form of 81. Similarly, 100 = 102, 64 = 43In 34, 3 is called the base. 4 is called the index.Thus, 34 means “3 is multiplied by itself 4 times.” 34 = 3 × 3 × 3 × 3 = 81 31 means 3 multiplied by itself once. Therefore, 31 = 3Any number having index (power) 1 gives the number itself.Solved ExamplesExample 1: Expand: (a) 43 (b) (– 2)3 (c) 344(d) ab3Solution: (a) 43 means, 4 is multiplied to itself 3 times. Therefore, 43 = 4 × 4 × 4 = 16 × 4 = 64 Similarly, (b) (– 2)3 = (– 2) × (– 2) × (– 2) = (+ 4) × (– 2) = – 8 (c) 344 = 34 × 34 × 34 × 34 = 3 × 3 × 3 × 34 × 4 × 4 × 4 = 81256(d) ab3= ab × ab × ab = a × a × ab × b × b = a3b3Example 2: Express 169 in the index form. Solution: 169 = 13 × 13 = 132Example 3: Write m × m × m × m × m × n × n in the index form. Solution: m × m × m × m × m × n × n = m5 × n2 = m5n2Example 4: Simplify: 573 × 752Solution: 573 × 752 = 5 × 5 × 57 × 7 × 7 × 7 × 75 × 5 = 5713 16913

Acme Mathematics 7 165Classwork1. Write the base and index (power) in the blanks. (a) 67 base = …….. and index = ………. (b) x4 base = …….. and index = ………. (c) (– 3)3 base = …….. and index = ………. (d) – (4)6 base = …….. and index = ………. (e) m2 base = …….. and index = ………. 2. Expand the following. (a) 63 (b) (– 2)5 (c) 35 (d) (– 2)4(e) 1052 (f) (– 11)4 (g) – 233(h) (mn)4Exercise 4.11. Write the following in the base and power form. (a) 5 × 5 × 5 × 5 × 5 (b) a × a × a × a (c) (– 4) × (– 4) × (– 4) × (– 4) × (– 4) (d) 2 × 2 × 2 × 2 × 2 × 2 (e) 278 (f) 6252401(g) 811000 (h) – 6492. Compare the following numbers and put the sign '<' or '>'. (a) 33 and 42 (b) 62 and 43 (c) 37 and 44(d) 112 and 63 (e) 53 and 44 (f) 38 and 923. Simplify: (a) 26 + 10 (b) 92 ÷ 34 (c) 53 ÷ 52(d) (– 5) × (49)2 ÷ 75 (e) (– 6)4 + 42 (f) 53 – 34 + 6 (g) (– 2)6 + (6)2 (h) – 234 × 33 (i) a7 ÷ a2 + 4a5

166 Acme Mathematics 7B Laws of indicesThere are certain laws in the indices. They are as follows, Law 1: For any rational number x and any integers m and n, xm × xn = xm + nFor examples 23 × 22 = 23 + 2 = 25x3 × x4 = x3 + 4 = x7Law 2: For any rational number xy and m, the positive integer, xym = xmymFor examples 253 = 2 × 2 × 25 × 5 × 5 = 2353ab4 = ab × ab × ab × ab= a4b4Law 3: For any rational number x and integers m and n, xm ÷ xn = xm – nFor examples 56 ÷ 54 = 5 × 5 × 5 × 5 × 5 × 55 × 5 × 5 × 5= 5 × 5 = 52 = 56 – 4 a3 ÷ a2 = a × a × aa × a= a = a1 = a3 – 2Law 4: For any rational number x and integers m and n, (xm)n = xmnFor examples, (23)2 = 23 × 23= 23 + 3 = 23 × 2 = 26(a2)3 = a2 × a2 × a2= a2 + 2 + 2 = a6Law 5: For any rational number x, x° = 1. For examples, 4° = 42 – 2= 42 ÷ 42= 4242= 4 × 44 × 4 = 1Law 6: For any rational number x and integer m and n if xm = xn then, m = n.

Acme Mathematics 7 167Classwork1. Complete the pattern.(a) 32 × 33 = 32 + 3 = 35 = 243 (b) 24 × 20 = ..................................(c) 43 ÷ 42 = .................................. (d) 52 ÷ 52 = ..................................(e) 78 ÷ 76 = ..................................2. Simplify and write the answer in exponential form. (a) 23 × 21 × 22 (b) 23 × 24 (c) 123× 122 × 124(d) (– 3)4 × (– 3)2 × (–3) (e) 341× 342 × 343(f) 124÷ 132 (g) (– 4)3 ÷ (4)2 (h) 25 ÷ 22 (i) 54 ÷ 54(j) x3 ÷ x3 (k) a7 ÷ a7Exercise 4.21. Simplify the following: (a) (2 × 3)3 (b) 454(c) (42)3 (d) 252 3(e) 122 6(f) 782 2(g) {(– 1)3}7 (h) – 56 2 4(i) – 32 3 22. Simplify: (a) (42 × 51) + (24 – 71) – 5° (b) (23)2 × 32 – 43 (c) {(5–2)– 1 × 52} ÷ 53(d) – 896÷ 894(e) 23 × 3234 × 24 (f) a6 × b4a3 × b23. Express as form of index exponential.(a) 8 (b) 27 (c) 128 (d) 243(e) 625 (f) 343 (g) 1100 (h) 15124. Simplify.(a) (x + y)2 × (x + y)4 (b) (x – y)7 ÷ (x – y)5 (c) (2xy)3 × (2xy)4(d) 9(xy)5 ÷ (3xy)2 (e) (9x3 × 4x5) ÷ 18x6

168 Acme Mathematics 75. Simplify:(a) xa – b × xb – c × xc – a (b) xa × xb × x–2a × x–2b × x2(a + b)(c) (ya)b – c × (yb)c – a × (yc)a – b6. If p = 1, q = – 1 and r = 2. Find the value of the following..(a) pq2 (b) qr3 (c) p2q2r2(d) 3p × 2r (e) 4q × 2p × r3 (f) (2 × 3 × 5)pqr7. Solve for x. (a) 23× 24= 2x (b) (–1)5× (– 1)4= (– 1)x (c) 122× 123 = 12x(d) (– 4)3× (– 4)2= (– 4)x(e) (23)4= 23x (f) (52)7= 5x + 10(g) ab4÷ ab4 = abx(h) 5310÷ 536 = 532x8. (a) Find the value of (a0 + 1)(b) Simplify : ax – y . ay – z . az – x(c) Find the value of x from the given figure.9. (a) What is the value of (2a)0.(b) Simplify : xa – b. xb – c. xc – a10. (a) Prove that : xya – b× xyb – c× xyc – a = 1(b) Simplify : 827– 2311. (a) What is the value of (7x)0.(b) Simplify : pa – b. pb – c. pc – a12. (a) What index of 'x' will be equal to 1?(b) Write the exponential form. xy × xy × xy × xyB CA8 cmx10 cm

Acme Mathematics 7 169Warm Up Test1. Make an algebraic expression using the sign +, – , × , ÷ between the given terms. (a) 2x and y (b) 4xy and 3z (c) x + z and 10y (d) 23 and x – y 2. Match the following (a) the sum of x and y (i) 4x3y(b) 10 times p (ii) 13 (3x + z)(c) 4x is divided by 3y (iii) x + y (d) one-third of sum of 3x and z (iv) x is a constant (e) the value of x is 3 (v) 10p 3. Fill in the blanks. (a) 4x + 5x = .................... (b) x + 3 + 4x = ...................... (c) 3x × 2y = .................... (d) 9x – 6x = .......................... (e) 6a2 + a2 = .................... (f) 10y2 – 7y2 = ...................... (g) If x = 3, x + 4 = .......... (h) If y = 3, 3y – 2 = .............. (i) 10x ÷ 2 = .................... (j) 12y3 ÷ 6y2 = ..................... 4. (a) If x = 5 and y = 6, find the numerical value of x2 + y2. (b) If a = 2, b = 3 and c = 4, find the numerical value of 5a – b – c. (c) If a = 2, b = 3 and c = 4, find the numerical value of 3a + 2bc . 5. Simplify: (a) 2x + 5y – 4y (b) a + 2b + c – (a + 3b + 4c) 6. Multiply: (a) (4x + 5y) by 3x (b) 7x by (2x + 7) 7. Divide: (a) 32x4y4 by 8xy (b) (10x3y2 + 5x3y3) by 5xy 8. Fill in the blanks: (a) x × x3 = ................. (b) x3 × x4 = .............. (c) am × an = ...............(d) x5 × x2 = ................ (e) x3 ÷ x = ............... (f) am ÷ an = ...........(g) x0 = ........................ (h) (x2)2 = ..................4.2 Algebraic Expressions

170 Acme Mathematics 7A. Revision of Algebraic Expressionsa. RevisionConstants and variables A quantity which has a fixed value is called a constant. 0, 1, 2, 3, …. are constants. A quantity which can take various values is called a variable. Algebraic expressions An algebraic expression is a combination of constants and variables by the signs +, –, × or ÷ . For example, 3x3y2 + 2xy3, 2ab + 9ab2 – 8ab3 , 6x2y and 4z + 6x5 , are algebraic expressions. Algebraic term The terms of an algebraic expression are parts of the expression separated by '+' or '–' signs. For example 2ab, 9ab2 and –8ab3 are terms of the expression 2ab + 9ab2 – 8ab3. An algebraic expression containing one term is called a monomial. e.g. 6x2y, –3a3 or 8 are monomials. An algebraic expression containing two terms is called a binomial. e.g. 4xy2 + 2xy3 or ab + 9 are binomials. An algebraic expression containing three terms is called a trinomial, e.g. –2ab + 9ab2 – 8ab3. In general, the algebraic expressions with two or more terms are called polynomial. A term in an algebraic expression which has no literal factor is called a constant term. In 3x2 + 2x – 3, 3 is the constant term. In an algebraic term, any one of the factors is the coefficient of the product of the other factors. e.g. in 7x2y, 7 is the coefficient of x2y; 7y is the coefficient of x2. Terms in an expression which have the same literal factors are called like terms. 3xy, 6xy and 9 xy are like terms. Terms which do not have the same literal factors are called unlike terms. 4xy, 5ab and 6x2y2 are unlike terms.b. Value of algebraic expressionValue of algebraic expression is the numerical expression. It is possible when value of variable is given. To find the numerical value, we substitute the given value of the variable in the expression and simplify it.

Acme Mathematics 7 171Solved ExamplesExample 1: If x = 1, y = 2 & z = 3, find the numerical value of the expression 2x + 3y + 4z. Solution: Here, 2x + 3y + 4z= 2 × 1 + 3 × 2 + 4 × 3= 2 + 6 + 12 = 20 Thus, the numerical value of 2x + 3y + 4z is 20. Example 2: Find the numerical value of 4x2 – 3y2 – 4z2 when x = 6, y = 5 and z = 2. Solution: Here, 4x2 – 3y2 – 4z2= 4 × 62 – 3 × 52 – 4 × 22= 4 × 36 – 3 × 25 – 4 × 4 = 144 – 75 – 16 = 144 – 91 = 53 Thus, the numerical value of 4x2 – 3y2 – 4z2 is 53. Example 3: If p = 23, q = 35, find the value of (5p2 × q) × (– 3pq2). Solution: Here, (5p2 × q) × (– 3pq2) = 5 × 23 2 × 35  × – 3 × 23  × 35 2 = 5 × 49 × 35 × – 3 × 23 × 925= 43 × – 1825 = – 2425 Thus, the numerical value of (5p2 × q) × (– 3pq2) is – 2425 .Classwork1. Calculate the numerical value for the following algebraic expressions when a = 1 and b = 2. (a) 12a – 3b2 (b) 8a2 + 5ab3 (c) 3b + 25a3(d) 32 a2b2 – 12 ab2 (e) – 4a2b2 – 1 + 12ab2 (f) 14ab3 – 12ab2 – a42. Calculate the numerical value for the following algebraic expressions. (a) 6x2 – 4x(x – 1) + x(4 + 5x) for x = –1. (b) 3x2 + 4xy for x = 34 and y = 3 (c) 3p6 + (– 4q2) for p = –1 and q = – 2 (d) (– 3x2) (4xy3) – 23 x3 for x = 34 and y = 43

172 Acme Mathematics 7Exercise 4.31. Calculate the numerical value for the following algebraic expressions when x = –1 and y = 2. (a) 4x3 + 3x4 – x5 (b) – x2 + 3x3 – 8x5(c) – 3xy – 2xy + 22 (d) – 23 xy2 + – 32 x2y – 112. Calculate the numerical value for the following algebraic expressions. (a) – 646 p3 + (2q2) – – 83 pq for p = 38 and q = 3 (b) 2x5 – (6xy2) + (9xy) for x = 2 and y = 4 (c) – 38 pq2 × 43 p2r + qr for p = 1, q = –1 and r = 2 (d) – 2c ab2 + 38 a2 × – 611 c2 for a = b = c = –1 3. Calculate the numerical value for the following algebraic expressions for x = –1, y = –2 and z = 1. (a) (x2 + 1) + (y2 – 2yz) (b) 3xy – (z2 + y2 + 4) (c) –12yz3 – (4yx2 + 34 y2z2) (d) – 47 x2z2 + 32 x2z – 94 xy2c. Addition and SubtractionTo add two or more algebraic expressions, group the like terms and add their constant coefficients. e.g (3x2 + 5y) + (4x2 – 2y) = (3x2 + 4x2) + (5y – 2y) = 7x2 – 3y To subtract an algebraic expression from another, change the sign of each term in the expression to be subtracted, and then add the expressions. e.g (7x2 + 5y) – (4x2 – 2y) = 7x2 + 5y – 4x2 + 2y = 7x2 – 4x2 + 5y + 2y= 3x2 + 7y Addition and subtraction can also be done by the vertical method, by writing the expressions one below the other so that all like terms are in the same column.Addition 3x2 + 5y + 4x2 – 2y 7x2 + 3ySubtraction 7x2 + 5y 4x2 – 2y – + 3x2 + 7y

Acme Mathematics 7 173Solved ExamplesExample 1: Add 5x2 – 4x + 3 , 2x2 + 6x + 8 and x2 + 2x – 1 Solution: (5x2 – 4x + 3) + (2x2 + 6x + 8) + (x2 + 2x – 1) = 5x2 – 4x + 3 + 2x2 + 6x + 8 + x2 + 2x – 1 = 5x2 + 2x2 + x2 – 4x + 6x + 2x + 3 + 8 – 1 [Collecting the like terms]= 8x2 + 4x + 10 Example 2: Subtract 2x2 – 3xy – 4ax2 from 8x2 + 7xy + 6ax2Solution: (8x2 + 7xy + 6ax2) – ( 2x2 – 3xy – 4ax2) = 8x2 + 7xy + 6ax2 – 2x2 + 3xy + 4ax2= 8x2 – 2x2 + 7xy + 3xy + 6ax2 + 4ax2 [Collecting the like terms]= 6x2 + 10xy + 10ax2Classwork1. Add: (a) 5x, 2x and 5x (b) 2x2, 8x2, x2 and 6x2y2(c) – 5x2y2, 7x2y2, 9x2y2 and 11x2y2 (d) 2x – 2y, 5y – 2x, 5x – y 2. Subtract: (a) x from 3x (b) – 7x2y2 from 2x2y2(c) 2x + y from x + 3y (d) x – y from x + y 3. Simplify : (a) 3x – 4y + 2x – 2y (b) (13x2 – 2b2) + (11x2 – b2) Exercise 4.41. Add: (a) 8x2 + 2y2, 3y2 + 2x2 and 5x2 – y2 (b) 2x2y – 4xy, 5x2y + 2xy, –x2y + 2xy (c) 2x2 – 5y2 + 4z2, 2y2 – x2 + 4z2, 2y2 – 5x2 + 4z2(d) 14x2 + 23 x + 1, – 54 x2 – x3 + 8 and 2x2 – 812 x – 75(e) 6a2 + 5b2 – 2c2, – 5a2 – 5b2 – 5c2 and 7a2 – 3b2 – 5c2(f) a + 2b – c + d, 2b – 3d + 5c + a, 5b – 2c + 5 – 2a 2. Subtract: (a) x2 + y2 – 2xy from x2 + y2 + 2xy (b) x2 + y2 from 2x2 – 6y2 + 3 (c) a2 – 2a + 9 from –a2 + 2a – 3 (d) 2x2 + 3y2 – 4z2 from 4z2 – 3x2 + 4y2(e) 2x2 + 3y2 – 4z2 from x2 + 9 (f) 11 – a2 + b2 from 20 + 3b2 – 4c2(g) 3 + 2x + 3y – z from 3 + 2x + 3y – z (h) 1 from 3 + 2x + 3y – z

174 Acme Mathematics 73. Simplify : (a) (6x2 – 7y2) – (2x2 – 3y2) (b) (6a2 – 2b2) – (3a2 – b2) (c) 8(x + y) – 6 (2 + x – 2y) (d) {5z – (4x + 2y)} – {3x – (4y – 2z)}(e) 67 a2 – 67 b2 + 7 – 37 a2 – 37 b2 + 9 (f) 13 x2 – 13 y2 + 13 z2 – 23 x2 – 53 y2 + 73 z24. Solve the following word problems. (a) Find the sum of 2x + 3y + 4z, 4x – 5y + 6z and –8x + 9y – 10z. (b) Subtract 3x + 4y – 2z from the sum of 3 + 4z + 3y – x and 3x + 2y – 3z + 2. (c) Subtract the sum of 4x + 5y – z and –2x + 3y – z from the sum of 5x – 2y and 3x – y + zB. Multiplication of Algebraic Expressions1. Multiplication Properties of an Algebraic ExpressionsMultiplication in algebra follows the same rules as multiplication of numbers and exponents. They are: (i) Commutative property → x × y = y × x (ii) Associative property → (x × y) × z = x × (y × z) (iii) Distributive property → x (y + z) = xy + xz (iv) Law of exponents → xm × xn = xm + n2. Multiplication of MonomialsLet us see how, using the above rules, we can multiply the monomials.Solved ExamplesExample 1: Multiply : 2x by 3y Solution: 2x × 3y = (2 × 3) × (x × y)= 6 × xy = 6xy Example 2: Multiply : 3x2y and 2xy. Solution: 3x2y × 2xy = (3 × 2) (x2 × x ) (y × y) = 6x3y2Remember! (a) we multiply the constant together, (b) we multiply the variable together, using the multiplication rule for exponents.We can use dot(.) in the place of 'x' sign in case of alphabets i.e. x × x = x.x = x2

Acme Mathematics 7 175Example 3: Multiply 4x2 , 5x and 2x. Solution: 4x2 × 5x × 2x = 4 × 5 × 2 × x2 × x × x = 40 × x4= 40x43. Multiplication of Binomials by MonomialsHow to multiply 3x + 4y by 4x ? We know that, the area of rectangle is the product of its length and breadth if its adjacent sides are ‘l’ and ‘b’. If the length of a rectangle is (3x + 4y) cm and the breadth is 4x cm, what is its area? Here, Area (A) = l × b = (3x + 4y) × 4x = (4x × 3x) + (4x × 4y)= 12x2 + 16xy Hence (3x + 4y) ×4x = 12x2 + 16xy™ We can multiply the binomial by a monomial by multiplying each terms of the binomial. ™ When a binomial is multiplied by a monomial, the distributive property is applied.Multiplication can be done in two ways. (a) By horizontal way (b) By vertical way4. Multiplication of Trinomials by MonomialsHow to multiply 3x + 4y + 3 by 5x ? We know that, the area of rectangle is the product of its length and breadth if its adjacent sides are ‘l’ and ‘b’. If the length of a rectangle is (3x + 4y + 3) cm and the breadth is 5x cm, what is its area? Here, Area (A) = l × b = (3x + 4y + 3) × 5x = 15x2 + 20xy + 15x Hence (3x + 4y + 3) × 5x = 15x2 + 20xy + 15xMultiplication Sign++––++–– ––++××××====3x 4y12x2 4x16xy3x 4y 315x2 5x20xy 15x

176 Acme Mathematics 7™ We can multiply the trinomials by a monomial by multiplying each terms of the trinomial. ™ When a trinomial is multiplied by a monomial, the distributive property is applied.Solved ExamplesExample 1: Multiply (4x + 5y) by 3x Solution: 3x × (4x + 5y) = 3x × 4x + 3x × 5y= 12x2 + 15xy This is the horizontal method of multiplication. We can also carry out the multiplication by the vertical method, as shown below. 4x + 5y × 3x 12x2 + 15xyExample 2: Simplify: 7x2 – 4x(x + 1) + x(4 + 5x) Solution: Here, 7x2 – 4x(x + 1) + x(4 + 5x) = 7x2 – 4x × x – 4x × 1 + 4 × x + 5x × x = 7x2 – 4x2 – 4x + 4x + 5x2= 3x2 + 5x2 = 8x2Classwork1. Multiply the following monomials. (a) 3 × 2a (b) 2y × 5y (c) – 6x × (2xy) (d) – 3x × (– 6x) (e) 12a × 3b2 (f) 8a2 × 5ab3(g) 3b × 5a3 (h) 7a2b2 × ab2 (i) – 4a2b2 × ab2(j) 14ab3 × ab2 × (– a4) (k) 4x3 × 3x4 × (– x5) (l) x2 × 3x3 × (– 8x5) (m) – 3xy × (– 2xy) (n) 9xy2 × 9x2y × 9x2y22. Find the product of the following algebraic expressions. (a) 4x(x – 1) (b) 3x2 (4xy + 3) (c) 3p6 × (pq – 4q2) (d) 3a(b – 3) (e) – 3x2(2x + 7) (f) 10y(2x2 – 5y)Exercise 4.51. Find the product of the following algebraic expressions. (a) 5a3(6a + 7b) (b) 2x3 (2x2 – 4y2) (c) x2 (x + y2 – 2xy)

Acme Mathematics 7 177(d) 3xy(x2 + y2 + z) (e) – 12yz3(4yz2 + 34 y2z2 + 8) (f) 5x2z2 12 x2z – 34 x + z2 (g) x2 (1 + y2 – 2xy) (h) 3xy(x2 + y2 + 4) (i) yz2(4xy – z2 + 34 y2z2) (j) 9xyz(x2 + y2 + z) 2. Find the area of the following rectangles.(a)x3(b) (c)3. Simplify: (a) 3x(x2 + y2) – 2x(x + y) + 20xy (b) x2(x + 1) – x(x2 + 1) – x3(c) x2(3x3 – 2x2 + 1) – x(3x3 – 2x2 + 1) (d) x3(x2 – c2) + 2b2(c2 – x3) + c2(x2 – b3)(e) a2b(b2 – 2a) + a(b – a2b) – b(a + 2a3) (f) 59 x(x2 – x3) – 13 x2 (x3 – x) + 13 x3 (1 – x2) (g) 2a2b(a3 – a + 1) – ab(2a4 – 2a2 + a) – ab(a3 + a – 1) (h) x3y(x2 – 2x + 5) + 2xy(x3 – 2x2 + 3y)5. Multiplication of Binomial by a BinomialLet two binomials are a + b and a – b. Let the length is (a + b) unit and the breadth is (a – b) unit of a rectangle. Now, Area of the rectangle (A) = length × breadth= (a + b) × (a – b) = a (a – b) + b (a – b) = a2 – ab + ab – b2= (a2 – b2) sq. unit Hence, (a + b) × (a – b) = a2 – b2Solved ExamplesExample 1: Multiply: (a + b ) by (a + b) Solution: (a + b) (a + b ) = a(a + b ) + b (a + b ) = a2 + ab + ab + b2= a2 + 2ab + b2Hence, (a + b) × (a + b) = a2 + 2ab + b2 x4xy 3x4xya – ba + ba + ba + b

178 Acme Mathematics 7Example 2: Multiply: a + b by a2 – b2Solution: (a2 – b2) × (a + b ) = a2 (a + b) – b2 (a + b) = a3 + a2b – ab2 – b36. Multiplication of Trinomial by BinomialExample 3: Multiply: (a2 – ab + b2) by (a + b) Solution: (a + b) × (a2 – ab + b2) = a (a2 – ab + b2) + b (a2 – ab + b2) = a3 – a2b + ab2 + a2b – ab2 + b3= a3 + b3Hence‚ (a + b) (a2 – ab + b2) = a3 + b3Example 4: Multiply: (a2 + ab + b2) by (a – b) Solution: (a – b) × (a2 + ab + b2) = a(a2 + ab + b2) – b (a2 + ab + b2) = a3 + a2b + ab2 – a2b – ab2 – b3 = a3 – b3Hence‚ a3 – b3 = (a – b) (a2 + ab + b2)Example 5: Multiply: (ab – bc – ca) by (a – b). Solution: (a – b) (ab – bc – ca) = a(ab – bc – ca) – b (ab – bc – ca) = a2b – abc – a2c – ab2 + b2c + abc = a2b – a2c – ab2 + b2cExample 6: Multiply: (a – b), (a – b) and (a – b). Solution: Here, (a – b) × (a – b) = a (a – b) – b (a – b) = a2 – ab – ab + b2= (a2 – 2ab + b2) Now, (a – b) (a – b) (a – b) = (a – b) (a2 – 2ab + b2) = a3 – 2a2b + ab2 – a2b + 2ab2 – b3= a3 – 3a2b + 3ab2 – b3Hence‚(a – b) × (a – b) × (a – b) = (a – b)3 = a3 – 3a2b + 3ab2 – b3(a2 – ab + b2)(a + b)

Acme Mathematics 7 179Example 7: Multiply: (a + b), (a + b) and (a + b). Solution: Here, (a + b) [(a + b) (a + b)] = (a + b) (a2 + 2ab + b2) [(a + b) (a + b) = a2 + 2ab + b2] = a(a2 + 2ab + b2) + b(a2 + 2ab + b2) = a3 + 2a2b + ab2 + a2b + 2ab2 + b3= a3 + 3a2b + 3ab2 + b3Hence‚ (a + b) × (a + b) ×(a + b) = (a + b)3 = a3 + 3a2b + 3ab2 + b3Remember !In case of multiplication of a binomial or a trinomial by a binomial or a trinomial each term of the binomial or the trinomial should multiply each term of the given binomial or the trinomial and if needed, simplification should be performed.CLASS WORK1. Multiply the following expressions using both the horizontal and the vertical methods. (a) a(c + d) + b(c + d) (b) (a + b) (c + d)(c) (2x + 3y) (3x – 2y) (d) (3a + 2b) (2a – b)(e) (3x – 2x )(x + 3) (f) (2a + 3b) (6a – 4b)(g) (6p – 5p) (4p + 7p) (h) (8ab + 5ab) (6ab – ab)(i) (3p – q2) (7q + 4p2) (j) (4a + 3ab) (4a + 3ab) (k) (3p – 2q) (3p – 2q) (l) (4x – 5y) (4x – 5y) 2. Multiply the following expressions. (a) (3x – 7y + 6) (7x + 2y) (b) (3q2 – 2p – 1) (5q2 – 1)(c) (x2 – y2 + xy) (x2 + y2) (d) (2p – q + pq) (2p – q) 3. Multiply the followings.(a) (x2 + 2y + 3) (x2 – 2) (b) (4x – 3y + 4) (2x – 3) Exercise 4.61. Multiply the following expressions. (a) (4a2 – 5b2 + ab) (a2 – 2b2) (b) (a2 – ab + 3b2) (3b – a) (c) (a2 – b + 5ab) ( a + b2) (d) (8x – 4y + z) (2y + z) (e) (x – 4y + 1) (3y – 2)

180 Acme Mathematics 72. Multiply the followings.(a) (x2 – xy + y2) (x + y) (b) (x2 + xy + y2) (x – y) (c) (2a3 – 3b2) (3a2 – 2b) (d) (3x2 – 2y) (4x2 – y2) (e) (3x + y) (–x + y2) (f) (4y – 3x2) (6x3 – y3) (g) (p2q2 + 3) (3p – 5q) (h) (a2 + ab) (b2 + ab) 3. Find the following products. (a) (a – 4) (a + 4) (a – 1) (b) (x – 5) (5 – x) (x + 1) (c) (x + 4) (x – 4) (x – 6) (d) (2x – y) (y – 2x) (x + y) 4. Simplify: (a) (2y + 3) (y – 2) – (5y + 3) (y – 2) (b) (3x – 4y) (x + y) + (5x + 7y) (x – y) (c) (3x + 4y) (3x – 4y) – (3x + 4y)2 (d) (5a – 3) (a + 4) – (2a + 5) (3a – 4)5. Calculate the area of given rectangles.(a) (b)(c) (d)6. The length and breadth of rectangular field are (a + 3b + c) m and (3c + a) m respectively. Calculate :(a) The area of field.(b) If a = 2 m, b = 3 m and c = 2 m. What is the actual area of that field?a + b2a – b(2a + b + c)2b + 2ca 2b 3cb2aaabb23

Acme Mathematics 7 181EvaluationTime: 43 minutes Full Marks: 181. Give the answer of the following questions. (a) Express as the index form: 5×5×5×5 [1](b) If a = 3, b = 1 and c = –4, find the value of a + b – c2ab . [2]2. (a) Find the value of 32×23 . [1] (b) If x = 2 , y = 3 and z = 4 then find the valu of (y + z)X . [2] 3. (a) Express as the index form: 625 [1](b) Express as numerical form: 81a4 + 72a2b2 + 16b4, when a = 1 and b =2. [2]4. (a) Find the product of (x –1) (x2 – 2x + 1), when x = 5. [1](b) If a + b + c = 0, find the value of Z?+? × Z?+? × Z?+?. [2]5. (a) Express as the continued multiplication form: (3x)4 [1] (b) Convert the algebraic expression 4x2 – 20xy + 25y2 to the square form. [2] (c) Find the area of given rectangle. Calculate it's actual perimeter when a=10.(d) Find the value of (a + b + c)100 when a = –1, b= –2 and c = 4.6. (a) Write the expanded form of (2x+3)2 [1](b) Calculate the volume of thhe given cuboid. [2](P c-a) cm(P b-c) cm(P a-b) cm(2a + 3) m(5a – 3) m

182 Acme Mathematics 77. Some Special ProductsSome binomials occur very frequently in algebra. It is therefore useful to remember their answers. Some of them are: (a) (a + b)2 (b) (a – b)2 (c) (a + b) × (a – b) (d) (a + b)3 (e) (a – b)3These results can be derived from geometry also. (a) Expansion of (a + b)2Geometric concept of (a + b)2. (a + b)2 is the area of a square of side (a + b). We can divide such a square into 4 parts as shown alongside. The areas of the 4 parts are as follows. Figure given alongside is a square ABCD. It is a square of side (a + b) unit. Now, Area of the square ABCD = (a + b) × (a + b) = (a + b)2..........(i) From the figure, Area of the square ABCD = Area of 4 parts = a2 + ab + ab + b2= a2 + 2ab + b2 .......(ii) From (i) and (ii), (a + b)2 = a2 + 2ab + b2 Hence, (a + b)2 = a2 + 2ab + b2(b) Expansion of (a – b)2Geometric concept of (a – b)2. Figure given alongside is a square ABCD. It is a square of side 'a' unit. Now, Area of the square ABCD = a2..........(i) Again, from the figure, Area of the square ABCD = (a – b)2 + b(a – b) + b(a – b) + b2= (a – b)2 + 2ab – b2 .......(ii) From (i) and (ii), (a – b)2 + 2ab – b2 = a2 or, (a – b)2 = a2 – 2ab + b2Hence, (a – b)2 = a2 – 2ab + b2BACb Dab a + bb2 baaa2aba + bBACG DF Eba–ba b2 ba–ba–bb(a–b)(a–b)2b(a–b)a

Acme Mathematics 7 183Solved Examples(c) Expansion of (a + b)3Example 1: Find the product of (a + b), (a + b) and (a + b) Solution: (a + b) × (a + b) × (a + b) = (a + b)3= (a + b) (a + b)2= (a + b) (a2 + 2ab + b2) = a(a2 + 2ab + b2) + b(a2 + 2ab + b2) = a3 + 2a2b + ab2 + a2b + 2ab2 + b3= a3 + 3a2b + 3ab2 + b3Hence, (a + b)3 = a3 + 3a2b + 3ab2 + b3(d) Expansion of (a – b)3Example 2: Find the product of (a – b), (a – b) and (a – b) Solution: (a – b) × (a – b) × (a – b) = (a – b)3= (a – b) (a – b)2= (a – b) (a2 – 2ab + b2) = a(a2 – 2ab + b2) – b(a2 – 2ab + b2) = a3 – 2a2b + ab2 – a2b + 2ab2 – b3= a3 – 3a2b + 3ab2 – b3Hence, (a – b)3 = a3 – 3a2b + 3ab2 – b3Hence, we conclude that: 1. (a + b)2 = a2 + 2ab + b22. (a – b)2 = a2 – 2ab + b23. (a + b)3 = a3 + 3a2b + 3ab2 + b3 or, a3 + 3ab(a + b) + b34. (a – b)3 = a3 – 3a2b + 3ab2 – b3 or, a3 – 3ab(a – b) – b3These are the basic formulae in algebra. By using these formulae, we can calculate the squares and the cubes of the algebraic expressions. Solved ExamplesExample 1: Find the product of (4 – x)2. Solution: Method 1: by direct multiplication (4 – x)2 = (4 – x) (4 – x) = 4 (4 – x) – x(4 – x) = 16 – 4x – 4x + x2= 16 – 8x + x2

184 Acme Mathematics 7Method 2: by using formula, (a – b)2 = a2 – 2ab + b2 (4 – x)2 = (4)2 – 2 × 4 × x + (x)2= 16 – 8x + x2Example 2: Expand: x + 1x2Solution: by using formula, (a + b)2 = a2 + 2ab + b2x + 1x2= x2 + 2 × x × 1x + 1x2= x2 + 2 + 1x2Example 3: Expand: (3x + 2y)3Solution: by using formula,(a + b)3 = a3 + 3a2b + 3ab2 + b3(3x + 2y)3 = (3x)3 + 3 × (3x)2 × 2y + 3 × (3x) × (2y)2 + (2y)3= 27x3 + 3 × 9x2 × 2y + 3 × 9x × 4y2 + 8y3= 27x3 + 54x2y + 108xy2 + 8y3(e) To find the product of (a + b) and (a – b) Example 4: Multiply : (a + b) and (a –b) Solution: (a + b) × (a – b) = a(a – b) + b(a – b) = a2 – ab + ab + b2= a2 – b2Hence, (a + b) × (a – b) = a2 – b2(f) Use of ( a + b)2 and (a – b)2Example 5: Find the square of 9 and 11. Solution: We have 9 = 10 – 1 So, 92 = ( 10 – 1)2= 102 – 2 × 10 × 1 + 12= 100 – 20 + 1 = 81 Thus, square of 9 is 81.

Acme Mathematics 7 185Similarly, we have 11 = 10 + 1 So, 112 = (10 + 1)2= 102 + 2 × 10 × 1 + 12 = 100 + 20 + 1 = 121 Thus, square of 11 is 121. Example 6: If a + b = 7 and ab = 10, find the value of a2 + b2. Solution: Here, a + b = 7 So, (a + b)2 = 72 [squaring on the both sides.]or, a2 + 2ab + b2 = 49 or, a2 + 2 × 10 + b2 = 49 or, a2 + 20 + b2 = 49 or, a2 + b2 = 49 – 20 or, a2 + b2 = 29 Thus the value of a2 + b2 is 29. (g) Use of ( a + b)3 and (a – b)3Example 7: If a + b = 3 and ab = 2, find the value of a3 + b3. Solution: Here, a + b = 3 So, (a + b)3 = 33 [Cubing on the both sides.]or, a3 + 3ab(a + b) + b3 = 27 or, a3 + 3 × 2 × 3 + b3 = 27 or, a3 + 18 + b3 = 27 or, a3 + b3 = 27 – 18 or, a3 + b3 = 9 Thus the value of a3 + b3 is 9. Example 8: If x – 1x = 4,prove that x3 – 1x3 = 76 Solution: Here, x – 1x = 4 x – 1x3 = 43 [cubing on the both sides.]or, x3 – 3 × x × 1xx – 1x – 1x3= 64 or, x3 – 3 × 4 – 1x3 = 64 or, x3 – 12 – 1x3 = 64 or, x3 – 1x3 = 64 + 12 = 76 Hence x3 – 1x3 = 76, proved.

186 Acme Mathematics 7Classwork1. Fill in the blanks. (a) (x + 4)2 = ................. (b) (x – 4)2 = .......................... (c) (x – 3)2 =.................. (d) x2 + 2xy + y2 = ..................... (e) a2 – 2ab + b2 = .............. (f) (a – b)3 = ............................ 2. Expand the following by direct multiplication. (a) (4 + 3y)2 (b) (3x + y)2 (c) (a – 2b)2(d) (5x – 2y)2 (e) (3p + 2q)3 (f) (3x – 5y)3(g) (2x – y) (2xz + y) (h) (x + 4y) (x – 4y) (i) (2x – 3y) (2x + 3y) 3. Find the product of the following expression by using the formula a2 – b2. (a) (x + 2) (x – 2) (b) (3x – 3) (3x + 3) (c) a – 23 a + 23(d) (a2 – b2) (a2 + b2) (e) a2 – 3ya2 + 3y (f) 12 – xy12 + xyExercise 4.71. Expand the following by using formula: (a) (2a + 3b)2 (b) (3x – 2y)2 (c) (p + 5q)3 (d) (3m – n)3(e) (5x + bc)3 (f) x2 – 142(g) 2x3 + 3y22(h) mn2 – 2mn22. Find the square of the following algebraic expressions: (a) ( x + y) (b) (a – 2b) (c) (2m + 3n) (d) 1x + x(e) x2y – y2x (f) x2 + y2 (g) 3ab + ab3 (h) xy – yx(i) (a + b + c)2 (j) (a 2 + b 2 + c 2)23. Simplify: (a) (x + y)2 + (x – y)2 (b) (x + y)2 – (x – y)2(c) (2a + 3b)2 + (2a – 3b)2 (d) (a + b)3 + (a – b)34. (a) If x + y = 8 and xy = 12, find the value of x2 + y2. (b) If x – 2y = – 2 and xy = 12, find the value of x2 + 4y2 . (c) If a + 1a = 4, find the value of a2 + 1a2.

Acme Mathematics 7 187(d) If x - 1x = 4 find the value of x2 + 1x2 . 5. Simplify: (a) (x + y)2 – (x2 + y2) (b) (x – 3)2 – (x – 5) (x + 3) (c) (a – b) (a2 + ab + b2) (d) (a + b) (a2 – ab + b2) (e) (a – b)2 – (a + b)2 + 4ab (f) (a + b)2 + (a – b)2 – 4ab 6. Find the square of the following numbers. [use (a + b)2 or (a – b)2] (a) 31 [Hint:31 = 30 + 1] (b) 29 [Hint:29 = 30 – 1] (c) 49 (d) 51(e) 99 (f) 101 (g) 999 (h) 1001 7. (a) If a + b = 5 and ab = 6, find the value of a2 + b2. (b) If x + y = 6 and xy = 9 find the value of x2 + y2. (c) If a – b = – 5 and ab = 6, find the value of a2 – b2. (d) If x – y = 1 and xy = 6 find the value of x2 + y2. (e) If a + 1a = 3, find the value of a2 + 1a2 .(f) If x - 1x = 3, find the value of x2 – 1x28. (a) If a + b = 2, find the value of a3 + b3 + 6ab. (b) If x – y = 2, find the value of x3 – y3. (c) If a – 1a = y , find the value of a3 – 1a3 . (d) If a – 1a = x, find the value of a3 – 1a3 . 9. Simplify: (a) (a + b)3 – a3 – b3 (b) (a – b)3 – (a3 + ab2) + (a2b + b3) (c) (x – y)3 – (x + y)3 + x2y (d) (x + y)3 – (x – y)3 – x2y10. Convert either to (a + b)2 or (a – b)2 form.(a) x2 + 4x + 4 (b) y2 – 14y + 49 (c) 9x2 + 2 + 19x2(d) 4a2 – 20ab + 25b2 (e) 9x4 + 24x2y2 + 16y4 (f) a2 + 2 + 1a2

188 Acme Mathematics 7C. Division of Algebraic Expressions1. Division of monomial by monomialLook at the given example and collect the idea about the division. Divide: 34x2 by 17 Here, 34x2 ÷ 17 17) 34x2 (2x2 – 34x20∴ 34x2 ÷ 17 = 2x2Alternative Method: 34x2 ÷ 17 = 34x2 × 117 = 2x2Remember ! In the above example, 34x2 is called Dividend. 17 is called Divisor. 2x2 is called Quotient. 117 is called the multiplicative inverse of 17.Division is the reverse process of multiplication. ™ Divide the coefficient of the dividend by the coefficient of the divisor™ While dividing a monomial by another monomial, we often write the multiplicative inverse of the divisor and put it in multiplication form with the dividend and simplify.Solved ExamplesExample 1: Divide: 20x2y2 by 5xy Solution: Method I20x2y2 ÷ 5xy5xy) 20x2y2 (4xy – 20x2y2 0∴ Quotient is 4xy. Method II 20x2y2 ÷ 5xy = 20x2y2 × 15xy = 2 × 2 × 5 × x × x × y × y5 × x × y = 4xy∴ Quotient is 4xy

Acme Mathematics 7 189Example 2: Divide: 30x4y4z4 by 6x2y2z2Solution: Method I30x4y4z4 ÷ 6x2y2z2 6x2y2z2) 30x4y4z4 (5x2y2z2 – 30x4y4z4 0∴ Quotient is 5x2y2z2Division of a Binomial or Trinomial by a MonomialThe process of division of a binomial by a monomial is similar to the division of a monomial by another monomial. Similarly, the process is same for the division of a trinomial by a monomial.Solved ExamplesExample 1: Divide (10xy + 15x2y2) by 5xy. Solution: Method I(10xy + 15x2y2) ÷ 5xy 5xy) 10xy + 15x2y2 (2 + 3xy – 10xy + 15x2y2 – 15x2y2 0Hence, the quotient = 2 + 3xy Method II Here, (10xy + 15x2y2 ) ÷ 5xy = (10xy + 15x2y2) × 15xy = 10xy5xy + 15x2y25xy = 2 + 3xyExample 2: The area of rectangle is (2xy + y2) cm2 and its breadth is y cm. Calculate its length.Solution: Here,Area of rectangle (A) = (2xy + y2) cm2Breadth of rectangle (b) = y cmLength of rectangle (l) = ?Method II30x4y4z4 ÷ 6x2y2z2 30x4y4z4 × 16x2y2z2= 2 × 3 × 5 × x2 × x2 × y2 × y2 × z2 × z22 × 3 × x2 × y2 × z2= 5x2y2z2∴ Quotient is 5x2y2z2Area = 10xy + 15x2y2 5xy?Area = (2xy + y)2 cm2 y?

190 Acme Mathematics 7Now, Area of rectangle = l × bor, l = Area of rectanglebreadthor, l = 2xy + y2y = 2xyy + y2y = (2x + y) cm.Hence, length of rectangle is (2x + y) cm.Classwork1. Divide the following expressions. (a) 4a2 ÷ 2a (b) 12y3 ÷ 3y (c) – 6x3 ÷ 2x (d) 18x ÷ (– 6x) (e) 12ab2 ÷ 3ab (f) 7a2b2 ÷ ab2(g) – 4a2b2 ÷ ab2 (h) 4xx (i) 6a22 (j) 10a3a (k) – 3x2x2 (l) – 12x2y3xy(m) – 25xy3– 5xy (n) 12a4b4– 3a2b3 (o) 50x7y8– 10x6y72. Find the quotient of: (a) (4x2 + 2x) ÷ 2x (b) (6y2 – 3xy) ÷ 3y (c) (18x2 + 6x) ÷ 3x (d) (10x2 + 15x ) ÷ 5xExercise 4.81. Find the quotient of: (a) (9a2 –15ab) ÷ 3a (b) (2a2b + 3ab2) ÷ ab(c) (3x2y – 2xy2) ÷ xy (d) (12a2b – 6ab2 ) ÷ 6ab (e) (14a2b + 16ab2) ÷ 2ab (f) (24a3b + 8ab2) ÷ 8ab(g) (12a2 + 14ab) ÷ 2a (h) (8a2b + 4b) ÷ 4b 2. Divide: (a) (4xy2 – 2xy + 6y) by y (b) (6x3 – 3x2 + 3x) by 3x(c) (25x4y4 – 20x3y2 – 5x2y) by 5x2y

Acme Mathematics 7 191(d) (15x5 + 25x4 – 5x3) by 5x3(e) (8x5y5 – 16x4y4 + 40x2y2) by (– 4xy)(f) (–10a3b3 – 5a2b2 + 50ab) by (–5a2b2) (g) (4xyz3 + 36xyz2 – xyz) by 4xyz (h) (14a3b3c3 + 35a2b2c2 – 28abc) by 7abc3. Calculate the unkonwn part of the following rectangle.(a) (b)(c) Area = (x2 + xy) square feet and length = y feet.(d) Area = (21xy + 12y2 + 9y) m2 and breadth = 3y m.Division of a trinomial by a binomialWhile dividing polynomial by another polynomial, at first we should arrange the terms inthe decreasing order of its power. Then we start the division by dividing the first term ofthe dividend by the first term of the divisor. See the solved examples.Solved ExamplesExample 1: Divide (x2 + 7x + 12) by (x + 3). Solution: Hare, x2 + 7x + 12 is a dividend and x + 3 is a divisor Now, x + 3) x2 + 7x + 12 (x + 4 x2 + 3x (–) (–) 4x + 12 4x + 12 (–) (–) 0Hence, (x + 4) is the quotient.Area : (x2 + 4x) m2?xArea : (x2 + 2xy) cm2 x ?Remember: ™ x2 ÷ x = x ™ x (x + 3) = x2 + 3x ™ x2 + 7x – x2 – 3x = 4x ™ 4x + 12 is new dividend ™ Continue the process

192 Acme Mathematics 7Example 2: Divide (a3 – b3) by (a – b). Solution: Hare, (a3 – b3) is a dividend and a – b is a divisor. Now,Hence, a2 + ab + b2 is the quotient.Example 3: The area of a rectangular ground is (x2 + 9x + 20)m2. If its length is (x + 5) m,calculate its length.Solution: Here, Area (A) = (x2 + 9x + 20) m2Length (l) = (x + 5) mBreadth (b) = ?Now, A = l × bSo, b = Alor, b = x2 + 9x + 20x + 5Dividing x2 + 9x + 20 by x + 5or, b = x + 4Hence, breadth of rectangle is (x + 4) ma – b) a3 – b3 (a2 + ab + b2 a3 – a2b (–) (+) a2b – b3 a2b – ab2 (–) (+) ab2 – b3 ab2 – b3 (–) (+) 0x + 5) x2 + 9x + 20 (x + 4 x2 + 5x (–) (–) 4x + 20 4x + 20 (–) (–) 0

Acme Mathematics 7 193Classwork1. Divide the following: (a) (x2 + 4x) by (x + 4) (b) (x2 – 6x) by (x – 6) (c) (10x3 – 5x2) by (2x – 1) (d) (a2b2 + b2) by (a2 + 1) 2. Find the quotient: (a) (x2 + 4x + 3) ÷ (x + 1 ) (b) (x2 + 5x + 6 ) ÷ ( x + 2 ) (c) (y2 + 9y + 20) ÷ ( y + 4) (d) (x2 – 5x +6) ÷ ( x – 3)Exercise 4.91. Find the quotient: (a) (x2 + 3x – 10) ÷ ( x + 5) (b) (x2 – 5x + 6) ÷ ( x – 2) (c) (a2 – 2a – 35) ÷ (a – 7) (d) (2x2 – 5x – 3) ÷ (2x +1) (e) (x2 + 5x + 6) ÷ (x + 3) (f) (x2 + 3x + 2) ÷ (x + 1) (g) (x2 + 10x + 21) ÷ (x + 7) (h) (6x2 – 13x + 6) ÷ (2x – 3) (i) (14x2 + 5x – 1) ÷ (7x – 1) (j) (x2 – y2) ÷ (x – y) (k) (x3 + y3) ÷ (x + y) (l) (4a2 – 1) ÷ (2a – 1) 2. Calculate the unknown sides of a rectangle.(a) Area = (x2 + 7x + 12) m2 Breadth = (x + 3) m(b) Area = (16x2 + 24xy + 9y2) m2 Breadth = (4x + 3y) m(c) Area = (a3 – b3) sq. ft. Breadth = (a – b) ft3. The product of two quantity is (10xy + 15x2y2). If one of them is 5xy, find the other.4. The product of two quantity is (a3 + b3). If one of them is (a + b) find the other.5. The area of my rectangular room is (8a2 – 14ab + 3b2 + 20a – 5b) square meter. If its length is (4a – b) m, (a) Calculate its breadth.(b) If a = 4 and b = 3, calculate its length, breadth and area.6. Divide the followings and verify that : [Dividend = Quotient × divisor + remainder](a) (6m2 + 5mn – 6n2) ÷ (2m + 3n)(b) (15m2 + 4mn – 4n2) ÷ (5m – 2n)(c) (2x2 + 17x – 32) ÷ (x + 10)

194 Acme Mathematics 7Project Work1. Objective : To make a square.2. Materials required :A cardboard paper, colour, glue, scissors and A4 sized paper3. Activities:” Make a 6 cm by 6 cm square piece from cardboard.” Make a 4 cm by 4 cm square piece from cardboard” Make 2 pieces of 6 cm by 4 cm square from cardboard .An Example is given below.'6 × 6 cm2C D6 × 4 cm26 × 4 cm2 4 × 4 cm2A B6 × 6 cm2A B6 × 4 cm2C6 × 4 cm2D4 × 4 cm2From the four pieces of cardboard by arranging them we made a square with dimension 10 cm by 10 cm.6 cm6 cm4 cm4 cm

Acme Mathematics 7 195EvaluationTime: 48 minutes Full Marks: 201. (a) The length and breadth of a rectangular field are (2x + 4y)m and (3x – y)m respectively. Find the area of field in terms of x and y. [2](b) If m + 1m = 5 then find the value of 1m m –2. [2]2. (a) DivideM (4a2 + 14a + 6) ÷ (2a + 6) [2](b) SimplifyM 3x4y5×4x2y72x3y6×3xy2 [2]3. (a) DivideM (x2 + 7x + 12) ÷ (x + 4) [2](b) MultiplyM (x + 2y) (3x – y) [2]4. (a) Find the area of given figure. [2] 3x 2x3(b) If p + 1p = 10, find the value of p2 + 1p2 . [2]5. Arectangular classroom isshown in the figure with itslength is(5a+b)m and breadthis (a+b)m. (a+b) m(5a+b) m(a) What is the area of classroom ? [2](b) If a = 2 and b = 3. Find the real area of classroom. [2]

196 Acme Mathematics 74.3 Equation, Inequality and GraphWarm Up Test1. Write True, False or Open statement in the blanks. (a) The sum of 3 and 4 is 7................. (b) The product of x and 8 is 80.............. (c) 3 is a factor of 20....................... (d) – 4 is a multiple of x........................ 2. Convert to the algebraic form. (a) x less than 20 is 30. (b) 5 times the number x is 10. (c) y is greater than 12. (d) z is not equal to 20. 3. Fill the value of unknown quantity in the blanks. (a) If x + 4 = 13, then x = .....................(b) If x – 8 = 2, then x = ........................ (c) If x4 = 1, then x = ........................ (d) If 2a = 20, then a = ...........................(e) If 3a – 2 = 10, then a = ........................ 4. Choose the unknown quantity as 'x' and make an equation for the following statements. (a) The sum of two numbers is 20. One of them is 12 (b) The difference of two numbers is 30. One of them is 40. (c) The product of two numbers is 12. One of them is 3. (d) The quotient of two numbers is 16. The greater number is 32. 5. Study the number line given below and make the inequality. Choose 'x' as variable.(a)–1 0 1 2 3 4 5 6 7(b)–3 –2 –1 0 1 2 3 4 5(c)–1 0 1 2 3 4 5 6 7(d)–3 –2 –1 0 1 2 3 4 56. Represent these inequalities in the number line. (a) x > 3 (b) x < 3 (c) x > 2 (d) x < 4 (e) 5 < x < 10 (f) 2 > x > – 1

Acme Mathematics 7 197A. Revision1. Equation in one VariableMathematical statementsStudy the given examples. (i) 2x = 20 (ii) y < 4 (iii) 21 > 3 These are called mathematical statements. The statements like x + 4 = 8 or, y > 6 are called Open statements.a. Linear equation in one variableLet us consider the following statements. x + 4 = 12 .......................(i) 2x – 6 = 20 .....................(ii) 4x = 48 ...........................(iii) x7 = 3 ...............................(iv) We observe that the symbol '=' appears in each of the statements. These are equations. A statement which contain variable and equality ('=' ) sign is called an equation. All equations are linear equations. If the exponent of the variable in the equation is 1, the equation is called a linear equation. In the above examples there is only one variable x so these are linear equation in one variable. Equation has two sides: Left hand side (LHS) and Right hand side (RHS) In equation (i), x + 4 is LHS and 12 is RHS.b. Solution of an equationConsider the equation x – 5 = 9. It is true when we replace x by 14 otherwise it is false. Here 14 is called the solution of the equation. The process of finding the solution of an equation is called solving the equation.c. Solving the equationMethods of solving the equations (i) Method of Balancing An equation may be compared with a balance used for weighing. Its sides are two pans and equality (=) tells us that the two pans are in equilibrium.

198 Acme Mathematics 7The following rules help us while solving the equation. Rule 1 We can add the same number to both sides. Rule 2 We can subtract the same number from the both sides. Rule 3 We can multiply both sides of the equation by the same non-zero number. Rule 4 We can divide both sides of the equation by the same non-zero number. Now we can solve the linear equations using the above four rules 1, 2, 3 and 4.Solved ExamplesExample 1: Solve: x – 9 = 1Solution: Here, x – 9 = 1 or, x – 9 + 9 = 1 + 9 → using rule 1or, x = 10 Hence the value of x is 10. Example 2: Solve: x + 15 = 19 Solution: Here, x + 15 or, x + 15 – 15 = 19 – 15 → using rule 2or, x = 4 Hence the value of x is 4. Example 3: Solve: y2 = 2 Solution: Here, y2 = 2 or, y2× 2 = 2 × 2 → using rule 3or, y = 4 Hence the value of y is 4. Example 4: Solve: 4x = 16 Solution: Here, 4x = 16 or,4x4 = 164 → using rule 4or, x = 4 Hence the value of x is 4.

Acme Mathematics 7 199Example 5: Solve 2(x + 1) = x + 14 Solution: Here, 2(x + 1) = x + 14 or, 2x + 2 = x + 14 or, 2x + 2 – 2 = x + 14 – 2 [Subtracting 2 from both sides]or, 2x = x +12 or, 2x – x = x – x + 12 [Subtracting x from both sides]or, x = 12 Hence the value of x is 12. Example 6: Solve : 5x – 76 = 3x – 54Solution: Here, 5x – 76 = 3x – 54or, 4(5x – 7) = 6(3x – 5) [by cross multiplication]or, 20x – 28 = 18x – 30 or, 20x – 28 + 28 = 18x – 30 + 28 [adding 28 on both sides]or, 20x = 18x – 2 or, 20x – 18x = 18x – 18x − 2 [subtracting 18x from both sides]or, 2x = − 2or, 2x2 = – 22 [dividing both sides by 2]or, x = − 1Hence the value of x is − 1.(ii) Method of transposition: It is mostly common method used for solving equation. The word transpose means 'changing the side'. Solved ExampleExample 1: Solve 2(x + 1) = x + 14 Solution: Here, 2(x + 1) = x + 14 or, 2x + 2 = x + 14 or, 2x– x + 2 = 14 [transposing x from RHS to LHS]or, x + 2 = 14 [transposing 2 from LHS to RHS]or, x = 14 – 2 or, x = 12 Hence, the value of x is 12.(Note: We put variable in LHS and constant in RHS while solving)(i) + becomes – on changing the side (ii) – becomes + on changing the side (iii) × becomes ÷ on changing the side (iv) ÷ becomes × on changing the side These are the method of transposition:

200 Acme Mathematics 7Classwork1. Choose the value of x from the set of numbers{1,2, 3,............, 10} and fill in the blanks. (a) In x + 3 = 7, the value of x is …… (b) In x – 4 = 3, the value of x is ……(c) In 5x = 10 , the value of x is ……. (d) In x3 = 2 , the value of x is ………(e) In 4x5 = 4 , the value of x is ……...2. Find value of x for the following statements. (a) x + 9 = 10 (b) x + 7 = 5 (c) x – 4 = 8 (d) x – 10 = 0 (e) 2x = 6 (f) 4x = 12 (g) x3 = 1 (h) x10 = 2 (i) 4x + 1 = 17 (j) 5x – 1 = 19 (k) 6x – 1 = 35 (l) 3x + 2 = 8 (m) 5x + 7 = 17 (n) 2x + 11 = 17 (o) x3 = 25Exercise 4.101. Solve for x. (a) 6(x – 1) = 2x + 2 (b) 5x – 2 = 2(x + 20) (c) 9(x – 1) = 3(x + 3) (d) 20(x – 1) + 5 = 14x +3 (e) 1 – 3(x + 2) = 9x + 4 (f) 8 – 6(x + 3) = 6x + 13 (g) (2 – x) + 12 = 2(x + 7) (h) 3( x + 6) – 8 = 2(x + 2) (i) 2(x + 5) + 8 = 7x + 8 (j) 4( x + 3) – 8 = 2(x +2) (k) 4(x +3) = 52 (x – 2) (l) 4(x + 3)3 – 3 = 1 2. Solve for x.(a) 7x – 14 = 15 (b) x2 + 4x2 = 16(c) x – 22 = x – 13 (d) x + 24 = x + 35(e) x – 32 = 3x – 14 (f) x + 53 = x + 44(g) x + 43 = x – 6 (h) 45 = 2x3 – 15