Acme Mathematics 8 101Evaluation Time: 60 minutes Full Marks: 251. 25 men earns Rs. 5,00,000 in the 30 days . (a) How much does a man earn in a day ? [1] (b) How many men will earn Rs. 5,00,000 in 10 days ? [2] (c) How many days are required for 25 men to earn 1,00,000 ? [1] (d) What is the earning of 5 men in 40 days ? [1] 2. A contractor employed 18 workers, to complete construction work in 30 days working 10 hrs a day.(a) In what proportion do the numbers of workers, working days and working hours lie? [1] (b) In how many days would 20 workers complete the construction working 6 hours a day? [2](c) How many workers are needed to complete the construction in 25 days working 8 hours a day? [2](d) How many hours per day should 10 workers work to complete the construction in 45 days? [1]3. Aman deposited 60,000 at rate 8% Pa in saving account after 5 years he withdraw rupees 40000 and total interest of 5 years. (a) If amount (A), Rate (R) and Time (T) are given write the formula to calculate the principal (P). [1] (b) Find the interest of 5 years. [2] (c) How long should be keep the remanining amount in the bank to get total interest Rs. 28800 from the beginning? [2] 4. Deeyana took a loan of Rs. 20,00,000 from a bank to go for abroad study at the rate of 12% p.a. simple interest. She cleared the loan after 10 years . (a) Define amount. [1] (b) What is the interest paid by her in 10 years ? [2] (c) If the interest rate was 10% p.a., what amount of money had she paid intotal ? [2] 5. Ruhi deposited some amount of money she has in Nepal Bank and Banijya Bank in the ratio 2:1. She deposited Rs. 30,000 in the Nepal Bank. (a) How much money Ruhi deposited in the Banijya Bank ? [1] (b) How much interest with Ruhi obtain in 2 years at the rate of 10% per annum from Banijya Bank ? [1] (c) How long will Ruhi have to go on depositing the amount in the Nepal Bank with the same rate to obtain the same amount of interest from the Banijya Bank ? [2]
AlgebraAcme Mathematics 8 1023UNIT Mensuration3.1 Area of TriangleActivity (Area of right angled triangle):Take rectangular sheet of paper having its length 'l cm' and breadth 'b cm' as shown in the figure.Now, From Figure (i).Area of rectangle = l × b square unit. Draw a diagonal in the rectangle as shown in the figure-(ii) .Fold the paper and cut the folded part. Now, the rectangle is divided into two right angled triangles. Put the cut parts one over the other triangle. It exactly fits to each other.Therefore, Area of rectangle = 2 × Area of right angled triangles . Area of right-angled triangle = 12× Area of rectangle.= 12× l × b square unit.We can find the area of triangles by using different formulae on the basis of the given measurements of various parts of the triangles.A. Area of a triangle in terms of its base and height In the adjoining triangle, AD is the height of ∆ABC, which is drawn from the vertex A on the base BC. Area of ∆ABC= 12 × base × height = 12 × BC × ADB. Area of Right Angled TriangleConsider a rectangle ABCD. AC is a diagonal of rectangle ABCD. Diagonals bisect the rectangle into 2 equal parts. They are ∆ABC and ∆ADC where ∆ABC = ∆ADC. Now, Area of rectangle = Area of ∆ABC + Area of ∆ADCAB D CADBCblFigure-(i)Breadth(b)Length(l)Figure-(ii)Breadth(b)Length(l)Figure-(iii)Breadth(b)Length(l)
Acme Mathematics 8 103or, l × b = Area of ∆ABC + Area of ∆ABCor, l × b = 2 × Area of ∆ABC or, Area of ∆ABC = 12 × l × b Area of triangle = 12 × base × height [For triangle : l = base and b = height]C. Area of Isosceles TriangleIn an isosceles triangle ABC, AB = AC = x and BC = a (say)BD = DC = a2 and AD = h. In right-angled ∆ABD, AD2 = AB2 – BD2 [Pythahoras Theorem]or, h2 = x2 – a22or, h2 = x2 – a24or, h2 = 4x2 – a24or, h = 12 4x2 – a2Now, area of an isosceles triangle ∆ABC = 12 × BC × AD= 12 × a × 12 4x2 – a2∴ Area of an isosceles ∆ABC = a4 4x2 – a2D. Area of Equilateral TriangleIn the equilateral triangle ABC, AB = BC = AC = a (say) BD = DC = a2 and AD = hIn right-angled triangle ABD, AD2 = AB2 – BD2or, h2 = a2 – a22or, h2 = a2 – a24or, h2 = 4a2 – a24or, h2 = 3a24or, h = a2 3B D CAah a aa2a2x xB D CAah a aa2a2
AlgebraAcme Mathematics 8 104Now, area of an equilateral ∆ABC = 12 × BC × AD = 12 × a × a2 3∴ Area of an equilateral ∆ABC = 34 a2Solved ExamplesExample 1 : Calculate the area of the given triangle. Solution : Area = 12 × base × height = 12 × BC × AD = 12 × 10 cm × 8 cm= 40 cm2Hence, area of the triangle is 40 cm2.3.2 Area of QuadrilateralWe know the, area of rectangle = l × b. Where l and b represent the length and breadth. (a) Area of a parallelogramA parallelogram is a quadrilateral in which both pairs of Opposite sides are parallel and equal to each other. Let us understand how to find the area a parallelogram.Draw a Parallelogram ABCD on the paper and cut it out as shown in the figure from D, draw DE ⊥ AB.Cut the parallelogram into two parts through DE. Pick up the triangular piece, ∆ADE and place it on right side of quadrilateral DEBC as shown in the figure.A E ED DBCE A ED DBCE RD CIn the figure parallelogram ABCD is converted to rectangle DERC, where length of rectangle ER = l and breadth of rectangle RC = b. Now, area of rectangle DERC = l × b. B D8 cmAC10 cmlbADBCA EDBC
Acme Mathematics 8 105Hence area of parallelogram ABCD = Area of rectangle DERC= length (DC) × height(RC), where RC = DE.(b) Area of quadrilateral.In the figure ABCD is a quadrilateral. Draw diagonal BD, and draw AF⊥BD, CE⊥BD. Now, Area of quadrilateral ABCD = Area of ∆ABD + Area of ∆BCD = 12 × BD × AF + 12 × BD × CE = 12 × BD (AF + CE) = 12 × diagonal × sum of lengths of perpendiculars Area of quadrilateral = 12 × d × (p1 + p2)∴ Where d = diagonal, p1 and p2 are perpendiculars. (c) Area of Rhombus: In the figure ABCD is a rhombus. The diagonals of rhombus bisect to each other at 90°. Let, diagonal AC = d1 and BD = d2. Now, Area of rhombus ABCD = Area of ∆ACD + Area of ∆ABC. = 12 × AC × OD + 12 × AC × OB = 12 × AC (OD + OB) = 12 × AC × BDArea of rhombus = 12 × d1 × d2) , where d1 and d2 are the diagonals.(d) Area of kite: In the figure ABCD is a kite. The diagonals of a kite intersect each other at 90°. BD is a Long diagonal and AC is a short diagonal. Let, diagonal AC = d1 and BD = d2. Now, Area of kite ABCD = Area of ∆ABD + Area of ∆BDC = 12 × BD × OA + 12 × BD × OC BCADFEP2P1ADOBCBDA CO
AlgebraAcme Mathematics 8 106= 12 × BD (OA + OC) = 12 × BD × AC Area of kites = 12 × d1 × d2) where, d1 and d2 are the diagonals. (e) Area of Trapezium:In the figure ABCD is a trapezium where AB||DC. Draw CE⊥AE and DF perpendicular to BF Let, CE = DF = h, AB = l1 and DC = l2Now, Area of trapezium = Area of ∆ABC + Area of ∆ADC = 12 × AB × CE + 12 × DC × DF = 12× CE (AB + DC) = 12 × h × (AB + DC) = 12 × distance between parallel lines × Sum of parallel sides. Here, Area of trapezium = 12 × h × (l1 + l2)Where, l1 and l2 are the lengths of parallel sides. Solved ExamplesExample 1 : Find the area of the given trapezium Solution : Area of trapezium ABCD = 12 × BC × (AB + DC) = 12 × 4 cm × (8 cm + 12 cm)= 2 cm × 20 cm = 40 cm2 Hence, the area of trapezium is 40 cm2. Example 2 : Calculate the area of the shaded part. AD EBC10 cm14 cm FD 12 cm4 cmCA 8 cm BA BD CF Eh h
Acme Mathematics 8 107Solution : Area of shaded part = Area of parallelogram ABCD – Area of triangle ABE. = AB × CF – 12 × AB × CF = 14 cm × 10 cm – 12 × 14 cm × 10 cm = 140 cm2 – 70 cm2 = 70 cm2This, area of shaded part is 70 cm2Classwork1. Study the given figure, and fill in the blanks.(a) AB is called ..................(b) CD is called ..................(c) AC, CD and DA are ...................... of the triangle ACD.(d) Area of triangle ACD = ...............................(e) Area of triangle ABD = ....................2. Tick () the correct answer.(a) The area of equilateral triangle whose each side 'x' unit is calculated as ........(i) 3x (ii) 3x2(iii) 34 x2 (iv) 4x2(b) The area of isosceles triangle whose equal sides are 'a' cm and third side 'x' cm is ....................(i) x4 4a2 – x2 cm2 (ii) 4xx2 – a2 cm2(iii) 4x a2 – x2 cm2 (iv) x4 (2a2 – x2) cm2(c) The area of given triangle ABC is .....................(i) 100 cm2 (ii) 50 cm2(iii) 75 cm2 (iv) 25 cm23. The area of quadrilateral is calculate as ..............(a) 12 × diagonal × Sum of length of perpendicular.(b) 2 × diagonal × Sum of length of perpendicular.(c) Diagonals × Sum of length of perpendiculars(d) None.AC B DAB D C 10 cm5 cm 5 cm
AlgebraAcme Mathematics 8 1084. The rhombus ABCD is given alongside. It's area is equal to.(a) 2 × AC × BC(b) AC × BC(c) 12 × AC × BD(d) All5. The area of trapezium ABCD is ...................(a) 2h(AD + BC) (b) h2(AD + BC)(c) 1h(AD × BC) (d) h2 × (AD ×BC)6. Perimeter of equilateral triangle is 12 m. Tick its area .............(a) 24 m2 (b) 4 3 m2(c) 8 m2 (d) 20 m2Exercise 3.11. In the trapezium given alongside, find :(a) BD and BC.(b) The area of trapezium ABCD. 2. (a) Find the area of an equilateral triangle each of whose sides measures 20 cm.(b) Find the area of an equilateral triangle whose one side is 8 3 cm.3. (a) If the perimeter of an equilateral triangle is 12 cm, find its area.(b) If the area of an equilateral triangle is 9 3 cm2, find its perimeter.4. (a) The base of an isosceles triangle is 12 cm and its perimeter is 32 cm. Find its area.(b) The base of an isosceles triangle whose area is 60 cm2 and the length of one of its equal sides 13 cm is 10 cm. Find the height of triangle.5. Calculate the area of the following triangles.(a)B 4 cm5 cmA3 cmC(b)Q 40 cm48 cm30 cm30 cmPR(c)D B 26 cm C30 cmA25 cmODACBB CA DhA BD 14 cm C9 cm12 cm
Acme Mathematics 8 1096. Calculate the area of the given squares and rectangles.(a)B 6 cmACD (b) AB 20 cmDC8 cm(c)QP5 cm R10 cmS7. Calculate the area of the following quadrilaterals.(a) 3 cm4 cm4 cm 3 cm(b)14 cm20 cm(c)12 cm16 cm(d)28 cm20 cm (e)5 cm10 cm 6 cm(f)5 cm4 cm(g)2 cm7 cm3 cm6 cm(h)2 cm2 cm4 cm 10 cm(i)7 cm2 cm5 cm8 cm(j)7 cm6 cm(k) 2 cm6 cm2 cm2 cm(l)10 cm 6 cm8. Calculate the area of the following shapes.(a) (b)6 cm8 cm27 cm3 cm3 cm3 cm3 cm4 cm12 cm4 cm7 cm6 cm5cm
AlgebraAcme Mathematics 8 1109. Find the area of the shaded parts in the following figures.(a) 10 cm 8 cm 6 cm2 cm(b)6 cm 2 cm12 cm(c)6 cm15 cm(d)7 cm5cm(e) 4 cm3.5 cm15 cm(f)3cm4cm10 cm2.5 cm(g)7 cm3 cm 3 cm(h)6 cm 4 cm8 cm10 cm(i) 1.5 cm7 cm10 cm10. Find the area of the plot of land with the following measurement.(Note: 16.9 m2 = 1 Dhur, 31.8 m2 = 1 Aana)AEF GD2 m 20 m12 m18 m18 m4m3m12 mB CJHI 9 m10 m8 cm12 cm7 cm 5cm4.8 cm
Acme Mathematics 8 111(a) Calculate the following area of land in 'Dhur'.(i) Area of land ABC.(ii) Area of land BCJI.(iii) Area of land GHJ.(b) Calculate the following area of land in 'Aana'.(i) Area of land AED.(ii) Area of land DEFG.(iii) Area of land FGH.Project WorkMeasure the dimension of bench, table, board and classroom. Calculate the area of all and present it in 'Math Corner'.
AlgebraAcme Mathematics 8 1123.3 Area of CircleDraw a circle of any radius. Divide it into 16 equal parts by drawing 8 diameters as shown in the figure. Arrange these sectors as shown:What will happen if the number of sectors is increased to 32 or 64 or 128 and so on. The figure ABCD becomes a rectangle whose length is equal to half the circumference of given circle and breadth is its radius. So, Area of the circle = Area of rectangle ABCD. = Length × Breadth = πr × r = πr2Hence, Area of the circle is, A = πr2.Solved ExamplesExample 1 : Find the area of a circle whose radius is 42 cm. Solution : Here, radius (r) = 42 cm Now, Area of circle = πr2 = 22 7 × (42)2 cm2= 22 7 × 42 × 42 cm2= 5544 cm2Hence, Area of a circle is 5544 cm2.Example 2 : Calculate the area of a circle whose circumference is 440 cm. Solution : Here, Circumference (C) = 440 cm or, 2πr = 440 cm or,2 × 22 7 × r = 440 cm or, r = 440 × 7 2 × 22 cm 12 4 6 8 10 12 14 163 5 7 8 9 10 11πrr rB πr CA D123 5 67891011 13 12 141516442 cmNote: Use π = 22 7 , in generalC = 440 cm
Acme Mathematics 8 113or, r = 70 cm Now, Area of the circle = πr2 = 22 7 × (70 cm)2 = 15400 cm2Hence, Area of the circle is 15400 cm2. Example 3 : Find the radius of the circular plate whose area is 1386 cm2. Solution : Here, Area of the circle = 1386 cm2or, πr2 = 1386 cm2or,22 7 × r2 = 1386 cm2or, r2 = 1386 × 7 22 cm2or, r2 = 441 cm2or, r = 441 cm2or, r = 21 cm Hence, radius of the circular plate is 21 cm. Example 4 : Find the area of the shaded part of the given figure where AB is its diameter. (π=3.14)Solution : Area of shaded part (A) = Area of circle – Area of ∆ADB.Now, Area of circle, (A1) = πr2= π × AB 22 = 3. 14 × AD 2 + DB24= 3.14 × 9 + 16 4= 3.14 × 6.25= 19.63 cm2Area of ∆ADB, (A2) = 1 2 × AD × DB = 1 2 × 3 cm × 4 cm= 6 cm2Hence, area of Shaded part (A) = A1 –A2= 19.63 cm2 – 6 cm2= 13. 63 cm2ACBD3 cm4 cm
AlgebraAcme Mathematics 8 114Example 5 : Find the area of the given window. Solution : Area of windows = Area of rectangle + Area of semi circle. Now, Area of rectangle = 90 cm × 150 cm = 13500 cm2Area of semi circle = 12 × π × r2= 12 × 22 7 × 902 cm2 [diameter = 90 cm]= 3182.14 cm2 Thus, area of windows = Area of rectangle + Area of semicircle. = 13500 cm2 + 3182.14 cm2= 16682.14 cm2Classwork1. Tick () for the true statement and cross () for the false statement.(a) Circumference of a circle is its perimeter(b) Circumference of a circle with radius 'r' is πr.(c) Area of circle increases as its perimeter increases.(d) Circumference of circle decreases as radius increases.(e) Area of circle with radius 'r' is πr2.(f) Diameter of circle and its radius are equal.2. Choose the correct answer.(a) The diameter of circle is 20 cm. Its radius is ...(i) 40 cm (ii) 2πr (iii) 10 cm (iv) π(b) The relation between diameter 'D' and radius 'R' of a circle is ........(i) 2D = R (ii) R = D (iii) 2R = D (iv) D = R2(c) Area of circle whose radius is 7 cm is ......(i) 49 cm2 (ii) 154 cm2 (iii) 7p (iv) 14 cm2(d) Area of circle when diameter 'd' is given is ........(i) πd2 (ii) π2d (iii) πd24 (iv) 4πd290 cm150 cm
Acme Mathematics 8 115Exercise 3.21. Find the area of the circle whose radii are given below. [Take π = 3.14](a) 20 cm (b) 35 cm (c) 9 cm (d) 8 m2. Find the area of the circles whose diameters are given below. [Take π = 3.14] (a) 28 cm (b) 50 cm (c) 12.5 cm (d) 4 m 3. Complete the table given below. [Take π = 3.14]S.N. Radius (r) Diameter (d) Circumference (C)Area (A)(a) 7 cm ................. ................ ...............(b) ................ 20 cm ................. ......................(c) ................ ...................... 612 cm ......................(d) ................... ................... .................... 154 cm2(e) ............... .................. ................. 2772 cm24. The diameter of a circular garden is 200 meters. On its outside, there is a road 6 meters wide. Find the area of the road.5. A rectangular field is 100 m long and 75 m broad. In the middle of it there is a well of radius 3 m find the area of the field. 6. Calculate the area of the shaded portion. [Take π = 3.14](a) (b) (c)(d) (e) (f) 28 cm28 cm 10 cm18 cm18 cm28 cm28 cm4 cm4 cm2 cm6 cm
AlgebraAcme Mathematics 8 1167. How many circles of radius 7 cm can be made from the rectangular metal sheet of 4.62 m long and 1 m wide ? Take π = 2278. Calculate the area of the given window.(a) (b)9. Find the area of the shaded portion in the following figures. [Take π = 3.14](a) (b) (c)(d) (e) (f)10. Kushal tied his puppy with a 10 m long rope to a pole in a garden. The puppy started to take circular rounds around the pole with straight rope. Calculate the area covered by the puppy in a round. [Take π = 3.14]Project WorkDraw two circles of different sizes and colour them with red and black colour. Which is bigger in area ? By how much ?3.8 cm7 cm6 ft.5 ft.14 cm14 cm14 cm14 cm14 cm14 cm5.1 cm4 cm3.2 cm6 cm 7 cm
Acme Mathematics 8 117Project 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 scissors3. 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.
AlgebraAcme Mathematics 8 1181. ABCD is a rectangle E and F are the mid points of AB and CD where BC = 16 cm and AB = 12 cm.(a) Write the formula to calculate area of rectangle.(b) Find the area of rectangle ABCD. (c) What is the area of quadrilateral EHFG? (d) If you need to paint quadrilateral EHFG at the rate of Rs. 3.25 per square cm. Calculate the cost of painting. 2. From the figure along side find:(a) Radius of outer circle. (b) Find the diameter of inner circle. (c) Write the formula of area of circle. (d) Find the area of shaded part. 3. Inside the park A there is swimming pool and inside the park B there is volleyball court.Park ASwimming pool100 m30 m10 m 50 mPark BVolleyball court70 m18 m9 m 50 m(a) Define perimeter. (b) Find the perimeter of park A. (c) Find the area of park B excluding the volleyball court. (d) How much more is the area of park A than the area of park B. 4 cm 6 cmE FA GB CDHMixed Exercise
Acme Mathematics 8 1194. Observe the following triangular shapes and answer the following questions.(a) Write the formula to find the area of a triangle having base 'b' and height 'h'.(b) Find the height of given triangle. (c) Calculate the area of given shapes. (d) Find the perimeter. 5. A circular plate of the diameter 14 cm rolls in the ground for 15 revolution about the circumference. π = 227(a) Find the radius of the plate. (b) Find the area of the plate. (c) Find the circumference of the plate. (d) Find the distance covered by it. 6. The trapezium shaped garden shown in the figure belongs to Sabina. She makes a square shaped workers' house with the length 2.5 m for the protection of her garden.(a) Write the formula to find out the area of the garden. (b) Find the area of the house. (c) What is the area without the house? 7. Observe the following figure.(a) Find the area of an equilateral triangle. (b) Find the area of circle. (c) Find the ratio of area of equilateral triangle and area of circle.8. A circular metal plate has a diameter of 21 cm.(a) Find the circumference of the plate. (b) Find the distance covered by the plate if it makes 12 revolutions about its circumference. (c) What would be the length of square if the perimeter of square is equal to the circumference of the plate? 9. (a) Write formula to find the area of parallelogram.(b) Draw a circle and show chord AB.AB D C10 cm6 cm12 cmA BD 20 m C2.5 m50 m30 m8 cmPQ RO13.7 cm
AlgebraAcme Mathematics 8 120(c) The area of circle is 50.24 cm2,(i) Find the radius of circle. (ii) Find the circumference of circle. 10. The figure along side is an umbrella made by stitching 8 equal triangular pieces of clothes of two different colours, each piece measuring 5 cm, 5 cm and 8 cm;(a) If length of a side of an equilateral triangle is x cm, then write the formula to calculate the area of the triangle. (b) Find the area of a piece of cloth. (c) What is the total area of the clothes required to make this umbrella.11. The area of circular pond is 616 m2.(a) What is the radius of pond?(b) How long wire is required to around the pond one time?(c) The cost to fencing the pond around is Rs. 250 per meter. What is the total cost of fencing surround the pond?(d) Which of the rectangular garden or swimming pool need more cost to fence it once? Find in percentage.8 cm5 cm
Acme Mathematics 8 121C BA4cm3cmP 24ftQR SM 46ft N30ftTEvaluationTime: 60 minutes Full Marks: 251. Observe the following triangular shapes and answer the following questions. BC = 4cm and AB =3cm (a) Write the formula to find the area of the given triangle. [1](b) What is the lengths of side AC [2](c) Find the perimeter. [1](d) Find the area of given traingle. [1] 2. Observe the adjoining figure in which ABCD is a parallelogram and PQRS is a square. (a) Write the formula to find the area of a rhombus. [1] (b) Find the area of the parallelogram ABCD. [1] (c) What is the area of the shaded portion in the figure ? Find it. [2] (d) Compare the area of the parallelogram ABCD and the square PQRS. [1] 3. The land shown in figure belongs to Shiva. She made the worker dig a well with diameter 3.5ft irrigate her land. (a) What is the formula to find the areas of lids of the well ? [1](b) Calculate the area of the lids of the well. [1](c) What is the area of the land except the well ? [2](d) Shiva wanted to fence the wall with wire at once on that land. If she asks you whether 300 ft wire is enough or not, write your answer with reason. [1]4. The radius of circular pond is 35m. There is a path 3.5m wide running outside the pond. (a) Find the area of the path. [2](b) Find the area of square whose length of each side is equal to the diameter of pond. [1](c) Find the 50% of the sum of area of square and area of path. [1](d) Write the formula to calculate the area of kite having length of diagonals d1cm and d2 cm. [1]5. The length and breadth of a rectangular floor are 11 and 10m respectively. It is covered with the square tile having the length of 20cm. (a) Find the area of the floor. [1](b) How much sq. cm space of floor does a tile cover ? [1](c) How many tiles will be needed to cover the floor of the room ? [2](d) Find the cost of tiling the room if the cost of one tile is Rs. 200. [1]APQ RSB C30m D20m 6m6m
122 Acme Mathematics 84UNIT Algebra4.1 IndicesA. Laws of indices There 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 = x(m + n)Law 2: For any rational number x, y and m, the positive integer, xym = xmymLow 3: For any rational number x and integers m and n, xm ÷ xn = x(m – n)Law 4: For any rational number x and integers m and n, (xm)n = xmnLaw 5: For any rational number x and integers m and n, (xy)n = xn × ynLaw 6: For any rational number x, x° = 1. Law 7: For any rational number x and any integers m, x– m = 1xm or, xm = 1x–m Solved ExamplesExample 1: Simplify: x3 × x4Solution : Here, x3× x4= x3 + 4= x7Example 2: Simplify: 2a3b4Solution : Here, 2a3b4= 2a3b × 2a3b × 2a3b × 2a3b= 16a481b4Example 3: Simplify: a6 ÷ a4Solution : Here, a6 ÷ a4= a6a4= a6 – 4= a2Example 4: Simplify: (4a2)3Solution : Here, (4a2)3= 43 × a2 × 3 = 64a6
Acme Mathematics 8 123Example 5: Simplify: (xa + b)a – b .(xb + c)b – c .(xc + a)c – a Solution : Here, (xa + b)a – b .(xb + c)b – c .(xc + a)c – a= x(a + b)(a – b) × x(b + c)(b – c) × x(c + a)(c – a)= x(a2 – b2) × x(b2 – c2) × x(c2 – a2)= xa2 – b2 + b2 – c2 + c2 – a2= x° = 1Classwork1. Tick () for the correct statement.(a) xm is read as x raised to the power m.(b) x3 is read as x cube.(c) The laws of exponents are ............(i) xa × xb (ii) xa + b (iii) xa ÷ xb = xa – b(iv) (xa)b (v) x0 = 0 (vi) y0 = 1(vii) xa = yb gives 'x' and 'y' are equal(viii) xa = xb gives 'a' and 'b' are not equal(ix) xm + xm = x2m (x) xm ÷ ym = xym(xi) 42 is greater than 24(xii) 163 and 47 are equal.2. Fill in the blanks.(a) 253 ÷ 523 = ................ (b) 435 × 575 = ............(c) 379 ÷ 374 = ............ (d) 792 × 975 = ............(e) (– 1)20 × (– 1)3 = ........... (f) – 342 = ............(g) (– 4)9 × (– 4)6 = ............Exercise 4.11. Simplify by using the law of indices.(a) x3 × x1 × x2 (b) x3 × x4 (c) – x23× – x22× – x24(d) (– 3y)4 × (– 3y)2 × (–3y) (e) x3 ÷ x3 (f) 10p5 ÷ 2p2
124 Acme Mathematics 8(g) (– 4a)3 ÷ (4a)2 (h) 1x4÷ 1x2. Simplify the following. (a) (x4)3 (b) 453(c) (y −3)3 (d) (− a5b2)4 (e) – 2x3z3 4(f) x2y3z52b2 2 (g) x2y2 3(h) {(–1)3}7 (i) – 326 1/23. Simplify: (a) (x2 × x3) + (x5 – 71) – x° (b) (x3)2 × x2 – x3 (c) (x–2)– 1 × x2} ÷ x3 (d) – xy6 ÷xy4 (e) a2 × b3 × c2a4 × b4 × c2 (f) xb – a× x a – b4. (a) If ax = 1, find the value of x ? (b) If 9x = 1, find the value of x ? (c) If a = 2, b = –3 and c = 4, then find the value of the following(i) 3a2b–2 (ii) (a2b)–2 × (ab3c2)25. Solve for p.(a) x3× x4= xp (b) (– x)5 × (– x)4 = (– x)p (c) (xp)4 = x3 (d) (x2)p = xp + 10 (e) xy4 ÷xy2 = xyp(f) 13z 10 ÷13z 6 = 1 3z2p6. Simplify: (a) xa + bxc + b × xc + dxd + a (b) xb x– a a – b . xc x– bb – c. xax – c c – a (c) xa – b xb – a a + b . xb – c xc – bb + c. xc – axa – c c + a(d) yp + q xrp – q. xq +r xp q – r. xr + pxq r – p(e) xa xb a + b × xb xc b + c × xc xa c + a (f) (xa + b)2 . (xb + c)2 . (xc + a)2(xa . xb . xc)4
Acme Mathematics 8 125Project WorkObjective : To verify that xm × xn = xm + n by paper foldingMaterials required :(i) Thin papers (e.g. tracing papers)Activity : Take some thin papers. We will verify that 23 × 24 = 23+4=27 experimentally. Fold the paper so as to obtain a 21 which divides the paper into two equal parts.Now fold the folded paper once again to obtain 22 equal parts on the paper.Fold the folded paper once again to obtain 23 equal parts on it.Next, take another paper and repeat the process of folding four times to obtain 24equal parts.On the first paper, we have 23 = 8 parts and on the second paper, we have 24 = 16 parts.∴ 23 × 24 = 8 × 16 = 128 partsNow, take a third paper and repeat the folding process 7 times to obtain 27 = 128 parts.Hence, it is verified that 23 × 24 = 128 = 27i.e. xm × xn = xm + n.
126 Acme Mathematics 81. Sristi multiplied 2 for 30 times and Barsha multiplied 3 for 20 times.(a) Compare whose number is greater. (b) Prove that : (xa + b ÷ xc + b) × (xc + d ÷ xd + a) = 1. 2. (a) What should be the power of y so that it's value will be 1?(b) Simplify: 113l + m × 113m + n × 113n + l 3. Answer the following questions.(a) Find the value of 7 – x0(b) Simplify : xa + bxc + b × xc + dxa + d4. (a) What is the value of 2x0 + 3?(b) Evaluate: 827 – 23 (c) Simplify: xmxnm + n × xnxpn + p × xpxm p + m 5. (a) Write the value of (x – y)0.(b) Simplify: (– 4ab2)3 (c) Solve it (xa – b)a + b × (xb – c)b + c × (xc – a)c + a 6. (a) Multiply: xa and xb.(b) Find the value of : (ii) (xa)b (ii) x0 (c) Simplify: (xa)b – c .(xb)c – a .(xc)a – b 7. (a) What is the power of x so that it becomes 1x2(b) Find the value of (8)2/3 8. Answer the following question.(a) If 3x = 1, what is the value of x?(b) Simplify: xbxab + a × xcxbc + b × xaxc a + c Mixed Exercise
Acme Mathematics 8 1279. Study the given table and write down the correct answer. Problem (Question) Answer(a) Simplify: xp × xq (a) .......................(b) What is the value of (x0 + 1)1 (b) .......................(c) Simplify: 32 × 34 × 3–7 (c) .......................(d) Evaluate: 34– 3× 34 3(d) .......................10. Simplify:(a) (x– a)b – c (x– b)c – a (x– c)a – b (b) x1xml + m . xmxnm + n . xnxl n + l 11. (a) Simplify (use law of indices) : 8x5 ÷ 2x2 (4x3)(b) Prove that: xaxba2 + ab + b2 × xbxcb2 + bc + c2× xcxa c2 + ca + a2 12. (a) What should be the power of x to make its value 1.(b) Simplify by using the law of indices : xm + n + 2 × xm + n + 2x2m + 2n + 213. (a) What is the value of 32a × 33a (b) Simplify: xaxbc × xbxca × xcxab 14. (a) Multiply: a2 × a3 (b) Simplify: xax– ba2 – ab + b2 × xbx– cb2 – bc + c2× xcx– a c2 – ca + a2 15. Answer the following questions:(a) If 52x = 1, what is the value of x? (b) Simplify: xa + bxa – b × xb + cxb – c × xc + axc – a
128 Acme Mathematics 84.2 Algebraic ExpressionActivityObserve the following algebraic models.The white and black colors of cards indicate positive and negative respectively. x –1–x–1–x–1–x–1x –x 2 Breadth = x unitLength = (x–4) unitFigure (ii)x 1x1x1x1x x 2 Breadth = x unitLength = (x+4) unitFigure (i)Discuss the following questions in your group in order to find the conclusion.(a) What is the total area of cards of figure no.(i)?(b) What is the total area of cards of figure no. (ii)?(c) What are the length and breadth of the rectangle formed by combining all the cards in figure no. (i)?(d) What are the length and breadth of the rectangle formed by combining all the cards in figure no. (ii)? In the figure-(i), Length of rectangular card = (x + 4) units Breadth rectangular of card = x units Area of card = (x + 4)× x square unit. Again, from the figure-(i), Total area of the cards = (x2 + x +x + x +x) square unit= (x2 + 4x) square unit. Hence, (x + 4) × x = (x2 + 4x) or, (x2 + 4x) = x × (x + 4) Here, x and (x + 4) are the factor of x2 + 4x. The process is called factirization. Now discus about the figure-(ii), in the group.
Acme Mathematics 8 129A. FactorizationWe know that x(x + 2) = x2 + 2x. In this multiplication x2 + 2x is called the product of x and (x + 2). x and (x + 2) are called the factors of x2 + 2x. The operation of expressing a polynomial as a product of two or more polynomials is called factorization. We have the following cases for the factorization. (a) When each of the terms contains a common monomial factor.Solved ExamplesExample 1: Factorize : 3x3 + 2x2 + x. Solution : In, 3x3 + 2x2 + x by inspection x is a common to all terms. So, 3x3 + 2x2 + x = x(3x2 + 2x + 1)Example 2: Factorize: 2x2y2 – 4x2y2z + 6x2y2z2 Solution : In 2x2y2– 4x2y2z + 6x2y2z2, 2x2y2 is a common. So, 2x2y2 – 4x2y2z + 6x2y2z2= 2x2y2(1 – 2z + 3z2)(b) When terms can be arranged in suitable group. When an expression is the product of two or more expressions, each of the expression is called a factor of the product. The process of writing a given expression in the product of two or more factors is called factorization.Solved ExamplesExample 1: Factorize x2 + xz + xy + yz. Solution : In this method, terms of the given expression are arranged in suitable groups, such that each group has a common factor.Here, x2 + xz + xy + yz = (x2 + xy) + (xz + yz) → by regrouping = x(x + y) + z(x + y) → x and z are common in each group. = (x + y) (x + z) → [(x + y) is common]Hence, x2 + xz + xy + yz = (x + y) (x + z)
130 Acme Mathematics 8Example 2: Factorize : xy + y + 2x + 2Solution : Here, xy + y + 2x + 2= (xy + 2x) + (y + 2) = x(y + 2) + 1(y + 2) = (y + 2) (x + 1) Hence, xy + y + 2x + 2 = (y + 2) (x + 1)Classwork1. Fill in the blanks.(a) In the expression a(a – b) = a2 – ab, (i) 'a' is called .........(ii) (a – b) is called ................(iii) a2 – ab is called ..............(b) 2 and 3 are factor of 6.(i) factors of 6 are ........... and ..............(ii) 6 is ............ of 2 and 3.(iii) every factors are less than their ..............(iv) product is grater than its all factors ...........2. Tick the correct answer.(a) Factor's form of – 7x2y2 is ........(i) – 7 × x × x × y × y (ii) 7 × (– 1) × xy2(iii) 7x × xy × y2 (iv) x × x × y × y × 7(b) Factors form of xy – x is ......(i) x(y – 1) (ii) xy (iii) – x (iv) – x2y3. Fill the factors in the blanks(a) 8x + 16 = ......................(b) 15 + 25y = ...................(c) 32 – 4a = ...................(d) 16x2 - 4x = ...................(e) 12z + 60z2 = ...................1 is common if no other term is common factor
Acme Mathematics 8 131Exercise 4.21. Find the common factors of the following terms.(a) 2x2 and 8ab (b) a3b2 and a2b3(c) 21ab2 and 42a2b (d) 5x2, 25x3 and –20x3(e) 6a2bc, 3a2b2c2 and 9abc (f) 11xy2, 12xy and 13yz2. Factorize: (a) 3xy (b) – 15xy2 (c) 120x3y3z(d) 10x – 5x2 (e) a2 – 2a (f) – 8a + 12ab(g) 3x – 9 (h) –20x3 + 25x2 (i) 6a3b2 – 9a2b3(j) 2a2bx – 4abcz (k) –x4y – x3y (l) 24x2y3 – 36x3y4 3. Factorize the following: (a) 3(a + 3) + x(a + 3) (b) x(x + 3) – 2(3 + x) (c) a(3 – b) – 4(b – 3) (d) m(a + b) + na + nb 4. Factorize each of the following: (a) 4x2 – 4y2 + 4z2 (b) x3 + x2 + x(c) 3x2 + 6x + 9 (d) 20 – 10x + 5x2(e) 16x4 – 20x3 + 12x2 (f) y3 – 2y2 – y (g) 2y3 – 6y4 – 10y2 (h) 15x2y2 + 12xy2 + 36x2y (i) 12x3y4 – 18x4y3 + 24x2y2 (j) 2x2y – 3xy2 + 5xyz + 4x5. Factorize each of the following by regrouping:(a) 2xy + 10x + y + 5 (b) 3x + y2 + 3y + xy(c) xy + x + y + 1 (d) xy2 + y2z + zr2 + xr2(e) 10xy + 5x + 2y + 1 (f) ax – ay + 5x – 5y(g) 4x3 + 8x2 – 5x – 10 (h) xyz2 + y2z + xy + x2z(i) x2 + yz – xz – xy (j) 3a2 + 2bc + 6ab + ac
132 Acme Mathematics 8(c) When terms can be put in the form a2 – b2. Geometrical Meaning : Take a square having its side a unit then its area = a2 square unit From a2 square unit take b2 square unit from its corner Cut b2 square unit from a2 square unit, the remaining area is a2 – b2 square unit Cut the remaining piece along a dotted line and join the parts and make a rectangleNow, Length of a rectangle = (a + b) unit Breadth of a rectangle = (a – b) unit Area of a rectangle = length × breadth= (a + b) × (a – b) But area of rectangle = a2 – b2∴ a2 – b2 = (a + b) (a – b) Look at the following examples: a2 – b2 = a2 – ab + ab – b2 = a(a – b) + b(a – b) = (a – b) (a +b) Thus, a2 – b2 = (a – b) (a + b) From this we conclude that a2 – b2 has two factors they are (a – b) and (a + b).Solved ExamplesExample 1: Factorize : x2 – 36Solution: Here, x2 – 36 = x2 – 62 = (x + 6) (x – 6) using a2 – b2 = (a + b) (a – b) Hence, x2 – 36 = (x + 6) (x – 6)Example 2: Factorize : 4x2 – 9y2Solution: Here, 4x2 – 9y2 = (2x)2 – (3y)2= (2x + 3y) (2x – 3y) Hence, 4x2 – 9y2 = (2x + 3y) (2x – 3y) b2bba2aa2aa – ba – ba – b1a – bbbaaArea = a2-b2 a – bbbaa1 2
Acme Mathematics 8 133Example 3: Find by factors, the value of (19)2 – (11)2 Solution: Here, (19)2 – (11)2 = (19 + 11) (19 – 11) = 30 × 8 = 240 Hence, (19)2 – (11)2 = 240 (d) When terms are perfect square or can be put in the form (a + b)2 or (a – b)2. Solved ExamplesExample 4: Factorize: x2 + 4x + 4Solution: Here, x2 + 4x + 4 = x2 + 2 × x × 2 + (2)2= (x + 2)2= (x + 2) (x + 2) Hence, x2 + 4x + 4 = (x + 2) (x + 2)Example 5: Factorize: 4x2 – 12xy + 9y2Solution: Here, 4x2 – 12xy + 9y2 = (2x)2 – 2 × (2x)× (3y) + (3y)2 = (2x – 3y)2= (2x – 3y) (2x – 3y) Hence, 4x2 – 12xy + 9y2 = (2x – 3y) (2x – 3y)Classwork1. Tick () the correct answer.(a) The product form of x2 – a2 is ...............(i) (x + a)(x – a) (ii) (x – a)(x – a)(iii) (x + a)(x + a) (iv) x2a2(b) (3x)2 – 16 is written as ...................(i) 48x2 (ii) (3x2 – 16)(iii) (3x – 4)(3x +4) (iv) 9x22. The simple form of 902 – 102 is................(a) 900 (b) 8100(c) 8000 (d) 803. To make x2 – 8x a perfect square ............ should be added to it.(a) 8 (b) 16(c) – 16 (d) 4I know(a + b)2 = a2 + 2ab + b2(a – b)2 = a2 – 2ab + b2
134 Acme Mathematics 84. ............ should be added to 9a2b2 – 12abc to make it a perfect square.(a) 4c2 (b) – 4c2(c) a2b2 (d) 1225. ...........should be added to a2 + ab + b2 to make it a perfect square.(a) 2ab (b) – ab(c) 3ab (d) ab6. ...................... should be added to a2 + 6ab to make it a perfect square.(a) 3ab (b) 9b2(c) 16b2 (d) a27. Fill in the blanks(a) (a – b)(a + b)(a2 + b2) = ..................................(b) (a + b )2 – (a – b)2 = ..............................(c) (80)2 – (20)2 = ....................................Exercise 4.31. Factorize (a) x2 – 25 (b) 4x2 – 1 (c) 49p2 – q2(d) 16x2 – 25y2 (e) a2 – (b – 3)2 (f) x4 – 256(g) 5x3 – 20xy2 (h) x4 – y4 (i) (b – c)2 – a2 2. Fill in the blank to make the given expressions a perfect square. (a) x2 + ...... + z2 (b) a2 – ...... + 16 (c) y2 + ....... + 100 (d) x2 – ...... + 1 (e) 9x2 + ...... + 16y2 (f) 4a2 + ...... + 25b2 (g) x2 + 2x + ...... (h) 4x2 + 4x + ...... (i) 16x2 + 40xy + .....(j) 4x2 – 8x + ...... (k) ...... – 2xy + y2 (l) ...... – 8ab + 163. Factorize the following. (a) x2 + 8x + 16 (b) x2 – 12x + 36 (c) 4x2 + 4x + 1(d) 64a2 – 48ab + 9b2 (e) 16x2 + 40xy + 25y2 (f) a2 – 22ab + 121b2(g) 144 + 24x + x2 (h) 121 – 66x + 9x2 (i) 49a2 + 28ab + 4b2 (j) 25a2 – 80ab + 64b2 (k) 9 + 6y + y2 (l) 16 – 8x + x2(m) 9 – 6a + a2 (n) a2 + 16 + 8a (o) x2 + 9 – 6x
Acme Mathematics 8 1354. Find the value of the following using a2 – b2 = (a + b) (a – b). (a) (49)2 – 12 (b) (121)2 – (21)2 (c) (75)2 – (25)2 (d) (100)2 – (10)2 (e) (99)2 – (1)2 (f) (505)2 – (5)2(g) 103 × 97 (h) 89 × 93 (i) 8.9 × 9.15. Calculate the area of the shaded part.(a) 2 cm2 cm4 cm4 cm(b)x cmx cm2x cm2x cm (c)4y4y2x2x6. Find the sum and difference of the radii, from the given figure. Also find the area of the shaded part. (use �=3.14)(a)18 cm9 cm(b)4 m8 m(c)7 cmProject WorkA square (8 cm × 8 cm) is cut out from the square (20 cm × 20 cm). Convert the remaining part to the rectangular shape and verify that a2 – b2 = (a + b) (a – b).
136 Acme Mathematics 8Project WorkObjective : To verify geometrically the algebraic identity a2 – b2 = (a + b) (a – b)Materials required :(i) Paper (ii) Geometry boxActivity : (a) Draw a rectangle ABCD of length 10 cm (let a = 6 cm and b = 4 cm so that a + b = 10 cm) and breadth 6 cm.(b) Take a point P on AB such that AP = a and PB = b.(c) Take a point Q on BC such that BQ = a – b and QC = b (so that BQ + QC = a – b + b = a).(d) Fold the paper to draw a crease PR || BC and a crease QS || CD. Let the point of intersection of these creases be T.ObservationArea of a rectangle ABQS = AB × BQ= (a + b) (a – b)Also, Area of rectangle ABQS = area of rectangles (ABCD – STRD – TQCR)= AB × BC – DR × SD – QC × CR= (a + b) a – a × b – b × b= a × a + b × a – ab – b2= a2 + ba – ab – b2= a2 – b2 ( ba = ab)Thus, the area of rectangle ABQS = (a + b) (a – b) = a2 – b2Conclusion: (a + b) (a – b) = a2 – b2 or a2 – b2 = (a + b) (a – b)a + baA BD Ca bba – bA P BCQDaab bbba – b a – bA PRBCT QDS
Acme Mathematics 8 137(e) When terms are trinomial x2 ± bx ± c, where b and c the integers.Solved ExamplesExample 1: Factorize: x2 + 6x + 8Solution: In the expression x2 + 6x + 8, integers are 6 and 8. Where, 8 = 1 × 8 and 1 + 8 = 9 → sum is not 6 8 = 2 × 4 and 2 + 4 = 6 → sum is 6.So, we choose, 8 = 2 × 4,Now, x2 + 6x + 8 = x2 + (2 + 4)x + 8= x2 + 2x + 4x + 8 = x(x + 2) + 4(x + 2) = (x + 2) (x + 4) Hence, x2 + 6x + 8 = (x + 2) (x + 4) Example 2: Factorize: x2 – 6x – 40Solution: In the expression x2 – 6x – 40 , integers are – 6 and – 40. Here, Sum = – 6 and Product = – 40 So, numbers are –10 and 4. Now, x2 – 6x – 40 = x2 – (10 – 4)x – 10 × 4 = x2 – 10x + 4x – 10 × 4 = x(x – 10) + 4(x – 10) = (x – 10) (x + 4) Hence, x2 – 6x – 40 = (x – 10) (x + 4)(f) When terms are trinomial ax2 ± bx ± c, a ≠ 1. We can factorize the trinomial ax2 + bx + c by grouping method. To factor ax2 + bx + c, remember the following steps. Multiply the coefficient of x2, + a and the constant term, + c. Try to factor the product ac so that the sum of the factors is b. i.e. find two integers p and q such that p + q = b and pq = c Regroup the middle term, b as the sum of factors Regroup the terms if needed and factor.Remember !The coefficient of x2 is 1It is called ac method
138 Acme Mathematics 8Solved ExamplesExample 1: Factorize: 3x2 + 5x – 2Solution: Here, 3x2 + 5x – 2 is given. The product is 3 × (– 2) = – 6 Possible factors area × c Sum of factors (b) = p + q1, – 6 – 5– 1, 6 + 5 → It is true for given expression2, – 3 – 1– 2, 3 + 1Now, 3x2 + 5x – 2 = 3x2 + (6 – 1)x – 2= 3x2 + 6x – x – 2= (3x2 + 6x) – (x + 2) = 3x(x + 2) – 1(x + 2) = (x + 2) (3x – 1) Hence, 3x2 + 5x – 2 = (x + 2) (3x – 1)Example 2: Factorize: 6x2 + 11x + 3Solution: Here, 6 × 3 = 18 → product of a and c. factors of 18 are 1, 2, 3, 6, 9, 18 by inspection 2 and 9 are possible factors,as 2 × 9 = 18 and 2 + 9 = 11 No, 6x2 + 11x + 3= 6x2 + (9 + 2)x + 3= 6x2 + 9x + 2x + 3 = (6x2 + 2x) + (9x + 3) → regrouping= 2x(3x + 1) + 3(3x + 1) = (3x + 1) (2x + 3)
Acme Mathematics 8 139Example 3: Factorize: 4x2 – 4x – 3Solution: Here, 4x2 – 4x – 3 = 4x2 – (6 – 2)x – 3= 4x2 – 6x + 2x – 3 = 2x(2x – 3) + 1(2x – 3) = (2x – 3) (2x + 1) Hence, 4x2 – 4x – 3 = (2x – 3) (2x + 1)Classwork1. Calculate the values of 'a' and 'b' in the following cases:(a) a + b = 10, ab = 24 (b) a + b = 10, ab = 9 (c) a + b = 7, ab = 12(d) a + b = 7, ab = 10 (e) a + b = –12, ab = 35 (f) a + b = –7, ab = –18(g) a + b = – 5, ab = –36 (h) a + b = 3, ab = – 4 2. Factorize the trinomials:(a) x2 + 7x + 10 (b) x2 + 6x + 8 (c) x2 + 6x + 5 (d) x2 + 7x + 12 (e) x2 + 10x + 24 (f) x2 + 10x + 9(g) x2 + 4x + 3 (h) x2 + 4x + 4 (i) x2 + 10x + 21(j) x2 + 12x + 20 Exercise 4.41. Factorize the given trinomials: (a) a2 – 6a + 9 (b) a2 – 37a + 36 (c) a2 – 16a + 39(d) a2 – 6a + 5 (e) a2 + a – 6 (f) a2 + 2a – 15 (g) a2 + ab – 12b2 (h) a2 + ab – 20b2 (i) a2 – a – 42(j) a2 – a – 56 (k) a2 – 12ab – 64b2 (l) a2 – 2ab – 15b2 2. Factorize: (a) 2y2 + 3y + 1 (b) 2y2 + 13y + 11 (c) 2y2 + 13y + 20(d) 4y2 + 15y + 9 (e) 12y2 + 11y + 2 (f) 5y2 – 18y + 16 (g) 7y2 – 50y + 7 (h) 9y2 – 24y + 16 (i) 3y2 + 17yz – 20z2(j) 10y2 + 59yz – 6z2 (k) 2y2 + 3yz – 2z2 (l) 3y2 – y – 4 (m) 6y2 – 7yz – 10z2 (n) 3y2 – y – 2 (o) 15y2 – 14yz – 8z2 (p) 2y2 – 3y – 54x (–3) = –12±1, ±2, ±3, ±4, ±6, ±12are possible factors
140 Acme Mathematics 8C. Highest Common Factor (HCF) and Lowest Common Multiple (LCM)(a) Highest common factor (HCF)Consider the expressions 3x2 and 6x2y. Here, 3x2 = 3 × x × x6x2y = 3 × 2 × x ×x × yAmong the factors of 3x2 and 6x2y, 3, x and x are the common factors. The product of common factors = 3 × x × x = 3x23x2 is the highest common factor. Hence, HCF = 3x2Solved ExamplesExample 1: Find the HCF of 4x + 2 and 6x2 + 3xSolution : Here, 4x + 2 = 2(2x + 1) 6x2 + 3x = 3x(2x + 1)Common factor = (2x + 1) Hence, HCF = (2x + 1) Note : HCF is a greatest common expression which divide the given expressions exactly.Example 2: Find the HCF of 3x2 + 6xy and x2 + 5xy + 6y2Solution : Here, 3x2 + 6xy = 3x(x + 2y)x2 + 5xy + 6y2 = x2 + (2 + 3)xy + 6y2 = x2 + 2xy + 3xy + 6y2= x(x + 2y) + 3y(x + 2y) = (x + 2y) (x + 3y) Hence, HCF = (x + 2y) Example 3: Find the HCF of x2 + 3x + 2 and x2 – x – 6. Solution : Here, x2 + 3x + 2= x2 + (1 + 2)x + 2 = x2 + x + 2x + 2 = x(x + 1) + 2(x + 1) = (x + 1) (x + 2) x2 – x – 6 = x2 – (3 – 2)x – 6 = x2 – 3x + 2x – 6= x(x – 3) + 2(x – 3) = (x – 3) (x + 2) Hence, HCF = (x + 2)
Acme Mathematics 8 141Classwork1. Fill in the blanks.(a) The factors of 3x2y are ..................(b) The factors of 6ab + 12bc = .......................(c) The common factors of 6x and 18x is ..............................(d) The greatest common factors of 18xy2 and 6xy = ..............................(e) The greatest common factors of 11x2 and 12y2 = ..................................(f) The common factors of 2xy, 4x2y2 and 6x3y3 = ...................................(g) (x – a) is a factor of ..................................(i) x2 – a2 (ii) x2 – ax(iii) 2x2 – 2ax (iv) all of above.2. Tick () the correct answer.(a) The HCF of 18a2 and 18a4 is ...............................(i) 9a (ii) 18a(iii) 18a2 (iv) 18a4(b) .................. is the heighest common factors of 3x2y and 12y2.(i) 3y (ii) 3xy(iii) 3y2 (iv) 12y2(c) ................... a common factor of a + b and a2 – b2.(i) a – b (ii) a + b(iii) a2 – b2 (iv) none(d) The HCF of 8a2, 6a2 and 4a2 is .....................(i) 8a2 (ii) 6a2(iii) 4a2 (iv) 2a2(e) 4xy is HCF of ............................(i) 8x and 16xy (ii) 6x2y2 and 4xy(iii) 4xy and 8xy (iv) 16x2y2 and 4x2y2
142 Acme Mathematics 8Exercise 4.51. Find the HCF of the following expressions: (a) 16x2y2 and 4xy (b) 8a2 and 2ab (c) 15a2b3 and 5a2b2 (d) 20x3, 25x2y and 30x2y2 (e) 5x3, 15y3 and 20z3 (f) 14xyz2, 21x2yz and 28xy2z2. Find the HCF of the given expressions: (a) 6x3 + 2x2 + 2x and 2x3 (b) 9x2 + 12x3 – 6x4 and 6x2 (c) 4x – 8 and x – 2 (d) x + 4 and x – 42(e) a2 – b2 and a + b (f) 2x + 10 and x2 – 25 (g) x3y + xy2 and x4 + x2y (h) x2 + 3x and x2 – 9 3. Find the HCF of the following expressions:(a) x + 2 and x2 + 4x + 4 (b) x + 5 and x2 + 11x + 30 (c) x – 7 and x2 – 10x + 21 (d) 2x – y and 6x2 + xy – 2y2 (e) x2 – 7x + 12 and x2 – 4x + 3 (f) 2x2 – 9x + 4 and 3x2 – 7x – 20 (g) x2 – 16 and x2 + 2x – 8 (h) (x + 1)2 and x2 + x (i) x2 – x – 2 and 3x2 – 13x + 14 (j) 2x2 – 7x + 6 and 3x2 – 7x + 2 (k) 3x2 + 5x – 12 and x2 + 2x – 34. Find the HCF of the following expressions:(a) x3 + 3x2 + 2x and x3 + 7x2 + 10x(b) 6x3 + 5x2 – 6x and 3x3 – 5x2 + 2x(c) x3 – x2 – 42x and x3 – 5x2 – 14x
Acme Mathematics 8 143(b) Lowest Common Multiple (LCM) Consider the expressions 3x2 and 6x2y. Here, 3x2 = 3 × x × x 6x2y = 3 × 2 × x × x × y Among the factors of 3x2 and 6x2y, 3, x and x are the common factors. 2 and y are the remaining factors. The product of common factors and remaining factors = 3 × x2 × 2 × y = 6x2y 6x2y is the lowest common multiple. Hence, LCM = 6x2y6x2y is exactly divisible by the term 3x2. 6x2y is exactly divisible by the term 6x2y. Note : LCM is a Least (smallest) expression which is exactly divisible by the given expressions exactly.Solved ExamplesExample 1: Find the LCM of x2 + 2x and x2− 42 Solution : Here, x2 + 2x = x(x + 2) x2 − 42 = (x + 2) (x – 2) Common factor = (x + 2) Remaining factors = x (x – 2) Hence, LCM = x(x + 2) (x – 2) Example 2: Find the LCM of 3x2 + 6xy and x2 + 5xy + 6y2Solution : Here, 3x2 + 6xy = 3x(x + 2y)x2 + 5xy + 6y2 = x2 + (2 + 3)xy + 6y2= x2 + 2xy + 3xy + 6y2= x(x + 2y) + 3y(x + 2y) = (x + 2y) (x + 3y) Common factor = (x + 2y) Remaining factors = 3x(x + 3y) Hence, LCM = 3x(x + 2y) (x + 3y)
144 Acme Mathematics 8Example 3: Find the LCM of x2 + 3x + 2 and x2 – x – 6. Solution : Here, x2 + 3x + 2 = x2 + (1 + 2)x + 2 = x2 + x + 2x + 2 = x(x + 1) + 2(x + 1) = (x + 1) (x + 2) x2 – x – 6 = x2 – (3 – 2)x – 6 = x2 – 3x + 2x – 6= x(x – 3) + 2(x – 3) = (x – 3) (x + 2) Common factors = (x + 2)Remaining factors = (x + 1) (x – 3)Hence, LCM = (x + 2) (x + 1) (x – 3)Example 4: Find the LCM of (x – 1) (x – 2) and (x + 1) (x – 1) Solution : Here,First expression = (x – 1) (x – 2) Second expression = (x + 1) (x – 1) LCM = (x – 1) (x – 2) (x + 1)Classwork1. Fill in the blanks.(a) The factors of a2b are ..................(b) The factors of 6ab + 12bc = .......................(c) The common factor of 6x and 18x is ................... and remaining factor are ..................(d) The common factor of 18xy2 and 6xy is ...................... and remaining factor are ..................(e) The common factor of 11x2 and 12y2 is .......................... and remaining factor are ..................(f) The common factor of 2xy, 4x2y2 and 6x3y3 is ........................ and remaining factor are ..................We take common if at least two expression has the same factor.
Acme Mathematics 8 145(g) (x – a) is a factor of ..................................(i) x2 – a2 (ii) x2 – ax(iii) 2x2 – 2ax (iv) all of above.2. Tick () the correct answer.(a) The LCM of 18a2 and 18a4 is ...............................(i) 9a (ii) 18a(iii) 18a2 (iv) 18a4(b) .................. is the least common multiple of 3x2y and 12y2.(i) 3y (ii) 3xy(iii) 3y2 (iv) 12x2y2(c) ................... a remaining factor of a + b and a2 – b2.(i) a – b (ii) a + b(iii) a2 – b2 (iv) none(d) The LCM of 8a2, 6a2 and 4a2 is .....................(i) 8a2 (ii) 6a2(iii) 24a2 (iv) 2a2(e) 16x2y2 is LCM of ............................(i) 8x and 16xy (ii) 6x2y2 and 4xy(iii) 4xy and 8xy (iv) 16x2y and 4x2y2Exercise 4.61. Calculate the LCM of the following expressions: (a) 2x2y3 and 4xy2 (b) 21x and 28x2(c) x2y2 and x3y (d) x4 and (x2y2 – x3y) (e) 14(x – y)2 and 7(x – y) (f) 6x2 and 3x2 + 6x(g) (x + 2) (x + 3) and (x + 3) (x – 1)(h) (x – 2) (x – 6) and (x – 2) (x + 1)(i) (x + 1) (x + 2) and (x + 1) (x – 2)
146 Acme Mathematics 82. Find the factors of the following expressions.(a) x2 – 2x + 1 and x – 1(b) x2 – 4x + 4 and x – 2(c) y2 + 6xy + 9x2 and (y + 3x)(d) x2 + 8x + 16 and x + 4(e) 25x2 + 10xy + y2 and 5x + y(f) 49x2 – 56xy + 16y2 and 7x – 4yNow, answer the following questionsi. List the common factors.ii. List the remaining factors.iii. Calculate the LCM.3. Find the LCM of: (a) x2 – x and 3x2 – 9x + 6 (b) 6x2 + 18x + 12 and 6(x2 – 1) (c) x – 2 and x2 – 4x + 4 (d) 2x – 3 and 6x2 – 17x + 12 (e) x2 – 4 and x2 + 3x + 2 (f) x2 – 25 and x2 – 9x + 20(g) x2 + 5x + 6 and x2 + x – 6 (h) x2 – 9x – 22 and x2 – 8x – 33 4. Find the HFC and LCM of the following expressons:(a) x2 – 1 and (x – 1)2 (b) x2 – y2 and 2(x + y) (c) x2 – 25 and x2 – 8x + 15 (d) 2x2 – 8 and x2 + x – 2(e) 2x2 + 3x – 2 and x2 – 4 (f) x2 + 2x – 8 and x2 + 3x – 4(g) x2 + 3x + 2 and x2 + 4x + 3(h) x2 + 6x + 8 and x2 + 7x + 105. Find LCM of:(a) x3 + 3x2 + 2x and x3 – x2 – 6x(b) 6x3 + 5x2 – 6x and 3x3 – 5x2 – 2x(c) x3 – x2 – 42x and x3 – 5x2 –14x
Acme Mathematics 8 147D. Like and Unlike FractionsThe algebraic fractions having the same denominator are called like fractions. For examples:2x3y, 3x + y3y , 5x – 23y and x2 – 2x + 13y are like fractions. The algebraic fractions having the different denominator are called unlike fractions. For examples.2x3y, 7xz , 9x + 10y– x and 10xy4p are unlike like fractions. E. Rational Expressions The number 23 is a rational number in the arithmetic. Similarly ab is a rational number (expression) in algebra, where b ≠ 0.x3, 3x + 1, x – 2x + 5 etc are rational expressions.In 3x + 1 if, x = – 1, then 3x + 1 is not rational algebraic expression with zero (0) as denominator is not defined. Rational algebraic expression can add, subtract, multiply, divide or simplify(a) Rational expression in the lowest term Look at the given solved examples carefully.Solved ExamplesExample 1: Convert to the lowest term: 12x2y224x3y2Solution : Here, 12x2y224x3y2= 12 × x × x × y ×y12 × 2 × x × x × x × y × y × y= 12xy Hence, 12x2y224x3y3 = 12xyExample 2: Simplify: x2 + 2x x2 + 5x + 6 Solution: Here, x2 + 2x x2 + 5x + 6 Find the factors and cancel the same factors from the Numerator and Denominator.
148 Acme Mathematics 8= x(x + 2)x2 + (2 + 3)x + 6= x(x + 2)x2 + 2x + 3x + 6= x(x + 2)x(x + 2) (x + 3)= xx + 3Classwork1. Circle the rational expressions.(i) ax (ii) x4 (iii) 1x + 1(iv) 5xx + 3 (iv) x – 3x + 2 (vi) 4aba – 32. The expression 7xyy + 4 is given. 7xy is its ....................... and (y + 4) is its .....................3. Tick the correct answer.(a) 2x is simplest form of .....................(i) 44x (ii) 36x(iii) 42x (iv) 2x4(b) 4x2 is the lowest term of .....................(i) 8x32x (ii) 10x26(iii) 14x2 (iv) 4x2Exercise 4.71. Separate the like and unlike fractions from each of the following.(a) xy, 4xz , 6xy , 10xyz , 4yz(b) xyz3a , pq10a, 2xyzb , xyb(c) xx + y, yy – x, 2xx + y, xyy – x
Acme Mathematics 8 1492. Which of the following fraction is equivalent to ab?(a) axbx, 3a3b, 5ax6b , 4a4b(b) a(a + b)b(a + b), 6a(x – y)6b (x – y), (3x + 2y)a(2y + 3x)b3. Reduce the following rational expressions to their lowest term: (a) 2x3y24x4y (b) 40x3y380xy (c) 2x3y24x4y(d) 19xy457x4y (e) a2 – b2a + b (f) 5x + 310x + 6(g) x2 – 4x – 2 (h) x2 + 2xx + 2 (i) x3 – 4xx3 + 2x24. Simplify the following expressions: (a) x2 + 3x + 2x(x + 1) (b) x2 + 10x + 214(x + 3) (c) 4x2 – 1(2x – 1) (x – 3)(d) 2x2 + x – 1 (x + 1)y (e) 8x2 + 6x4x2 + 7x + 3 (f)x2 – y2x2 – 3xy + 2y2(g) 12 + a – a224 – 10a + a2 (h) x2 + 6x + 8x2 – 2x – 8 (i)x2 – 12x + 32x2 – 7x + 12(j) x4 – x2y2y4 – x2 y25. For what value of x, the following rational expressions are not defined ? (a) 10x – 1 (b) yz x + 5 (c) 1x + 7 (d) x – y 3 – y (e) x2 – y2– 4 + x2 (f) 100x2 – 36. Prove the following: (a) 40x – y = – 40y –x (b) 7 – xx – 4 = x – 74 – x (c) –(x –1)x +y = 1 – xx + y (d) 2x + 3y3y – 4x = – 2x + 3y4x – 3y
150 Acme Mathematics 8(b) Addition of rational expressions Rational expressions can be added in the same way as rational numbers.Solved ExamplesExample 1: Add: xx + 5 and yx + 5 Solution : Here, the denominator (x + 5) is sameSo, xx + 5 + yx + 5= x + yx + 5Example 2: Simplify 4x + 1 + xx + 2Solution : Here, 4x + 1 + xx + 2 = 4 (x + 2) + x (x +1)(x + 1) (x + 2)= 4x + 8 +x2 +x(x+ 1) (x + 2)= x2 + 5x + 8 (x + 1) (x + 2)(c) Subtraction of rational expressions Rational expressions can be subtracted in the same way as rational numbers.Solved ExamplesExample 1: Subtract 32x from 62xSolution: Here, 62x – 32x= 6 – 32x = 32xExample 2: Simplify: 2 x – 2 – 3x + 2Solution : Here, 2 x – 2 – 3x + 2= 2(x + 2) – 3(x – 2) (x – 2) (x + 2) = 2x + 4 – 3x + 6 (x – 2) (x + 2) = – 3x + 2x + 4 + 6(x – 2) (x + 2) = – x +10 (x – 2) (x + 2) It is common denominatorIt is LCM of denominatorsIt is common denominator