Mathematics Standard Eight PDF

STANDARD EIGHT 5) ) (x - x-7 10 x2- ( x2 - 3x - 9x +1 (x +...

STANDARD EIGHT 5) ) (x - x-7 10 x2- ( x2 - 3x - 9x +1 (x + 4 ) 2) -2 (x 20 % OFF NO PLASTIC PLEASE < 48.00  The Constitution of India Chapter IV A Fundamental Duties ARTICLE 51A Fundamental Duties- It shall be the duty of every citizen of India- (a) to abide by the Constitution and respect its ideals and institutions, the National Flag and the National Anthem; (b) to cherish and follow the noble ideals which inspired our national struggle for freedom; (c) to uphold and protect the sovereignty, unity and integrity of India; (d) to defend the country and render national service when called upon to do so; (e) to promote harmony and the spirit of common brotherhood amongst all the people of India transcending religious, linguistic and regional or sectional diversities, to renounce practices derogatory to the dignity of women; (f) to value and preserve the rich heritage of our composite culture; (g) to protect and improve the natural environment including forests, lakes, rivers and wild life and to have compassion for living creatures; (h) to develop the scientific temper, humanism and the spirit of inquiry and reform; (i) to safeguard public property and to abjure violence; (j) to strive towards excellence in all spheres of individual and collective activity so that the nation constantly rises to higher levels of endeavour and achievement; (k) who is a parent or guardian to provide opportunities for education to his child or, as the case may be, ward between the age of six and fourteen years. The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4 Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 29.12.2017 and it has been decided to implement it from the educational year 2018-19. Mathematics Standard Eight Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune - 411 004 Sachin Mehta Preface  English Mathematics - Standard VIII Suggested Pedagogical Processes Learning Outcomes The learner may be provided opportunities The learner — in pairs/groups/ individually and encouraged to — 08.71.01 generalises properties of addition, subtraction, multiplication and explore examples of rational numbers with division of rational numbers through all the operations and explore patterns in patterns. these operations. explore patterns in square numbers, square 08.71.02 finds out as many rational numbers roots, cubes and cube roots of numbers and as possible between two given form rules for exponents as integer. rational numbers. provide situations that lead to simple equations and encourage them to solve using 08.71.03 finds squares, cubes and square roots and cube roots of numbers suitable processes. using different methods. multiply/experience two algebraic expressions and different polynomials may be 08.71.04 solves problems with integral provided based on their previous knowledge exponents. of distributive property of numbers and generalise various algebraic identities using 08.71.05 solves puzzles and daily life problems concrete examples. using variables. factorise algebraic expressions using relevant 08.71.06 multiplies algebraic expressions. activities based on previous knowledge of factorising two numbers. For example, expands (2x-5)(3x2+7). situation may be provided that involve the use of 08.71.07 uses various algebraic identities in percentages in contexts like discount, profit and solving problems of daily life. loss, simple and compound interest, etc. provide various situations to generalise the 08.71.08 applies the concept of percent in formula of compound interests through profit and loss situation in finding repeated use of simple interest. discount and compound interest. a number of situations may be given where one quantity depends on the other, the 08.71.09 calculates discount percent when marked price and actual discount are quantities increase together, or in which given or finds profit per cent when while one increases the other decreases. For cost price and profit in a transaction example, as the speed of a vehicle increases are given. the time taken by it to cover the distance decreases. 08.71.10 solves problems based on direct and measure the angles and sides of different inverse proportions. quadrilaterals and let them identify patterns in the relationship among them, let them make 08.71.11 solves problems related to angles hypothesis on the basis of generalisation of a quadrilateral using angle sum of the patterns and later on verify through property. examples. 08.71.12 verifies properties of parallelograms verify the properties of parallelograms and and establishes the relationship apply reasoning by doing activities such as between them through reasoning. constructing parallelograms, drawing their diagonals and measuring their sides and 08.71.13 constructs different quadrilaterals angles. using compasses and straight edge. verifies Euler’s relation through pattern.  demonstrate the construction of various 08.71.15 estimates the area of shapes like quadrilaterals using geometric kit. trapezium and other polygons by sketch the figure of trapezium and other using square grid/graph sheet and polygons in the given graph paper and verifies using formulas. ask students to estimate their areas using counting of unit square. 08.71.16 finds the area of a polygon. derive the formula for calculating area of 08.71.17 finds surface area and volume of trapezium using the areas of triangle and cuboidal and cylindrical object. rectangle. (square) 08.71.18 draws and interprets bar charts. identify that surfaces of various 3-D objects like cubes, cuboids and cylinder. 08.71.19 verifies the properties of angles formed derive formulae for surface area of cubes by the transversal of two parallel lines. and cuboids using the formulae for areas of 08.71.20 Uses, SSS, SAS, ASA, Hypo-side tests rectangles, squares and circles. of congruence of triangles. demonstrate to find volume of a given cube and cubiod using unit cubes. 08.71.21 estimates the area of closed figures using graph paper or grid paper. collect data, organise it into groups and represent it into bar graphs/ pie chart. 08.71.22 computes mean of the data used in establish congruence criterion and later on day-to-day life. verify the property by superimposing one above the other. 08.71.23 constructs parallel line to the given line. find a representative value of data, for example, mean, mode or median of ungrouped data. Encourage them to arrange it in a tabular form and represent it by bar graphs. Guidlines for Teachers It is expected that the text book of standard VIII should be used to establish dialogue with students. Tools such as question-answers, discussions, activities, etc. should be used to serve the purpose. This will be possible by reading the book throughly. While reading, underline the important sentences. Read the books of previous and next standards and other books also for reference. The matter on Q. R. code will also be useful. In the book attempt is made to correlate mathematics with other subjects such as Environment, Geography, Science, Economics etc. Bring the fact to the notice of students. Encourage students to work out projects, activities and practicals. This will help students to understand the use of mathematics in practical life. In the text book, mathematical concepts are explained in simple language. It is expected that teachers should construct examples similar to those in practice sets and ask the students to solve them. Encourage the students also to construct and solve examples of their own. The star-marked questions are a little challenging. The matter given under the head ‘For more information’ will definitely be useful to students for further studies. We hope, you will definitely appreciate the book.  Index Part 1 1. Rational and Irrational numbers........................ 01 to 06 2. Parallel lines and transversals........................... 07 to 13 3. Indices and Cube root...................................... 14 to 18 4. Altitudes and Medians of a triangle.................. 19 to 22 5. Expansion formulae.......................................... 23 to 28 6. Factorisation of Algebraic expressions............. 29 to 34 7. Variation............................................................ 35 to 40 8. Quadrilateral : Constructions and Types........... 41 to 50 9. Discount and Commission................................ 51 to 58 Miscellaneous Exercise 1.................................. 59 to 60 Part 2 10. Division of Polynomials.................................... 61 to 66 11. Statistics............................................................ 67 to 74 12. Equations in one variable.................................. 75 to 80 13. Congruence of triangles.................................... 81 to 87 14. Compound interest............................................ 88 to 93 15. Area................................................................... 94 to 105 16. Surface area and Volume................................... 106 to 113 17. Circle : Chord and Arc...................................... 114 to 118 Miscellaneous Exercise 2.................................. 119 to 120  Part 1 1 Rational and Irrational numbers Let’s recall. We are familiar with Natural numbers, Whole numbers, Integers and Rational numbers. Natural numbers Whole numbers Integers 1,2,3,4,... 0,1,2,3,4,......,-4,-3,-2,-1, 0, 1, 2, 3,... Rational numbers −25 , 10 , -4, 0, 3, 8, 32 , 67 , etc. 3 −7 3 5 m Rational numbers : The numbers of the form are called rational numbers. n Here, m and n are integers but n is not zero. We have also seen that there are infinite rational numbers between any two rational numbers. Let’s learn. To show rational numbers on a number line 7 −2 Let us see how to show , 2, on a number line. 3 3 Let us draw a number line. -3 -2 -1 0 1 2 3 4 −2 2 7 3 3 3 · We can show the number 2 on a number line. 7 1 · = 7 ´ , therefore each unit on the right side of zero is to be divided in three 3 3 7 7 1 equal parts. The seventh point from zero shows ; or =2+ , hence 3 3 3 1 7 the point at rd distance of unit after 2 shows. 3 3 1  −2 2 · To show on the number line, first we show on it. The number to the left of 3 3 −2 0 at the same distance will show the number. 3 Practice set 1.1 1. Show the following numbers on a number line. Draw a separate number line for each example. 3 5 3 7 −2 −4 −5 11 13 −17 (1) , , − (2) , , (3) , (4) , 2 2 2 5 5 5 8 8 10 10 2. Observe the number line and answer the questions. A B E O F C D -3 -2 -1 0 1 2 3 (1) Which number is indicated by point B? 3 (2) Which point indicates the number 1 ? 4 5 (3) State whether the statement, ‘the point D denotes the number ’ is true or false. 2 Let’s learn. Comparison of rational numbers We know that, for any pair of numbers on a number line the number to the left is smaller than the other. Also, if the numerator and the denominator of a rational number is multiplied by any non zero number then the value of rational number a ka does not change. It remains the same. That is, = , (k ¹ 0). b kb 5 2 Ex. (1) Compare the numbers and. Write using the proper symbol of. 4 3 5 5´ 3 15 2 2´ 4 8 Solution : 4 = 4´3 = 12 3 = 3´ 4 = 12 15 8 5 2 12 > 12 \ > 4 3 2 Ex. (2) Compare the rational numbers −7 and 4. 9 5 Solution : A negative number is always less than a positive number. 7 4 Therefore, - <. 9 5 To compare two negative numbers, let us verify that if a and b are positive numbers such that a < b, then -a >-b. 2 < 3 but -2 > -3 5 < but 7 4 −5 4 > 4 −7 } Verify the comparisons using a number line. 4 −7 −5 Ex. (3) Compare the numbers and. 3 2 7 5 Solution : Let us first compare and. 3 2 7 7´ 2 14 5 5´ 3 15 14 15 = 3 ´ 2 = 6 , = 2´ 3 = and 6 < 6 3 2 6 7 5 −7 −5 \ < \ > 3 2 3 2 3 6 Ex. (4) and are rational numbers. Compare them. 5 10 3 3´ 2 6 3 6 Solution : = = \ = 5 5´ 2 10 5 10 The following rules are useful to compare two rational numbers. a c If and are rational numbers such that b and d are positive, and b d a c (1) if a × d < b × c then < b d a c (2) if a × d = b × c then = b d a c (3) if a × d > b × c then > b d Practice Set 1.2 1. Compare the following numbers. −9 −5 (1) -7, -2 (2) 0, (3) 8 , 0 (4) , 1 (5) 40 , 141 5 7 4 4 29 29 17 −13 15 7 −25 −9 12 3 −7 −3 (6) - 20 , 20 (7) , (8) , (9) , 5 (10) , 12 16 8 4 15 11 4 3  Let’s learn. Decimal representation of rational numbers If we use decimal fractions while dividing the numerator of a rational number by its denominator, we get the decimal representation of a rational number. For example, 7 = 1.75. In this case, after dividing 7 by 4, the remainder is zero. Hence the process 4 of division ends. Such a decimal form of a rational number is called a terminating decimal form. We know that every rational number can be written in a non-terminating recurring decimal form. 7 · 5 · For example, (1) = 1.1666... = 1.16 (2) = 0.8333... = 0.8 3 6 6 −5  (3) 3 = -1.666... = 1. 6 22 23 (4) 7 = 3.142857142857... = 3.142857 (5) = 0.2323... = 0.23 99 Similarly, a terminating decimal form can be written as a non-terminating 7 · recurring decimal form. For example, = 1.75 = 1.75000... = 1.75 0. 4 Practice Set 1.3 1. Write the following rational numbers in decimal form. 9 18 9 −103 11 (1) (2) (3) (4) (5) − 37 42 14 5 13 Let’s learn. Irrational numbers In addition to rational numbers, there are many more numbers on a number line. They are not rational numbers, that is, they are irrational numbers. 2 is such an irrational number. We learn how to show the number 2 on a number line. · On a number line, the point A shows the number 1. Draw line l perpendicular to the number line through point A. Take point P on line l such that OA = AP = 1 unit. · Draw seg OP. The D OAP formed is a right angled triangle. 4  By Pythagoras theorem, l OP = OA + AP 2 2 2 = 12 + 12 = 1 + 1 = 2 OP2 = 2 P \ OP = 2...(taking square roots on both sides) R O A Q · Now, draw an arc with centre O and - 2 0 1 2 radius OP. Name the point as Q where the arc intersects the number line. Obviously distance OQ is 2. That is, the number shown by the point Q is 2. If we mark point R on the number line to the left of O, at the same distance as OQ, then it will indicate the number - 2. We will prove that 2 is an irrational number in the next standard. We will also see that the decimal form of an irrational number is non-terminating and non-recurring. Note that - In the previous standard we have learnt that p is not a rational number. It means 22 it is irrational. For calculation purpose we take its value as or 3.14 which are 22 7 very close to p ; but and 3.14 are rational numbers. 7 The numbers which can be shown by points of a number line are called real numbers. We have seen that all rational numbers can be shown by points of a number line. Therefore, all rational numbers are real numbrs. There are infinitely many irrational numbers on the number line. 2 is an irrational number. Note that the numbers like 3 2, 7+ 2, 3- 2 etc. are also irrational numbers; because if 3 2 is rational then 3 2 should also 3 be a rational number, which is not true. We learnt to show rational numbers on a number line. We have shown the irrational number 2 on a number line. Similarly we can show irrational numbers like 3 , 5... on a number line. Practice Set 1.4 1. The number 2 is shown on a number line. Steps are given to show 3 on the number line using 2. Fill in the boxes properly and complete the activity. 5 Activity : · The point Q on the number line shows the number....... R · A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on O Q C -1 0 1 2 3 the line. · Right angled D ORQ is obtained by drawing seg OR. · l (OQ) = 2 , l(QR) = 1 \ by Pythagoras theorem, [l(OR)]2 = [l(OQ)] 2 + [l(QR)] 2 2 2 = + = + =   \ l(OR) = Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number 3. 2. Show the number 5 on the number line. 3«. Show the number 7 on the number line. ÑÑÑ Answers Practice Set 1.1 −10 2. (1) (2) C (3) True 4 Practice Set 1.2 −9 8 −5 1 40 141 1. (1) -7 < -2 (2) 0 > (3) >0 (4) < (5) < 5 7 4 4 29 29 −17 −13 15 7 −25 −9 12 3 (6) < (7) > (8) < (9) > 20 20 12 16 8 4 15 5 −7 −3 (10) > 11 4 Practice Set 1.3 (1) 0.243 (2) 0.428571 (3) 0.6428571 (4) -20.6 (5) -0.846153 6  2 Parallel lines and transversal Let’s recall. The lines in the same plane which do not intersect each other are called parallel lines. l ‘ Line l and line m are parallel lines,’ is written as m ‘ line l || line m’. Let’s learn. l Transversal A In the adjoining figure, line l intersects line m m and line n in two distinct points. line l is a B n transversal of line m and line n. If a line intersects given two lines in two distinct points then that line is called a transversal of those two lines. Angles made by a transversal In the adjoining figure, due to the transversal, there are two distinct points of intersection A namely M and N. At each of these points, four Q M angles are formed. Hence there are 8 angles in P all. Each of these angles has one arm on the R N S transversal and the other is on one of the given T lines. These angles are grouped in different pairs of angles. Let’s study the pairs. · Corresponding angles · Interior angles If the arms on the transversal of a A pair of angles which are on the same pair of angles are in the same direction side of the transversal and inside the and the other arms are on the same side given lines is called a pair of interior of the transversal, then it is called a pair angles. of corresponding angles. 7  pairs of corresponding angles in the pairs of interior angles in the given given figure - figure - (i) ÐAMP and ÐMNR (i) ÐPMN and ÐMNR (ii) ÐPMN and ÐRNT (ii) ÐQMN and ÐMNS (iii) ÐAMQ and ÐMNS (iv) ÐQMN and ÐSNT · Alternate angles Pairs of angles which are on the opposite sides of transversal and their arms on the transversal show opposite directions is called a pair of alternate angles. In the figure, there are two pairs of interior alternate angles and two pairs of exterior alternate angles. Interior alternate angles Exterior alternate angles (Angles at the inner side of lines) (Angles at the outer side of lines) (i) ÐPMN and ÐMNS (i) ÐAMP and ÐTNS (ii) ÐQMN and ÐRNM (ii) ÐAMQ and ÐRNT Practice Set 2.1 1. In the adjoining figure, each angle is shown by a letter. Fill in the boxes with the help of the figure. Corresponding angles. (1) Ðp and (2) Ðq and p q s r (3) Ðr and (4) Ðs and w x Interior alternate angles. z y (5) Ðs and (6) Ðw and 2. Observe the angles shown in the figure and write the following pair of angles. a b e f d c h g (1) Interior alternate angles (2) Corresponding angles (3) Interior angles 8  Let’s learn. Properties of angles formed by two parallel lines and a transversal Activity (I) : As shown in the figure (A), draw two parallel lines and their transversal on a paper. Draw a copy of the figure on another blank sheet using a trace paper, as shown in the figure (B). Colour part I and part II with different colours. Cut out the two parts with a pair of scissors. II I a b d c e f h g (A) (B) Note that the angles shown by part I and part II form a linear pair. Place, part I and part II on each angle in the figure A. Which angles coincide with part I ? Which angles coincide with part II ? We see that, Ðb @ Ðd @ Ð f @ Ðh, because these angles coincide with part I. Ða @ Ðc @ Ðe @ Ðg, because these angles coincide with part II. (1) Ða @ Ðe, Ðb @ Ð f, Ðc @ Ðg, Ðd @ Ðh (These are pairs of corresponding angles.) (2) Ðd @ Ð f and Ðe @ Ðc (These are pairs of interior alternate angles.) (3) Ða @ Ðg and Ðb @ Ðh (These are pairs of exterior alternate angles.) (4) m Ðd + mÐe = 180° and mÐc + mÐ f = 180° (These are interior angles.) Let’s discuss. When two parallel lines are intersected by a transversal eight angles are formed. If the measure of one of these eight angles is given, can we find measures of remaining seven angles ? 9  Let’s learn. (1) Property of corresponding angles Each pair of corresponding angles A formed by two parallel lines and their transversal is of congruent angles. P M Q In the adjoining figure line PQ || line RS. Line AB is a transversal. R N S Corresponding angles ÐAMP @ ÐMNR ÐPMN @ ÐRNB B ÐAMQ @ ÐMNS ÐQMN @ ÐSNB (2) Property of alternate angles (3) Property of interior angles Each pair of alternate angles Each pair of interior angles formed formed by two parallel lines and their by two parallel lines and their transversal transversal is of congruent angles. is of supplementary angles. Interior alternate Exterior alternate Interior angles angles angles mÐPMN + mÐMNR = 180° ÐPMN @ ÐMNS ÐAMP @ ÐSNB mÐQMN + mÐMNS = 180° ÐQMN @ ÐMNR ÐAMQ @ ÐRNB Solved Examples L Ex. (1) In the adjoining figure line AB || line PQ. A O B Line LM is a transversal. P N 70° Q mÐMNQ = 70°, then find mÐAON. M Solution : Method I Method II mÐMNQ = mÐONP = 70°...(Opposite mÐMNQ = 70° angles) \mÐNOB = 70°...(Corresponding mÐAON + mÐONP = 180°...(Interior angles) angles) mÐAON + mÐNOB = 180° \ mÐAON = 180° - mÐONP \ mÐAON + 70° = 180° = 180° - 70° \ mÐAON = 110° = 110° (The above example can be solved by another method also.) 10 Ex. (2) In the adjoining figure line m || line n l line l is a transversal. b a c d m If mÐ b = (x + 15)° and f e n mÐ e = (2x + 15)°, find the value of x. g h Solution : Ð b @ Ð f.... (corresponding angles) \ mÐ f = mÐ b = (x + 15)° mÐ f + mÐ e = 180°.......... (Angles in linear pair) substituting values in the equation, x + 15 + 2x + 15 = 180° \ 3x + 30 = 180° \ 3x = 180°- 30° = 150°......... (subtracting 30 from both sides) 150 o x=......... (dividing both sides by 3) 3 \ x = 50° Now I know. When two parallel lines are intersected by a transversal, the angles formed in each pair of · corresponding angles are congruent. · alternate angles are congruent. · interior angles are supplementary. Practice Set 2.2 1.Choose the correct alternative. p (1) In the adjoining figure, if line m || line n 3x m and line p is a transversal then find x. x n (A) 135° (B) 90° (C) 45° (D) 40° (2) In the adjoining figure, if line a || line b a b and line l is a transversal then find x. 4x (A) 90° (B) 60° (C) 45° (D) 30° 2x l t s 2. In the adjoining figure line p || line q. 40° p Line t and line s are transversals. y Find measure of Ð x and Ð y 70° q using the measures of angles given x in the figure. 11 3. In the adjoining figure. line p || line q. p q line l || line m. Find measures of 80° a l Ð a, Ð b, and Ð c, using the measures of given angles. Justify c m b your answers. a b 4«. In the adjoining figure, line a || line b. 105° x line l is a transversal. Find the z l measures of Ð x, Ð y, Ð z using y the given information. p 40° 5. In the adjoining figure, line p || line l || line q. « Find Ð x with the help of the measures given l x in the figure. q 30° For more information : If a transversal intersects two coplaner lines and a pair of - corresponding angles is congruent then the lines are parallel. - alternate angles is congruent then the lines are parallel. - interior angles is supplementary then the lines are parallel. To draw a line parallel to the given line Construction (I) : To draw a line parallel to the given line through a point outside the given line using set - square. Method I : Steps of the construction (1) Draw line l. (2) Take a point P outside the line l. (3) As shown in the figure, place two set - P m squares touching each other. Hold set - squares A A and B. One edge of set - square A is close B to point P. Draw a line along the edge of B. l (4) Name the line as m. (5) Line m is parallel to line l. 12 Method II : Steps of construction (1) Draw line l. Take a point P outside the line. P Q m (2) Draw a seg PM ⊥ line l. (3) Take another point N on line l. (4) Draw seg NQ ⊥ line l. l such that l(NQ) = l(MP). M N (5) The line m passing through points P and Q is parallel to the line l. Construction (II) : To draw a parallel line to a given line at a given distance. P Q Method : Draw a line parallel to line l at a distance 2.5 cm. 2.5 cm 2.5 cm Steps of construction : (1) Draw line l. (2) Take two points A and l A B B on the line l. (3) Draw perpendiculars to the line l from points A and B. (4) On the perpendicular lines take points P and Q at a distance of 2.5cm from A and B respectively. (5) Draw line PQ. (6) Line PQ is a line parallel to the line l at a distance 2.5cm. Practice Set 2.3 1. Draw a line l. Take a point A outside the line. Through point A draw a line parallel to line l. 2. Draw a line l. Take a point T outside the line. Through point T draw a line parallel to line l. 3. Draw a line m. Draw a line n which is parallel to line m at a distance of 4 cm from it. ÑÑÑ Answers Parctice Set 2.1 1. (1) Ð w (2) Ð x (3) Ð y (4) Ð z (5) Ð x (6) Ð r 2. (1) Ð c and Ð e, Ð b and Ð h (2) Ð a and Ð e, Ð b and Ð f, Ð c and Ð g, Ð d and Ð h (3) Ð c and Ð h, Ð b and Ð e. Parctice Set 2.2 1. (1) C (2) D 2. Ð x = 140°, Ð y = 110° 3. Ð a = 100°, Ð b = 80°, Ð c = 80° 4. Ð x = 105°, Ð y = 105°, Ð z = 75° 5. Ð x = 70° 13  3 Indices and Cube root Let’s recall. In earlier standards, we have learnt about Indices and laws of indices. · The product 2 ´ 2 ´ 2 ´ 2 ´ 2, can be expressed as 25, in which 2 is the base, 5 is the index and 25 is the index form of the number. · Laws of indices : If m and n are integers, then (i) a m ´ a n = a m+n (ii) a m ¸ a n = a m-n (iii) (a ´ b)m = a m ´ b m (iv) a  0 = 1 m m m am b (vii)    m (viii)   1 a a (v) a -m = (vi) (a ) = a m n mn   am b b b a  · Using laws of indices, write proper numbers in the following boxes. (i) 35 ´ 32 = 3 (ii) 37 ¸ 39 = 3 (iii) (34)5 = 3 1 (iv) 5-3 = (v) 50 = (vi) 51 = 5 3 3 3 3 5   (ix)   5 (vii) (5 ´ 7) = 5 2 ´7 (viii)   = 3 =   7 7   Let’s learn. Meaning of numbers with rational indices 1 (I) Meaning of the numbers when the index is a rational number of the form. n Let us see the meaning of indices in the form of rational numbers such as 1 1 1 1 , , ,...,. 2 3 5 n To show the square of a number, the index is written as 2 and to show the 1 square root of a number, the index is written as 2. For example, square root of 25, is written as 25 using the radical sign ‘ ’. 1 1 Using index, it is expressed as 25. \ 25 = 25. 2 2 In general, square of a can be written as a 2 and square root of a is written as 2 a 1 or a or a 2. 1 3 3 Similarly, cube of a is written as a and cube root of a is written as a or a. 3 14  For example, 43 = 4 ´ 4 ´ 4 = 64. 1 1 \ cube root of 64 can be written as 3 64 or  64  3. Note that, 64 3 =4 3 ´ 3 ´ 3 ´ 3 ´ 3 = 35 = 243. That is 5th power of 3 is 243. 1 1 Conversely, 5th root of 243 is expressed as  243 5 or 5 243. Hence,  243 5 = 3 1 In genral n th root of a is expressed as a n. 1 1 For example, (i) 128 7 = 7th root of 128, (ii) 900 12 = 12th root of 900, etc. 1 Note that, If 10 5 = x then x5 = 10. Practice Set 3.1 1. Express the following numbers in index form. (1) Fifth root of 13 (2) Sixth root of 9 (3) Square root of 256 (4) Cube root of 17 (5) Eighth root of 100 (6) Seventh root of 30 2. Write in the form ‘nth root of a’ in each of the following numbers. 1 1 1 1 1 1 (1)  81 4 (2) 49 2 (3) 15  5 (4)  512  9 (5) 100 19 (6)  6  7 m (II) The meaning of numbers, having index in the rational form. n We know that 82 = 64, 1 1 Cube root at 64 is =  64  3   82  3  4 \ cube root of square of 8 is 4.......... (I) 1 Similarly, cube root of 8 = 8 = 2 3 2  1 \ square of cube root of 8 is  8 3  = 22 = 4..........(II)   From (I) and (II) 2 =  8 3 . 1 1 cube root of square of 8 = square of cube root of 8. Using indices,  8  2 3   The rules for rational indices are the same as those for integral indices 2 1  13  2 \ using the rule (a ) = a , we get m n mn   8 2 3 = 8  = 83.   2 From this we get two meanings of the number 8 3. 1 (i) 8 3 =  82  3 i. e. cube root of square of 8. 2 2 2  13  (ii) 8 = 3  8  i. e. square of cube root of 8.   15  4 1 Similarly, 27 =  27 5  4 5 means ‘fifth root of fourth power of 27’, 4 4  51  and 27 = 5  27  means ‘fourth power of fifth root of 27’.   m Generally we can express two meanings of the number a n. m 1 a n = a  m n means ‘nth root of mth power of a’. m m  n1  a n =  a  means ‘m power of n root of a’. th th   Practice Set 3.2 1. Complete the following table. Sr. No. Number Power of the root Root of the power 3 (1)  225  2 Cube of square root of 225 Square root of cube of 225 4 (2)  45  5 6 (3)  81 7 4 (4) 100 10 3 (5)  21 7 2. Write the following numbers in the form of rational indices. (1) Square root of 5th power of 121. (2) Cube of 4th root of 324 (3) 5th root of square of 264 (4) Cube of cube root of 3 Let’s recall. · 4 ´ 4 = 16 implies 42 = 16, also (-4)2 = 16 which indicates that the number 16 has two square roots ; one positive and the other negative. Conventionally, positive root of 16 is shown as 16 and negative root of 16 is shown as - 16. Hence 16 = 4 and - 16 = -4. · Every positive number has two square roots. · Square root of zero is zero. 16  Let’s learn. Cube and Cube Root If a number is written 3 times and multiplied, then the product is called the cube of the number. For example, 6 ´ 6 ´ 6 = 63 = 216. Hence 216 is the cube of 6. To find the cube of rational number.   2 Ex. (1) Find the cube Ex. (2) Find the cube of Ex. (3) Find the cube of   5 .   of 17. (-6).  2 3  2  2 173 = 17 ´ 17 ´ 17  2 (-6)3 = (-6) ´ (-6) ´ (-6)  5  =  5  ´   5  ´  5         = 4913 = -216 8 = − 125 Ex. (4) Find the cube of (1.2). Ex. (5) Find the cube of (0.02). (1.2) = 1.2 ´ 1.2 ´ 1.2 3 (0.02)3 = 0.02 ´ 0.02 ´ 0.02 = 1.728 = 0.000008 Use your brain power. In Ex. (1) 17 is a positive number. The cube of 17, which is 4913, is also a positive number. In Ex. (2) cube of - 6 is -216. Take some more positive and negative numbers and obtain their cubes. Find the relation between the sign of a number and the sign of its cube. In Ex. (4) and (5), observe the number of decimal places in the number and number of decimal places in the cube of the number. Is there any relation between the two ? To find the cube root We know, how to find the square root of a number by factorisation method. Using the same method, we can find the cube root. Ex. (1) Find the cube root of 216. Solution : First find the prime factor of 216. 216 = 2 ´ 2 ´ 2 ´ 3 ´ 3 ´ 3 Each of the factors 3 and 2, appears thrice. So let us group them as given below, 216 = (3 ´ 2) ´ (3 ´ 2) ´ (3 ´ 2) = (3 ´ 2)3 = 63 1 \ 3 216 = 6 that is  216  3 = 6 17 Ex. (2) Find the cube root of -1331. Ex. (4) Find 3 0.125. Solution : To find the cube root of -1331, 125 Solution : 3 0.125 = 3 1000 let us factorise 1331 first. 3 125 m 1331 = 11 ´ 11 ´ 11 = 113 =... a   am 3 1000 b bm   -1331 = (-11) ´ (-11) ´ (-11) 3 53 = (-11)3 = 3 10 3 \ 3 −1331 = -11 5 1 = 10 = 0.5...   am m a Ex. (3) Find the cube root of 1728. Solution : 1728 = 8 ´ 216 = 2 ´ 2 ´ 2 ´ 6 ´ 6 ´ 6 \ 1728 = 23 ´ 63 = (2 ´ 6)3........ am ´ bm = (a ´ b)m 3 1728 = 2 ´ 6 = 12 (Note that, cube root of - 1728 is -12.) Practice Set 3.3 1. Find the cube roots of the following numbers. (1) 8000 (2) 729 (3) 343 (4) -512 (5) -2744 (6) 32768 27 16 2. Simplify : (1) 3 (2) 3 54 3. If 3 729 = 9 then 3 0.000729 =? 125 ÑÑÑ Answers 1 1 1 1 1 1 Practice Set 3.1 (1) 135 (2) 96 (3) 256 2 (4) 17 3 (5) 100 8 (6) 30 7 2. (1) Fourth root of 81. (2) Square root of 49 (3) Fifth root of 15 (4) Ninth root of 512 (5) Nineteenth root of 100 (6) Seventh root of 6 Practice Set 3.2 1. (2) 4th power of 5th root of 45 ; 5th root of 4th power of 45. (3) 6th power of 7th root of 81 ; 7th root of 6th power of 81. (4) 4th power of 10th root of 100 ; 10th root of 4th power of 100. (5) 3rd power of 7th root of 21 ; 7th root of 3rd power of 21. 5 3 2 2. (1) 121 2 (2)  324  4 (3)  264  5 (4) 33 3 Practice Set 3.3 1. (1) 20 (2) 9 (3) 7 (4) -8 (5) -14 3 2 (6) 32 2. (1) (2) 3. 0.09 5 3 18  4 Altitudes and Medians of a triangle Let’s recall. In the previous standard we have learnt that the bisectors of angles of a triangle, as well as the perpendicular bisectors of its sides are concurrent. These points of concurrence are respectively called the incentre and the circumcentre of the triangle. Activity : Draw a line. Take a point outside the line. Draw a perpendicular from the point to the line with the help of a set - square. Let’s learn. Altitude The perpendicular segment drawn from a vertex A of a triangle on the side opposite to it is called an altitude of the triangle. In D ABC, seg AP is an altitude on the base BC. B C P To draw altitudes of a triangle : Y 1. Draw any D XYZ. 2. Draw a perpendicular from vertex X on the side YZ using a set - square. Name the point where it R meets side YZ as R. Seg XR is an altitude on side YZ. Z X 3. Considering side XZ as a base, draw an altitude YQ on side XZ. seg YQ ⊥ seg XZ. 4. Consider side XY as a base, draw an altitude ZP Y on seg XY. seg ZP ⊥ seg XY. seg XR, seg YQ, seg ZP are the altitudes of R D XYZ. P Note that, the three altitudes are concurrent. O The point of concurrence is called the orthocentre Z X of the triangle. It is denoted by the letter ‘O’. Q 19 The location of the orthocentre of a triangle : A Activity I : D Draw a right angled triangle and draw all its altitudes. Write the point of concurrence. B C Activity II : P Draw an obtuse angled triangle and all its altitudes. B A Do they intersect each other ? Q R Draw the lines containing the altitudes. Observe that these lines are concurrent. C O A Activity III : Draw an acute angled D ABC and all its altitudes. F Observe the location of the orthocentre. E O B D C Now I know. The altitudes of a triangle pass through exactly one point ; that means they are concurrent. The point of concurrence is called the orthocentre and it is denoted by ‘O’. · The orthocentre of a right angled triangle is the vertex of the right angle. · The orthocentre of an abtuse angled triangle is in the exterior of the triangle. · The orthocentre of an acute angled triangle is in the interior of the triangle. Let’s learn. Median The segment joining the vertex and H midpoint of the opposite side is called a median of the triamgle. D In D HCF, seg FD is a median on the C base CH. F 20 To draw medians of a triangle : 1. Draw D ABC. C 2. Find the mid-point P of side AB. Draw seg CP. Q R 3. Find the mid-point Q of side BC. Draw seg AQ. G 4. Find the mid-point R of side AC. Draw seg BR. B P A Seg PC, seg QA and seg BR are medians of D ABC. Note that the medians are concurrent. Their point of concurrence is called the centroid. It is denoted by G. Activity IV : Draw three different triangles ; a right angled triangle, an obtuse angled triangle and an acute angled triangle. Draw the medians of the triangles. Note that the centroid of each of them is in the interior of the triangle. The property of the centroid of a triangle : · Draw a sufficiently large D ABC. · Draw medians ; seg AR, seg BQ and seg CP of D ABC. · Name the point of concurrence as G. Measure the lengths of segments from the figure and fill in the boxes in the following table. l(AG) = l(GR) = l(AG) : l (GR) = : l(BG) = l(GQ) = l(BG) : l (GQ) = : l(CG) = l(GP) = l(CG) : l (GP) = : Observe that all of these ratios are nearly 2:1. Now I know. The medians of a triangle are concurrent. Their point of concurrence is called the Centroid and it is denoted by G. For all types of triangles the location of G is in the interior of the triangles. The centroid divides each median in the ratio 2:1. Let’s discuss. As shown in the adjacent figure, a student A drew D ABC using five parallel lines of a note G book. Then he found the centroid G of the triangle. How will you decide whether the B C location of G he found, is correct. 21  Practice Set 4.1 L 1. In D LMN,...... is an altitude and...... is a median. (write the names of appropriate segments.) M N X Y 2. Draw an acute angled D PQR. Draw all of its altitudes. Name the point of concurrence as ‘O’. 3. Draw an obtuse angled D STV. Draw its medians and show the centroid. 4. Draw an obtuse angled D LMN. Draw its altitudes and denote the orthocentre by ‘O’. 5. Draw a right angled D XYZ. Draw its medians and show their point of concurrence by G. 6. Draw an isosceles triangle. Draw all of its medians and altitudes. Write your observation about their points of concurrence. 7. Fill in the blanks. A Point G is the centroid of D ABC. (1) If l(RG) = 2.5 then l(GC) =...... R Q (2) If l(BG) = 6 then l(BQ) =...... G (3) If l(AP) = 6 then l(AG) =..... B C and l(GP) =..... P Try this. (I) : Draw an equilateral triangle. Find its circumcentre (C), incentre (I), centroid (G) and orthocentre (O). Write your observation. (II): Draw an isosceles triangle. Locate its centroid, orthocentre, circumcentre and incentre. Verify that they are collinear. ÑÑÑ Answers Practice Set 4.1 1. seg LX and seg LY 7. (1) 5, (2) 9, (3) 4, 2 22  5 Expansion formulae Let’s recall. We have studied the following expansion formulae in previous standard. (i) (a + b)2 = a2 + 2ab + b2, (ii) (a - b)2 = a2 - 2ab + b2, (iii) (a + b) (a - b) = a2 - b2 Use the above formulae to fill proper terms in the following boxes. (i) (x + 2y)2 = x2 + + 4y2 (ii) (2x - 5y)2 = - 20xy + (iii) (101)2 = (100 + 1)2 = + + 12 = (iv) (98)2 = (100 - 2)2 = 10000 - + = (v) (5m + 3n)(5m - 3n) = - = - Let’s learn. Activity : Expand (x + a)(x + b) using formulae for areas of a square and a rectangle. x b x x2 xb b x ax ab = x x+a + x+ a a x b x (x + a)(x + b) = x2 +ax + bx + ab (x + a)(x + b) = x2 + (a +b)x + ab (I) Expansion of (x + a) (x + b) (x + a) and (x + b) are binomials with one term in common. Let us multiply them. (x + a)(x + b) = x (x + b) + a(x + b) = x2 + bx + ax + ab = x2 + (a + b)x + ab \ (x + a)(x + b) = x2 + (a + b)x + ab 23 Expand Ex. (1) (x + 2)(x + 3) = x2 + (2 + 3)x + (2 ´ 3) = x2 + 5x + 6 Ex. (2) (y + 4)(y - 3) = y2 + (4 - 3)y + (4) ´ (-3) = y2 + y - 12 Ex. (3) (2a + 3b)(2a - 3b) = (2a)2 + [(3b) + (- 3b)]2a +[3b ´ (-3b)] = 4a2 + 0 ´ 2a - 9 b2 = 4a2 - 9 b2  3  1 3 1 3 1 3 Ex. (4) m 2 m 2 = m2 +    m+ ´ = m2 + 2m +      2 2 2 2 4 Ex. (5) (x - 3)(x - 7) = x2 + (-3 - 7)x + (-3)(-7) = x2 - 10x + 21 Practice Set 5.1 1. Expand. (1) (a +2)(a - 1) (2) (m - 4)(m + 6) (3) (p + 8)(p -3) (4) (13 + x)(13 - x) (5) (3x + 4y)(3x + 5y) (6) (9x - 5t)(9x + 3t)  2  7  1  1 1  1  (7) m 3 m 3 (8) x x x  (9)   4   9        x y  y  Let’s learn. (II) Expansion of (a + b)3 (a + b)3 = (a + b) (a + b) (a + b) = (a + b) (a + b)2 = (a + b)(a2 + 2ab + b2) = a(a2 + 2ab + b2) + b(a2 + 2ab + b2) = a3 + 2a2b + ab2 + ba2 + 2ab2 + b3 = a3 + 3a2b + 3ab2 + b3 \ (a + b)3 = a3 + 3a2b + 3ab2 + b3 Let us study some examples based on the above expansion formula. (x + 3)3 Ex. (1) We know that (a + b)3 = a3 + 3a2b + 3ab2 + b3 In the given example, a = x and b = 3 24  \ (x + 3)3 = (x)3 + 3 ´ x2 ´ 3 + 3 ´ x ´ (3)2 + (3)3 = x3 + 9x2 + 27x +27 (3x + 4y)3 = (3x)3 + 3(3x)2(4y) + 3(3x)(4y)2 + (4y)3 Ex. (2) = 27x3 + 3 ´ 9x2 ´ 4y + 3 ´ 3x ´ 16y2 + 64y3 = 27x3 + 108x2y + 144xy2 + 64 y3 3 3 2 2 3  2m n   2m   2m   n   2m   n   n  Ex. (3)  n  2m    n   3  n   2m   3  n   2m    2m               8m3  4 m 2   n   2m   n 2  n3   3  2    3  2   n3  n   2m   n   4 m  8m 3 8 m 3 6 m 3n n3     n3 n 2m 8m3 Ex. (4) (41)3 = (40 + 1)3 = (40)3 + 3 ´ (40)2 ´ 1 +3 ´ 40 ´ (1)2 + (1)3 = 64000 + 4800 + 120 + 1 = 68921 Practice Set 5.2 1. Expand. (1) (k + 4)3 (2) (7x + 8y)3 (3) (7 + m)3 (4) (52)3 3 3 3  1  1  5x y  (5) (101) x   3 (6) (7)  2m  5  (8)    x    y 5x  Activity : Make two cubes of side a and of side b each. Make six parallelopipeds; three of them measuring a ´ a ´ b and the remaining three measuring b ´ b ´ a. Arrange all these solid figures properly and make a cube of side (a + b). Let’s learn. (III) Expansion of (a - b)3 \ (a - b)3 = (a - b) (a - b) (a - b) = (a - b)(a - b)2 = (a - b)(a2 - 2ab +b2) = a(a2 -2ab +b2) - b(a2 - 2ab + b2) 25  = a3 - 2a2b + ab2 - a2b + 2ab2 - b3 = a3 - 3a2b + 3ab2 - b3 \ (a - b)3 = a3 - 3a2b + 3ab2 - b3 Ex. (1) Expand (x - 2)3 (a - b)3 = a3 - 3a2b + 3ab2 - b3 Here taking , a = x and b = 2, (x - 2)3 = (x)3 - 3 ´ x2 ´ 2+ 3 ´ x ´ (2)2 - (2)3 = x3 -6x2 + 12x - 8 Ex. (2) Expand (4p - 5q)3. (4p - 5q)3 = (4p)3 - 3(4p)2(5q) + 3(4p)(5q)2 - (5q)3 (4p - 5q)3 = 64p3 - 240p2q + 300pq2 - 125q3 Ex. (3) Find cube of 99 using the expansion formula. (99)3 = (100 - 1)3 = (100)3 - 3 ´ (100)2 ´ 1 + 3 ´ 100 ´ (1)2 - 13  = 1000000 - 30000 + 300 - 1 = 9,70,299 Ex. (4) Simplify. (i) (p + q)3 + (p - q)3 = p3 + 3p2q + 3pq2 + q3 + p3 - 3p2q + 3pq2 - q3 = 2p3 + 6pq2 (ii) (2x + 3y)3 - (2x - 3y)3 = [(2x)3 + 3(2x)2(3y) + 3(2x)(3y)2 + (3y)3] - [(2x)3 - 3(2x)2(3y) + 3(2x)(3y)2 - (3y)3] = (8x3 + 36x2y + 54xy2 + 27y3) - (8x3 - 36x2y + 54xy2 - 27y3) = 8x3 +36x2y + 54xy2 + 27y3 - 8x3 + 36x2y - 54xy2 + 27y3 = 72x2y + 54y3 Now I know. (i) (a + b)3 = a3 + 3a2b + 3ab2 + b3 = a3 + b3 + 3ab(a + b) (ii) (a - b)3 = a3 - 3a2b + 3ab2 - b3 = a3 - b3 - 3ab(a - b) 26  Practice Set 5.3 1. Expand. (1) (2m - 5)3 (2) (4 - p)3 (3)(7x - 9y)3 (4) (58)3 3 3 3  1   1  x 3 (5) (198) 1  a  3 x 3 (6) 2p  (7) (8)  2p      2. Simplify. (1) (2a + b)3 - (2a - b)3 (2) (3r - 2k)3 + (3r + 2k)3 (3) (4a - 3)3 - (4a + 3)3 (4) (5x - 7y)3 + (5x + 7y)3 Let’s learn. (IV) Expansion of (a + b + c)2 (a + b + c)2 = (a + b + c) ´ (a + b + c) = a (a + b + c) + b (a + b + c) + c (a + b + c) = a2 + ab + ac + ab + b2 + bc + ac + bc + c2 = a2 + b2 + c2 + 2ab + 2bc + 2ac \ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac. Ex. (1) Expand: (p + q + 3)2 = p2 + q2 + (3)2 + 2 ´ p ´ q + 2 ´ q ´ 3 + 2 ´ p ´ 3 = p2 + q2 + 9 + 2pq + 6q + 6p = p2 + q2 + 2pq + 6q + 6p + 9 Ex. (2) Fill in the boxes with appropriate terms in the steps of expansion. (2p + 3m + 4n)2 = (2p)2 + (3m)2 + + 2 ´ 2p ´ 3m + 2 ´ ´ 4n + 2 ´ 2p ´ = + 9m + 2 + 12pm + + Ex. (3) Simplify (l + 2m + n)2 + (l - 2m + n)2 = l2 +4m2 + n2 + 4lm + 4mn + 2ln + l2 + 4m2 + n2 - 4lm - 4mn + 2ln = 2l2 + 8 m2 + 2n2 + 4ln 27  Practice Set 5.4 1. Expand. (1) (2p + q + 5)2 (2) (m + 2n + 3r)2 (3) (3x + 4y - 5p)2 (4) (7m - 3n - 4k)2 2. Simplify. (1) (x -2y + 3)2 + (x + 2y -3)2 (2) (3k -4r -2m)2 - (3k + 4r - 2m)2 (3) (7a - 6b + 5c)2 + (7a + 6b - 5c)2 ÑÑÑ Answers Practice Set 5.1 (1) a2+ a - 2 (2) m2 + 2m - 24 (3) p2 + 5p - 24 (4) 169 - x2 (5) 9x2 + 27xy + 20y2 (6) 81x2 - 18xt - 15t2 5 14 1 1 5 (7) m2 - 3 m - (6) x2 - (9) − − 36 9 x2 y2 y Practice Set 5.2 (1) k3 + 12k2 + 48k + 64 (2) 343x3 + 1176x2y + 1344xy2 + 512y3 (2) 343 + 147m + 21m2 + m3 (4) 140608 (5) 1030301 3 1 12 m 2 6 m 1 (6) x3 + 3x + + (7) 8m3 + + + x x3 5 25 125 125 x 3 15 x 3 y y3 (8) + + + y3 y 5 x 125 x 3 Practice Set 5.3 1. (1) 8m3 - 60m2 + 150m - 125 (2) 64 - 48p + 12 p2 - p3 (3) 343x3 - 1323x2y + 1701xy2 - 729y3 (4) 1,95,112 3 1 (5) 77,62,392 (6) 8p3 - 6p + − 3 2p 8p 3 3 1 x3 9 27 (7) 1-  2 3 (8) x  3 a a a 27 x x 2. (1) 24a2b + 2b3 3 (2) 54r + 72 rk 2 (3) -288a2 - 54 (4)250x3 + 1470 xy2 Practice Set 5.4 1. (1) 4p2 + q2 + 25 + 4pq +10q + 20p (2) m2 + 4n2 + 9r2 + 4mn + 12nr + 6mr (3) 9x2 + 16y2 + 25p2+ 24 xy-40py-30px (4) 49m2 + 9n2 + 16k2 - 42mn + 24nk - 56km 2. (1) 2x2 + 8y2 + 18 - 24y (2) 32rm - 48kr (3) 98a2 + 72b2 + 50c2 - 120bc 28  6 Factorisation of Algebraic expressions Let’s recall. In the previous standard we have learnt to factorise the expressions of the form a x + a y and a2 - b2 For example, (1) 4xy + 8xy2 = 4xy(1 + 2y) (2) p2 - 9q2 = (p)2 - (3q)2 = (p + 3q)(p - 3q) Let’s learn. Factors of a quadratic trinomial An expression of the form ax2 + bx + c is called a quadratic trinomial. We know that (x + a)(x + b) = x2 + (a + b)x + ab \ the factors of x2 + (a + b)x + ab are (x + a) and (x + b). To find the factors of x2 + 5x + 6, by comparing it with x2 + (a + b)x + ab we get, a + b = 5 and ab = 6. So, let us find the factors of 6 whose sum is 5. Then writing the trinomial in the form x2 + (a + b)x + ab, find its factors. x2 + 5x + 6 = x2 + (3 + 2)x + 3 ´ 2.......... x2 + (a + b)x + ab = x2 + 3x + 2x + 2 ´ 3........ multiply (3 + 2) by x, make two groups of the four terms obtained. = x (x + 3) + 2(x + 3) = (x + 3)(x + 2) Study the following examples to know how a given trinomial is factorised. Ex. (1) Factorise : 2x2 - 9x + 9. Solution:First we find the product of the coefficient of the square term and the constant term. Here the product is 2 ´ 9 = 18. Now, find factors of 18 whose sum is -9, that is equal to the coefficient of the middle term. 2x - 9x + 9 2 18 = (-6) ´ (-3) ; (-6) + (-3) = -9 = 2x2 - 6x - 3x + 9 Write the term -9x as -6x - 3x = 2x (x - 3) - 3(x - 3) = (x - 3)(2x - 3) \ 2x2 - 9x + 9 = (x - 3)(2x - 3) 29 Ex. (2) Factorise : 2x2 + 5x -18. Ex. (3) Factorise : x2 - 10x + 21. -36 +21 2x2 + 5x - 18 Solution : solution: x2 - 10x + 21 +9 -4 = x2 - 7x - 3x + 21 -7 -3 = 2x2 + 9x -4x- 18 = x(2x + 9) -2(2x + 9) = x(x - 7) -3(x - 7) = (2x + 9)(x - 2) = (x - 7)(x - 3) Ex. (4) Find the factors of 2y2 - 4y - 30. Solution : 2y2 - 4y - 30 = 2(y2 - 2y - 15)........ taking out the common factor 2 -15 = 2(y2 - 5y + 3y - 15)........ = 2 [y(y - 5) + 3(y - 5)] -5 +3 = 2(y - 5)(y + 3) Practice Set 6.1 1. Factorise. (1) x2 + 9x + 18 (2) x2 - 10x + 9 (3) y2 + 24y + 144 (4) 5y2 + 5y - 10 (5) p2 - 2p - 35 (6) p2 - 7p - 44 (7) m2 - 23m + 120 (8) m2 - 25m + 100 (9) 3x2 + 14x + 15 (10) 2x2 + x - 45 (11) 20x2 - 26x + 8 (12) 44x2 - x - 3 Let’s learn. Factors of a3 + b3 We know that, (a + b)3 = a3 + 3a2b + 3ab2 + b3, which we can write as (a + b)3 = a3 + b3 + 3ab(a + b) Now, a3 + b3 + 3ab(a + b) = (a + b)3.......... interchanging the sides. \ a3 + b3 = (a + b)3 - 3ab(a + b)= [(a + b)(a + b)2] - 3ab(a + b) = (a + b)[(a + b)2 - 3ab] = (a + b) (a2 + 2ab + b2 - 3ab) = (a + b) (a2 - ab + b2) \ a3 + b3 = (a + b) (a2 - ab + b2) 30  Lets us solve some examples using the above formula for factorising the addition of two cubes. Ex. (1) x3 + 27y3 = x3 + (3y)3 = (x + 3y) [x2 - x(3y) + (3y)2] = (x + 3y) [x2 - 3xy +9y2] Ex. (2) 8p3 + 125q3 = (2p)3 + (5q)3 = (2p + 5q) [(2p)2 - 2p ´ 5q + (5q)2] = (2p + 5q) (4p2 - 10pq + 25q2) 1 3  1    1  1  1   2 Ex. (3) m3 + = m  3 =  m  4 m   m2  m  64 m3     4m      4 m  4 m    1   m2  1  1  =  m  4 m   4 16 m 2    Ex. (4) 250p3 + 432q3 = 2(125p3 + 216q3) = 2 [(5p)3 + (6q)3] = 2(5p + 6q) (25p2 - 30pq + 36q2) Practice Set 6.2 1. Factorise. (1) x3 + 64y3 (2) 125p3 + q3 (3) 125k3 + 27m3 (4) 2l3 + 432m3 1 8 q3 (5) 24a3 + 81b3 (6) y3 + (7) a 3 + (8) 1 + 8y3 a3 125 Let’s learn. Factors of a3 - b3 (a - b)3 = a 3 - 3a 2b + 3ab2 - b3 = a3 - b3 -3ab(a - b) Now, a3 - b3 - 3ab (a - b) = (a - b)3 \ a3 - b3 = (a - b)3 + 3ab (a - b) = [(a - b)(a - b)2 + 3ab (a - b)] = (a - b) [(a - b)2 + 3ab] = (a - b) (a 2 - 2ab + b2 + 3ab) = (a - b) (a 2 + ab + b2) \ a3 - b3 = (a - b)(a 2 + ab + b2) 31  Lets us solve some examples using the above formula for factorising the difference of two cubes. Ex. (1) x3 - 8y3 = x3 - (2y)3 \ x3 - 8y3 = x3 - (2y)3 = (x - 2y) (x2 + 2xy + 4y2) Ex. (2) 27p3 - 125q3 = (3p)3 - (5q)3 = (3p - 5q) (9p2 + 15pq +25q2) Ex. (3) 54p3 - 250q3 = 2 [27p3 - 125q3] = 2 [(3p)3 - (5q)3] = 2(3p - 5q)(9p2 + 15pq +25q2) 1  1  2 1  Ex. (4) a3 − = a  a  a  1 a 2  a3     Ex. (5) Simplify : (a - b)3 - (a3 - b3) Solution : (a - b)3 - (a3 - b3) = a3 - 3a2b + 3ab2 - b3 - a3 + b3 = - 3a2b + 3ab2 Ex. (6) Simplify : (2x + 3y)3 - (2x - 3y)3 Solution :Using the formula a3 - b3 = (a - b) (a2 + ab + b2) \(2x + 3y)3 - (2x - 3y)3 = [(2x + 3y) - (2x - 3y)] [(2x + 3y)2 + (2x + 3y) (2x - 3y) + (2x - 3y) 2] = [2x + 3y - 2x + 3y] [4x2+12xy+ 9y2 + 4x2-9y2+ 4x2-12xy+ 9y2] = 6y (12x2 + 9y2) = 72x2y + 54y3 Now I know. (i) a3 + b3 = (a + b) (a2-ab + b2) (ii) a3 - b3 = (a - b) (a2 + ab + b2) Practice Set 6.3 1. Factorise : (1) y3 - 27 (2) x3 - 64y3 (3) 27m3 - 216n3 (4) 125y3 - 1 27 128 (5) 8 p3 − (6) 343a3 - 512b3 (7) 64x3 - 729y3 (8) 16a 3 − p3 b3 2. Simplify : (1) (x + y)3 - (x - y)3 (2) (3a + 5b)3 - (3a - 5b)3 (3) (a + b)3 - a3 - b3 (4) p3 - (p + 1)3 (5) (3xy - 2ab)3 - (3xy + 2ab)3 32  Let’s learn. Rational algebraic expressions A If A and B are two algebraic expressions then is called a rational algebraic B expression. While simplifying a rational algebraic expression , we have to perform operations of addition, subtraction, multiplication and division. They are similar to those performed on rational numbers. Note that, the denomintors or the divisors of algebraic expressions are non-zero. (1) Simplify : a 2  5a  6  a2  4 2 Ex. 7 x 2  18 x  8 14 x  8 Ex. (2)  a  a  12 a  4 49 x 2  16 x2 a 2  5a  6 a  4 7 x 2  18 x  8 14 x  8 Solution:  Solution :  a 2  a  12 a 2  4 49 x 2  16 x2  a  3  a  2    a  4  7 x  4   x  2 2 7 x  4  = =  7 x  4   7 x  4    x  2   a  4   a  3  a  2   a  2  1 = 2 = a −2 x2 − 9 y2 Ex. (3) Simplify : 3 x − 27 y 3 x2  9 y2   x  3y  x  3y  x  3y  Solution: x  27 y 3 3   x  3 y  x 2  3 xy  9 y 2  x  3 xy  9 y 2 2 Practice Set 6.4 1. Simplify : (1) m2  n2 m 2  mn  n 2 (2) a 2  10a  21 a 2  1 (3) 8 x 3 − 27 y 3    m  n m3  n3 2 a 2  6a  7 a 3 4x2 − 9 y2 x 2  5 x  24 x 2  64 3x 2  x  2 3x 2 − 7 x − 6 (6) 4 x  11x6 2 (4)  (5) ¸  x  3  x  8   x  8 2 x 2  7 x  12 x2 − 4 16 x  9 2 a 3  27 a 2  3a  9 1 2x  x2 1 x  x2 (7) 5a 2  16a  3 ¸ 25a 2  1 (8)  1 x3 1 x ÑÑÑ 33  Answers Practice Set 6.1 1. (1) (x + 6) (x + 3) (2) (x - 9) (x - 1) (3) (y + 12) (y + 12) (4) 5(y + 2) (y - 1) (5) (p - 7) (p + 5) (6) (p + 4) (p - 11) (7) (m - 15) (m - 8) (8) (m - 20) (m - 5) (9) (x + 3) (3x + 5) (10) (x + 5) (2x - 9) (11) 2(5x - 4) (2x - 1) (12) (11x - 3) (4x + 1) Practice Set 6.2 1. (1) (x + 4y) (x2 - 4xy + 16y2) (2) (5p + q) (25p2 - 5pq + q2) (3) (5k + 3m) (25k2 - 15km + 9m2) (4) 2(l + 6m) (l2 - 6lm + 36m2)  1  2 1 1  (5) 3(2a + 3b) (4a2 - 6ab + 9b2) (6)  y   y   2   2y  2 4y   q  q q  2  2  4  (7)  a    a 2  2  2  (8)  1    1     a  a   5   5 25  Practice Set 6.3 1. (1) (y - 3) (y2 + 3y + 9) (2) (x - 4y) (x2 + 4xy + 16y2) (3) 27(m - 2n) (m2 + 2mn + 4n2) (4) (5y - 1) (25y2 + 5y + 1)  3  2 9  (5)  2 p   4 p  6  2  (6) (7a - 8b) (49a2 + 56ab + 64b2)  p  p   2  2a 4  (7) (4x - 9y) (16x2 + 36xy + 81y2) (8) 16  a    a 2   2   b  b b  2. (1) 6x2y + 2y3 (2) 270a2b + 250b3 (3) 3a2b + 3ab2 (4) -3p2 - 3p - 1 (5) -108x2y2ab -16a3b3 Practice Set 6.4 1 4 x 2 + 6 xy + 9 y 2 1. (1) (2) a + 1 (3) m+n 2x + 3y (4) 1 (5) ( x  1)( x 2 2)( x  2) ( x  3) ( x  4 ) x2 1 x (6) (7) 5a + 1 (8) 4x  3 1 x 34  7 Variation Let’s recall. If the rate of notebooks is ` 240 per dozen, what is the cost of 3 notebooks? Also find the cost of 9 notebooks ; 24 notebooks and 50 notebooks and complete the following table. Number of notebooks (x) 12 3 9 24 50 1 Cost (In Rupees) (y) 240 20 From the above table we see that the ratio of number of notebooks (x) and 1 their cost (y) in each pair is. It is constant. The number of notebooks and their 20 cost are in the same proportion. In such a case, if one number increases then the other number increases in the same proportion. Let’s learn. Direct variation The statement ‘x and y are in the same proportion’ can be written as ‘x and y are in direct variation’ or ‘there is a direct variation between x and y’. Using mathematical symbol it can be written as x a y. [a(alpha) is a greek letter, used to denote variation.] x a y is written in the form of equation as x = ky, where k is a constant. x x = ky or = k is the equation form of direct variation where k is the y constant of variation. Observe how the following statements are written using the symbol of variation. (i) Area of a circle is directly proportional to the square of its radius. If the area of a circle = A, its radius = r, the above statement is written as A a r 2. (ii) Pressure of a liquid (p) varies directly as the depth (d) of the liquid ; this statement is written as p a d. To understand the method of symbolic representation of direct variation, study the following examples. Ex. (1) x varies directly as y, when x = 5, y = 30. Find the constant of variation and equation of variation. Solution: x varies directly as y, that is as x a y 35  \ x = ky......... k is constant of variation. when x = 5, y = 30, is given 1 \ 5 = k ´ 30 \ k = 6 (constant of variation) y \ equation of variation is x = ky, that is x = 6 or y = 6x Ex. (2) Cost of groundnuts is directly proportional to its weight. If cost of 5 kg groundnuts is ` 450 then find the cost of 1 quintal groundnuts. (1 quintal = 100 kg) Solution: Let the cost of groundnuts be x and weight of groundnuts be y. It is given that x varies directly as y \ x a y or x = ky It is given that when x = 450 then y = 5, hence we will find k.