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Maths 8

Maths class 8

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36 views197 pages

Maths 8

Maths class 8

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maymunaxmed092
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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TABLE OF CONTENTS PREFACE ACKNOWLEDGMENTS ... NUMBERS 1.1 READING AND WRITING NUMBERS..... 1.2 SQUARE ROOT... 1.3 CUBE OF ANUMBER 1.4 CUBE ROOTS. 1.5 BINARY NUMBERS 1.6 CONVERSION OF BINARY AND DECIMAL NUMBERS 1.7 PRODUCT RULE .... 1.8 QUOTIENT RULE .... 1.9 OTHER RULES OF INDICES 1.10 LOGARITHMS... 1.11 SCIENTIFIC NOTATION 1.12 SETS 1.13 SUBSETS. 1.14 UNION AND INTERSECTION OF SETS(1).. 1.15 UNION AND INTERSECTION OF SETS (2) 1.16 VENN DIAGRAM... 1.17 UNIVERSAL SET MEASUREMENT 2.1 Conversion of m, dm, sm and mm 2.2 Conversion of m, dam, hm, and km 2.3 PERIMETER AND AREA OF A SQUARE 2.4 PERIMETER AND AREA OF A RECTANGLE. 2.5 AREA OF PARALLELOGRAM 2.6 AREA OF TRAPEZIUM.. 2.7 AREA OF A TRIANGLE... 2.8 CONVERSION OF CAPACITY UNIT! 2.9 VOLUME OF A CYLINDER. 2.10 VOLUME OF A SPHERE 2.11 VOLUME OF CONE GEOMETRY 3.1 CONGRUENT TRIANGLES.. 3.2 PYTHAGORAS THEOREM 3.3 ANGLES OF ATRIANGLE .. 3.4 ANGLES OF QUADRILATERALS 3.5 POLYGONS .... 3.6 EXTERIOR ANGLES OF A POLYGON 3.7 NETS OF SOLIDS STATISTICS... 4.1 MEAN, MODE AND MEDIAN, 4.2 BAR GRAPH 4.3 PIE CHART . 4.4 LINE GRAPHS PROBABILITY... 5.1 PROBABILITY TERMINOLOGIES: 5.2 CALCULATING PROBABILITIES ALGEBRA 6.1 MONOMIAL AND POLYNOMIAL EXPRESSIONS .....:0:0 6.2 ADDITION AND SUBTRACTION OF POLYNOMIALS ... 146 6.3 EVALUATING ALGEBRAIC EXPRESSIONS... 149 6.4 MULTI AND DIVISION OF ALGEBRAIC EXPRESSIONS 151 6.5 SOLVING SIMULTANEOUS EQUATIONS .155 6.6 ELIMINATION METHOD... . 158 6.7 SUBSTITUTION METHOD. 161 6.8 PROBLEMS SIMULTANEOUS EQUATIONS .. 164 6.9 BINOMIAL PRODUCTS . 168 6.10 SOLVING QUADRATIC EQUATION BY FACTORIZATION METHOD..... 171 6,11 SOLVING QUADRATIC EQUATION BY FORMULA...... 174 TRIGONOMETRY... a 79 7.1 TRIGONOMETRIC RATIOS... 179 GENERAL EXERCISE 185 8.1 Unit 1 review questions . 8.2 Unit 2 review questions . 8.3 Unit 3 review question 8.4 Unit 4 review questions . 8.5 Unit 5 review questions . 194 8.6 Unit 6 review questions . .195 8.7 Unit 7 review Questions ....sssssssesssessssueeanensssessseenneeeanseeses 196 185 187 190 192 ee ee ee 1.1 READING AND WRITING NUMBERS Place value is the value of each digit in a number. To read large numbers easier, we put them into groups of three digits starting from the right. liana three digits consist of ones, tens, and hundreds. = | "PLACE VALUE Ts Pe TS THOUSANDS J g 3 oe ee Ty E Eo ore 3 3 Er Fe = Eke) — — 987, 654,321 Nine hundred and eighty-seven million, six hundred and fifty-four thousand, three hundred and twenty-one. levels Write the following numbers in words: a. 12,436,789, b. 35,766 2,009,752 a. 12,436,789, before writing in words look at the table below. This table shows the basic concept of place value. [Millions |Thousands [Units T O° H T iH |T " / | [a 2 4° (13 | So, twelve million, four hundred thirty- six thousand, seven hundred eighty-nine. a. (Thirty-five thousand, seven hundred sixty-six). b. 2,009,752 (two million, nine thousand, seven hundred fifty-two) Write the following numbers in symbols a. Thirty five thousand six hundred forty five b. Three million four hundred sixty six thousand five hundred c. Forty eight thousand five hundred thirty three a. 35,645. b. 3,466,500. c. 48,533. 1. Write the following numbers in words d. 988,458. e. 5.107,362. f. 64,187. g. 7,988 453. h. 6,567,000. i. 56,971,023. j. 53,000,534. 2. Write the following in sym bols a. Nineteen million, two hundred seventy five thousand, eight hundred thirty one. b. Seventy one million, five hundred sixty four thousand, one hundred ninety eight. c. Forty eight million, three hundred twelve. d. Fifty one million, and sixty. e. Sixty five million, four hundred ninety thousand, seven hundred thirty two. Twenty nine million, nine hundred ninety nine thousand. Write the place value of the underlined digit. g. 47,093,156 e) 30,248,916 h. 87,524,397 f) 68,959,854 i. 4,560,635 g) 6,454,874 j. 25,935,714 h) 87,546,894 4. Write the following numbers in expanded form: a. 834,345. d) 9,321,006 b. 47,926,135 e) 60,010,025 c. 7,346,291 f) 24,619,658. 1.2 SQUARE ROOT Square root is anumber multiplied by itself. For instance ‘a’ is written as . So square root of a number is the opposite of the square of a number. There are two methods of finding a square root. 1. Prime factorization method. 2. Division method. Prime factorization method is to express a given number as a product of a prime factor. The following steps must be followed © Step 1: Divide the given number by the smallest prime number. © Step 2: Again, divide the quotient by the smallest prime number. © Step 3: Repeat the process, until the quotient becomes 1. Step 4: Finally, multiply all the prime factors taking one from any same two factors . Find the prime factorization of 576. Square root of 576 = V576=24 wn Set a ul; WwWwNnNNNNNN wl olelwl ate ol als! es e The square root of 324 by prime factorization, we get: ty | w v & iB The square root of 324 = ¥324 = 18 |] |es|bo lolsle vfol Sle | - Example 3 Find the square root of 484 by the prime factorization method. 2 | 484 The square root of 484 = V484 = 22 Find the square root of each of the following numbers by using prime factorization method: 1.81 2.100 3.729. 4.169. 5.225. 6.441. 7.324 “Division method: To solve the square root by division method: Step 1: Divide the given number by divisor by identifying the suitable integer. —_, Step 2: Multiply the divisor and integer (quotient) to get the number to be subtracted from the dividend. Step 3: Subtract the number from the dividend. Step 4: Bring down the remainder and another digit (if any) from the dividend. Square root of 529 = V529 = 23 Find the square root 484 by using the division method. Square root of 484 = V484 = 22 2 a 4 0 2 84 84 0 Find the square root 1024 by using the division method Sa 32 The square root of 1024 = 3 | 1024 -9 V1024 = 32 ae 62 | 124 -124 aes ° Find the square root using the long division method: 1.V144 6.V81 2.V361 7.V324 3.¥8100 8. ¥1444 4.V4489 9. ¥900 5.V9216 10. ¥2025 1.3 CUBE OF ANUMBER Cube of anumber is anumber multiplied by itself three times. Thus, 3 is the power of x ¥? — cu be and is read as “x cubed”. 1. 1%= 1xixt=1 2. 23= 2x2x2 =8 3. 4° = 4x4x4 = 64 4. 69 = 6x6x6 = 216 Thus 1, 8, 64, 216, etc. are perfect cubes. Calculate the cubes 2 3 5 wee 1,.2x2x2=2%=8. (23is read as two cubic). 2, 39 = (3x 3x 3) = 27. (3?is read as 3 27. 3.59=5x5x5 = 125. Thus, cube of 5 is 125. Find the difference between 9° and 8° P=9xXIXI=729 8°=8x8x8=512 Therefore 9° - 8° = 729 - 512 = 217. elle) (es The diagram shows a cube of side 3 cm V=LxLxL=L3. V=39=3x3x3= V =27 cm? Pika Calculate the cube of the following numbers : a5 d.6 g. 7 bg e 8 h. 13 c. 10 f.12 i, 14 2. Calculate the difference: a) 15*9-12? b) 11°-10° cc) 173-153 d) 119-93 3. Find the volume of a cube box of side length 17 cm. 1.4 CUBE ROOTS Cube root is an inverse operation of the cube of a number. To find the cube root of a number, the number should be expressed in terms of its prime factors. The factors should be numbers repeated three times or in powers of 3 numbers. The radical sign isused as a cube root symbol for any number with a small 3 written on the top left of the sign. Look at the diagram below Cubes and Cube Roots Cube ——— 4 64 Cube root Find the cube root of the following numbers a) 64 b) 27 c) 721 a. *M64= 3V4x4x4 = 349-4 b. 27 = 3x3x3 = 3° 327 = 3 c. 271=9x9x9 =9? [ 3271 = 9 Find the cubic root of 216 using prime factorization. 3¥216= *V(29x3? )= 2 x3=6 Find the cube roots of the following numbers 3V125 3343 3V27 3¥512 3¥1000 3V¥1331 1. A cube water tank has a volume of 3375 cm®. Find the length of the side of the tank. 2. The volume of a cube room is 2744 m*. What are the measurements of the room? 1.5 BINARY NUMBERS The total number of symbols used in a particular number system is called base or radix and each symbol is called a digit. The decimal number system has base 10 and the digits are O, 1, 2, 3,4, 5, 5,6, 7,8, 9: To binary number system have base 2 and the binary digits are Oand 1. Tips to Remember « Abinary number consists of two numbers Os and 1s. © Os has no value as usual « 1s have values according to the place value of the 1. From right to left, first 1 = 1 Second 1 = 2, third 1 = 4, fourth 1 = 8 and so vaues 1 in binary system is powers of 2 : for example : the value of 1 from right to left is: Value of 8or2? 4or 22 2or 2? |10r 2° each digit | ‘Number 1 0 ja : ‘4 So, the value of 1011=8+0+2+1=11. What is the value of the bolded 1 in the following binary numbers? a. 100010 d) 1001 b. 11000 e) 1010001 c. 1101 1.6 CONVERSION OF BINARY AND DECIMAL NUMBERS To convert binary into decimal is very simple and can be done as shown below: Say we want to convert 10011101 into a decimal value, we can use a formula table like that below To convert, you simply take a value from the toprowwherever there is a 1 below and then add the values together. If you have 0 then it will not have a value but if you have 1 take the value above it. The values have these positions and cannot change. Changes binary to decimal Example 1 Change into decimal 1001 ao 1001 = 1x 29+0x 2?+0x2!+1x 2° =8+0+0+1=9 Therefore, 1001 is 9 as decimal system. Example 2 Find the decimal value of 1110. 1110 =1x 29 + 1x2? +1x2?+0x2° =8+44+2+0 Therefore, (1110), is (14),, Find the decimal value of 11001. rare 11001 =1x 24 + 1x29 +Ox2? +0x2!?+1x2° = 16+8+0+0+1 = 25 How to convert decimal to binary Conversion steps: 1. Divide the number by 2. 2. Get the integer quotient for the next iteration. 3. Get the remainder for the binary digit. 4. Repeat the steps until the quotient is equal to 0. 5. Write remainders starting from the last and ending with the first. Seno Convert 13 to binary Quotient | Remainder | Division by2 6 13 is 1101 as binary. Convert 30 to a binary number form 2 7 2 oO 2 ? 1 2 1 1 (30),, = (1110), Example 6 Express 174 to binary ey Quotient | Remainder 17412 87 0) 87/2 43 1 32,21 4 | 10 1 0) 4) | 0) 1 } A. Express the following binary numbers to decimal system 1. 11001011. 6. 00000100. 2. 00110101. 7.00010010. 3. 10000011. 8.00111111. 4. 10001111. 9. 10101010. 5. 11100011. 10. 01010101. B. Express the following decimal system to binary system. 1. 213. 6. 143. rane: 7. 6. 8. 1. mmr a7 10. 252. 1.9 OTHER RULES OF INDICES Here we will be learning three laws of index and their rules power zero rule: Rule a°=1 Any number with power zero is equal to 1. =a™"= g? Therefore a°=1 Example 1 Any number with a negative power number is equal to the reciprocal of the number with positive power. Simplify = x4 Power raised to another power (am) = a™® or am When a power is raised by another power multiply the powers. Simplify the following a. (x*)3 = x4%3 = x12, . (2y3)3 = 23y3%3 = gy? . (Bxty?)? = 33x4x3y2x3 — 27x22y6_ . (26)? + (24)? = 212 . 28 = 212-8 24, (x°)? = x0%3 = x9 4, Simplify the following 1) (v7)? 6) (y*)* 2) (b°)° 7) x7? 3) (a* x a5)? 8) (2x3y-3)2 4) (2a*)? 9) (x7)* 5) (x?) 10) 2-3 1.10 LOGARITHMS logarithm, is the exponent or power to which a base must be raised to yield a given number. We have learnt that when a number is written in the form . ais the base while b is the index. This is index notation of a number. There is also logarithmic notation of a number such as and read as; the logarithm of 25 to base 5 is 2. index notation logarithm notation 57225 log, 25=2 In general, If then Common Logarithms: Base 10: The logarithm of anumber to the base 10 is known as common logarithm. Sometimes a logarithm is written without a base, like, log,,? is written as Log,. The base 10 in common logarithm is usually omitted. (10)°=1 Therefore, —_ log,, 1=0 or log1=0. (10)*=10 Therefore, log,, 10=1 or log10=1 (10)?=100 Therefore, log,, 100=2 or log100=2. (10)=1,000 Therefore, log,, 1000=3 or log100= 3. Express the following as logarithm notation: 1.29=8 2.37=9 3.5%=125 1.23=8 — log, 8=3 2.37=9 — log,9=2 3.59=125 — log,125=3 Express the following as index notation: a. log, 512 =9 b. log, 36 =2 c. log, 81=4 re la a. log, 512=9 — 2°=512 b.log, 36=2 -— 67=36 c.log, 81=4 —- 3*=81 Evaluate d. log, 32 e. log, 81 f. log, 625 a. Let Log,32 = x First, write as index notation 2*= 32 (Make same base with both sides 2=28 x=5 log, 32=5 — ——— 1. Express the following in logarithmic notation a) 5* = 625. d) 53 = 125. b) 10? = 1,000. e) 8? = 64. c) 117=121. f) 367 =6. 2. Express the following in index notation a) logyo 1000 = 3 d) logyy = 2 b) log; 10 =1 e) logi4 = —2 2 c) log, 27 =3 f)log.64 = 6 3. Evaluate a) log19100 d) log, 128 b) log. 16 e) log; 10000 c) logy)1000 f) log; 27 1.11 SCIENTIFIC NOTATION Scientific Notation (also called Standard Form ) is a special way of writing numbers. Numbers such as 10, 1000, 10 000, 1 000 000 can be expressed as powers of 10. Example 1 b. 1,000 = 10°. c. 100 = 10%. d. 1,000,000 = 10°. e. 10,000 = 10*. Any number can be written in scientific notation as a x 10" ais anumber between 1 and 10 where 1 < a< 10 nis an integer. These numbers are written in a standard form. a. 500 000 5 x 10° b. 58,753 5.8753 x 10%. c. 8,900 8.9 x 10° When the number is 10 or greater, the decimal point has to be moved to the left, and the power of 10 is positive SIE Write the following numbers in standard form. a. 82,000 8.2x104 b. 1264 1.26410? c. 34 3.4x10 d. 10,000,000= 1x107 e. 100 = 1x10? Ae Write the following numbers in ordinary form. a. 8.2x10*= 8.2x10x 10x 10x10 = 8.2x10000 = 82,000. b. 7.76x10°= 7.76x10x 10x10 = 7760. c. 6x105 = 6x10x10x10x 10x10 = 600,000. d. 10° = 10x10x10 = 1,000. e. 107 = 10x10x 10x 10x10x10x10 = 10,000,000. 1. Write these numbers in standard form a. 80,000 h) 100 000 b. 300,000 i) 11000 000 - 630,000 j) 358.74 |. 70 k) 18,000,000 - 9,000,000 |) 856,086 f. 590 m) 12,763 g. 3600 n) 230,000,000 Write these numbers in ordinary form. a. 5x10? h) 7.6 x 10%. b. 9x 10° i) 1.1 10. c. 3.7 x 10% j) 3x 10°. d. 6.2 x 107 k) 2x 102, e. 4.5534 x 10%. !) 8.8 x 10® f. 3.3x 10. m) 4.3 x 10°. g. 7x 10°. n) 6x 104. Decimals in Scientific Notation (Standard Form) Decimal numbers less than 1 can be expressed as powers of 10. Express the following decimals in standard form a. 0.0002. b. 0.00001. c. 0.000005. a) 0.0002 = — 1 =— b) 0.00001 = 255 = ios c) 0.000005.=5 x 107°. d) 0.3876 = 3.876 x 107+. ae ae Write the following as ordinary form: - 10% = 0.001. b. 10%= 0.000001. c. 4x10%= 0.0004, d. 3.654x10% =0,003654, e. 7.3x107 =0.00000073. When the number is smaller than 1, the decimal point has to be moved to the right, and the power of 10 is negative. 1. Express these decimal numbers in standard form. g) 0.1 h) 0.0000001 i) 0.05 j) 0.1435 k) 0.00009. 1) 0.0000787. a) 0.003 b) 0.4456 c) 0.0051 d) 0.00051 e) 0.0083 f) 0.00001 Write the following numbers as ordinary form. h) 3x10. i) 8.2x 10°. j) 1.1.x 107. d. 8 x10°. k) 3.3x 10°. e. 4.2x 10%, 1) 6x10 f. 2.3x 10%. m) 1.3 x10. a.5x107, b. 3.2x 10°. c. 6.3 x 10, g. 2.4x 10%. n) 2.1.x 10%. 1.12 SETS A set is a collection of objects, numbers, symbols, etc. The different objects, numbers, symbols and so on in the set are called the elements or members of the set. Capital letters are - commonly used to name sets. Example 1, set C = {1, 2, 3, 4, 5, 6,...}. Similarly, B = {a, b, c,d, e, ......} Notation Description Listing elements | N Set of natural|N={1,2,3,4,5,6,7,8,9 numbers | Zz The set of integers. R- Set of real numbers 1. Write the elements of each of the following sets. a. The set of continents of the world. b. The set of whole numbers less than seven. c. The set of odd numbers between 6 and 16. d. The set of the letters of the word SOMALIA

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