INDEX Signs and Symbols_3 Alphabet Numbers---4 Multiplication Table-4 To Multiply-4, 5, 6, 7 To Square numbers-7 Special Rule-8 To Measure Round Timber-8 To Measure a Round Log-9 Paying Lumber-9 Buying by the Ton-9 To Measure -foists-10 Buying and Selling by the 100-10 Measuring Land-10 Measuring Girfani in the Bin-11 Measuring CoWon the Cob-11 To change lord tons to short tons 11 To find No tons of iron in a R. R. 12 Selling by doz. to gain 20 per ct_-12 •To Measure Stone-12 Arithmetical Progression-13 Geometrical Progression-13 To find the Area of a Circle-14 Allegation- 14 Interest---14 15 16 Equation ofPaymet^s-.— 16 Per centage and Annual Interest-17 fo square No. & square root—*.18 Cube Root_-—19, 20 To Cube Numbers-20 Time Rule_ 21, 22 Common Multiple---—22 Common Divisor-22 Greatest Common Divisor_23 Fractions-23, 24 Decimals_24, 25 Weights and Measures-25, 26. 27,28 Aliquot parts of 100 and 1000_28 Geometry and Trigonometry_29 Terms and Axioms_29 Geometrical Formulas_30, 31,32 The Cylinder___3: The Frustrum of a Pyramid_33 The Spheie-33 The Sphereoid-34 The round Iron Ring_ The Circular Ring_ The Circle and Triangle. The Circle and Square_36 The No. degrees in the Arc_36 To find the length of the Arc_3* The Quadrant_3 The Grave Yard and Oblique Triangle_38 The Equilateral Triangle_—39 How the width of a st. was found 4f To measure a Tree_41 The hardest of all FA)blems_41 To divide any Triangle_42, 43 The inverted Cone-_44, 4£ The Wine Glass-45 A Rectangle within a Rectangle-46 The prismatic Solid_4f Test Problems___46, 47, 4.' ■ggsss, PROF. S. KING'S EQatlieinatical Instructor, -AND- *■ Ss. AGGOUNTANT'S GUIDE/ Embracing Geometry and Trigonometry. PUBLISHED BY PROF. R. S. KING, AUTHOR, PREFACE. A successful attempt has been made by the Author of this book to so simplify the science of numbers so that the most common mind can understand it. This new and long felt need for mathematical work is in¬ tended to 5heet the wants of every person seeking a thorough knowledge of simple and practical mathe¬ matics that can ->e acquired without spending half a lifetime in the school-room. The rules in this system are more simple and easier understood, than in any other mathematical work ever published and is the result of over ten years hard study. This work is published by none but the Author and is above price to the most learned scholar. Respectfully, R. S. KING. Prof. King’s Mathematical Instructor. Science is knowledge reduced to a system. Art is knowledge applied to practical purposes. A unit is anything regarded simply as one; and numbers are repetitions of a unit. Quantity vis anything which admits of being measured; a line is a quantity, and we express its measure in saying it is so many feet or inches long. Every number expresses thfc measure of itself in units, and numbers are us^d. to exp;fess all other quantities. Number has two terms—the unit and the unity terms; the unit term represents t$jinumerator and Jhe unity term the denominatpr. ’ Arithmetic is the science of number, and when practically .applied, the art of calculation. Its chief signs and symbols are as follows : X + is plus, or the symbol of Addition, and when placed between two numbers they are to be added together. The sign — minus, or Subtraction, when placed between two numbers denotes one is to be taken from the other. The sign X (into) between two numbers denotes they are to be multiplied together. ^ The sign -f- when placed between two numbers denotes that one is to be divided by the other. The sign = equality, placed between two numbers denotes that they are equal to each othei. The sign (.) is the decimal point and denotes the number on the left is the whole number and the 4 Prof. King’s Mathematical Instructor. right is tenths, hundredths, thousandths and so on and is called a decimal fraction. 1234567890 iiiiiiiiii number. is the proper alphabet of To read numbers point them off in periods of three figures each, commencing at the decimal point from right to left and left to right, subtract two from the number of periods; the remainder is the name of the last period. 45G ia 674 | | | 111 C'J t n 438 1567 432 376 849 The first lin this table is the numerical alpha¬ bet, the second is twice the first, the third is three times the first, and so on. 1 2 3 4 5 6 7 8 9 0 1 1 1 1 1 1 1 1 1 2 4 6 8 10 12 14 16 18 0 3 6 9 12, 15 18 21 24 27 0 4 8 12 16 20 24 28 32 36 0 5 10 15 20 25 30 35 40 45 0 6 12 18 24 30 36 42 48 54 0 7 14 21 28 35 42 49 56 63 0 8 16 24 32 40 48 56 64 72 0 9 18 27 36 45 54 63 72 81 0 The product of any two numbers is written in one line by the following Rule: Units, tens and hundreds of the Multiplicand and Multiplier is the first, second and third term. ft- Pbof. King’s Mathematical Instructor. 5 Multiply the first term by the first, then add the products of the first by second and second by first, then first by third, second by second and third by first, the second by third and third by second, then third by third. Solution —Multiply. 321 by 456. 231 321 456 - K* 146376 First term X by first =6X1—6. -f second X first is 6X2+5X1 — 1 7 - the answer and the 1 ova* th^ hundreds column. Now firstXthird=6X3+ second X fcecond=5X2+ thirdXhrst=4X i=32+the carrying 1=^3. Write the 3 in the answer and the carrying 3 over the up¬ per number. Now multiply second by third+third X second =£X3+4X2=23+the carrying 3—26. Write the 6 and :arry the two. Now third Xthird= 4X3 + 2 the carry figure=i4. -'Write the whole 14 in the answer. For any two figures by two as 33X22. 33 22 726 FirstXfirst = 3X2=6. Now first X second+sec- ondXfirst 3X2+2X3=12. Write the 2 in the answer and carry the 1 to the product of the second Xsecond=3X2+i=7, which gives the entire ans¬ wer. These rules are very valuable to every accoun¬ tant, and are easier understood than any other rules. Fi*St X second Write the 7 in 6 Prof. King’s Mathematical Instructor. When the multiplicand is all the same the product is found as follows : 4 4 4 4 4 3 2 1 19202524 The first answer figure equals 4X 1=4; the second answer figure equals 1+2X4=12 \ the third answer figure equals 1+2+3X4=24+1 (the carrying fig¬ ure) =2^; the fourth answer figure equals 1+2+3 +4X4=40+2 (carrying figure) =42; the fifth ans¬ wer figure equals 2+3+4X4=36+4 (the carrying figure) =40; the sixlh answer figure equals 4+3X4 =28+4 (the carrying figure) =32; the seventh answer figure equals 4X4+3 (the carrying figure) = 19, which completes *jf.he answer. This rule holds good when the multiplicand is all the same numbers, and the multiplier-anything. 21, Rule .To* two figures by two—1st term 21, by 1st, 1st by 2d and 2d by 1st, and - 2d by 2d. 441 t i The language of the rule for any three 32 t figures by any other three proceeds 321 from the number 321 by 321: thus 103041 Rule —1st term by the 1st; 1st by 2d and 2d by 1st, 1st by 3d and 2d by 2d and 3d by 1st, now 2d by 3d and 3d by 2d, and 3d by 3d. N. B.— Chang¬ ing the figures does not change the terms in any ex¬ ample. Pbof. King’s Mathematical Instructor. 7 The language of the law of any four figures by any other four figures, proceeds from the number 4321 by 4321, thus: 4321 thus reaching more than millions and 4321 bringing every figure representing the -- product in the right place while the 18671041 highest thought in the operation is only twenty-seven. 454321 54321 54321 Rule.— 1 st term by 1 st; 1 st by 2 d and 2 d by 1 st; 1 st by 3 d, 2 d by 2 d and 3 d by 1 st; 1 st by 4 th, 2 d by 3 d, 20co*771 oai 3 d by 2d ar ‘ d 4 th by ist; istby 5 th, 95 77 4 2 d ^ gd by 3d, 4 th by 2d and 5 th by 1 st; now, 5 th by 2 d, 4 ^h by 3 d, 3 d by 4 th and 2 d by 5 th; now, 5 th by 3 d, 4 th by 4 th and 3 d by 5 th; now, 5 th by 4 th and 4 th by 5 th, ar /2 5 th by 5 th, &c, for any number of terihs. 23321 54321 3142 170676582 To square numbers by my compliment and sup¬ plement method: Decrease the number by its complement, multiply it by the base and add the square of the complement; or, increase the number by its supplement, multiply it by the base and add the square of the supplement. Take the most convenient number for the base. Square 99. In this case 100 is the base, 99—1 the complement =98 X 100=9800the square of the complement 1=9801 the answer. Square 18^. Here take 20 for the base. The complement is \% or f, and 18^—1^=17^, and 1 7/4X20, the base, equals 350, and 350-j-f squared ~ 35 I xV answer. This rule applies to any case. * 8 Prof. King’s Mathematical Instructor. Square 22 by the supplement. Here 20 is the base and 22+2 the supplement =24, and 24X20 the base =480-}- the square of the supplement 2=4 gives 484 the answer. A SPECIAL RULE TO MULTIPLY FRACTIONAL NUMBERS, When the whole numbers are equal, and the sum of the fractional parts make one, increase the whole number one and multiply by the whole number and annex the product of the fractional part. SysXS 2 /s. CaUthe 8=9X8=7«+#XJ$=7*f i2£Xi2^. Call the 12=13X^2=156-(-£xJ= 156^4 the answer. * When the sum of che units equals ten, and the tens are equal, increase the tens figure one, and mult ply it by tfa? *ens, and annex the product of the units. 42x4k- 48-1-10=^=58 and 58X42=2016. 8X2 =16 and 5X4=20, ?pd 20 prefixed to 16=2016 the answer. 34 Say 6 tirhes 4= 24 36 and 3 times 4=1-2 1224 1224 answer. RULE TO MEASURE SQUARE AND ROUND TIMBER. Take twelve times twelve for the unity term and the indicated product of the area of one end in inches and the length in feet for the unit term. How many cubic feet in a stick 12X18 inches and 36 feet long? The form of statement in all cases is 18X12=216, the area of one end multiplied by the length of the stick, which is 36 feet, and divided by 144 equals answer. Form of any statement. 2 - i 6X3 6 cu bi c feet. 2X12 Prop. King’s Mathematical Instructor. 9 A RULE TO MEASURE A ROUND LOG. The square of the diameter, minus four inches, multiplied by the length of the log in feet, divided by sixteen, is the number of feet of board measure in the log. How many feet of boards in a log 36 feet long and 24 inches in diameter? Subtract 4 inches front the diameter 24 leaves 20, and 20X20=400, and 400X 36, the length of the log, ^nd divided by 16 equals the number of feet in the log wjiich is 900 feet. Form of statement.. = 900 feet. l6 ^ BUYING AND SELLING LUMBER. Remove the decimal point \>ree places to the left in any number of feet, and multiply by the price of one thousand feet in all examples. What will 859 feet of lumber cost at $12 per thous- and feet ? Remove the point three places to the left in 859=.8S9, and .8s9X$i2=$io.3o8 answer. BUYING AND SELLING BY THE TON. Remove the decimal point three places to the left, and multiply by one-half the price per ton in all ex¬ amples. What will 1799. pounds of hay cost at $10 per ton? 1799 lbs with the point removed equals 1.799, and 10 Prof. King’s Mathematical Instructor. 1.799X5, one half the price per ton, is $8,995 answer. Every case is worked the same. To measure any kind of Sawed Lumber or Joists or pieces, whose thickness is more than one inch, as follows; How many feet of lumber in a piece of timber 2 inches thick, 9 inches wide and 12 feet long ? This is the form of statement in any case : The area of one end in inches multiplied by the length in feet and divided by 1.2 equals the answer. i8y !2 12 = 18 feet a BUYING AND SELLING BY THE HUNDRED. Remove the decimal point two places to the left, and multiply by the price of one hundred in all ex¬ amples. What will 39 cigars cost at $2.00 per hundred? 39. with the point removed equals .39 .39X^2^78 cents the answer. RULE FOR MEASURING LAND. Remove the decimal point two places to the left in the number of square rods, divide by 8 and multi¬ ply by 5 and you have the number of acres. How many acres in a piece of land 80 rods long and 40 rods wide ? 80X40=3200. 3200 with the point removed equals 32.--8=4. X5=20 acres. Pbof. King’s Mathematical Instructor. II RULE FOR MEASURING GRAIN. Eight-tenths of the number of cubic feet, plus four and one half for every one thousand bushels is the number of bushels. sllifiiyilliilllii 25 feet How many bushels of grain in'a bin 25 feet long 5 feet high and 10 feet wide? '?5XSX 10=1250 cu¬ bic feet in the bin. 1J50. with the point removed equals i25.X8=iooo, and iooo-}- 4/4 bushels = 1004^4 bushels the answer. Every case is worked the same. * , * RULE TO MEASURE CORN ON THE COB. Remove the decimal point one place to the left in the number of cubic feet, and divide by two and mul¬ tiply by nine, and you have the number of bushels of shelled corn. How many bushels of shelled corn in a crib 5 feet wide, 10 feet high and 20 feet long? 5X10X20 =1000. With the point removed equals 100. ioo-|-2 =5 oX9 =45 0 bushels of shelled corn, exactly (if good dry corn.) RULE TO CHANGE LONG TONS TO SHORT TONS. Multiply the expression 1.12 by the number of long tons and the result is the number of short tons. UL Of SLL Lift, €2 Prof. King’s Mathematical Instructor. How many short tons in 632 long tons? Simply multiply 632 by 1.12 equals 701.84 tons the answer. RULE TO FIND THE NUMBER OF TONS OF IRON IT TAKES TO BUILD ANY NUMBER OF MILES OF RAILROAD. The number of miles multiplied by eleven-sevenths and this product by the weight of one yard is the answer in any case. How many tons of iron will it take to build 140 miles of railroad, each yard of rail weighing 56 pounds ? Form of any statement: il22llil^ll^=i232o tons. 7 SELLING ARTICLES BOUGHT BY THE DOZEN TO GAIN 20 PER CENT. < Remove the decimal point *>ne place to the left in the price per dozen. Increase or diminish to suit all other per cents. > At what price must y hats that costs $2.25 per dozen foe sold to gain 20 per cent? Simply 82.25 with point removed equals .225. RULE TO MEASURE STONE. Remove the decimal point-two places to the left in the number of cubic feet. Multiply by four and add to the result one hundredth, part and you have the number of perch. How many perch of stone in 480 cubic feet? 480. with the point removed equals 4.80 4 .80X4= part of 19.20=19.392. Peof. King’s Mathematical Instructor 13 ARITHMETICAL PROGRESSION. Arithmetical Progression is a series of quantities which constantly increase, or decrease by a common difference, as i, 3, 5, 7, 9 is a series in which 2 is the common difference. The first and last terms are called the extremes, and the intermediate terms the means. The last term of increasing arithmetical pro¬ gression is equal to the first, plus the product of the common difference, multiplied by the number of terms less one, and in a decreasing progression it is equal to the first term minus the same product. The common difference of the terms in arithmetical progression is equal to the difference of the extremes divided by the number of terms less one. ( THE SUM OF A SERIES OF NUMBERS. Increasing by a common difference is equal to half the sum of the first and last terms multiplied by the number of terms. How many times does a clock strike in 12 hours ? i2-J-i=i3-f-2=6|Xi2=78 times. A Geometrical progression is a series of numbers increasing by a constant multiplier, as 1, 2, 4, 8, 16 and so on. The last term of a geometrical progression is equal to the first term multiplied by that power of the ratio which is expressed by the number of terms less one. The last term of a geometrical progression divided by the first term gives the power of the ratio which is expressed by the number of terms less one. A geometrical mean or a mean proportional be¬ tween two quantities is equal to the square root of the product of those quantities. 14 Prof. King’s Mathematical Instructor. The sum of all the terms of a geometrical pro¬ gression is equal to the difference between the first term and the product of the last term multiplied by the ratio, divided by the difference between the ratio and a unit. THE CIRCUMFERENCE OF ANY CIRCLE IS equal to the diameter, multiplied by 3.1416-]- What is the circumference of a circle 10 feet in diameter? Simply multiply 3.1416 by 10 equals 31.416 feet. 3.1416 is the ratio of the diameter to the circumference of any circle. ONE-FOURTH OF THE DIAMETER OF ANY CIRCLE Multiplied by the circumference, gives the area. What is the area of a circle 10 feet in diameter and 31.416 feet in circumference? 31.416X2!* one-fourth of the diameter, =7^.54 square feet the answer. RULE TO FIND HOW MUCH OF ; AN EQUAL QUANTITY OF SEVERAL ARTICLES AT DIFFERENT PRICES CAN BE BOUGHT FOR A UNIT OF MONEY. The sum of the different costs inverted gives the quantity that can be bought for one unit of money. How much tea, worth 37^ cents, 62J cents and $1.00 per pound, can be bought for $1.00. Solution: 374+62^+1.00=82.00, and $2 inverted equals \ a pound of each kind can bejxmght for 81.00. Instantaneous Interest Rule —Invert the rate, annex ciphers and prefix points. Inverting the rate demonstrates the time it takes $1.00 to earn one cent, and one hundredth part of any principal is the interest for that time, and the lines pass down through all sums of money represent¬ ing the decimal point. Pbof. King’s Mathematical Instructor. 15 Practical Rule —Take for the unity term of number the time it takes a dollar to earn one cent, and for the numerator the principal multiplied by the whole number of days, and the complete analysis stands before you. What is the interest on $400 for 90 days at 8 % ? This is the proper form of statement in any case, which gives the interest in cents: j 4 Q °X 9 ° —^3 oq answer. .45 When the time is in days then the time it takes one dollar to earn one cent must be in days. When the time is in months, then the time it takes one dollar to earn one cent must be in months and fraction of a month. $ 400X3 months _ $8 op ans> 1.5 months When the time is in years months and days, reduce the time to months or days, whichever is the easiest. The time it takes one dollar to earn one cent is found by dividing 12 months or 360 days by the rate per cent. - p • EXAMPLES IN INTEREST. What is the interest on $5 for 9 days at 8 % ? simply *SX2=:i cent ans. 45 This 45 is }£ of a year. -* What is the interest on $135 for 90 days at S% ? simply ^35 X 9 g 2 45 What is the interest on $327 for 160 days at 9% ? statement $ 33 ? ^ - l - ~ - $13.08 ans. 4 ° This 40 is £ of a year. 16 Prof. King’s Mathematical Instructor. What is the interest on $973 for 180 days at 4 % ? statement .46 a ns. 90 This 90 is of a year. Any rate per cent inverted indicates the part of one year; that it takes $1 to earn one cent as of 360=45 days. ^ of 360=40 days. ^ of 360=90 days, and so on in all cases. RULE FOR EQUATION OF PAYMENTS. Divide the sum of the interest on all the payments from the date they become due to the date the prin¬ cipal is due, by the interest on the principal for one month, at the given rate-, and subtract this quotient from the date the principal becomes due, and you have the equated time of payment, without loss to either party. A owes B $1800, to be paid in 18 months at 6 % per annum, as follows : $600 in 6 months and $600 in 12 months. What .is the equated time of settle¬ ment of the entire amount ? Solution: The interest on $6go for 12 months is $36 The interest on $600 for 6 months is $18 $54 Now the interest on $1800 for one month is $9 and $54^9=6, which mustJbe subtracted from the most remote date of maturity, which gives 12 months, the equated time of payment of the entire amount. All cases under this Rule are worked the same. Prof. King's Mathematical Instructor. 17 PERCENTAGE. A man buys a lot of goods for $325. By paying cash he gets them 20% off. 20% off leaves 80% to pay or y 8 ^, and by removing the point one place to the left in $325=32.5> and by multiplying by 8=^260 the amount to pay. In every case this Rule applies the same. The present worth of any sum is found by multi¬ plying the amount by 100 and dividing that product by the amount of $1.00 for the given time and rate. What is the present worth of $1650 without interest for eight months, money being worth 6 % ? $1650 X|^|=$is 86.53, present worth. This $1.04 is the amount of $i.co for 8 months at 6%. ANNUAL INTEREST. One of any number of equal annual installments is found by dividing the product of the interest for one year, and the amount of one dollar for the given time and rate at compound inter^t by the compound interest of one dollar for the given time and rate. A buys a farm for $6000, to be paid in five years, annual interest at 10%. What is one of the equal annual installments that will discharge the debt, principal and interest ? Every case of this kind is worked as follows : The compound interest on one dollar for five years at 10 % is .61051, and the amount of one dollar for five years at compound interest is $1.61051. The form of any statement is as follows: The interest on $6,000 at 10% for one year is $600 $600X1-6105.^ 8 g +AnSt .6lOSl IS Prof. King’s Mathematical Instructor. cs RULE TO FIND THE SQUARE OF ANY NUMBER. Is the square of the first, plus twice the first, mul¬ tiplied by the second plus the square of the second. 32 2 . Units and tens are the first and second terms. The square of the first is 2X2 or 4, which is th 6 unit figure in the answer. Twice the first, 2=4 multiplied by the second, 3=12, 2 being the second figure in the answer and 1 the carrying figure; the square of the second is 3X3 ::: =9 +i, the carrying figure equals 1024 the answer. RULE FOR SQUARE ROOT. Point the number off in periods of two figures each and find the greatest square in the first left-hand period. Subtract it from this period. To the re¬ mainder bring down the next period for a new divi¬ dend. The sum of two sides of the square (on which you are making the addition) with one cipher affixed is the trial divisor by which you find each root figure. Each root figure is -added to each trial divisor for .each true divisor, etc. What is the squark root of 1089 ? Pointed off in two periods equals 10.89 The greatest square in ten is 9.00 (the root of which is 30) - New dividend 1.89 1.89 The square is 30 on each side, which doubled gives 60, the trial divisor, by which we find the next root iigure to be 3, which, added to 60, equals 63, the complete divisor. Now multiplying 63 by 3 we find there is no remainder. Consequently the square root of 1089 is 33. If there was a remainder for a new dividend, the next trial divisor is found by doubling the 33, which 30 3 Prof. King's Mathematical Instructor. 19 is 66, with one cipher affixed equals 660, the trial divisor, with which you proceed as before until there is no remainder. CUBE ROOT, The surface of three sides of a complete cube is the trial divisors. To find each true divisor add to the trial divisor three times the surface of one side of one parallelopipede and one side of the small cube. To find each trial divisor after the first, add to the last true divisor three times the surface of one side of one parallelopipede and two sides of the small cube, and annex two ciphers. Always point the number off in periods of three figures each, commencing at the right hand and find the greatest cube that does not exceed the first left-hand period. Fig. 1 . Ffg.X The surface of three sides of a complete cube is the trial divisor in all cases and is the dark surface of this cube. The dark surface of this cube la the true divisor in all cases, and the dark surface added to what appears vacant is the second trial divisor by annexing two ciphers.^ 20 Prof. King’s Mathematical Instructor. 64,000 3Mi3 27,125 What is the cube root of 95,443,993 ? The greatest cube in the 95 is 64, 95 , 443,993 4° the cube root of which is 4, 64,000 5 which I call 40 for convenience. Now the cube is 40 feet ^ach way. The surface of the three sides is 4o 2 .X3 equals the first trial divisor, 48 oo+( 5 X 4 °X 3 ) 4 , 3 lS > 99 ^ equals the surface of one side 4,318,993 of each corner piece which equals -- 600+5X5=25, one side of small cube, equals 5425 . First true divisor . 5425 One side of three corner pieces 600 Two sides of small cube 50 Affix 00—second trial divisor 607500 The cube now, is 450 each way. The thickness of the second addition equals 7 feet, and 45 °X 7 . ec l ua l s surface of one side of one parallelopipede; multiplying by 3 equals surface of one side of 3 parallelopipede, is equal to 9450, which added to the second trial divisor and one side of the small cube or 49 equals 616999, the second true divisor, and multiply this divisor by 7 , the last root figure leaves no remainder. If there was a remainder then we would make another trial divisor which is equal to 616,999, second true divisor, +9450, one side of three corner pieces, + 98, two sides of small cube, and affixing 00 gives the trial divisor 62654700, by which you proceed as be¬ fore.' THE CUBE OF ANY NUMBER Is the cube of the first, plus three times the square of the first multiplied by the second, plus three times the first, multiplied by the square of the second plus the cube of the second. Pbof. King’s Mathematical Iksthuctob. 21 33 Z = 3 X> 3 >< 3 = 2 7 > 3 2 z = 9 s < 3 X 3=% 1 +2 the car¬ rying figure; 3X3=9X3 2:: =8i-f8 the carrying figure; 3X3X3+8 the carrying figure equals the answer, 35937 - A BRIEF ANft MENTAL RULE TO FIND THE DAY OF THE WEEK THAT ANY DATE OCCURS FOR 3,000 YEARS. The excess of sevens in the sum of the ratio of the century, year, month and date of the month is the day of the week. Sunday is the first day of the week and Saturday the last. The ratio of the year is the excess of sevens in the last two figures of the year multiplied by one and one fourth as follows: 1848 ; last two figures are 48; 48Xi/^=60) and 60+7=8 and 4 excess, which is the ratio of 48. The ratio of any year is found in the same way. Reject fractions as they occur, for a fraction of a day does not change the day of the week. The ratio of the months are as follows, and must be memorized by the student to become expert in han¬ dling the Rule: 1 2 3 4 5 6 0 September December April July January October May August Febuary March November June In leap year the ratio of January and February is one less (2 and 5), but in any other year remains un¬ changed. What day of the week was the 4th of July 1836 ? 3 6 XiX— 45*r7—6 and 3 over; this 3 is the ratio of 1836. The ratio of July is 2, and the fourth day equals 4; the sum of 3+24-4=9; 9+7=1 and 2 over, which is the second day of the week or Mon¬ day Every example is worked the same. When the century has a ratio it must be added in, but in the 19th century the ratio is a cipher and does not count. 22 Prof. King’s Mathematical Instructor. The ratios of the centuries are as follows, and are found by subtracting any century from 16, increase the remainder by 1 and divide by seven; the remain¬ ing number is the ratio. This rule holds good from the i st to the 16 century, but not in the other centuries above 16, it being the starting place: Ratio HNeC^iOOOHWeO^>f5«CO Century — rtrtcCrtrtctictfrtrtoSrtcGctcflrfctfc'JrtoSrtoJRlrtrttdrtrtolrtcd 333333333333333333333333333333 :r;J*c^a , o , ucra , G , c < ccr l cro , cro , c7 , a , cro‘o , o , o , o‘o‘crcro , c^'o , yUWyUWlHHJWUUtUWWUUlHJJMMIWUCtWUO The next starting place is 17, which equals 4 and every 4th century from 17 in the future equals four. Then 18 equals 2 and every 4th century in the future from 18 equals 2. Then 19 equals o and every 4th century in the future from 19 equals o. Then 20, which is 5, and every 4th century in the future from 20 equals 5. These ratios repeat without end in the future. A composite quantity is one which is the product of two factors, each differing from unity, and a prime quantity is one which is not the product of such factor. Common Multiple. A Common Multiple of two or more numbers is any number that will contain each of them without a remainder. What is the common multiple of 3 , 6 , 7, 4? It is 84. The Least Common Multiple is the least number that will contain each of them without a remainder. Common Division. A common divisor of two or more numbers is any number that will divide each of them without a rem¬ ainder, thus 2 is a common divisor of 16, 20, 24, 28, and so on. Pkof, King’s Mathematical Instructor. 23 The Greatest Common Divisor of two or more numbers is the greatest number that will divide each of them without a remainder. Thus 4 is the greatest common divisor of 16, 20, 24 28 and so on. When two numbers have no common divisor but unity they are said to be prime to each other as 3, 5, ' 7, 9 and so on. Fractions. The Reciprocal of a number is a unit divided by the number. The reciprocal of a fraction is the frac¬ tion inverted. To reduce two or more fractions to a common denominator, multiply all the denominators except its own for a new numerator and multiply all the denominators together for a common denomina¬ tor. ADDITION OF FRACTIONS. If the fractions have not a common denominator, reduce them to a common denominator, then add the numerators and place the sum over the common denominator. 232 40+45+24 109 49 Add --t-{-- =-=-= 1— 3 4 5 60 60 60 Here 3X4X5 or 60 is the common denominator. SUBTRACTION OF FRACTIONS. Reduce the fractions to a common denominator, subtract the numerators and place their difference over the common denominator. Subtract 2 /z from Here the common denominator is 3X4=12, and the difference of the numerators 1, hence ^ is the ans- MULTIPLICATION OF FRACTIONS. Multiply the numerators for a new numerator and the denominators for a new denominator. 24 Prof. King’s Mathematical Instructor. Multiply ^ixH—'A == i an swer. 3X4 is the new denominator, and 2X3 is the new numerator. DIVISION OF FRACTIONS. Invert the terms of the divisor and proceed as in multiplication. Inverting the divisor tells how many times the div¬ isor is contained in one, as ^ inverted equals the number of times ^ is contained in 1. Multiplication of Decimals. Decimals are expressed in tenths, hundredths and thousands, multiply as in whole numbers, and from the right hand of the product point off as many fig¬ ures as there are decimal places in both multiplicand and multiplier. DIVISION OF DECIMALS. When the dividend and divisor contain the same number of decimal places, the quotient will represent how many times the divisor is contained in the div¬ idend. In addition and subtraction of decimals, the decimal point is brought perpendicular down in the sum or difference. When the divisor has more deci¬ mal places than the dividend, annex as many ciphers to the dividend as there axe decimal places in the divisor, thus 63-h.3, now annex one cipher to 63= 630-^3=2 10 answer. When the dividend has more decimal places than the divisor, then point off in the quotient as many figures for decimals as those in the dividend exceed those in the divisor, thus .63-^3= 21, the number of places in the dividend is two and the divisor one, consequently point off one place in 21=2.1 When the dividend and divisor are abstract Prof. King’s Mathematical Instructor. 25 numbers, the quotient will be abstract as .8~.8=i, this one is used without reference to any particular thing or quantity, the expression .8 is contained in .8 the same expression, one time, but .8 of a bushel -f- .8=1, and by pointing the quotient off properly this or ,oi of a bushel. Tenths divided by tenths gives hundredths. Any fraction whatever divided by the same fraction, as * 4 -r} 6 =i, but this i is of a denomination equal to the product of the denominators which is -fa of whatever concrete number we talk about, if we talk about $i it is -fa of $i and so on in any case. TABLE OF WEIGHTS AND MEASURES. 12 inches make 1 foot 3 feet U | yard Si yards it . rod 320 rods a mile 144 sq. in. a sq. foot 9 sq. feet < ( . sq. yard 3 °i sq. yds a . sq. rod ll60 rods a j acre 2150.4 cub in a j bushel 1760 yards “ I mile SURVEYORS' MEASURE. 7.92 inches i link 100 links 1 chain 80 chains 1 mile Dry Measure. 4 gills t pint 2 pints 1 quart 8 quarts 1 peck 4 pecks 1 bushel 26 Prof. King’s Mathematical Instructor. Long Ton Measure. 28 pounds 1 quarter 4 quarters 1 cwt 20 cwt or 2 24olbs 1 ton TROY WEIGHT. 24 grains 1 pwt 20 pwts 1 ounce f 12 ounces 1 pound APOTHECARIES WEIGHT. 20 grains 1 scruple 3 scruples 1 dram 8 drachms 1 ounce 12 ounces 1 pound TROY WEIGHT FOR WEIGHING GOLD, DIAMONDS, ETC. 20 mites 1 grain 20 grains 1 pwt 20 pwts 1 ounce 12 ounces 1 pound LIQUID MEASURE. 1 gallon ale 282 cubic inches 1 gallon wine 25* cubic inches 1 bushel 2150.4 cubic inches. I APOTHECARIES FLUID MEASURE. 60 minims 1 fluid drachm 8 fluid drachms 1 ounce troy 16 ounces troy 1 pint 8 pints 1 gallon i^rof. .King's Mathematical Instructor. 27 LIQUID MEASURE. 4 g'lls 2 pints 4 quarts 31^ gallons 42 gallons 63 gallons 2 hogshead 2 pipes 1 pint 1 quart 1 gallon 1 barrel 1 tierce 1 hogshead 1 pipe or butt 1 ton ANGULAR MEASURE. 60 seconds 1 minute. 60 minutes 1 degree. 30 degrees 1 sign. 90 degrees 1 quadrant. 360 degrees 1 circumference of any circle. 24,85 5 miles the circumference of the earth nearly 0 MEASURE OF TIME. 60 seconds 1 minute. 60 minutes 1 hour. 24 hours 1 day. 7 days 1 week. 52 weeks 1 year. 365 days 1 common year. 366 days 1 leap year. 365X days 1 Julian year. 365 days, 5 hours, 48 minutes, 48 seconds, 3 solar year. 12 months 1 year. 28 days 1 lunar month. 29, 30, or 31 ; 1 calendar month. 1728 cubic inches 1 cubic foot. 27 cubic feet 1 cubic yard. 8 cubic feet, 1 cord foot. 28 Prof. Ring’s Mathematical Instructor. 16 cord feet, or (128) cubic feet, 1 cord wood 24^ cubic feet i perch of stone. To multiply any number by the following numbers: To X by 25 —annex two ciphers and -f- by 4. To X by 2 ) 4 —annex one cipher and 4 - by 4. To X by 125 —annex three ciphers and -f- by 8. To X by 12^4—annex two ciphers and -f- by 8 . To X by i }^—annex one cipher and -f- by 8. To X by 62^ —annex three ciphers and by 16. To X by 6J^—annex two ciphers and 4 - by 16. To X by 31^—annex three ciphers and 4- by 32. To X by 333 / 4 —-annex three ciphers and -7- by 3. To X by 33 / 4 —annex two ciphers and 4- by 3. To X by 3 j 4 —annex one cipher and 4" by 3. To X by 50 —annex two ciphers and 4 “ by 2. To X by 66 ^ 4 —annex three ciphers and 4 “ by 15. To X by 6 ^ 4 —annex two ciphers and -f- by 15. To X by 833^4—annex four ciphers and ~ by 12. To X by 83^4—annex three ciphers and -f- by 12.* To X by 8 j 54 —annex two ciphers and -f- by 12. To X by 166^4—annex three ciphers and -f- by 6 . To X by 16^4-^annex two ciphers and 4- by 6. To X by i^ 4 —annex one cipher and ~ by 6. To X by 37^ —annex two ciphers and X by To X by 87 J —annex two ciphers and X by ^j. To X by ii, as 34X11, add the three and 4 together and set the sum between the 3 and 4=374, the ans¬ wer. Reason means hook to hold on by. Please find the reason in each case. Prop. King’s Mathematical Instructor. 2 $ PROF. R. S. KING’S NEW SYSTEM OF GEOMETRY AND TRIGONOMETRY* Definitions of Terms, An axiom is a self-evident truth. A demonstration is a train of logical arguments brought to a conclusion. A theorem is a truth which becomes evident by means of a demonstration. A problem is a question proposed, which requires a solution. Axioms. • All right angles are equal to each other. From one point to another only one straight line can be drawn. A straight line is the shortest distance between two- points. A triangle is a figure having three sides and three angles. The altitude of a triangle is the perpendicular distance from the side assumed as its base to the vertex of the opposite angle. Rule .—To find the surf ace of any triangle, multiply the base by half the altitude. A right-angle-triangle is a triangle having a right angle. 30 Prof. King’s Mathematical Instructor Lines are parallel when they lie in the same direc¬ tion. A parallelogram is a four-sided figure having its opposite sides parallel. A trepizoid is a four-sided figure, having two of its sides parallel. A polygon is a figure bounded on all sides by straight lines. Similar figures are those which have the same shape, and their corresponding sides proportional. The base of a figure is the side on which it is sup¬ posed to stand. The altitude of a rectangle, a parallelogram or a trepizoid, is the perpendicular distance between its parallel basis. Geometrical Formulas BY WHICH ALL SURFACE AND SOLIDS ARE MEASURED. The circumference of a circle equals the diameter multiplied by 3.1416, the ratio of the circumference to the diameter. The area of a circle equals the square of the radius multiplied by 3.1416. The area of a circle equals one quarter of the diameter multiplied by the circumference. The radius of a circle equals the circumference multiplied by o. 15915 5. The radius of a circle equals the square root of the area multiplied by 0.56419. The diameter of a circle equals the circumference multiplied by 0.31831. The diameter of a circle equals the square root of the area multiplied by 1.12828. The side of an inscribed equilateral triangle equals the diameter of the circle multiplied by 0.86. Prof. King’s Mathematical Instructor 31 The side of an inscribed square equals the diameter multiplied by 0.7071. The circumference of a circle multiplied by 0.282 equals one side of a square of the same area. The side of a square equals the diameter of a circle of the same area multiplied by 0.8862. The area of a triangle equals the base multiplied by one half of its altitude. The area of an eclipse equals the product of both diameter and .7854. The solidity of a sphere equals its surface multi¬ plied by one-sixth of its diameter. The surface equals the product of the diameter and circumference. The surface of a sphere equals the square of the diameter multiplied by 3.1416. The surface equals the square of the circumference multiplied by 0.3183. The solidity of a sphere equals the cube of the diameter multiplied by 0.5236. The diameter of a sphere equals the square root of the surface multiplied by 0.56419. The square root of the surface of a sphere multi¬ plied by 1.772454 equals the circumference. The diameter of a sphere equals the cube root of its solidity multiplied by 1.2407. The circumference of a sphere equals the cube root of its solidity multiplied by 3.8678. jThe side of an inscribed cube equals the radius multiplied by 1.1547. The length of the arc of a sector of a circle multi¬ plied by its radius, equals twice the area of the sector. The area of the segment of a circle equals the area of the sector, minus the area of a triangle whose ver¬ tex is the centre, and whose base equals the chord of the segment. Prof. King’s Mathematical Instructor. 32 The solidity of a spherical zone equals the sum of the squares of the radii of its two ends, and one-third the square of its height multiplied by the height and by 1.5708. The square of the diameter of a cylinder multiplied by its length, and divided by any other required length, the square root of the quotient equals the diameter of the other cylinder of equal contents or capacity. Mensuration of tlie Cylinder. The curved surface of a cylinder is equal to the circumference at the base multiplied by the altitude. To find the solidity of a cylinder multiply the area of the base by the altitude. The convex surface of a cone is equal to the cir¬ cumference at the base multiplied by one-half the slant height. The convex surface of a frustrum of a cone is equal to the slant height multiplied by one-half the sum of the circumference at both ends. The solidity of a cone is equal to the area of the base multiplied by one third the altitude. A cylinder, sphere, and cone resting on the same base and having equal altitudes have their volumes in proportion to one, two and three. One half of the solidity of the cylinder equals the salidity of the sphere. One third of the solidity pf the cylinder is equal to the salidity of the cone in any case. Pbof. King’s Mathematical Instruoto*. S 3 THIS FIGURE Ifi THE FRUSTRUM OF A PYRAMID. How many square feet of inch boards in a wagon tongue that is 2 by 2 inches square at one end and 3 by 3 inches at the other end and 12 feet long? Solution, 22 =4 3 2 =9 2X3=6 r 9 —X4Xi*=6>$ft. RULE. The sum of the areas of both ends added to the square root of the product of the areas of both ends and multiplied by ^ of the height and by 12 in all cases gives the number of square feet. There are 12 square feet in 1 cubic foot. THE SPHERE. V Solution. 5 2 X3=75 75+9=84 84X3X5236 equals 131.9472 RULE FOR ALL EXAMPLES. Three times the square of the radius at the base plus the square of the height, multiplied by the height and .5236 will give the solidity in any case. To find the solidity of a spherical seg- a ment with one base (as a. e. b.) whose height is 3 feet and base 10 feet. 34 Prof. King’s Mathematical Instructor. rule 2nd. To find the surface of a spherical segment as (a, e, b.) Multiply the height of the segment by the circumfer¬ ence of the sphere of which it is a part and you have the surface. THE SPHEROID. To find the solidity of a prolate spher¬ oid. a, b = 6 feet, c, d =10 feet. What is the solidity of a prolate spheroid whose perpendicular diameter is 6 feet and its horizontal diameter is io feet ? SOLUTION. .a, b 2 =36Xc, dor io—360X,5236=188496 ans. If it is of a spheroid then the square of the base multiplied by the height and .5236 will give the solidity. THE ROUND IRON RING. How many cubic inches in a round iron ring 6 inches thick and with its inner diameter 24 inches. Solution. 64-24X9— 2 7 ° 270 X.9 8696 is 266.4792. an. BULB. The sum of the inner diameter and the thickness of the ring multiplied by the square of half the thick- J ness of the ring and by 9.8696 will give the cubical 1 contents of any ring. Prof. King’s Mathematical Instructor. 35 TO FIND THE AREA OF A CIRCULAR RING. Rule — Multiply ^-product will be the the difference of the f/^~^\\ are a of the ring, squares of the inside/ f j—j Solution—Suppose and outside diame-V \_/ /the inside diameter is tersby .7854 and the 5 feet and the out¬ side is 9 feet, then g 2 —5*X-7854=the area of the ring which is 43.9824 sq. in. Rule 2 D. —The product of the sum and difference of the diameters and .785! gives the area between the two circles. MENSURATION OF THE CIRCLE AND TRIANGLE. 1st.—The diameter of any circle multiplied by ,86 will give one side of an equilateral triangle as a, a, a. 2nd.—One side of the triangle divided by .86 will give the diameter of a circumscribed circle in any case. 3rd.—The shortest distance from the center of any equilateral triangle to the center of any of its sides is \ of its altitude, or y of the diameter of a circum¬ scribed circle. 86 Prof. King’s Mathematical Instructor. If the diameter of a circle is 10 feet, what is one side of the greatest square that can be inscribed in the circle. Rule.—T he diam¬ eter multiplied by .7071 equals one side or 7.07i ) or v / i o 2 —2 equals 7.071 one side of the square or the line c. Given the chord and versed sine to find the diam¬ eter of any circle. a, a=chord, b=versed sine. Rule in all cases is: The sum of the squares of half the chord and versed sine divided by the versed sine gives the diameter. Suppose a, a to be 12 feet and, b, to be 3 feet, then ^ of 12=6 and 6 2 —36 and 36-4-9=45 and 454-3—15 feet the diameter of the circle. To find the length of the arc when the number of degrees in the arc and the diameter of the circle are given. If the diameter of a circle is 10 feet and the num¬ ber of degrees in the arc is 50, what is the length of the arc in feet ? RULE FOR ALL EXAMPLES Multiply the number of degrees in the arc by the diameter of the circle and by .008- 727 and the pro- d 50° duct will be the ^ length of the arc ^ in feet. SOLUTION. a, d = 50° 5g°XioX-oo 8727=4.3635 feet the length of the arc. a, d is the arc and b f b is the diameter, 10 feet Prof. King’s Mathematical Instructor. 37 To find the length of the arc of a circle when the chord and versed sine are given. a,a, is the chord. ^d,is °hord of c, is the versed sine. b, b, is the arc. half the arc, SOLUTION. (7.2i 11 X&)—(i2)=i 5.229 the length of the arc. 3 RULE FOR ALL EXAMPLES. From 8 times the chord of half the arc subtract the chord of the whole arc and divide the remainder by 3 and the quotient will be the length of the arc. TO FIND THE GREATEST CIRCLE THAT CAN BE IN* SCRIBED IN A QUADRANT. What is the diameter of the greatest circle that can be inscribed in a quadrant, or a right triangle whose base and perpendicular are equal ? b Solution. v /82_j_g 2 = I 1 -3 1 37 or the line a, a, and 11 - 3 1 37"r* 2 =5.6568 or the line b, b, and b, b, is equal to c, c, now 8 ft.—5.6568=0, e, or 2.3432 and 38 Prof. King's Mathematical Instructor. is the radius of the circle, now by multiplying the radius by 2 we have the diameter which is 4.6864 feet the ans. c, c,=b, b, and e, e, are equal. If the circle sought in this case was supposed to be in a quadrant, the calculation would be made pre¬ cisely the same. If in fencing a grave yard it was found that 200 feet square would take in all the graves, but one. which ex¬ tends over the line so that it was necessary to make a curve in the fence 2 feet from the straight line at the center marked (a.), if the fence was built on this curve until both ends met, what would be the diame¬ ter of the circle enclosed ? TO FIND THE AREA OF AN OBLIQUE TRIANGLE. Add the three sides together and divide by two, from this half sum subtract each side separately; multiply together the three remainders and the half sum and extract the square root of the product, the root found will be the area, then by dividing the area by half the length of the longest side will give the altitude in any case. PjEior King’s Mathematical Instbuotob* 89 Problem. An equilateral triangle 10 ft. on each side has a 15 foot pole standing in its center, what is the length of a hipothenuse that will reach from the top of the pole to the center of any one of the sides, or what is the length of the line, a ? Solution.—The diameter of the circle is equal to 10 feet -f-.86=11.627 9 and 11.6279-^ -4—2,90695 which is the line b, now v' 2.90695 2 +15 2 = *he line (a) which is 15.2788 ans. The altitude of an equilateral triangle is 4 rods. What are its sides a, a, a ? a Solution.—4 rods multiplied by f and .86 gives one side in this case which is 4.58-1 and by multiply¬ ing by 3 gives the three sides which is 13.76 rods ans. 40 Pbof. King’s Mathematical Instructor. RULE FOR ALL EXAMPLES. The altitude of any equilateral triangle multiplied by | and .86 gives one side in the same denomina¬ tion as the altitude. How the width of a street was found. If two ladders, one 90 feet and the other 80 feet long be placed in the middle of the street and lean¬ ing against a building it is found that the perpendic¬ ular distance between the upper ends is 20 feet, what is the width of the entire street ? go 2 =8ioo and 202=400 8100—400=7 7 00 802=6400 and 20X2=40 7700—6400=1300 1300-^-40=32^ the line a, b. ^8o 2 _32^2=73.101 which is b, c, and 73.101X2 =146.202 feet, the width of the street b, c. d. The number of revolutions a wheel makes in going one mile is equal to the number of inches the wheel is in circumference; what is the circumference of the wheel? ^63360=251.69+ Ans. Prof. King’s Mathematical Instructor. 41 TO MEASURE A TREE. Solution.- (The square of the base _ divided by the 1 length of the tree © added to the length of the tree)divided by 2 gives the part that breaks off in all cases. In this case it is .151^- feet THE HARDEST OF ALL PROBLEMS Is to divide a cone or pyramid perpendicular to its altitude so that the solidity of each part shall be equal to each other. * RULE FOR ALL EXAMPLES. Multiply the altitude by 2, and cube the product; then divide this product by 2, and extract the cube root of the quotient, and divide the root found by 2, and the last quotient will be the distance from the small end to divide the cone or pyramid so that the solidity of each part will be equal to each other. This rule holds true in all cases. A tree 300 feet high is broken off in a storm, the butt resting on the stump and the top on the ground 30 feet from the root of the tree. What is the length of the part broken off? 30 feet 42 Prof. King’s Mathematical Instructor. TO DIVIDE ANY TRIANGLE. What distance from each end must a plank be sawed in two, that is io feet long and 8 inches wide at one end, and runs to a point at the other end, so that the area of each part shall be equal to each other, RULE FOR ALL EXAMPLES. First multiply the altitude by 2; then square the product and divide this square by 2; extract the square root of the quotient and divide this root by 2, and the last quotient will be the distance from the small end to measure to get half the area. REASON OF THE RULE, d a, b=io feet. 10X 2—20 or c, d, c, d 2 =40o and 400-7-2=200 c a To divide a piece of land that does not run to a point. First find its altitude if it did go to a point; then multiply this altitude by 2 and square the pro¬ duct ; and from this square subtract the square of 2 times the imaginary altitude; then divide the remain¬ der by 2, and to this half add the square of 2 times the imaginary altitude; extract the square root of this sum; then divide the root found by 2; and trom this quotient subtract the imaginary altitude and the remainder will be the distance to measure down the altitude from the narrow end of the lot to get half its area. This rule will give the correct answer in all cases and is of great importance to surveyors in dividing triangular lots, etc. y/ 2oo= =I 4‘i4 2I 3 14.14213=0, f. 14.14213-^2= 7.071065 answer. Peof. King’s Mathematical Instructor. 43 What is meant by the imaginary altitude in the explanation, is the difference between the altitude if it did run to a point and the altitude of the lot. To find the altitude of any right triangle that does not go to a point (if it did go to a point) first X the part of the altitude given by the greatest base, and divide the product by the difference of the two bases, and the quotient will be the entire altitude if it did run to a point. In mowing one swarth $}£ feet wide around a square field it was found that one-half the surface was mowed. What was the dimensions of the field ? s All questions of this kind are easily solved by forming a proportion between the similar lines in the ng. on page 42 and these given in the question. In the fag. on page 42, the line c, d=20, and the line a, v is 2.928935, and in this fig. the line, r, s is 2 %, hence from these three quantities we can find the width of the field as follows : c, d : a, v : : (’) : r, s —width of the field, 56.344 ft, or (20 X^)t ( io —v 50) =56.344 ft, the width of the field. All problems of this kind are solved in the same manner. If a cone 4 inches in diameter at its mouth and its altitude 10 inches is placed with its mouth down, it is found that the altitude of a quantity of liquid in it is one inch (as in Fig. 1); then suppose the cone is inverted (as in Fig. 2) what will be the altitude of the liquid ? 44 Pbof. Kikg’s Mathematical Instbuotob. SOLUTION. jThe capacity of the entire cone is 41.888 cubic inches, the diameter at (a) is 3.6 inches, and is found by forming a proportion between the altitude and the radius at the base of the entire cone, and the altitude of the vacant space in (Fig. r) as follows : 2 | 10 || - (’ ) | 9=1.8 inches; 1.8 inches X 2 = 3.6 inches the diameter at a. Now the solidity of the entire cone 41.888 minus the solidity of the blank space in (Fig. i) 30.536352 will give 11.351648 cubic'inches oi liquid m (Fig 1.) From now on I apply the rule for dividing a cone into two equal parts on page 41 of this book, which is expressed in this statement: / H351S48 ., 3^ \41.888000 X 20 ) 1 by 2 gives 6.475 inches the altitude of the liquid in (Fig. 2.) The expression f. M - 35 i 643 ) i s the ratio of ^41.888000 ' Tugf. King’s Mathematical Instructor. 45 the solidity of the liquid to the capacity of the entire cone. This expression (20 3 ) means double the alti¬ tude of the entire cone cubed. The whole expres¬ sion means that the answer sought is equal to one- half the cube root (of the ratio of the solidity of the liquid to the solidity of the entire cone multiplied by Rouble the altitude of the entire cone cubed) which gives 6.475 inches the answer. The diameter at c, in (Fig. 2) is 2.59 inches and is found by the following proportion 2 | 10 || (’) | 6.475. This 2, is £ the diameter at the mouth of the cone. This io, is the altitude of the entire cone. This 6.475 * s l ^ e altitude of the liquid in (Pig. 1.^ THE WINE GLASS PROBLEM. The perpendicular depth of a wine glass is 6 in. and the diameter at its mouth is 4 in. Suppose a billiard ball 2 in. in diameter is dropped into the glass, how far will the center of the ball be from the bottom of the glass ? SOLUTION. 6 in.is the altitude | of the glass, z—half the diam¬ eter of the mouth of the glass. * | 6 || x | ( ) or 3 the distance from d, to the bottom of the glass. Now \/$*' 7 *= 3162 the distance from the center of the ball to the bottom of the glass. 46 ^Pbof. King’s Mathematical Instr ctor. TO FIND THE SOLIDITY OF A PRISMATIC SOLID. How many cubic inches in a piece of timber 4 in. wide at each end and no thicknes and 12 in. long ? 4 in TEST PROBLEMS. What will $10,000 amount to in 50 years, annual interest 10% ? SOLUTION. The interest on $10,000 for one year is $1,000 and for 50 years is $50,000; this $1,000 draws interest 49 . 48, 47 . 46, 45, 44, 43 , 42, 4 i, 4°, 39 , 38, 37, 36,. 35, 34, 33, 3 2 > 3 1 , 3°, z 9, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 years and is 1225 years, and $100 ^the interest on $1,000 tor one year) multiplied by 1225 equals $122,500, and $i22,5oo+$5°)0°o+$i of six what is \ of 20? Ans. 3^. What is the diameter of a circular field that cor tains 785,400 sq. rds ? Ans. 10 ids. What is the diameter of a sphere that contair 1 oco cubic inches? Ans. 12,407. What is the interest on $6 for 17 months at 6% ? Solution. —$6X17=51 cents answer. This 2 is ) of 12 months. 2