QUESTIONS IN MENSURATION. 1. THE largest of the Egyptian pyramids is square at the base, and measures 693 feet on a side: how much ground does it cover? 696 ×393 272.25 1764 1764 poles, and =11 acres and 4 poles, Ans. 2. What difference is there between a floor 20 feet square, and two others, each 10 feet square? 20×20-10×10+10×10=200 feet, Ans. 3. There is a square of 2500 yards in area what is each side of the square, and the breadth of a walk along one side and one end, which may take up just one half of the square? 2500 √2500=50 yards, each side. ✔ 35.35, and 50-35.35 2 14.65 yards, breadth of the walk, Ans. 4. A pine plank is 16 feet and 5 inches long, and I would have just a square yard slit off: at what distance from the edge must the line be drawn? A square yard=1296 inches, and 16 feet 5 inches 197 inches. 5. If the area of a triangle be 900 yards, and the perpendicular 40 yards required the length of the base? 6. If the three sides of a plane triangle be 24, 16, and 12 perches required its area? 24+16+12 12 2 4.6+ =26; 26-24-2; 26-16-10; 26-12-14, and √26×14×10×2=85.32 perches,=area. Again, as 24: 16+12 :: 16-12: 4·6+, the difference of the segments of the base; then, 9.6, and 12×12-9-6×96-7.11 the perpendicular on the longest side; whence 24÷÷2×7·11=85-32, area as above. 7. Required the area of a circular garden, whose diameter is 12 rods? 12×12X 7854-113 0976 poles, Ans. 8. The wheel of a perambulator turns just once and a half in a rod what is its diameter ? 16.5×3 11 circumference, and 11X 31831-31 feet, Ans. 9. Agreed for a platform to the curb of a round well, at 74d. per square foot the inward part, round the mouth of the well, is 36 inches diameter, and the breadth of the platform was to be 151 inches: what will it come to? 36+15.5×2 67the greatest diam.; 67×67×·7854-36x36x7854 2507.8722 144 17.4157 square feet, at 74d. per foot,=10s. 10, d. [Aus. 10. Required the difference between the area of a circle, whose radius (or semidiameter) is 50 yards, and its greatest inscribed square? 50x2100 the diameter, and 100x100x 7854-7854 the area of the circle; then, 50×50×2=5000 the area of the greatest inscribed square, and 7854-5000=2854, Ans. 11. There is a section of a tree 25 inches over; I demand the difference of the areas of the inscribed and circumscribed squares, and how far they differ from the area of the section? 25×25—12 5×12·5×2=312·5 the difference of the squares. 25×25 -25×25×·7854=134.125 the circumscribed square, more than the section, and 25X25X 7854-12·5×12-5×2=178-375 inscribed square, less than the area of the section. 12. Four men bought a grindstone of 60 inches diameter : how much of its diameter must each grind off, to have an equal share of the stone, if one first grind his share, and then another, till the stone is ground away, making no allowance for the eye? RULE. Divide the square of the diameter by the number of men, subtract the quotient from the square, and extract the square root of the remainder, which is the length of the diameter after the first man has ground his share; this work being repeated by subtracting the same quotient from the remainder, for every man, to the last; extract the square root of the remainders, and subtract those roots from the diameters, one after another; the several remainders will be the answers. 13. If a cubick foot of iron were hammered, or drawn, into a square bar, an inch about, that is, of an inch square required its length, supposing there is no waste of metal ? 12×12×12 : 25X25X46912 inches, 576 feet, Ans. 14. Required the axis of a globe, whose solidity may be just equal to the area of its surface? -7854×4 ⚫5236 6 inches, Ans. 15. A joist is 7 inches wide, and 24 thick; but I want one just twice as large, which shall be 33 inches thick: what will be the breadth ? 7.5x2.25×2 16. I have a square stick of timber 18 inches by 14; but one of a third part of the timber in it, provided it be 8 inches deep, will serve how wide will it be? 18X14 3 8-10 inches, Ans. 17. A had a beam of oak timber, 18 inches square throughout, and 25 feet long, which he bartered with B, for an equilateral tri angular beam of the same length, each side 24 inches required the balance at 1s. 4d. per foot? 18X18X25 : 56.95, solidity of the square beam. The perpendicular let fall on one of the sides of the trianguJar beam is 20-7846 inches, and the half perp.=10-3923; then 10.3923 x 24 144 =1.732 foot, area at the end, and 1·732×25=43 3 feet, solidity of the triangular beam; therefore 56-25-43·3=12.95 feet, at 1s. 4d. per foot=17s. 3.2d. balance due to A, Ans. 18. What is the difference between a solid half foot, and half a foot solid? 12x12x6 6×6×6 4, therefore, one is but of the other. 19. A lent B a solid stack of hay, measuring 20 feet every way; sometime afterward, B returned a quantity measuring every way 10 feet what proportion of the hay remains due ? : 20×20×20-10×10×10=7000 feet=1, Ans. 20. A ship's hold is 75 feet long, 18 wide, and 7 deep how many bales of goods 3 feet long, 24 deep, and 23 wide, may be stowed therein, leaving a gang way the whole length, of 31 feet wide? 75.5X18.5X7·25-75.5×7.25×3-25 3.5×2.25X2.75 385 44 bales, Ans. 21. If a stick of timber be 20 feet long, 16 inches broad, and 8 inches thick, and 34 solid feet be sawed off one end: how long will the stick then be? 1728X3.5 201 - 16X8 =16 feet, 63 inches, Ans. 22. The solid content of a square stone is found to be 136 feet; its length is 9 feet: what is the area of one end? and if the breadth be 3 feet 11 inches, what is the depth? 136.5×1728 9.5×12 inches, Ans. =area 2069 0526 inches, and 2069.0526 =44-022 23. I would have a cobick box made capable of receiving just 50 bushels, the bushel containing 2150-425 solid inches: what will be the length of the side? 3 2150 4×50=47.55 inches. 24. A statute bushel is to be made 8 inches high, and 18 inches diameter, to contain 2176 cubick inches; (though the content of the dimensions is but 2150 425 inches) I demand what the diameter of the bushel must be, the height being 8 inches; and what the height, the diameter being 18 inches, to contain 2176 cubick inches? Solidity. Height 8)2176 and Area 272 272×1·273=18 6 diameter. 18.5×18.5 X 7854268-80315 area, and the solidity 2176-268.88·0956 inches, height. 25. There is a garden rolling stone 66 inches in circumference, and 34 cubick feet are to be cut off from one end, perpendicular to the axis: where must the section be made? 1728×3.5 Area 412.5 14.65 inches from one end, Ans. 26. I would have a syringe of 14 inch diameter in the bore, to hold a quart, wine measure what must be the length of the piston, sufficient to make an injection with? 1.5X1.5X7854=1·76715, and 231÷÷4=57-75 the cubick inches in a quart, then 57.75 1.76715 32 679 inches, Ans. 27. If a round pillar, 9 inches diameter, contain 5 feet: of what diameter is that column, of equal length, which measures 10 times as much? As 5: 9x9 :: 5×10 : 810, and ✅/810=28·46 inches, Ans. 28 There is a square pyramid, each side of whose base is 30 inches, and whose perpendicular height is 120 inches, to be divided by sections parallel to its base into 3 equal parts: required the perpendicular height of each part? 30×30×40=36000 the solidity in inches, now thereof is 24000, and is 12000. Therefore, As 36000 120X120X120 :: { 1152000=104.8 Also, 576000-83-2. Then, 120-104-8 15.2 length of the thickest part, and 104.8-83-2=21.6 length of the middle part, consequently 83 2 is the length of the top part. 29. Suppose the diameter of the base of a conical ingot of gold in be 3 inches, and its height 9 inches; what length of wire may be expected from it, without loss of metal, the diameter of the wire being one hundredth part of an inch? 3×3×·7854×3=21.2058 the solidity of the cone. 21.2058 01X 01× 7854 =270000 inch. 4 miles, and 460 yards, Ans. 30. Suppose a pole to stand on a horizontal plane 75 feet in height above the surface at what height from the ground must it be cut off, so as that the top of it may fall on a point 55 feet from the bottom of the pole, the end, where it was cut off, resting on the stump, or upright part? As the whole length of the pole is equal to the sum of the hypotenuse and perpendicular of a triangle, (the 55 feet on the ground being the base) this, as well as the following question, may be solved by this RULE. From the square of the length of the pole (that is, of the sum of the hypotenuse and perpendicular) take the square of the base; divide the remainder by twice the length of the pole, and the quotient will be the perpendicular, or height at which it must be cut off. 31. Suppose a ship sails from latitude 43°, north, between north and east, till her departure from the meridian be 45 leagues, and the sum of her distance and difference of latitude to be 135 leagues : I demand her distance sailed, and latitude come to? 135X135 -45X45 135×2 =60 leagues, and 60×3=180 miles=3 de grees the difference of latitude, 135—60=75 leagues the distance. Now as the vessel is sailing from the equator, and consequently the latitude is increasing: Therefore, To the latitude sailed from Add the difference of latitude 43°,00' N. 3,00 And the sum is the latitude come to 46,00 N. |