14757120 : 112 13176000 26352000 13176 13176 14757120)1475712000(100 Ans. 100lb. of copper in the composition. and 12lb. of tin T8 MISCELLANEOUS QUESTIONS. 1. WHAT difference is there between a floor 48 feet long, and 30 feet broad, and two others each of half the dimensions? Ans. 720 feet. 2. From a mahogany plank 26 inches broad, a yard and a half is to be sawed off; what distance from the end must the line be struck? Ans. 6.23 feet. 3. A joist is 8 inches deep, and 3 broad; what will be the dimensions of a scantling just as big again as the joist, that is 44 inches broad? Ans. 12.52 inches deep. 4. A roof is 24 feet 8 inches by 14 feet 6 inches, and is to be covered with lead at 8lbs. to the foot; what will it come to at 188. per cwt.? Ans. 221. 19s. 104d. 5. What is the side of that equilateral triangle, whose area cost as much paving at 8d. per foot, as the palisading the three sides did at a guinea per yard? Ans. 72.746 feet. 6. The two sides of an obtuse-angled triangle are 20. and 40 poles; what must the length of the third side be that the triangle may contain just an acre? Ans. 58.876, or 23.099. 7. If two sides of an obtuse-angled triangle, whose area is 603, be 12 and 20; what is the third side? Ans. 28. 8. If an area of 24 be cut off from a triangle, whose three sides are 13, 14, and 15, by a line parallel to the longest side; what are the lengths of the sides including that area? Ans. 14, 214, and 14. 9. The distance of the centres of two circles, whose diameters are each 50, is equal to 30; what is the area of the space inclosed by their circumference? Ans. 559.115. 10. There is a segment of a circle the chord of which is 60 feet, and versed sine 10 feet; what will be the versed sine of that segment of the same circle whose chord is 90 feet? Ans. 28.2055. 11. The area of an equilateral triangle, whose base falls on the diameter, and its vertex in the middle of the arc of a semicircle, is equal to 100; what is the diameter of the semicircle? Ans. 26.32148. 12. The four sides of a field, whose diagonals are equal to each other, are 25, 35, 31, and 19 poles, respectively; what is the area?* Ans. 4 ac. 1 ro. 38 poles. * Construction. In this question the sums of the squares of the opposite sides of the trapezium being equal, the figure may be constructed as follows: Draw AB and AE at right angles, and each equal to one of the given sides (35); join BE, and from the points E and B, with radii equal to the two remaining opposite sides (25 and 31) respectively, describe arcs intersecting in C on the farther side of BE; join AC, and draw BF at right angles to it. With the centre C, and radius equal to the remaining side (19) describe an arc cutting BF produced in D. Join AD and CD, then will ABCD be the figure required. Demonstration.-By the question AB2 + CD2 = BC2 + CE and since BD and AC cross each other at right angles, (47.1) AB2 + CD2 = BC2 + AD2; wherefore AD2=E02, or aD =EC. 13. A cable which is 3 feet long, and 9 inches in compass, weighs 22 pounds: what will a fathom of that cable weigh whose diameter is 9 inches? Ans. 434.25 lbs. 14. A circular fish-pond is to be dug in a garden that shall take up just half an acre: what must the length of the chord be that strikes the circle? Ans. 27.75 yards. 15. A carpenter is to put an oaken curb to a round well, at 8d. per foot square; the breadth of the curb is to be 7 inches, and the diameter within 3 feet: what will be the expense? Ans. 58. 24d. Hence, in the two triangles ABD and EAC, we have the two sides BA, AD equal the two AE, EC, each to each, and the angles ABD and EAC equal (each being the complement of BAF) and EC and AD, similarly situated; wherefore BD-AC.. Q. E. D. Calculation.-On BE let fall the perpendiculars co and AH; now BE2=AB2+AE2=352 × 2; BE =√352 × 2 =35 √2=49.4975; and AH=BH=BE=24.7487; again (13.2) BC2+BE2-CE2 312+2 × 352-252 2786 BG= = 2BE 2 X 49.4975 28.1428; GH = BG BH 28.1428-24.7487=3.3941; CG= √✓ BC2— BG2= √✓ 312 — 28.14282= √168.98280816 = 12.9993. By sim. triangles AH+CG(37.748) : GH(3.3941) :: AH (24.7487): HI = 2.2253; AI = √ AH2 + HI2 = √24.74872 + 2.22532 = √612.49815169+4.95196009= √617.45011178=24.8485. Again, by sim. triangles, HI (2.2253): HG(3.3941):: AI(24.8485): Ac=37.8997 BD; now, by Problem V. Superficies, the area of the trapeAC X BD -AC2 37.89972 ACX BF + FD 2 zium ABCD = Note. This method is applicable to all questions of the kind wherein the diagonals cross each other at right angles; that is, when the sums of the squares of the opposite sides are equal, but a general solution to the question, without this, would involve an equation of the higher order. 16. Suppose the expense of paving a semicircular plot, at 28. 4d. per foot, amounted to 107. what is the diameter of it? Ans. 14.7739. 17. Seven men bought a grinding stone of 60 inches in diameter, each paying part of the expense; what part of the diameter must each grind down for his share? Ans. The 1st, 4.4508; 2d, 4.8400; 3d, 5.3535; 4th, 6.0765; 5th, 7.2079; 6th, 9.3935; and the 7th, 22.6778. * This problem may be thus constructed: On the radius AC describe a semicircle; also divide AC into as many equal parts CD, DE, EF, &c. as there are shares, and erect the perpendicular DL, EM, FN, &c. meeting the semicircle described on AC in L, M, N, O, P, Q. Then with the centre C and radii CL, CM, CN, &c. describe circles, and the thing is done. A B I HGFED C For, by the nature of the circle, the square of the chords or radii CL, CM, CN, &c. are as the cosines CD, CE, CF, &c. |