Thus, if the length of the arc AB be 15; then, by reafon of the fimlarity of the arcs AB, DE, it will be 10 6 15: 3×39, the arc DE. Whence (15+9) × (10 −6) 12 X2 24, the area of the part ABEDA. RULE III. Multiply the perpendicular breadth of the ring, that is, the difference of the radii, by the circumference RST (or part Rs for the part ABEDA) having the fame center with, and equally diftant from the bounding arcs.* Taking, ftill, the fame example; it will be circumferences. And CRCD + DR = CD+AD=3+1=4, the radius of the middle arc. Hence 31416 x 8251328 the middle circumference. Therefore 25.1328×250 2656, the area the fame as before. SR And for the part ABEDA, we have CA: CR :: AB : 15×4 5 =3×4=12. Then ADXRS 2 X 12 24, the area. PRO DEMONSTRATION. For this circumference, being equally distant from the other two, will be equal to half their fum. Wherefore &c. Corol. Hence the whole ring, or any part of it ABEDA, included between two radii, is equal to a parallelogram on the fame base AD, and whofe altitude is equal to Rs the middle cir cumference. PROBLEM XIII. To find the Areas of Lunes. B D. A Lunes, lunule, or little moons, are spaces included between the interfecting arcs of two excentric circles, as the lune ABCDA; which is, evidently, equal to the difference of the fegments ABCA, ADCA. If ABC be a triangle right-angled at c, and if femicircles be defcribed on the three fides as diameters; then the triangle T (ABC) will be equal to the fum of the two lunes L1, L2.-For the greatest femicircle is equal to the fum of the other two; from the greatest femicircle take the fegments s I, s2, and there will remain the ▲ T; from the two less femicircles take alfo the fame two fegments s 1, s2, and there will remain the two lunes L1, L2; wherefore T=LI+L2. 1 Whence, if the two fides Ac, CB, of the triangle be equal to each other; the two lunes will, alfo, be equal, and each lune L1 equal to the ▲ ACD; and therefore the fegment s 1 femicircle AQCA-AACD double the feg. AWDA, or double the feg. DVCD. SECT. II. A PROMISCUOUS COLLECTION OF QUESTIONS CONCERNING AREAS. QUESTION I. N the trapezium ABCD are given AB IN 61, BC= 153, CD 12, and DA 9, alfo в a right angle; to find the area of the trapezium. A D B First √AB2 + BC2 = √(23)2 + X 169 = AC; and √379 × 41 × 1379 × I 2 2 2 19'9 2 ✓ 37′9 × 4°1 × 13′9 × 19′9 = 4√42982•4279 518305098 the area of the triangle ADC. And Some of the questions in this fection are taken from other books; but the methods of folution are, generally, different from thofe ufed in the books from which they were taken, they being there mostly folved by an analytical procefs. And I have conftructed thofe of which the conftructions do not appear to be felf-evident. the triangle ABC. 20 The fum is 102.5305098, the area of the trapezium required. QUEST. 2. To find the area of a trapezium, the length of its fides being as in the margin, and the fum of the two oppofite angles B and D equal to 180 degrees. Chains AB = 15.6 BC 132 CD 10°O DA 26.0 By rule 5 of prob. 3 of fect. 1. (32.4 being half the perimeter) √(32·4 — 15·6) × ( 32°4 — 13′2) × (32·4—10) × ( 32'4—26)= 16.8 X 19.2 X 22.4 X 6.4 215'04 fquare chains 21504 acres 21 a. 2r. o 64 perches, the area required. Note. A construction of this problem may be seen in Simpfon's Select Exercifes, page 135. QUEST. 3. In the pentangular field ABCDE are given AB 14, BC7, CD10, DE 12, EA5, and the diagonal AC 17 chains, alfo £ a right angle; required the area. D Firft ADDE2 + EA2 = = √122 √ 122 + 52 = 13. Then AE XED = 5×6=30 AADE. And20X3X7X10=10√42=64.807407=4ADC. Alfo✓19X2X5X12=2/570=47*749345=▲ ABC. The fum of all three is 142.556752 fquare chains 14a. 10227 r. the area required. QUEST. QUEST. 4. Given the bafe Ac=32, AD=5, EC9, the perpendicular EF4, and the perpendicular DG3; required the area of the triangle Draw GH parallel to BC. Then, by the fimilar triangles GDH, FEC, we have FE 4: EC 9:: GD 3: DH; and hence AH 47. Again, from the fimilar triangles AGH, ABC, we have A AC 32:: DG 3 the perpendicular IB = 35. 384 47 4 Hence ACXIB 16X38413024130723447 = is the area of the triangle required. QUEST. 5. What is the fide of that equilateral triangle, whofe area coft as much paving at 8d a foot, as the pallifading the three fides did at guinea a yard? The fides are 75. a foot, and the areas. a foot. And that the produces may be equal, the quantities muft be inversely as the prices; but, by rule 2, page 113, BC2/3 is the area; therefore:7:3BC:BC2√3 3×4X7X3 :: 3x4: BC= 3×4 √3 213 = 42√3 72 7461339, the fide required. B a A P C QUEST. 6. Surveying a quadrangular field, I found the four fides to be 10, 9, 7, and 6 chains, in a fucceffive order: I likewife, at the two extremes of the longeft fide, took the bearings of the oppofite angles, which were N. E. by E. and N. w. Hence the content of the field is required. From |