20. Hydraulic mean depth of a rectangular water-course is found by dividing the area by the wetted perimeter; i. e., = area divided by the sum of 2 A C + C B. 21. When the breadth is to the depth as 1: V2, i. e., as 1 : 1,4152, the rectangular beam will be the strongest in a circular tree. 22. When the breadth is to the depth as 1: √3, i. e., as 1 : 1,732, the beam will be the stiffest that can be cut out of a round tree. 23. Rhomboid. (Fig 4.) (Fig 4.) In a rhomboid the four sides are parallel. Area = longest side by the perpendicular height = A BXCE=AB XACX nat. sine < C A B. 24. Trapezoid. In a trapezoid only two of its sides are parallel to one another. Let A D E B (fig. 4) be a trapezoid. Area = † (C D + A B) × by the perpendicular width C E. OF THE TRIANGLE. 25. Let A B C (Fig. 5) be a triangle. A B 26. If one of its angles, as B, is right-angled, the area XB C 2 27. Or, area = ABX tangent of the angle B A C. 28. When the triangle is not right-angled, measure any side; A Cas a base, and take the perpendicular to the opposite angle, B; then the area= · † (A C × E B.) In measuring the line A C, note the distance from A to E and from E to C, E being where the perpendicular was erected. 29. Or, area = АСХАВ Xnat. sine of the angle C A B. When the perpendicular E B would much exceed 100 links, and that the surveyor has not an instrument by which he could take the perpendicular E B, or angle C A B, his best plan would be to measure the three sides, A B = a, B C = b, and A C: Then the area will be found as follows: C. 30. Add the three sides together, take half their sum; from that half sum take each side separately; multiply the half sum by the three differences. The square root of the last product will be the area. 82. Let s equal half the sum of the three sides then to the logarithm of half the sum add the logs of the three differences, divide the sum by 2, and the quotient will be the log of the required area. 1 34. Or, to the log of A C add the log of A B and the log sine of the contained angle C A B. The number corresponding to the sum of these three logs will be double the area, i. e., Log a+ log clog sine angle C A B double the area. 35. Or, by adding the arithmetical compliment of 2, which is 1,698970, we have a very concise formula, Area log a Example. Let A B log clog sine angle C A B + 1,698970. a = 18,74, and A C tained angle C A B — 29° 43′ Log 18,47 chains, Log 16,95 chains, C = 1695 and the con 1,2664669 1,2291697 9,6952288 1,6989700 11,8898354 10 1,8898354 The natural number corresponding to this log will be the required area 77,5953 square chains, which, divided by 10, will give the area 7,75953 acres. 35a. In Fig. 5, let the sides A C and B C be inaccessible. Measure a2 sine A X sine B, A Ba; take the angles A and B, then the area which, in words, is as follows: 2 sine C Multiply together the square of the side, the natural sines of the angles A and B; divide the contained product by twice the sine of the angle C. The quotient will be the required area. Or thus: Add together twice the log of a, the log sine A, and the log sine B; from the sum subtract log 2 + log sine C. The difference will be the log of the area. Example. Let the <A 50°, angle B 70°; and let A B & From the sum A subtract the sum B, the difference, having rejected 10 from the index will be the log of the natural number corresponding to the area 141,198 square chains, which divided by 10 gives the area 1410000 acres. Or thus: By using the table of natural sines. Having used Hutton's logs, we will also use his nat. sines. See the formula (34) a2 20 × 20, Nat. sine 50° nat. sin.<A Product, 400 ,7660444 306,4177600 356. If on the line A B the triangles A C B, A D B, A E B, etc., be described such that the difference of the sides A C and C B, of A D and D B, and of A E and E B is each equal to a given quantity, the curve passing through the points C, D and E is a hyperbola. 36. If the sum of each of the above sides A C+CB, AD + D B, A'E + E B is equal to a given quantity, the curve is an ellipses. 37 In the AC B, (Fig. 5,) if the base C E is of the line A C, the ACE B will be of the A A C B, and if the base A C be n times the base C E, the A A C B will be n times the area of the ACE B. 38. From the point P in the AC B, (Fig. 11,) it is required to draw a line P E, so that the AA PE will be the area of the A A C B. Divide the line A B into 4 equal parts, let A D one of these parts, join D and C and P and C, draw D E parallel to P C, then the AA E P will be of the AC B; for by Euclid I. 37, we find that the A EOC AOD P.. the ▲ A E P 39. ACD the From the A A C B, required to cut off a AADE ▲ ACB by a line D E parallel to B C. 5 AC B, Q.E.F. to of the By Euclid VI. 20, ▲ A DE: ▲ ACB :: A D2: A B2; therefore, in this case, divide A B into two such parts, so that A D2 the square of A B. Let D be the required point, from which draw the line D E parallel to B C, and the work is done. ▲ADE: 40. In the last case we have A DE: AACB:: A D2: A B2; Generally, 1:n:: A D2: A B2; and by i. e., 1 : 5 :: A D2: A B2. 41. If D be a point in the A A C B, (Fig 13,) through which the line PE is drawn parallel to C B, make C E to meet CB in G, then the line F D G triangle. E F, join F D, and produce it will cut off the least possible 42. By Euclid VI. 2, F D = D G, because F E = E C. 43. To bisect the A Cb, BC = a, C P AC B (Fig. 16,) by the shortest line P D. Let x, and C D=y, ^\ CPD=▲ A C B, condi tions which will be fulfilled when x= ་་ a b =CP=V and y=CD=1 a b 2 Hence it follows that CPC D. (See Tate's Differential Calculus, p. 65.) 44. The greatest rectangle that can be inscribed in any A A C B, is that whose height n m, is the height n C of the given triangle (see Fig 14,) A B C. Hence the construction is evident. Bisect A C in K. draw K L parallel to A B, let fall the perpendiculars K D and L I, and and the figure K L ID will be the required rectangle. 45. The centre of the circumscribing circle A C B, (Fig. 7,) is found by bisecting the sides A B, A C, and C B, and erecting perpendiculars from the points of bisection; the point of their bisection will be the required centre. (See Euclid IV. 5.) 46. The centre of the inscribed circle (Fig. 6,) is found by bisecting the angles A, B and C, the intersection of these lines will be the required centre, O, from which let fall the perpendicular O E or O D, each equal to the perpendicular O F to the required radius. 47. Let R = radius of circumscribing circle and r radius of the inscribed circle, and the sides A B =a, BC b, and A Cc of the 48. To find r, the radius of the inscribed circle in (Fig. 6,) 49. Radius of the circumscribing circle is equal to the product of the three sides divided by 4 times the area of the triangle, and substituting the formula in ? 31 for the area of the triangle, we have 51. The area of any least line KL, when C K required area, then x = CKL (Fig. 14,) will be subtended by the C L. Let x = C K = C L, and A the 2 A nat. sine <C 52. Of all the triangles on the same base and in the same segment of a circle, the isoceles contains the greatest area. 53. The greatest isoceles in a circle will be also equi-lateral and will have each side =r √/3 where r = radius of the given circle. 54. In a right-angled, when the hypothenuse is given, the area will be a maximum when the ▲ is isoceles; that is, by putting h for the h hypothenuse the base and perpendicular will be each √2 h 1.414313 55. The greatest rectangle in an isoceles right-angled ▲ will be a square. 56. In every triangle whose base and perpendicular are equal to one another, the perimeter will be a maximum when the triangle is isoceles. 57. Of all triangles having the same perimeter, the equi-lateral ▲ contains the greatest area. 58. In all retaining walls (walls built to support any pressure acting laterally) whose base equals its perpendicular, or whose hypothenuse makes an angle of 45° with the horizon, will be the strongest possible. 61. Example. Let d = 46, then 46 × 3,1416 = 144,5136 = circumference; or, by logarithms, |