4. The circumference of a circle is 64'4: what is the diameter ? Ans. 20:49916. 5. Two chords whose sum is 21 and difference 1, and the rectangle of the segments of either is 24, cut each other at right angles: it is required to find the circumference of the circle, and the distance of their point of intersection from the centre. Ans. circumf. = 5π/5, dist. = √29. 6. Let C be the circumference of a circle, d its diameter, and c the chord of an 16 ad (Ca) arc, a: show that c = 502 4a (Ca) nearly. 7. If a be an arc of a circle to radius 1, a1=5.482. 3+ cos a 4 costa 237-cosa nearly. 124 costa 8. Show that 3a=tan a + 2 sin a, very nearly, when a is small. PROBLEM VIII. To find the length of an arc of the circle. Multiply the number of degrees and parts expressing the arc by 0174533, and by the radius of the circle; or, take out the numbers corresponding to the given number of degrees, minutes, and seconds successively from Hutton's Tables, p. 360; and then their sum, so multiplied, will be the length required. 1. Find the length of 30° in a circle whose radius is 9ft. Ans. 4.712388. 2. To diameter 10ft find the length of 12° 10′ 15′′. Ans. 4 248422. 3. What portion of the arc of a circle is equal to the radius ? Ans. 57° 2957795. ... PROBLEM IX. To find the area of a circle. No. 12, and scholium, page 477. Let r be the radius, d the diameter, c the circumference, and A the area: then A is found from any one of the following equations. A = r2 = r. c = {dc = d2. 0795775c2. π 1. Find the area of a circle whose circumference is 3141593. Ans. 78.5398. Ans. 38 484501. 2. Find the area of a circle whose diameter is 7. 4. Required the area of a circle whose circumference is 12ft. Ans. 1069014. Ans. 11-45916. 5. Find the area of a circle, the difference of whose diameter and circumference is 1056 64 ft. Ans. 191105 4ft. 6. If the centre of a circle whose diameter is 20 be in the circumference of another whose diameter is 40, what are the areas of the three included spaces? Ans. 173 852, 140 308, and 1116.332. 7. What is the diameter of that circle which contains an acre? Ans. 78 yds. 8. If the area of a circle be 100, find the sides of the inscribed square, pentagon and hexagon. PROBLEM X. To find the area of an annulus. 1. The diameters of two concentric circles are 10 and 6: required the area of the annulus. Ans. 50 26552. 2. The bounding circles are 10 and 20 in diameter: what is the area of the annulus ? Ans. 235 61947. 3. The circumference of a ring is 161in, and its width lin: required its internal diameter and area. Ans. int. diam. = 49.248in, and area = 157·8634. 4. The radii of two concentric circles are in the ratio of 10 to 9, and the area of the ring is 375-562feet: find the diameters of the circles. Ans. 50 1624, and 45.1462. 5. Let c be the outer circumference, and b the breadth of a ring: show that its area will be (c — bπ) b. PROBLEM XI. To find the area of a sector of a circle. (1) Multiply the radius by half the length of the arc. 1. Find the area of a sector of 18° to a diameter of 3ft. Ans. 35343. 2. The radius is 10ft, and the arc 20ft: find the area and angular measure. Ans. 100 and 114° 35' 29". 3. A sector of 147° 29′ 18′′ has a radius of 25ft: what is its area? Ans. 804 3986. 4. Find the area of a sector whose radius is 50 and arc 56° 30'. Ans. 1232 6387. 5. The area of a sector is 100, and the length of its arc 20: what is the angle of the sector? Ans. 114° 35' nearly. PROBLEM XII. To find the area of a segment of a circle. No. 13, page 477. (1) Find the difference or sum of the sector having the same arc and the triangle formed by the chord and the radii bounding the sector, according as the arc is less or greater than a semicircle. (2) Segment = r2 {a sin a}, which is only the same rule in another form. 1. The chord AB is 12, and the radius AE is 10: what is the area of the segment ? sin AED, or sin AEC= = AD ='6 sin 36° 52' 11"-2. AE Hence the arc ACB 73° 44' 224; and we have 360° 73° 44′ 22" 4 :: 102. π: 64.3504 area of sector. Also, for the triangle AEB, we have AD2=√100 — 36 = = 8, triangle AEB = 16·3504. and the area of the triangle is AD.DE 6.8 48. Whence segment ACB = sector AEB · Or again, by the formula: we have as before arc ACB 73° 44′ 22′′-4; and by Hutton's Tables, p. 360, arc 73° = 1.2740904 sin 73° 44′ = '9599684 and greater segment = r2 {a + sin a} = 112-349765. 2. The height CD is 18, and radius CE is 50: what is the area of the segAns. 961-3532. 3. Required the area of each of the segments where the chord is 16, the diameter of the circle being 20. Ans. 44 728 or 269-432. ment? Scholium. It occurs in many problems that we have to find the arc of the sector from knowing the area, together with the radius of the circle, or some other given line in the circle: or in other words, to solve the equation. For this purpose no method seems so generally and easily applicable as the method of trial and error. One example is annexed: viz. to find from the equation sin 02:0943951. Take as a conjectural approximation = 150°: then Taking next = 149° 16' we have a second correction. PROBLEM XIII. To measure any long and irregular figure. No. 14, page 478. 1. The breadths of an irregular figure at five equidistant places are 8-2, 7·4, 9.2, 102, and 86, and the whole length is 39: what is the area? Ans. 343.2. 2. The length of the figure is 84, and the six equidistant ordinates are 174, 20 6, 142, 16 5, 201, and 24:4: what is the area? Ans. 1550 64. 3. The distances of seven points in the long side of a figure formed by straight lines from the first extremity of that side were 108, 104, 25, 106, 18·5, 56·2, and the ordinates of the angular points were 0, 12, 18, 25, 18, 5, 15, and 0: what was the area? 3. In the vertical section of a rampart AS is the horizontal base, and the horizontal distance, in feet, of the several angular points of the work reckoned on this line, together with the heights of those points above it, are ranged below, from which to find the area of the section and construct the figure: viz.Distance on AS, AB = 16, BD = 18, DH = 2, HK = 3, KL = 2, LP = 12, PS 10. Height above AS, BC = 12, DE = 12}, HG = 131, KI = 134, LO = 18, B H ER S D P 4. Let ABC be the profile, or perpendicular section of a breast-work, and EP that of the ditch. Now, suppose the area of the section ABC is 88 feet, the depth of the ditch RD 6 feet, ER SO = 3 feet; what is the breadth of the ditch at top when the sections of the ditch and the breast-work are equal; that is, when the earth thrown out of the ditch is sufficient to make the breast-work? = 5. And what must be the breadth of the ditch at top, the depth and width at bottom remaining the same, when the profile of the breast-work remains the same, and the earth, in consequence of removal, occupies th more space than it did before it was taken out of the ditch? MENSURATION OF SOLIDS. THE cube described upon the linear unit is always taken as the unit of volume. It will hence follow in precisely the same manner as for superficial measure, that the volume of a rectangular prism or cylinder is the product of the numerical measures of the base and altitude. For any two rectangular parallelopipedons are to one another in a ratio compounded of that of their bases and that of their altitudes; and the bases are compounded of the ratios of their sides containing one of the right-angles: hence they are compounded of the ratios of the three edges of the one to the three edges of the other, each to each respectively. So far as the surfaces of solids bounded by plane faces are concerned, the surfaces are computed by the rules appropriate to them, as already explained in the mensuration of planes: but the surface of the sphere and its segments or zones will form a part of this section. I. THE SURFACES AND VOLUMES OF FIGURES OF THREE DIMENSIONS. 1. The volume of a rectangular parallelopipedon is expressed by the continued product of these edges which meet at one of the solid right-angles, as is evident from the foregoing introductory remarks. 2. The volume of any parallelopipedon, whose plane angles forming one of the solid angles are a, ß, y, and whose opposite edges are a, b, c, is expressed by 2abc/sin o sin (σ—a) sin (σ—ß) sin (σ—y) where 20 = a + ß+y. Let Q be the solid angle, the three edges AQ, BQ, CQ, of which are denoted by a, b, c; and the three angles CQB, CQA. AQB are a, ẞ, y. From A draw the perpendicular AP to the plane CQB, and from P draw PD, PE perpendicular to QC, QB, and join AE, AD, and PQ, Then (Geom. Planes, th. 7), AEQ, ADQ, are right-angles, and AQP is the inclination of the line AQ to the plane CQB. Denote this by 0: then we have E R whence, transposing, squaring, and performing obvious reductions, it becomes at once sin 0 sin a = √1 — cos3a — cos 3-cos y + 2 cos a cos ß cos y. = 2 sino sin (o - a) sin (σ — (3) sin (σ — y). But the parallelopipedon QR is compounded of the base QS and altitude AP, and hence we have vol be sin a. a sin 0 abc sina sin 0. = 3. The volume of any prism is the product of one end into the distance of the two parallel ends. (Geom. Pl. and Sol., th. 30, p. 366). 4. The volume of a cylinder is expressed by the product of its base and altitude; and that of a cone by one-third of that product: as is clear from Geom. of Pl. and Sol., theorems 30, and Cor. 34. 5. The curve surface of a right-cylinder is the product of the perimeter of the base into the length of the axis, or by 2rhя. For that surface is composed of an infinite number of infinitely narrow rectangles, all of the same length. When the two ends are also required in the expression, the entire surface is expressed by 2r (r + h) π. For 2r2 = area of the base, and 2rhπ = curve surface; hence the whole surface 2r2 + 2rh = 2r (r + h) x. 6. The curve surface of a right-cone is half the product of the perimeter of the base into one of the slant sides or edges. The reason is similar to the last. Cor. The curve surface of a frustum of a cone whose bounding sections have a and b for radii, and their distance reckoned along the side of the cone is d, is (a + b) dл. 7. The volumes of a cone and pyramid are one-third of the volumes of the cylinder of a prism respectively of the same bases and altitudes. 8. The volume of a truncated pyramid or cone is expressed by1h {a2+ab+b2}, where a and b are the areas of the two ends, and h is their distance. For, let ABCD be the pyramid of which BCDEFG is the frustum, a2 the area of BCD, b2 the area of EFG, and h the height Denote AI by c: then by similar figures HI of the frustum. 3 a- -b 3 a-b The same demonstration applies to the conic frustum. Cor. If D, d, be the corresponding linear dimensions of the ends, m their appropriate multiplier, (so that mD2, md2, are their areas,) and 2 the difference of those areas: then the volume is expressed by } mh {3Dd + 82}. 9. The curve surface of a sphere is equal to four times the area of one of its great circles. |