To find the area of a circle. RULE. · Multiply the square of the diameter by .7854, (.785398); or, multiply the square of the circumference by .07958; or, multiply half the circumference by half the diameter. In either case the product is the area. the chord a b, was EXAMPLE.-A line, stretched on the curve of a circular railway, 46 feet, and the versed sine of half the arc, p d, was found to be 18 feet; required the radius of the curve, the diameter, the circumference, and the area of the circle. P 46 ÷ 2 = 23, and 232 +1.82÷ 1.8295.69 ft. = diameter. Ans. RULE. = 147.845 feet 68669.56 square feet, or To find the length of an arc of a circle. Ans. 1. From eight times the chord of half the arc, subtract the chord of the whole arc, and divide the remainder by 3; the quotient will be the length of the arc, nearly a trifle too small. - 2. Multiply the number of degrees in the arc by .0174533,* and the product by the radius of the circle; the last product will be the length of the arc. Or, -3. Multiply ten times the square of the versed sine of half the arc by twice the chord of half the arc, and divide the product by fifteen times the square of the chord of the whole arc, +33 times the square of the versed sine of half the arc; to the quotient add twice the chord of half the arc, and the sum will be the length of the arc. EXAMPLE.Required the length of the arc a d b, the chord of half the arc, a d, being 10.5981 feet, and the chord of the whole arc, a b, being 20 feet. 770.22 X .0174533 X 16 21.6038 feet. Ans. By Rule 3. 202 X 15 = 6000 + 12.32 X 33 = 6406.56, 12.32 X 10 X 21.1962 = 2611.37 6406.56 .407621.1962= 21.6038. Ans. *Ratio of diameter to circumference. .0174533 X 180 = 3.141592. To find the area of a sector of a circle. and RULE. - 1. Multiply the length of the arc by half the radius of the circle, by half the length of one of the sides of the sector, the product is the area of the sector. Or, - 2. As 360 is to the degrees in the arc of the sector, so is the area of the circle to the area of the sector. EXAMPLE.-The length of the arc, C E, of the sector CE D, is found to be 328 inches, and the radius, CD, is 21 inches; required the area of the sector. 32.8 X 10.666 = 349.8448 square inches. 360 Ans. 88°09 :: 1429.76: 349.85. Ans. To find the area of a segment of a circle. RULE. —1. Find the area of a sector whose arc is equal to that of the given segment, and then find the area of the isosceles triangle formed by the chord of the segment and the two radii of the sector; if the segment is less than a semicircle, subtract the area of the triangle from the area of the sector, and the remainder will be the area of the segment; if the segment is greater than a semicircle, subtract the area of the less segment from that of the circle, and the difference will be the area sought. Or,- 2. To once and one third the chord of half the arc add the chord of the whole arc, and multiply the sum by the versed sine of half the arc, or height of the segment, and four tenths of the product will be the area of the segment, very nearly, - a mere fraction too small. Or, -3. Multiply the chord of the arc by the height of the scgment, divide the product by 3, and multiply the quotient by 2; cube the height, divide the cube by twice the length of the chord, and add the quotient to the product before obtained, and the sum will be the area of the segment, nearly, a fraction too small. EXAMPLE. The length of the arc a db is 32.8 feet, and the radii, ca, cb, are each 21 feet 4 inches; required the area of the segment, ab d. a 32.8 X 21.333= 349.86 sq. ft. area of sector, and angle; hence, 349.8612-227.41785 122.44335 sq. ft., area of seg. Ans. By Rule 2. By Rule 3. RULE. ad = pd 16 3 5.333, and 5.333 + 16 +29.66+ X 6 = = 63 = 216 ÷ 29.663856 × 2 = 3.640792+, and 29.663856 × 6 ÷ 3 = 59.327 + X 2 =118.655+3.64 + 122.296. Ans. = To find the area of a zone. Find the area of the circle containing the zone and the areas of the segments each side of the zone, and from the area of the circle subtract the sum of the areas of the segments, and the difference will be the area required. To find the area of a crescent. RULE. Find the areas of the two segments, formed by the two arcs of the crescent and their chord; subtract the less area from the greater, and the difference will be the area of the crescent. To find the side of a square that shall contain an area equal to that of a given circle. RULE. - Multiply the diameter of the circle by .886227; or, multiply the circumference of the circle by .282094; in either case the product will be the side sought. EXAMPLE. The diameter of a circle is 16 feet, its area, consequently, is 201 square feet; required the side of a square that shall contain the same area. 16 X.88622714.18 feet. Ans. Proof, 14.18 X 14.18 = : 201. To find the diameter of a circle that shall contain the same area as a given square. RULE.-Multiply the side of the given square by 1.12838, and the product will be the diameter of the required circle. As 8 is to 7.09 so is diam. of circle to side of equal square, nearly. As 7.09 is to 8 so is side of square to diam. of equal circle, nearly. To find the side of a square inscribed in a given circle. A square is said to be inscribed in a circle when all its corners touch the circumference. Its area is precisely half that of a circumscribing square, of the same circle. RULE. Multiply the diameter of the circle by .707106; or, multiply the circumference of the circle by .225079; in either case the product will be the side of the square inscribed. The square root of half the square of the diameter is the side. EXAMPLE. A log is 28 inches in diameter at the smaller end, required the side of the greatest square that can be hewn from it. = 19.80 inches. Ans. 28 X .7071 As 14 is to 9.9 so is diam. to side of inscribed square, nearly. As 9.9 is to 14 so is side of inscribed square to diam. of circle, nearly. To find the diameters of three equal circles, the greatest that can be inscribed in a given circle. RULE. Divide the diameter of the given circle by 2.155, and the quotient will be the diameter of either the inscribed circles. To find the diameters of four equal circles, the greatest that can be inscribed in a given circle. RULE. Multiply the diameter of the circumscribing circle by .4161, and the product will be the diameter of either the inscribed circles. EXAMPLE. The diameter of a circle is 24 inches; what is the diameter of each of the four greatest circles that may be inscribed therein ? 24 X.416110 inches, nearly. Ans. To find the area of the space contained between two concentric circles. RULE. Multiply the sum of the two diameters, (that of the outer circle and that of the inner,) by their difference, multiplied by .7854, and the product will be. the area of the ring or space referred to. EXAMPLE. The diameter of the larger circle, A B, is 36 inches, and the diameter of the less circle, CD, is 30 inches; required the area of the ring or space between the two circles. 363066, and 36 - 30= 6; then, 66 X 6 X .7854 311 square inches. Ans. RULE. = OF THE CONIC SECTIONS. To find the area of an ellipse. Multiply the transverse axis by the conjugate, and that product by .7854; the last product is the area. EXAMPLE. The transverse axis, a b, of the ellipse a c b d, is 30 inches, and the conjugate axis, c d, is 16 inches; required the area of the figure. = 25 square feet. Ans. To find the area of an elliptic segment. RULE. Consider the segment as that of a circle, and the axis of the ellipse (whether major or minor) vertical, or at a right angle with the base of the segment, as the diameter of the circle, and find the area as for the segment of a circle; then, as the vertical axis of the ellipse is to that parallel to the base of the segment, so is the area found to the area required. EXAMPLE. The major axis, a b, of the ellipse a c b d, is 30 inches, the minor axis, c d, 16 inches, and the height of the segment, p b, 10 inches; required the area of the segment, e ƒ b. diam. 2 radius; radius - height = cosine; radius2 — cosine2 = sine2; sine2X2=base of seg. = 28.284+; sinc2 + height? = ch'd of half the arc2, and ch'd arc2 = ch'd of half the arc = 17.32+; then, 17.323= 5.77317.32 + 28.284 X 10 X 4 ÷ 10 = =205.51= area of seg. of circle; and 30 16 205.51 RULE. 109.61 sq. in., area of elliptic seg. Ans. To find the circumference of an ellipse. Add the square of the transverse axis and the square of the conjugate together, and divide the sum by 2; find the square root of the quotient and multiply it by 3.1416, and the product is the circumference. EXAMPLE. The axes of an ellipse are 30 inches and 16 inches; required the circumference of that ellipse. 302+162÷2=578 = 24.04 × 3.141675.52+ inches. Ans. To find the length of an ordinate of an ellipse. RULE.-Multiply one abscissa by the other, find the square root of the product, multiply it by the axis parallel to the ordinate, and divide the product by the axis converted into abscissæ; the quotient will be the length of the ordinate. EXAMPLE. In the ellipse a c bd, the transverse axis, a b, is 30 inches, the conjugate, c d, 16 inches, and the abscissa, p b, 10 inches; required the length of the ordinate, e p. RULE.Subtract the square of the ordinate from the square of the semi-conjugate, find the square root of the difference, and add the semi-conjugate thereto; then, multiply the greater abscissa by the conjugate, and divide the product by the sum last obtained. Or, subtract the square root of the difference of the square of the ordinate and semi-con |
