given square. Or, if the side of the square be divided by 886, the quotient will be the diameter. Or, as 12 to 13.54, so is the side of any square to the diameter of an equal circle. EXAMP. The side of a square being 10-635, to find the diameter of a circle equal to that square? 10.635×1.128=12 nearly. Or, 10·635÷÷÷886=12=diameter. Or, as 12: 13.54 :: 10·635 : 12 nearly. ART. 21. The Side af a Square being given, to find the circumference of a Circle equal to the given Square. RULE. Multiply the given side by 3.545 and the product will be the circumference required. Or, divide it by 282, and the quotient will be the circumference. EXAMP. The side of a square being 10-631, to find the circumference of a circle equal to that square. 10.631×3 545=37-586-circum. Or, -282) 10.631(37-698 circum. ART. 22. To find the Area of a Semicircle, the Diameter being given. RULE. Find the area of the circle by Art. 15, and take the half of it. In the same manner may the area of a quadrant, or a quarter of a circle, be found, by taking a fourth part of the area of the whole circle. But with regard to measuring a sector, or a segment of a circle, it will be necessary first to show how to find the length of the arch line of a sector, and the diameter of the circle to a given segment. ART. 23. A Segment of a Circle being given, to find the length of the Arch Line RULE. Divide the segment into two equal parts; then measure the chord of the half arch, from the double of which subtract the chord of the whole segment; and one third of that difference, being added to the double of the chord of the half arch, will give the length of the arch line. ART. 24. The Chord and versed Sine of a Segment being given, to find the Diameter of a Circle. RULE. Multiply half the chord by itself, and divide the product by the versed sine; then add the quotient to the versed sine, and the sum will be the diameter of the circle. Definition. A sector is a part of a circle, contained between an arch line, and two radii or semidiameters of the circle. RULE. Find the length of half the arch by Art. 23: Then multiply this by the radius or semidiameter, and the product will be 1 144.16 length of the arch 5207.76 area. [ABC, by Art. 23. 72.33 EXAMP. 2. In the sector ABCD, greater than a semicircle, given the radius AE or ED=112, the chord BD (of half the arch ABD)=204, and the chord BC (of half the arch BCD)=120, to find the area of the sector. 120 BC. 2 240 204 subtract. 3)36 12 240 Add. 252= 120 204 252 balf the arch ABD. 504 252 252 28224 area of the sector. Length of the arch ART. 26. To find the Area of a Segment of a Circle. Definition. A segment of a circle is any part of a circle cut off by a right line drawn across the circle, which does not pass through the centre, and is always greater or less than a semicircle. EXAMP. 1. To find the area of the segment ABC, whose chord AC is 172, the chord of half the arch ABC, viz. BC=104, and the versed sine BD=58.48. RULE. By Art. 23, find the length of the arch line ABC, and by Art. 24, the diameter FB; then multiply half the chord of the arch ABC by half the diam-" eter, and the product will be the area of the sector ABCE: then find the area of the triangle AEC, whose base AC is 172, and perpendicular height 34, found by subtracting the versed sine BD from half the diameter; and the area of the triangle AEC, being subtracted from the area D ΕΙ F of the sector ABCE, will leave the area of the segment ABC. EXAMP. 2. In the segment ABCD greater than a semicircle, given the chord of the whole segment AD=136, the chord AC of half the arch ACD=146, the chord AB or BC one fourth of the arch ACD =86, and the radius AE or ED= 80, to find the area of the segment ABCD. B First find the area of the sector ABCDE, by Art. 25, at the second Example; then find the area of the triangle AED, by Art. 6, and, adding the area of the triangle to the A area of the sector, you will have the area of the segment. 86 2 chord AB. 146 C E 136 8.666 172 146 chord AC, subtract, 3)26 180.666 arch line ABC. 11453.280 area of the sector. Carried over. Brought over. 68 half the base AD. 42 perpendicular E 136. - 136 272 2856 area of the triangle AED. 14453.28 area of the sector, [add. 17309-28 area of the segment. C Note 1. The area of a Lune or Crescent, is calculated by the preceding rule. A Lune is a figure made by two circular arcs, which intersect each other, as ACBD. The area of the Lune is the difference of the two segments, which are contained by the arcs and the chord. Thus the A difference of the segments ACBE and ADBE is the area of the crescent ACBD. Note 2. A Circular Zone is a figure contained between two parallel chords. If the chords be equal, it is called a middle zone, as A ABCD. The area of a zone is evidently the difference between the area of the circle and D the areas of the two segments. E D B B C ART. 27. To find the Area of an Ellipsis. Definition. An ellipsis, or oval, is a curve which returns into itself like a circle, but has two diameters, one longer than the other, the longest of which is called the transverse, and the shortest the conjugate diameter. RULE. Multiply the two diameters of the ellipsis together; then multiplying the product by 7854, this last product will be the area of the ellipsis. EXAMP. In the ellipsis ABCD, the fransverse diameter AC is 38, and the conjugate diameter BD is 72, to find B Mensuration of Superficies is easily applied to Surveying thus, take the angles of the plot with a good compass, then measure the G |