ART. 18. The Area of a Circle being given, to find the Diameter. RULE. Multiply the given area by 1-2732, and the product will be the square of the diameter; then, extracting the square root of the product, you will have the diameter.* EXAMP. The area of a circle being 113-09, to find the diameter. ART. 19. RULE. root of the 10187 remainder. The Area of a Circle being given, to find the circumference. Multiply the given area by 12.566, and extract the square product, which root will be the circumference required. EXAMP. The area of a circle being 113 03 to find the circumfe rence. ART. 20. The Side of a Square being given, to find the Diameter of a circle equal to the Square, whose Side is given. RULE. Multiply the given side by 1.128, and the product will be the diameter of a circle, whose area is equal to the area of the *As the area of a circle, whose diameter is 1, is 7854, the area divided by 7854 must give the square of the diameter; but as 1.2732 is the reciprocal of 7854, the rule is evident. 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, 10.635886=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=37-686-circum. Or, 282) 10.631(37-693 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. EXAMP. In the segment ABCD, the whole chord ADC is 216, and the chord AB or BC 126, to find the arch line ABC. 120 216 B D 252 double of AB. 12=1 difference added. 264 length of the arch ABC. 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. EXAMPLE. In the segment ABCD, the chord AC is 1869-5, A 1869,5 and the versed sine BD 423-5, to find the diameter. E 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 the area. D 144.16-length of the arch 5207-76=area. [ABC, by Art. 23. 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 146 C E 136 8.666 172 double of AB, add. 180.666 arch line ABC. 80=radius. 11453.280 area of the sector. Carried over. |