I. EXAMPLES. Find the area of a circle whose diameter is 56. 2. Find the area of a circle whose diameter is 13 ft. 4 in. 3. Find the area of a circle whose diameter is 24 poles. 4. 5. is 47. 6. Find the area of a circle whose diameter is 34ft. 10 in. Find the circumference of a circle whose diameter Find the circumference of a circle whose diameter 8. The radius of a circle is 9 feet; find the length of an arc which contains 24°. Ans. 333 ft. 9. The radius of a circle is 3 feet; find the length of an arc which contains 36° 17'. Ans. 19 ft. IO. links. Find the area of a circle whose diameter is 476 II. Find the diameter of a circle whose area is 386. o. 10 square feet. Ans. 22 ft. 2 in. 12. Find the diameter of a circle whose area is II a. 3 r. 287 P. 13. Find the circumference of a circle whose diameter is 12 ft. 6 in. 14. Find the area of a circular ring whose inner and outer diameters are 13 ft. 8 in. and 7 ft. 6 in. Ans. 102.6.8 feet. 15. The radius of a circle is 12; find the area of a sector which contains 15o 20'. Ans. 19.274. 16. The radius of a circle is 9 feet; find the area of a sector which contains 35°. Ans. 2423 feet. 17. The radius of a circle is 37 ft. 4 in.; find the area of a sector which is of a quadrant. Ans. 136 feet. 18. Find the area of a circular ring whose inner diameter is 6 chains 25 links, and outer diameter 10 chains 35 links. Ans. 5 a. Ir. 15 p. 19. The radius of a circle is 17 ft. 3 in.; find the area of a sector which contains 24° 25'. Ans. 63 425 feet. 20. The area of a circular sector is 343, and it contains an angle of 11° 15'; find the radius. Ans. 183. Let ACB be a semicircle, of which AB is the diameter, and let CD be perpendicular to AB. Then AD is called the abscissa, and CD the ordinate of the point C; and by a property of the circle, AD × DB = CD3, or CD= √(AD × DB). Ex. The diameter of a semicircle is 40, which is divided into 8 equal parts. first three points of division. The distance between the ordinates is 5; and each pair of abscissæ are 5, 35; 10, 30; 15, 25. Hence the ordinates are √ (5 × 35) = √(175) = 13°228, √(10 × 30)=√(300) = 17°320, √(15 × 25) = √(375) = 19′365. EXAMPLES. I. The diameter of a semicircle is 35, which is divided into 7 equal parts; find the ordinates at the first three divisions. Ans. 12 247, 15.811, 17°320. 2. The diameter of a semicircle is 146, and it is divided into two parts, 25 and 121, by a perpendicular ordinate. The segment 25 is divided into 5 equal parts. Calculate the ordinates at the points of division. Ans. 265518, 36.8781, 44°3282, 50*1995. THE ELLIPSE. Let S and Hbe two fixed points, and let P be a point such that the sum of SP and HP is constant; the locus of Pis an ellipse; that is, if a piece of thread be attached to two pins fixed at S and H, and kept tight by a pencil at P, the point P will trace out a curve called the ellipse. Let the ellipse cut the straight line SH produced in A, A'; bisect AA' in C, and through C draw BB at right angles to AA', meeting the ellipse in B, B. Then AA' is the transverse diameter of the ellipse, On AA' describe a semicircle, and from any point D in AA' draw the ordinate DEF at right angles to AA' meeting the ellipse in E, and the semicircle in F Then DE is a fourth proportional to AC, BC, DF; that is, AC: BC: DF : DE. Also, if def be any other ordinate; AC BC area FDdf: area EDde. XVIII. To find the area of an ellipse. Multiply the transverse by the conjugate diameter; then multiply by 11 and divide by 14. Ex. I. Find the area of an ellipse whose transverse diameter is 18 ft. 1 in., and conjugate 12 ft. 5 in. Ex. 2. The transverse and conjugate diameters of an ellipse are 12 and 8; find the ordinates which divide the transverse diameter into 6 equal parts. The segments of the transverse diameter are 2 ; hence the abscissæ are 2, 10; 4, 8; 6, 6; 8, 4; 10, 2. The middle ordinate will be equal to the semi-conjugate diameter, and the last two the same as the first two. The ordinates in the semicircle are √ (2 × 10) = √ (20) = 4°4721, No↓ (4 × 8) = √ (32) = 5·6568. .. 6: 44°4721 : 2.981, :: 56568 3771. Ans. 2'981, 3771, 4, 3'771, 2·981. EXAMPLES. I. Find the area of an ellipse whose transverse and conjugate diameters are 16 ft. 4 in. and 13 ft. 5 in. Ans. 172.2. 2 feet. |