Or, again (which comes to A the same thing), if we square the radius, we get the area of EFBG, and it is evident that the area of the circle is not so much as four times this; the truth is that it is exactly 3.14159 times the square of the radius, or, as it is more commonly expressed, area of circle = r2. D G E These two latter rules are sometimes stated in words thus: To find the area of a circle 1. Multiply the square of the diameter by 7854; or, 2. Multiply the square of the radius by 3.14159. While dealing with the subject of the circle, it will be as well to discuss the method for find ing the area of a ring, i.e., of the space included between the circumference of two concentric circles. It is evident that the area of the ring may be found by subtracting that of the smaller circle from that of the larger. This, however, may be put in a simpler form thus: Let R denote the radius of the larger circle, Then area of larger circle and area of smaller circle = smaller circle. TR2, r2; Ꭱ T The rule may be expressed in words thus: - To find the area of a ring, multiply the sum of the radii by their difference, and this product by 3.14159. EXERCISES (H). Find the circumference of each of the circles whose diameters are respectively 1. 28 yds. 2. 120 ft. 3. 24.25 ft. 4. 7912 miles. 5. 3 ft. 9 in. 8. 7958 miles. 9. 483 ft. 10. 126 ft. Find the diameter of each of the following circles whose circumferences are respectively 11. 144 miles. 13. 9 ft. 6 in. 18. 64 ft. 19. 483 ft. 20. 55 in. Find the area of each of the circles whose dimensions are as follows: 25. How many sq. yds. of carpet will be required to cover a circular floor 73.25 ft. across? 26. A coach wheel turns round 250 times in travelling half a mile; what is its diameter? 27. A locomotive engine is travelling at the rate of 60 miles an hour, and the radius of its driving wheel is 3 ft.; how many times does it turn round in a second? 28. To enclose a circular garden there are required 2351⁄2 yds. of wire fencing; what is the area of the garden? 29. A circular field contains exactly one acre; find the length of the hedge which encloses it. 30. A circular meadow is 525 ft. in diameter; what would be the cost of making round the outside of it a road 21 ft. 6 in. wide, at 2s. 6d. per square yard? 31. A square is inscribed in a circle whose radius is 10 ft.; find the side of the square. 32. A large circular iron plate, 12 ft. in radius, weighs 15 lb. for every square inch in its surface; find the weight. 33. A circular gravel walk contains exactly 100 sq. yds., and its inner edge is 14 yds. long; what is the length of its outer edge? 34. The circumferences of two concentric circles are 62.832 ft., and 37 6992 ft.; find the area of the ring enclosed between them. 35. The inner diameter of a circular ring is 6 chains 25 links, and its outer diameter 10 chains 35 links; find its area. 36. The radius of a carriage wheel is 1 ft. 9 in.; how many times will it turn round in going 2 miles? LESSON XII.-SECTORS AND SEGMENTS OF CIRCLES. Η G K A portion (HGK) of the circumference of a circle is called an arc. A figure which is bounded by two radii and an arc of a circle is called a sector, as AOB, FOE, GFOA. A figure which is bounded by a straight line drawn through a circle, and the part of the circumference which it cuts F off is called a segment of the circle, as HKG. It will be seen that the sector FAO of the circle is something like a triangle, but E D B A having a curvilinear instead of a rectilinear base. The arc ABCDEF may, however, be regarded as composed of a great number of very short straight lines, and the whole sector, as thus composed, of a great number of triangles. Now the area of each triangle is half the product of the base and the perpendicular height; but if we regard the number of triangles as very great, the bases would practically coincide with the arc, and the perpendicular with the radii. We may, therefore, regard the whole arc as corresponding to the base of the triangle, and the radius of the circle as representing the height. The rule may then be stated thus: To find the area of the sector of a circle, multiply half the length of the arc by the radius. A more satisfactory proof of this rule, which some of you perhaps will understand, is as follows: Sectors of a circle are to each other as the arcs on which they stand. (Euclid VI., 33). .. Sector quadrant :: arc of sector : circumference. The rule for finding the area of a segment may easily be deduced from this: If the segment is less than a semicircle, ABC, its area may be found by subtracting from the sector AOCB the area of the triangle AOC; and if the segment be greater than a semicircle, as ADC, its area may be formed by adding to the area of the sector ADCO that of the triangle AOC. In the same way the area of a lune or crescent (i.e., the space included between the arcs of two eccentric circles, as ADBC) is evidently the difference between the areas of the segments ADB and ACB. Eccentric circles are those which have not the same centre. EXERCISES (I). 1. The diameter of a circle is 3 yds., and the arc of a sector of it contains 96 degrees; find the area of the sector. 2. Find the area of a sector the radius of which is 10 ft. and the arc 20 ft. 3. The radius of a sector is 8 yds. 1 ft., andi ts are contains 147° 29'; find its area. 4. A sector greater than a semicircle has arc 14 yds., and the radius 2 yds.; find its area. 5. The radius of a circle is 32 yds., and the arc of a sector of it 187° 37'; find the area. 6. A sector of a circle contains 240 sq. ft., and its arc 150°; find the radius. 7. The radius of a circle is 2 yds., and its arc contains 57° 30'. 8. The radius of a sector and the length of its arc are respectively 12 ft. and 16 ft.; find the area. 9. The radius of a sector of a circle is 25 ft., and the arc contains 218° 54'; find its area. 10. The radius of a quadrant is 27 ft.; find its area. LESSON XIII.—THE ELLIPSE. An ellipse, or oval, is like a circle flattened in one direction and elongated in the other. You will not fully understand this lesson until you have gone farther than at present in your mathematical studies; but it will help you at once if you read carefully what is said about the cone in the mensuration of solids farther on in this book.. An ellipse is made by cut ting through a cone obliquely. The base of a cone is circular, and every horizontal section of it (ie., every section which is parallel to the base, as in fig. A) is circular also, the circle becoming larger the nearer the section is made to the base of the cone. But an oblique section (as in fig. B) forms an ellipse; and the nearer the section is to being B perpendicular, the narrower is the ellipse; while the nearer |