238 



THE POPULAR EDUCATOR 



2. The radius of a circle is 14, and the lengths of the parallel 

 chords of a zone of that circle are 22 and 28. Eequired the 

 area of the zone. 



PROBLEM XIX. To find the area of a circular ring, A B c D, 

 A' B' c' D' (Fig. 25), that is, of the space included between two 

 concentric circles. 



Eule : Find the area of the interior circle, which subtract 

 from the area of the outer circle. 



This simple problem and its rule is so self-evident as scarcely 

 to need an example. We refer our reader to Problem XIV. for 

 all the information requisite. We give one example for practice. 



EXAMPLE. The diameter of the earth's orbit being (approxi- 

 mately) 180,000,000 miles, and that of the earth being 7,912 

 miles, what part of the superficial area of the orbit is occupied 

 by a plane passing through the diameter of the earth and 

 bounded by its circumference P Ans. 517,574,000. 



Fig. 25. 



Fig. 26. 



PROBLEM XX. To find the area of a lune, as A B c D (Fig. 26). 



Rule : Find the area of the two segments, and their difference 

 will be the area of the lune. 



EXAMPLE. What ia the area of a lune, the chord AC of 

 which is 48, and the height- or versed sine of the two arcs 10 

 and 7 respectively? Ans., about 103. 



PROBLEM XXI. To find the superficial area of a sphere. 



Rule 1 : Multiply the circumference ef the sphere by its 

 diameter.* 



Rule 2 : Multiply the square of the diameter by 3'1416. 



EXAMPLE. The diameter of the earth being 7912 miles, 

 what is its superficial area, supposing it a perfect sphere ? 



By Rule 2, d 2 ir = 62,599,744 X 3'1416 = about 196,663,356 



mile3 - EXERCISE 24. 



1. What is the superficial area of a sphere whose radius is 1*5 ? 



PROBLEM XXII. To find the surface of a regular solid ring. 



Rule : Find the length of the ring by adding together the 

 exterior and interior diameters, and by multiplying half their 

 sum by TT. This product, multiplied by the circumference of 

 the cross section of the ring, will give its superficies. 



The formula is- X * X c area, in which D = exterior 



m 



diameter, d = interior diameter, and c = circumference of cross 

 section of ring. 



EXAMPLE. The inner and outer diameters of a ring are 8 

 and 12 ; what is its superficial area ? Ans., about 197'4. 



PROBLEM XXIII. To find the area of an ellipse or oval. 



Rule : Multiply the product of the axes by -. 



4 



Note. An ellipse is a figure formed by a plane cutting a cone 



in a direction parallel to neither side of the cone, nor yet to its 



base. It has two axes, the long and the short, as A B, c D 



(Fig. 27). A/f - 7,6 



c 



Fig. 27. c 



Fig. 28. 



EXAMPLE 1. The major axis of an ellipse is 10, and the 

 minor axis 7 ; what is its area ? 



10 x 7 X -. = 70 X 

 4 



7854 = about 54-98. 



EXERCISE 25. 



1. What is the area of an ellipse whose axes are respectively 

 12 and 9 ? 



Having thus briefly glanced at the more important problems 

 in the measurement of surfaces, we shall occupy the attention 

 of the student for a very short space in noticing the question 

 or the contents of solids. 



* The superficial area of a sphere is equal to four times the area of 

 a ~ iaiic massing through its diameter. 



As a general definition, a solid may be regarded as a body 

 having length, breadth, and thickness, and in this sense it of 

 course includes liquids. It is, in fact, anything which is 

 boundc' 5 by surfaces in all directions. The measurement of a 

 solid is called its cubical content, and it involves three separate 

 acts of multiplication. If the figure be a cube, its content i 

 measured by the cube of one of the lines which connect any 

 two adjacent angles. Thus, in Fig. 28 the contents will 

 be represented by A B 3 , or A c 3 , or o u 3 . Suppose, however, that 

 the length of A B = 1, then I 3 = 1 ; that is, the solid content of 

 a cube whose side is unity is also unity, the difference being 

 between lineal and solid measure. 



EXAMPLE. The solid content of a cube is required whose 

 side measures 10. 



10 X 10 X 10 (or 10) = 1,000, the solid measure. 



EXERCISE 26. 



1. What is the solid content of a cube whose side measures 

 2 ft. 6in. P 



In order to find the length of the side of a cube whose solid 

 content is known, extract the cube root of the contents. 



In order to find the solid content of any other form of paral- 

 lelopipedon than a cube, multiply the length by the breadth, 

 and that product by the height. 



EXAMPLE 1. Required the c>-:aent of a parallelopipedon 

 whose length, breadth, and height are respectively 12, 6, and 3. 



12 X 6 = 72 ; 72 X 3 = 216. Ans. 



EXAMPLE 2. What is the content of a parallelopipedon 

 whose length is 3 ft. 2 in., its breadth 2 ft. 3 in., and its height 

 1 ft. 4 in. ? Ans., 9 cub. ft. 864 cub. in. 



The solid content of a prism or cylinder is found by multi- 

 plying the area of the end by the length. 



EXAMPLE 1. A hollow cylinder is 12 inches in diameter 

 inside and 12 inches high ; how many cubic inches of water 

 will it contain ? 



12 2 x '7854 = 113-1, nearly = area of base. 

 113-1 X 12 = 1357-2 cub. inches. 



EXERCISE 27. 



1. Required the solid content of a triangular prism whose 

 height is 3 feet, and the breadth of each side 6 inches. 



The solidity of a cone is determined by multiplying the area 

 of the base by one-third the height. 



The cubic content of a sphere or globe is ascertained by 

 multiplying the cube of the diameter by '5236 ; this number 



7T 



being - . Another rule is to multiply one-third of the surface 



by the radius. 



EXAMPLE 1. What is the content of a globe whose diameter 

 is 10 ? iQ3 = 1,000 ; 1,000 X -5236 = 523'6. 



EXERCISE 28. 



1. The mean diameter of the earth being 7,912 miles, what is 

 its cubic content, supposing it a regular sphere P 



To find the solid content of a segment of a sphere, add the 

 square of its height to three times the square of the radius of 

 the base. Multiply the sum by the height, and that product 



ky ~6 EXAMPLE. 



What is the content of the segment of a sphere whose height 

 is 2, and the diameter of the base 8 ? 



By rule, 2 2 + (3 X 4 2 ) = 52 ; 



52 X 2 X '5236 



= 54-4544. Ans. 



The solid content of a regular solid ring is found by multi- 

 plying the area of the cross section of the ring by its length, 

 the length being found by multiplying the mean diameter (that 

 is, half the sum of the inner and outer diameters) by it. 



We need scarcely observe that there are many more problems 

 in connection with the measurement of the content of solid 

 bodies as, for instance, of a circular spindle ; of a spheroid, or 

 the segment of a spheroid ; of a paraboloid, or the frustrum of 

 a paraboloid ; of a hyperboloid, or its segment ; and so on. But 

 our subject is directed principally to a consideration of the areas 

 ef flat surfaces, so as to apply the rules to the measurement of 

 land. We have merely and briefly called the student's attention- 



