THE POPULAR EDUCATOR. 



The seal is set. Now welcome, thou dread power, 

 Nameless, yet thus omnipotent, which here 

 Walkest iu the shadow of the midnight hour." Byron. 



Power is the antecedent, and wliich the relative : the sense 

 chows that the relative ivhich has reference to the antecedent 

 power. 



This and that, demonstrative pronouns, differ thus : this points 

 out the person or thing nearer to the speaker, that the more 

 remote ; as 



" Bring me that book." " This book ?" " No, that hook which stands 

 on the farthest part of the shelf." 



LESSONS IN MENSURATION. III. 



PROBLEM IV. Given the chord B E (Fig. 16) and the height 

 or versed sine c D of an arc of a circle, to find the diameter of 

 the circle. Rule : Divide the square of half the chord by the 

 height ; the quotient will be the diameter, 

 less the height. 



EXAMPLE. The chord being 8, and the 

 height 4, what is the diameter of the circle 

 of which the arc is a part ? 



\ 



Fig. 16. 



= 16 ; 

 \2/ 



= 4 = diameter, less height. 



.-. diameter = 4 + 4 = 8. 

 Note. In this example the correctness of 

 the rule is proved, since the diameter = the chord B E, and half 

 of this = radius = 4 ; i.e., the height of c D. 

 EXERCISE 8. 



1. The chord of an arc is 18'8, and the height is 8 ; what is 

 the radius ? 



2. In levelling for a canal, a certain allowance must be 

 made for the curvature of the earth, since the line of sight 

 is absolutely horizontal, in taking the level. This allowance 

 it is found necessary to make to the extent of 8 inches per 

 mile. Hence, what is the diameter of the earth ? 



PROBLEM V. Given the height of the arc, or the Versed 

 sine c D (Fig. 16), and the chord of half the arc B c, to find the 

 radius. Rule : Divide the square of the chord of half the arc by 

 the height of the arc ; the quotient will be the diameter, which 

 halve for the radius. Therefore the height of the arc is the 

 square of the chord of half the arc divided by the diameter ; 

 and the chord of half the arc is the square root of the diameter 

 multiplied by the height of the arc. 



EXAMPLE. The height of an arc is 16 feet, and tho chord 

 of half the arc is 32 feet ; what is the diameter of the arc ? 



32 2 = 1024 -r 16 = 64 = diameter. 

 EXERCISE 9. 



1. The neight of an arc is equal to half the chord of the whole 

 arc, and this is 25 feet. What is the chord of half the arc ?* 



2. The circular arch of a bridge rises 12 feet above the 

 water, whose level touches the spring of the arch. The radius 

 of the arch is 100 feet. How far is it in a direct line from the 

 spring of the arch to its crown ? 



3. The arch of a bridge forms a part of a circle. The river 

 beneath rises to the spring of the arch, and is 80 feet wide ; 

 and a boat's mast, 16 feet high, can just pass clear of the arch 

 when in mid-stream. With what radius was the arch struck ? 



PROBLEM VI. The circumference of a circle being given, to 

 find the length of an arc of it, the number of degrees, etc., it 

 contains being known. 



This is a simple and obvious case of proportion. Hence the 

 Rule : As 360 (the number of degrees in the whole circle) is 

 to the number of degrees, etc., in the arc, so is the length of the 

 whole circumference to the length required. 



EXAMPLE. The circumference of a circle measures 31,416 

 yards; what is the length of an arc of that circle containing 

 90? 



360 : 90; 

 4 1 



: 31,416 yards . r^? =7,854 yards. 

 4 



* The equality of the two given elements in this example shows 

 them to be the radii of a circle ; hence the chord of half the arc, being 

 the hypothenuse, can be found by Euc. I. 47. 



EXERCISE 10. 



1. What is the length of a degree at the earth's equator, 

 supposing the circumference of the earth at that part is 24,900 

 miles ? 



2. The radius of a circular plot of ground is 28 feet. How 

 many degrees will be contained in a portion of the circum- 

 ference which measures 29'3216 feet ? 



PROBLEM VII. To find the length of an arc, when the chords 

 of the whole and of half the arc respectively are given, or when 

 any lines in Fig. 16 are given by which these can be ascertained 

 by previous rules. Rule : Subtract the chord of the whole arc 

 from 8 times the chord of half the arc, and divide the remainder 

 by 3. 



EXAMPLE. The chord B E of the whole arc (see Fig. 16) is 

 20, and the radius A c is 14 ; what is the length of the arc ? 



To find B c, the chord of half the arc A B, the radius, and 



(BEN 

 x- ) being known proceed by Euc. I. 47. 

 i- / 



Thus, A D = A/A B 2 - B r> 2 = A/196-100 = A/96 = about 9'8 ; 

 D c A c - A. D = 14 - 9-8 = 4-2 ; 



2 = A/100 + 17-64 = A/117-64 = about 



Then, by the rule, l^ 85x8 " 20 - ?|? = 22'26 = length of 



arc, about. 



EXERCISE 11. 



1. The chord of the whole arc is 36 feet 9 inches, and the 

 chord of half the arc is 23 feet 3 inches. What is the length 

 of the arc ? 



2. The span of a circular arch is 48 feet, and the length from 

 the spring of the arch to its crown is 30 feet in a direct line. 

 How many stones of 9 inches each compose the arch ? 



There might follow several other problems having reference to 

 the mensuration of lines, such as the finding of the length of 

 the diameter of a circular zone that is, of the circle of which it 

 forms a part ; of the component parts of an ellipse, some of the 

 parts being given ; the same of a parabola ; but all these, 

 although interesting, are not of such importance as those we 

 have given, and we shall therefore at once proceed to the con- 

 sideration of the next part of our subject, namely, the mensura- 

 tion of surfaces that is, of spaces entirely enclosed by lines. 



The necessity for considering the measurement of lines as 

 introductory to that of superficies will be apparent from a con- 

 sideration of the fact that the area of surfaces are necessarily 

 dependent upon the lines which enclose them, and that, there- 

 fore, a knowledge of the one gives us at once the key to a 

 knowledge of the other. 



Take, for instance, our first problem under this head : 



PROBLEM VIII. To find the area of a rectangular parallelo- 

 gram that is, of a four-sided figure, both 

 pairs of its opposite sides being parallel, and 

 its angles right angles. Rule : Multiply the 

 lengths of any two adjacent sides together; 

 the product is the area. 



Let ABCD (Fig. 17) be a parallelogram; c Fig. 17. 

 then A B X A c = area. Hence, if A B and A c 

 be equal, the area is represented by A B 2 . If the figure be not 

 rectangular, as E F G H (Fig. 18), find the perpendicular E K, as 

 explained in our last lesson, by mul- 

 tiplying the hypothenuse E G by the 

 natural sine of the angle E G K, and 

 proceed by the above rule. 



EXAMPLE 1. The two sides of a 

 rectangular parallelogram are 5 and 

 6; what is its areaP Ans. 5 X 6 = 30. 



EXAMPLE 2. The side of a square is 10 ; what is its area? 

 Ans. 10 X 10 = 100. 



EXAMPLE 3. The sides of an oblique-angled parallelogram 

 (Fig. 18) are 15 and 10, and the angle at G is 30; what is its 

 area ? 



Side E G = 10 ; natural sine of angle EGK or 30= '5. Hence 

 E K = IP X '5 = 5 ; and area = E F X E K, or 15 X 5 = 75. 



EXERCISE 12. 



1. The sides of a rectangular parallelogram are respectively 

 25 and 4 ; what is its area ? 



Fig. 18. 



