LESSONS IN MT 





4. The area of a circle is 18 ft. 142 in. (square measure) ; what 

 in itfl null . 



i.'.-nliir plot of ground contain* one acre; what u its 



We call the attention of the reader to Example! 1 and 2 

 Problem XIV. It will be there observed that when the 

 <iiametor is 1 the area ia '7854, and when it is 2 the area U 

 : 1 U i! that ia to say, double the diameter produces fowr times 

 t he area. Thin we might expect. But by further comparison 

 I >o soon that in either case the area is equivalent to the 

 f tlu> diameter multiplied by the same figures, '7854. 

 Henoo wo obtain 



Kule 2 : The area of a circle is equal to D* X '7854. 

 EXAMPLE 1. The radius of a circle ia 1 ; what is its area? 

 (See Example 2 under last rule.) 



H = 1, .-. i) = 2. And D 2 X '7854 = 4 X '7854 = 3'1416, 

 which corresponds with the answer obtained by Rule 1. 



EXERCISE 19. 



1. A circular table ia 59 inches in diameter ; what ia its area 

 by Eu 



2. Find tho area of a circle whose diameter ia 78 1 yards. 



PROBLEM XV. The circumference only of a circle being 



given, to find its area. 



Rule : Multiply tho square of the circumference by '07358.* 

 EXAMPLE. The circumference of a circle is 3 1-116 ; what ia 



itsareaP 



3-1416 3 X "07958 = -7854, nearly. 



This example proves tho correctness of the rule, because the 



circumference being = IT, D = 1, and by previous rule D 2 X 



7851 is iu this case '7854. 



EXERCISE 20. 



1. The circumference being 6, what is the area of the circle tf 



2. The circumference of a circular plot of ground is 246 yds. 

 1 foot 10| inches ; what is its area ? 



3. How many square yards are contained in a circular table 

 whose circumference is 11 feet? 



PROBLEM XVI. To find the area of the sector of a circle. 

 (See Definition in Geometry, Vol. L, p. 53.) 



Bule 1 : Multiply the radius by half the length of the arc of 

 the sector. 



Rule 2 : As 360 degrees is to the number of degrees in the 

 given arc, so is the whole area of the circle to the area of the 

 sector. 



EXAMPLE 1. The radius of a circle is 6, and the length of 

 the arc is 12 ; what is the area of the sector ? 



ByRulel. X 



- 6 X 6 = 86. 



By Rule 2. 360 : 114-6 : : 113'1 (whole area of circle) : 36. 



Note. A sector may assume the form of Fig. 22, as well as 



of Fig. 21. 



Fig. 23. 



Fig. 21. 



Fig. 22. 



EXERCISE 21. 



1. Find by both rules the area of a sector, the length of the 

 radius being 5 feet, and the length of the arc 20 feet. 



2. The arc of a circle contains 36 42' 16'', and the length of 

 the radius is 4 ; required the area of the sector. 



3. The diameter of a circle is 578 feet, and the number of 

 degrees in the arc is 93 48' 30" ; what is the area of the sector 

 in acres, roods, etc. ? 



PROBLEM XVII. To find the area of a segment of a circle. 



Rule 1 : When the number of degrees in the segment is 

 given. Find the area of the sector A c B E (Fig. 23), and then 

 the area of the triangle ABE. If the segment is less than a 



* This number, '07058, is produced by dividing uuity by *; .., 



4 w -= '07958. 



t The fact that the area appears less than the circumference ia 

 because the latter is square measure and the former lineaL 



semicircle, the difference of theae will be the ana of the M* 

 inent ; if it be greater, the area i* the earn. 



Rule 2 : Prom the arc of the segment subtract Mi tiM. and 

 multiply the remainder by half the radios. 



Bole 3: AreaJ *X V * . .,, whiob * i* the height 



8 4 10 



of the segment, or its versed aioa, and c U the chord. 



Not. Thia role ia approximate only. 



EXAMPLE 1. What ia the area of the segment of aotalt 

 the number of degrees being 27 and the length of the radio* tOf 



By Role 1. To find the area of the Motor AC (Fig. If). 

 Aa 860 : 27 : : 1256*6 (whole area of circle) > 94 24, area of 

 sector. 



Again : To find the area of the triangle ABB (Pig. 28). 



Because ^IABB = 27, .'. ^. A B D = 13'5, and ^ A D B is) a> 

 right angle. Hence DK A E X nat. sine ^ D A B. 



But D A E 90 - 13'5 = 7'5 ; 



.-. D B = 20 X "9724 = 19-448. 



Then in the right-angled triangle A D E, 



4-66. 



':-. 



AD= ^/AB 8 DB* = ^400 878-22i 



We have now a triangle in which the base A B and the alti- 

 tude D E is known ; hence it* area = A B x or 9*32 (4'66 X 2) 



X 19 ' 448 = 90.6. 

 2 



Finally, area of segment A c B D = 94*24 (area of sector) 

 90-6 (area of triangle) = 3*64. Ant. 

 Solve the same question by Rule 3. 



Area =-'552 (c E - DB) X </* 

 3 4 



= 736 X J 2i-7 X -122 = '736 X 4 78 = 3*518. Ant. 

 It will be seen, by comparing the two answers solved diffe- 

 rently, that they do not agree. A closer approximation would 

 result by carrying the decimal places further, bot the first answer 

 is nearest the truth. 



EXERCISE 22. 



1. Required the area of the segment of a circle, the number 

 of degrees in the arc being 107 30', and the length of the radio* 

 12-41. 



2. What is the area of a segment, the length of arc being 

 46'58, and the whole circumference being 156 P 



PROBLEM XVIII. To find the area of a circular 

 A B c D E F (Fig. 24). 



Draw the straight lines A E, B D ; the ,_. ?- ^ 



zone is then divided into a trapezoid, 

 A B D E, and two segments, A F E, BCD. 



Rule 1 : Find the area of the trapezoid 

 by rule to Problem XI., and also the area 

 of the two segments, A F E, BCD. The 

 sum of these areas will be the area of the 

 zone. Xx --' 



Rule 2 : Find the area of the two Beg- H 



ments AQB, EHD, which subtract from r . _. 



the area of the whole circle. 



EXAMPLE 1. The radius of a circle is 5. A tone of that 

 circle has one of its parallel chords passing through the centre 

 of the circle, and the other chord equals the radios. What 

 is the area of the zone ? 



By Rule 2. Area of whole circle = d 1 X ^ =100x *785478 54 



78 - 54 

 Area of greater segment = 5 (aenacircle) = 39*27. 



To find area of smaller segment. 



The chord of the arc of this segment being - radius, it 

 forms the base of an equilateral triangle, each _ being-" 



60. Hence number of degrees in arc of segment = (30. 



3 



Then by Tablea. Segment of 60 = *0906, and area = "0906 

 X r 3 (or 25) = 2*265. Then area of zone = 78 54 (area of whole 

 circle) less areas of segments (39*27 + 2*265), or 41*535 = 37*005, 

 Am. 



EXERCISE 23. 



1. The parallel chorda of a circular zone are 5 and 6. and the 

 diameter of the circle is 20 : what ia the area of the aone t 



