arc is 64 ft.; find the diameter of the circle, and also the length of the arc. (15) The span of a bridge, the form of which is an arc of a circle, is 200 ft.; its height is 42 ft.; find the length of the arch of the bridge. (16) The rise or height of a bridge, the form of which is an arc of a circle, is 12 ft., and its span is 78 ft.; find the length of the arch of the bridge. (17) The span of each of the three arches of a bridge, the form of each being the arc of a circle, is 240 ft., and the height of the crown in each above the stone piers is 24 ft.; find the length of each arch. XV. THE SECTOR OF A CIRCLE. Definitions.-The sector of a circle is the figure bounded by two radii and the part of the circumference between them. Thus AOCBA is the sector of a circle, being bounded by the two radii oa and oc, and the arc ABC between them; and is less than a semicircle. AOC is the angle of the sector AOCBA, or the angle subtended by the arc ABC. Also ADCOA is the sector of a circle greater than a semicircle, being B C D bounded by the two radii OA and oc, and the arc ADC. The straight line AC is the chord of the sector. RULES. (1) To find the area of a sector, when the radius of the circle and angle of sector are given. Find the area of the circle by rules in Prob. XII. Then we have the following proportion for finding the area of the sector : 360° number of degrees in angle of sector :: area of circle required area of sector. (2) To find the area of a sector, when the radius of the circle and the arc are given. Multiply the length of the arc by the radius, and divide the product by 2. Note 1.-To find the radius of the circle, when the area of the sector and the number of degrees in the angle of the sector are given.-The area of the circle must first be found from the following proportion : Number of degrees in the angle of the sector: 360° :: area of sector: required area of circle. Then, having found the area of the circle, its radius can be found by Note 1, Prob. XII. Note 2.-If the sector be greater than a semicircle, as, for instance, the sector ADCOA, then it will be better to find, first, the area of the sector ABCOA, which is less than a semicircle, by either of the preceding rules. Then the sector ADCOA Example 1.-Find the area of circle-sector AOCBA. area of the sector of a circle whose radius is 10 inches, and whose arc subtends at the centre an angle of 18°. Now, area of circle = 22 × 102=2200 sq. in. Then, by Rule 1: 360° 18° :: 2200 sq. in.: required area of sector. From this proportion, we find that 15 sq. in. is the area of the sector. Example 2.-The radius of a circle is 4 ft.; the arc of a sector of the circle is 3 ft. 4 in.; find the area of the sector. By Rule 2: Area of sector arc × radius=×40 × 48 =960 sq. in.=6 sq. ft. 96 sq. in. Example 3.-The chord of a sector is 56 in.; the radius of the circle is 35 in.; find the area of the sector. In this example, we must first find the chord of half the arc, which is 31.304 in. Then, by Rule 3, Prob. XIV.: Length of the arc (8x31.304-56)=64.81 in. Therefore, area of sector= (arc × radius) = × 64.81 × 35 =1134 175 sq. in. Formulæ II. 360° number of degrees in angle of sector area of circle required area of sector. EXAMPLES. (1) The radius of a circle is 14 in.; the angle which the arc subtends at the centre is 45°; find the area of the sector. (2) The radius of the circle is 2 ft. 11 in.; the angle of the sector is 24°; find the area of the sector. ft.; the (3) The radius of a circle is 56 ft.; the angle subtended by the arc at the centre is 224; find the area of the sector. (4) The area of the sector of a circle is 385 sq. angle of the sector is 36°; find the radius of the circle, and also the whole perimeter of the sector. (5) The area of a sector is 48.125 sq. ft.; the angle of the sector is 18°; find the radius. (6) The chord of a sector is 30 ft.; the radius of the circle is 25 ft.; find the area of the sector. (7) The chord of an arc is 40 ft.; the radius of the circle is 25 ft.; find the area of the sector. (8) The arc of a circle greater than a semicircle is 86 ft.; the radius of the circle is 21 ft.; find the area of the sector. (9) The chord of a sector is 20 ft.; the radius of the circle is 14.5 ft.; find the area of the sector. (10) The chord of the sector of a circle is 56 ft.; the radius of the circle is 53 ft.; find the area of the sector. 78 XVI. THE SEGMENT OF A CIRCLE. Definitions.-The segment of a circle is any part of a circle bounded by an arc and its chord. ABCA is a segment less than a semicircle, being bounded by the arc ACB and the chord AB. ABDA is a segment greater than a semicircle, being bounded by the arc ADB and the chord AB. N.B.-Segment ABCA = sector AOBCA-triangle AOB. A Ꭰ RULES.-(1) To find the area of the segment of a circle. Find the area of the sector AOBCA, having the same arc as the segment, by Prob. XV., and subtract from it the area of the triangle AOB, formed by the radii and the chord. (2) To find the area of the segment of a circle, when the chord of the arc and its height are given. To of the product of the chord and height of the segment, add the cube of the height divided by twice the chord; the sum is the area of the segment, nearly. Note 1.-To find the area of the segment AD BA, greater than a semicircle.-In this case, it will be better to find the area of the segment ABCA, less than a semicircle, by preceding rules. Then, the area of the segment ADBA area of circle-area of segment ABCA. Example 1.-The radius of a circle is 35 ft., and the chord of a segment is 42 ft.; find the area of the segment. Now, area of segment sector AO BCA-triangle AOB. By Prob. XIII., the chord of half the arc=22·13 ft.; and by Prob. XIV., the arc 45'01 ft., nearly. = Therefore, area of sector (45.01 x 35)=787.67 sq. ft. And area of triangle, having its sides 35 ft., 35 ft., and 42 ft respectively, is 588 sq. ft. Therefore, area of segment=787·67-588-199·67 sq. ft., nearly. Example 2.-Find the area of the segment of a circle whose chord is 35 ft., and the height of the segment is 9.6 ft. By Rule 2: Area=3(35 × 9·6) + =x336+ Formulæ. 9.63 35 x 2 884-736 70 =224+12-639-236 639 sq. ft., nearly. height3 I. Segment sector-triangle. EXAMPLES. [Work Examples 1 and 2 by Rule 1; the rest by Rule 2.] (1) Find the area of a segment, when the chord of the arc is 56 ft. and the radius of the circle is 35 ft. (2) The chord of a segment is 30 ft., and the radius of the circle is 17 ft.; find the area of the segment. (3) The chord of a segment is 24 ft., and its height is 9 ft.; find the area of the segment. (4) The chord of a segment is 40 ft., and its height is 15 ft.; find the area of the segment. (5) The chord of a segment greater than a semicircle is 24 ft.; the diameter of the circle is 26 ft.; find the area of the segment. |