RULE. Find the area of a sector which has the same arc as the seg-ment; also the area of the triangle formed by the chord of the segment, and the radii of the sector. Then take the difference of these areas when the segment is less than a semicircle, and the sum when it is greater. EXAMPLES. 1. Find the area of a segment of a circle, the chord being 12 feet, and radius 10 feet. Ans. 16.3504 sq. ft. 2. Find the area of a segment of a circle whose arc contains 10°, the radius being 10 feet. Ans. 6.336 sq. ft. PROBLEM XIII. To find the area of a circular ring; or the area included between the circumferences of two circles which have a common centre. is is or, Let R denote the radius of the greater circle. Let r denote the radius of the less circle. Then (Geom., B. IV., th. 17) the area of the larger πηλ ..πR2—π2-area of the circular ring; π(R2—Î2), or π(R+r) (R—r). Whence we derive the following RULE. Multiply the sum of the two radii by their difference, and that product by 3.1416. EXAMPLE. What is the area of the space between two concentric circles whose radii are 10 and 6? Ans. 201.0624, MENSURATION OF SOLIDS. THE mensuration of solids consists of two parts: SURFACES OF SOLIDS. PROBLEM I. To find the convex surface of a right prism. Let ABCDE-K be a right prism. Then, since the planes FE, FB are perpendicular to the base, their line of inter- F section FA is also perpendicular to the base (Geom., B. V., Prop. XVII.), and therefore the angle FAB is a right angle. Hence the area of ABGF is ABX AF. Similarly for all the other parallelograms. A Hence the area of the parallelograms taken together is AFX (AB+BC+CD+DE+EA). From which we derive the following RULE. H Multiply the perimeter of the base by the altitude, and the product will be the convex surface. When the entire surface is required, we must add the area of the two bases to the convex surface. EXAMPLE. What is the convex surface of a cube, the length of each side being 20 feet? Ans. 2400 sq. feet. PROBLEM II. To find the convex surface of a right cylinder. Since the cylinder is the limit of the convex surface of the inscribed prism, whatever may be the number of its sides, therefore the convex surface of a cylinder is obtained by multiplying the circumference of its base by the altitude. EXAMPLE. What is the convex surface of a cylinder whose length is 20 feet, and the circumference of its base 3 feet? PROBLEM III. Ans. 60 sq. feet. To find the convex surface of a right cone, or right pyramid. Let BCDE-A be any right cone, BCDE its circular base. In the circle inscribe any regular polygon, BCDEFG, and join AC, AD. Then will these be the edges of the inscribed pyramid. Bisect any side, CD in H; join AH; then AH is perpendicular to CD (Geom., B B. I., th. 3, Cor. 1); and therefore the area of the triangle ACD=CD × AH. A Now the line joining A with the point of bisection of any other side of the polygon is equal to AH. Hence the convex surface of the inscribed pyramid is =†AH (BC÷CD+DE+, &c.), AHxperimeter of the base. Now this is true whatever be the number of sides of the polygon forming the base, and, hence, is true in the limit. If, therefore, we increase the number of the sides indefinitely, the polygon will become the circumference of the circle, and AH the slant height of the cone. Therefore, The convex surface of a right pyramid or a right cone is equal to half the slant height multiplied by the perimeter, or circumference of its base. Rem. If we suppose the surface of the cone to be un wrapped, it is evident that its surface will be a sector of a circle whose radius is the slant height of the cone, and the arc of the same length as the circumference of the base of the cone. It is plain that the area of this sector is, as it should be, the same as that of the convex surface of the cone. PROBLEM IV. To find the convex surface of a frustum of a right pyramid. Since the section is similar to the base (Geom., B. VII., Prop. VII.), and the base is a regular polygon, it follows that the sides of the section are all equal to each other. Hence the convex surface is composed of equal trapezoids, whose common altitude is equal to the slant height of the frustum. But the area of any one of these trapezoids is equal to half the sum of the parallel sides multiplied by the altitude (Prob. VI.). Hence the convex surface of a frustum of a right pyramid is obtained by multiplying half the sum of the perimeters of the two bases by the slant height of the frustum. EXAMPLE. How many square feet are in the surface of a frustum of a square pyramid whose slant height is 10 feet; also each side of the greater base is 3 feet 4 inches, and each side of the less base is 2 feet 2 inches? Ans. 110 sq. feet. B and if its base PQ-circumference BE, we have seen that the area of the sector OPQ is the same as the convex surface of the cone ABE (Prob. III., Rem.). D Take Op Ab, and describe the arc pq; then, as before, the area of Opq=convex surface of the cone Abc; and, therefore, the convex surface of the cone is equal to pqQP. Let the angle POQ=0. OP=OQ=R. ᎡᎯ but is equal to the circumference of the lower base; 180° 180° -circumference of the upper base (Prob. XI., Mens. Surfaces); and R-r-slant height of the frustum. Therefore, to find the convex surface of a frustum of a cone, we have the following RULE. Multiply half the sum of the circumferences of the two bases by the slant height. Cor. If we bisect pP in t', and describe the arc t't, then Therefore the convex surface of a frustum of a cone is equal to the product of the slant height, and the circumference of a section at equal distances between the two bases. EXAMPLE. What is the convex surface of a cone whose slant height is 10 feet, and the circumferences of its bases are 6 feet and 4 feet? Ans. 50 sq. feet. PROBLEM VI. To find the surface of a spherical zone, and also that of a sphere. |