The tetraedrons O-AES and J-AES have a common base, AES; and, therefore, are to each other as their altitudes. Denoting the tetraedrons by V and V', and (2) V: V' OY: JX OS: JS. = The tetraedrons S-EAO and S-AEJ have equal altitudes (?). These volumes are V and V'(?). Therefore, (3) V: V' EAO: EAJ (?). Complete. = 12. Two tetraedrons, which have a triedral angle of the one equal to a triedral angle of the other, are to each other as the products of the three edges of the equal diedrals. The bases are to each other as BD × BC is to BD' X BC' (?). The altitudes are to each other as BA is to BA'(?). 13. What is the entire surface of a cone whose slant is 50 feet, and whose base is 10 feet in diameter? 14. What is the volume of a cone having half the altitude of the above and double its base? 15. The dimensions of a bushel measure are 18 inches in diameter and 8 inches deep. What would be the dimensions of a gallon measure of the same shape? Ques.-In what ratio are similar solids to each other? 16. The altitude of a right pyramid is 2 feet, and the base is a regular hexagon whose sides are each 8 inches. Find the entire surface. 17. What is the edge of a cube equal in volume to the pyramid given in the 16th Ex.? SECTION XVIII.-THE SPHERE. DEFINITIONS. 1. A sphere is a solid which may be described by the revolution of a semicircle around its diameter as a fixed axis. The semi-circumference describes the convex surface. The center is the middle point of the axis. The radius is a straight line from the center to any A C point of the surface; and it is equal to the radius of the semicircle. A diameter is a double radius. Sch. As the semicircle ADB revolves about the axis AB, a perpendicular, as DC, let fall on the axis, from any point in the circumference, will describe a circle. Cor. 1.-All points on the surface of a sphere are equally distant from the center (?). Cor. 2.-Every plane section of a sphere is a circle (?). Ques. 1.-What shape is the horizon when viewed at sea, or upon any other level extent of surface; and at different distances from the surface? Ques. 2.-What does this prove as to the shape of the earth? 2. A great circle on a sphere is one whose plane passes through the center. Its radius is the same as the radius of the sphere. Its circumference is also called the circumference of the sphere. Any other circle on the sphere is called a small circle. Cor. 1.-All great circles of a sphere are equal (?). Cor. 2.-Two great circles of a sphere are never parallel (?). Cor. 3.-Every great circle of a sphere bisects the surface of the sphere (?). Cor. 4.-Two great circles of a sphere mutually bisect (?). Ques. 1.-Why does the intersection of the planes of two great circles pass through the center? Ques. 2.-If it passes through the center, what is it? Cor. 5. Of two small circles, the less is the farther from the center (?). 3. A portion of a sphere cut off by a plane, or included between two parallel planes, is called a segment. 4. The curved surface of a segment is called a zone. The altitude of the segment is the altitude of the zone. Ques. How is a zone bounded? 5. A frustum of a cone is said to be inscribed in a sphere, when the circumferences of its bases lie in the surface of the sphere. THEOREM XXVII. If a frustum of a cone be inscribed in a sphere, its convex surface will be equal to the altitude of the frustum multiplied by the circumference of a circle whose radius is a perpendicular from the center of the sphere to the slant height of the frustum. Let CD and GH be both perpendicular to the axis AB. Then, as the semicircle revolves about the axis, describing the sphere, the trapezoid DCGH will describe a frustum of a cone, which will be inscribed in the sphere. From the center K let fall the E A D F H K Hence, GC CI perpendicular KE, on the chord CG; then E will be the middle point of CG (Theo. XXXIII, Book I). Draw EF perpendicular to AB, and CI perpendicular to GH. Now, since the triangles GCI, KEF, have the sides of the one respectively perpendicular to the sides of the other, they are similar (Theo. XII, Book II). KE: EF. Multiplying an extreme and a mean equally (Theo. V, Book II. GC: CI 2 KE × 3.14159: 2 EF × 3.14159. The first term of this proportion is the slant height of the frustum; the second is its altitude; the third is the circumference of a circle whose radius is KE (Cor. 1, Theo. XXI, Book II); the fourth is the circumference of a circle whose radius is EF. Therefore, multiplying extremes and means, we have the product of the slant height into the circumference of a middle section of the frustum, equal to the product of its altitude into the circumference of a circle whose radius is the perpendicular from the center to the slant height. But the former product is equal to the convex surface of the frustum (Sch., Theo. XXVI); consequently, the latter product is equal to the same. That is, if a frustum of a cone be inscribed, etc. Sch. If the chord CG be one side of a regular inscribed polygon, is is evident that KE will be its apothegm (Book I, Theo. XXIX, Def. 2). Cor. If a square be rotated upon its diagonal as an axis, the convex surface of the double cone thus generated will be equal to the circumference of the circle inscribed in the square multiplied by the diagonal of the square. Similarly, rotate an octagon, and examine the surface of the figure. THEOREM XXVIII. The surface of a sphere is equal to the product of its diameter by its circumference. B G H K F Let ABCDEF be a semicircle, having the half of a regular polygon inscribed in it. As the semicircle revolves about the axis AF, describing the sphere, each of the trapezoids GBCH, HCDI, etc., will describe a frustum of a cone, which will be inscribed in the sphere. The convex surface of each of these frustums will be equal to its altitude. multiplied by the circumference of a circle whose radius is the apothegm of the polygon (Sch., Theo. XXVII). Therefore, the sum of their convex surfaces will be equal to the circumference of such a circle multiplied by the sum of the altitudes GH, HI, etc.; that is, multiplied by AF, the diameter of the sphere. Now, if the number of sides of the semi-polygon be indefinitely increased, its perimeter will ultimately coincide with the semi-circumference, and its apothegm with the radius, of the sphere. Then, the sum of the convex surfaces of the frustums will be equal to the surface of the sphere, and the circumference of which the apothegm is radius will be the circumference of the sphere (Def. 2). Therefore, the surface of a sphere is equal, etc. Cor. 1. Since the circumference is equal to 3.14159 × D (Cor. 1, Theo. XXI, Book II), it follows that the surface is equal to 3.14159 × D2 π D2 = 4R2. R2π is the area of a circle whose radius is R. Therefore, the surface of a sphere is four times the area of one of its great circles. Cor. 2. The area of a zone is equal to the product of its altitude by the circumference of a great circle. |