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Ex. 1776. Through a sphere whose diameter is 10 in. a cylindrical hole of 6 in. diameter is bored. Find the volume of the solid if the axis of the cylinder passes through the center of the sphere.

Ex. 1777. The radius of a sphere is r, the area of a small circle a. Find its distance from the center.

Ex. 1778. The volume of a sphere is V. Find the surface of an equilateral spherical triangle whose angle is equal to 100°.

Ex. 1779. Find the area of the torrid zone if its altitude is of the radius of the earth.

Ex. 1780. How many miles from the surface of the earth can of the surface be seen?

Ex. 1781. The average pressure of the atmosphere upon each square inch of the surface is 15 lb. Find the total weight of the atmosphere.

Ex. 1782. How many bullets + of an inch in diameter can be made from a cubic foot of lead ?

Ex. 1783. A coffee pot is 8 in. high, 4 in. in diameter at the top, and 5 in. in diameter at the bottom. How many cups of coffee, each having a capacity of 10 cu. in., could the pot hold ?

Ex. 1784. The places, A, B, and C, on the surface of the earth determine a spherical triangle ABC, whose angles are: A = 50°, B = 61°, C=71°. Find the area of triangle ABC.

APPENDIX TO SOLID GEOMETRY

PROPOSITION I. THEOREM

798. A truncated triangular prism is equivalent to the sum of three pyramids whose common base is the base of the prism and whose vertices are the three vertices of the inclined section.

[blocks in formation]

Given ABC-HKG, a truncated triangular prism of which ABC is the base.

To prove ABC-HGK = H-ABC + K-ABC + G-ABC.

Proof. Pass planes through H, B, C, and through H, K, C, forming the pyramids H-ABC, H-BCK, and H-CKG.

H-ABC is evidently one of the required pyramids.

H-BCK may be read C-HBK.

C-HBK = C-AKB.

C-AKB may be read K-ABC.

(628)

.. H-BCK is equivalent to the second of the required

pyramids.

H-CKG = H-GBC.

H-GBC or B-HGC = B-AGC or G-ABC.

Hence H-CKG = to the third of the required pyramids. ... ABC-HKG = H-ABC + K-ABC + G-ABC.

(628)

Q. E. D.

799. COR. 1. The volume of a truncated right triangular prism is equal to the product of its base by one third the sum of its lateral edges.

800. COR. 2. The volume of any truncated triangular prism is equal to the product of its right section by one third the sum of its lateral edges.

HINT. The right section divides the truncated prism so that two truncated right prisms are formed.

Ex. 1785. Find the volume of a truncated right triangular pyramid, if the lateral edges are 4, 5 and 6 and the edges of the base are 6, 8, and 10.

Ex. 1786. The sides of the base of a truncated triangular prism are 13, 14, and 15 and the lateral edges are 6, 8, and 10. Find the volume if the inclination of the edges to the base is 30°.

PROPOSITION II. THEOREM

801. Two similar polyhedrons can be decomposed into the same number of tetrahedrons similar each to each

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Given R and S, two similar polyhedrons; A and A', homologous

vertices.

To prove R and s can be decomposed into the same number of tetrahedrons, similar each to each, and similarly placed.

Proof. From A and A' draw all diagonals of the two solids. If we consider every face of R and S, excepting those that pass through a or A', as the base of a pyramid whose vertex is A or A', each solid is decomposed into as many pyramids as we have assumed bases.

By drawing in the homologous bases the homologous diagonals and passing planes through these diagonals and A or A' respectively, we decompose homologous pyramids into the same number of triangular pyramids.

Hence, Rand s can be decomposed into the same number of tetrahedrons.

In tetrahedrons A-KFG and A'-K'F'G',

[blocks in formation]

The tetrahedrons A-KFG and A'-K'F'G' have all homologous

trihedral angles equal.

(552)

... tetrahedron A-KFG is similar to tetrahedron A'-K'F'G'.

(633)

When A-KFG and A'-K'F'G' are removed, the remaining polyhedrons are similar, for their faces remain similar, and the polyhedral angles are equal.

In like manner any other two homologous tetrahedrons can be proved similar.

Hence Rand s can be decomposed into the same number of tetrahedrons similar each to each and similarly placed. Q. E. D.

802. COR. 1. The homologous edges of similar polyhedrons are proportional.

803. COR. 2. Any two homologous lines in two similar polyhedrons have the same ratio as any two homologous edges. 804. COR. 3. Two homologous faces of similar polyhedrons are to each other as the squares of any two homologous edges.

805. COR. 4. The total areas of two similar polyhedrons are to each other as the squares of any two homologous edges. 806. COR. 5. The volumes of two similar polyhedrons are to each other as the cubes of any two homologous edges.

HINT. If an edge of the first polyhedron is to the homologous edge of the second polyhedron as 1:n, then any edge of the second polyhedron is n times the homologous edge of the first polyhedron. Hence any tetrahedron which forms a part of the second polyhedron is no times the homologous tetrahedron of the polyhedron. Hence by addition we obtain the conclusion.

Ex. 1787.

Prove that all regular dodecahedrons are similar. Ex. 1788. If R and Sare the volumes of two similar polygons, and rands two homologous edges, find S, if R = 100 cu. in., r = 5, and s = 4.

Ex. 1789. The volumes of two similar polyhedrons are 216 cu. in. and 343 cu. in. If an edge of the first figure equals 12 in., find the corresponding edge of the second figure.

PROPOSITION III. THEOREM

807. In any convex polyhedron, the number of edges increased by two, equals the sum of the number of vertices plus the number of faces. (Euler's Theorem.)

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