Elements of Geometry |
From inside the book
Results 6-10 of 27
Page 131
... prism GHF - BCD , are equal to the three planes , which form the solid angle A in the prism ABD - H'E'F ' , each to each ; they are morcover simi- larly disposed in the two cases ; therefore these two prisms are equal , and being ...
... prism GHF - BCD , are equal to the three planes , which form the solid angle A in the prism ABD - H'E'F ' , each to each ; they are morcover simi- larly disposed in the two cases ; therefore these two prisms are equal , and being ...
Page 132
... prism ( 368 ) ; and this prism is a right prism , since the side BF is perpendicular to the plane of the base . This being premised , if the right prism Bh be divided by the the plane BFHD into two right triangular prisms a B d - h e F ...
... prism ( 368 ) ; and this prism is a right prism , since the side BF is perpendicular to the plane of the base . This being premised , if the right prism Bh be divided by the the plane BFHD into two right triangular prisms a B d - h e F ...
Page 133
... prism AEM be applied to the prism BFL , the base AEI being placed upon the base BFK , these two bases , being equal , will coincide ; and , since the solid angle E is equal to the solid angle F , the side EH will fall upon its equal FG ...
... prism AEM be applied to the prism BFL , the base AEI being placed upon the base BFK , these two bases , being equal , will coincide ; and , since the solid angle E is equal to the solid angle F , the side EH will fall upon its equal FG ...
Page 135
... prism MIKL , parallel to the base , is equal to this base ( 395 ) , and equal altitudes , because the altitudes are the divisions themselves Ax , xy , yz , & c . Now , of these 15 equal parallelopipeds 8 are con- tained in AL ...
... prism MIKL , parallel to the base , is equal to this base ( 395 ) , and equal altitudes , because the altitudes are the divisions themselves Ax , xy , yz , & c . Now , of these 15 equal parallelopipeds 8 are con- tained in AL ...
Page 138
... prism whatever , is equal to the product of its base by its altitude . Demonstration . 1. A parallelopiped of whatever kind is equiv- alent to a rectangular parallelopiped having the same altitude and an equivalent base ( 401 ) . But ...
... prism whatever , is equal to the product of its base by its altitude . Demonstration . 1. A parallelopiped of whatever kind is equiv- alent to a rectangular parallelopiped having the same altitude and an equivalent base ( 401 ) . But ...
Other editions - View all
Common terms and phrases
ABC fig adjacent angles altitude angle ACB angle BAD angles equal base ABCD bisect centre chord circ circular sector circumference circumscribed common cone consequently construction convex surface Corollary cube cylinder Demonstration diagonals diameter draw drawn equal and parallel equiangular equilateral equivalent faces figure four right angles frustum Geom gles greater hence homologous sides hypothenuse inclination inscribed circle intersection isosceles join less let fall line AC manner mean proportional measure the half meet multiplied number of sides oblique lines opposite parallelogram parallelopiped perimeter perpendicular plane MN polyedron prism proposition quadrilateral radii radius ratio rectangle regular polygon right angles right-angled triangle Scholium segment semicircumference side BC similar solid angle sphere spherical polygons spherical triangle square described straight line tangent THEOREM three plane angles triangle ABC triangular prism triangular pyramids vertex vertices whence
Popular passages
Page 65 - The square of the hypothenuse is equal to the sum of the squares of the other two sides ; as, 5033 402+302.
Page 21 - If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are congruent.
Page ii - Co. of the said district, have deposited in this office the title of a book, the right whereof they claim as proprietors, in the words following, to wit : " Tadeuskund, the Last King of the Lenape. An Historical Tale." In conformity to the Act of the Congress of the United States...
Page 63 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.
Page 22 - CIRCLE is a plane figure bounded by a curved line, all the points of which are equally distant from a point within called the centre; as the figure ADB E.
Page ii - States entitled an act for the encouragement of learning hy securing the copies of maps, charts and books to the author., and proprietors of such copies during the times therein mentioned, and also to an act entitled an act supplementary to an act, entitled an act for the encouragement of learning by securing the copies of maps, charts and books to the authors and proprietors of such copies during the times therein mentioned and extending the benefits thereof to the arts of designing, engraving and...
Page 80 - The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their areas are to each other as the squares of those sides (Prop.
Page 164 - If two triangles have two sides and the inchtded angle of the one respectively equal to two sides and the included angle of the other, the two triangles are equal in all respects.
Page 24 - In the same circle, or in equal circles, equal arcs are subtended by equal chords ; and, conversely, equal chords subtend equal arcs.
Page 153 - XVII.) ; hence two similar pyramids are to each other as the cubes of their homologous sides.