The New Practical Builder and Workman's Companion, Containing a Full Display and Elucidation of the Most Recent and Skilful Methods Pursued by Architects and Artificers ... Including, Also, New Treatises on Geometry ..., a Summary of the Art of Building ..., an Extensive Glossary of the Technical Terms ..., and The Theory and Practice of the Five Orders, as Employed in Decorative ArchitectureThomas Kelly, 1823 - 596 páginas |
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Página 7
... several modes are brought forward , for the first time , interesting , both with respect to the disposition and joining of the timbers ; and the examples which are given will be found of the greatest utility to the practical PREFACE . 7.
... several modes are brought forward , for the first time , interesting , both with respect to the disposition and joining of the timbers ; and the examples which are given will be found of the greatest utility to the practical PREFACE . 7.
Página 20
... join CG ; and because the two triangles CAG , DEF , have an angle of the one equal to an angle of the other , and the sides ДА B which contain these angles are equal , CG shall be equal to EF ( theorem 5 ) . Now there may be three cases ...
... join CG ; and because the two triangles CAG , DEF , have an angle of the one equal to an angle of the other , and the sides ДА B which contain these angles are equal , CG shall be equal to EF ( theorem 5 ) . Now there may be three cases ...
Página 21
... join CD ; the B angle DBC is , by hypothesis , equal to the angle ACB , and the two sides DB , BC , are equal to the two sides AC , CB ; therefore the triangle DBC is equal to the triangle ACB , the less to the greater 3 . F ELEMENTS OF ...
... join CD ; the B angle DBC is , by hypothesis , equal to the angle ACB , and the two sides DB , BC , are equal to the two sides AC , CB ; therefore the triangle DBC is equal to the triangle ACB , the less to the greater 3 . F ELEMENTS OF ...
Página 22
... join FC ; and , because AB is equal to BF , and BC is common to the two triangles ABC , FBC , and the angles ABC and FBC are equal ; the angle ACB is equal to FCB ( theorem 5 ) ; therefore AC and CF must be a continued line ( theorem 2 ) ...
... join FC ; and , because AB is equal to BF , and BC is common to the two triangles ABC , FBC , and the angles ABC and FBC are equal ; the angle ACB is equal to FCB ( theorem 5 ) ; therefore AC and CF must be a continued line ( theorem 2 ) ...
Página 24
... join AG . The triangles ABG and DEF , having AB equal to DE , and BG equal to EF , by hypothesis , and also having the angle ABG equal to DEF , they will be equal ( theorem 5 ) : therefore AG is equal to DF ; but DF is equal to AC ...
... join AG . The triangles ABG and DEF , having AB equal to DE , and BG equal to EF , by hypothesis , and also having the angle ABG equal to DEF , they will be equal ( theorem 5 ) : therefore AG is equal to DF ; but DF is equal to AC ...
Términos y frases comunes
ABCD abscissa adjacent angles altitude angle ABD annular vault axes axis major base bisect called centre chord circle circumference cone conic section conjugate contains COROLLARY 1.-Hence cutting cylinder describe a semi-circle describe an arc diameter distance divide draw a curve draw lines draw the lines edge ellipse Engraved equal angles equal to DF equation equiangular figure GEOMETRY given straight line greater groin homologous sides hyperbola intersection join joist latus rectum less Let ABC line of section meet multiplying Nicholson opposite sides ordinate parallel to BC parallelogram perpendicular PLATE points of section polygon PROBLEM produced proportionals quantity radius rectangle regular polygon ribs right angles roof segment similar triangles square straight edge subtracted surface Symns tangent THEOREM timber transverse axis triangle ABC vault vertex wherefore
Pasajes populares
Página 27 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Página 20 - 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. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.
Página 51 - The area of a parallelogram is equal to the product of its base and its height: A = bx h.
Página 15 - AXIOMS. 1. Things which are equal to the same thing are equal to one another. 2. If equals be added to equals, the wholes are equal. 3. If equals be taken from equals, the remainders are equal.
Página 15 - LET it be granted that a straight line may be drawn from any one point to any other point.
Página 28 - ... angles of another, the third angles will also be equal, and the two triangles will be mutually equiangular. Cor.
Página 81 - C' (89) (90) (91) (92) (93) 112. In any plane triangle, the sum of any two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference.
Página 80 - The sine of an arc is a straight line drawn from one extremity of the arc perpendicular to the radius passing through the other extremity. The tangent of an arc is a straight line touching the arc at one extremity, and limited by the radius produced through the other extremity.
Página 28 - After remarking that the mathematician positively knows that the sum of the three angles of a triangle is equal to two right angles...
Página 22 - The perpendicular is the shortest line that can be drawn from a point to a straight line.