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 Architecture |
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Page 30
... intersection , and shall divide it into two equal parts . CH B For the line AB , which joins the points of intersection , being a common chord to the two circles ; if , through the middle of this chord , a perpendicular be drawn , it ...
... intersection , and shall divide it into two equal parts . CH B For the line AB , which joins the points of intersection , being a common chord to the two circles ; if , through the middle of this chord , a perpendicular be drawn , it ...
Page 60
... intersection to each ex- tremity of the one chord , is equal to the rectangle of the two distances from the point of intersection to each extremity of the other chord . Let AB and CD be two chords , and let them be produced to meet in O ...
... intersection to each ex- tremity of the one chord , is equal to the rectangle of the two distances from the point of intersection to each extremity of the other chord . Let AB and CD be two chords , and let them be produced to meet in O ...
Page 65
... intersection of these two lines , will be the centre of the inscribed circle : draw ED perpendicular to AB , cutting AB in D ; from E , with the radius ED , describe the circle DFG , which will 8 . R ELEMENTS OF GEOMETRY , & c . 65.
... intersection of these two lines , will be the centre of the inscribed circle : draw ED perpendicular to AB , cutting AB in D ; from E , with the radius ED , describe the circle DFG , which will 8 . R ELEMENTS OF GEOMETRY , & c . 65.
Page 66
... intersection D , with the radius DA or DC , describe the circle ACE , which is that required . PROBLEM 18 . 191. In a given circle , ABCD , ( fig . 13 , pl . II , ) to inscribe a square . Draw the diameters AC and BD at right angles ...
... intersection D , with the radius DA or DC , describe the circle ACE , which is that required . PROBLEM 18 . 191. In a given circle , ABCD , ( fig . 13 , pl . II , ) to inscribe a square . Draw the diameters AC and BD at right angles ...
Page 89
... intersection of the tangent . For the triangle CNT ( see the preceding diagram ) is right - angled at N , since PN meets the circle at N ; and , because PN is perpendicular to TC , we shall have , by the similar triangles PCN , NCT ...
... intersection of the tangent . For the triangle CNT ( see the preceding diagram ) is right - angled at N , since PN meets the circle at N ; and , because PN is perpendicular to TC , we shall have , by the similar triangles PCN , NCT ...
Common terms and phrases
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
Popular passages
Page 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.
Page 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'.
Page 51 - The area of a parallelogram is equal to the product of its base and its height: A = bx h.
Page 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.
Page 15 - LET it be granted that a straight line may be drawn from any one point to any other point.
Page 28 - ... angles of another, the third angles will also be equal, and the two triangles will be mutually equiangular. Cor.
Page 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.
Page 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.
Page 28 - After remarking that the mathematician positively knows that the sum of the three angles of a triangle is equal to two right angles...
Page 22 - The perpendicular is the shortest line that can be drawn from a point to a straight line.