First lessons in Plane Geometry. Together with an application of them to the solution of problems, etcCarter and Hendee, 1830 - 12 sider |
Inni boken
Resultat 1-5 av 50
Side 30
... EFD , the angles AEF and CFN will also be equal ; because EFD and CFN are opposite an- gles at the vertex ; and , in the same manner , if the angles BEF and EFC are equal , MEA and EFC will also be equal ; because MEA and BEF 30 GEOMETRY .
... EFD , the angles AEF and CFN will also be equal ; because EFD and CFN are opposite an- gles at the vertex ; and , in the same manner , if the angles BEF and EFC are equal , MEA and EFC will also be equal ; because MEA and BEF 30 GEOMETRY .
Side 32
... manner , that the sum of the two interior angles CBH and BHE ( Fig . I. ) , or DBH and BHE ( Fig . II . ) , is less than two right angles , what will then be the case with the two straight lines CD , EF ? Fig . I. NCE Fig . II . CN ...
... manner , that the sum of the two interior angles CBH and BHE ( Fig . I. ) , or DBH and BHE ( Fig . II . ) , is less than two right angles , what will then be the case with the two straight lines CD , EF ? Fig . I. NCE Fig . II . CN ...
Side 34
... manner by the parallel lines AB , CD therefore we have a side and the two adjacent angles in the triangle MPO , equal to a side and the two adjacent angles in the triangle MPI ; consequently these two triangles must be equal ; and the ...
... manner by the parallel lines AB , CD therefore we have a side and the two adjacent angles in the triangle MPO , equal to a side and the two adjacent angles in the triangle MPI ; consequently these two triangles must be equal ; and the ...
Side 35
... manner I can prove that RS is equal to MI , and conse- quently equal to OP ; and so I might go on , and shew that every perpendicular , dropped from the line AB to the parallel line CD , is equal to RS , MI , OP , & c . The two parallel ...
... manner I can prove that RS is equal to MI , and conse- quently equal to OP ; and so I might go on , and shew that every perpendicular , dropped from the line AB to the parallel line CD , is equal to RS , MI , OP , & c . The two parallel ...
Side 43
... manner that a c will fall upon AB , and ab upon AC ( for you will still have two sides and the angle which is included by them in the one , equal to two sides and the angle which is included by them in the other ) ; and therefore the ...
... manner that a c will fall upon AB , and ab upon AC ( for you will still have two sides and the angle which is included by them in the one , equal to two sides and the angle which is included by them in the other ) ; and therefore the ...
Andre utgaver - Vis alle
First lessons in Plane Geometry. Together with an application of them to the ... Francis Joseph Grund Uten tilgangsbegrensning - 1830 |
First Lessons in Plane Geometry: Together with an Application of Them to the ... Francis Joseph Grund Ingen forhåndsvisning tilgjengelig - 2009 |
First Lessons in Plane Geometry: Together with an Application of Them to the ... Francis Joseph Grund Ingen forhåndsvisning tilgjengelig - 2009 |
Vanlige uttrykk og setninger
adjacent angles angle ABC angle ACB angle opposite angle x basis and height bisected called centre chord circum circumference circumscribed circle consequently construct the triangle DEMON diagonal diameter draw the lines equal angles exterior angle figure ABCDEF found by multiplying fourth term geometrical figures geometrical proportion given angle given circle given straight line given triangle hypothenuse isosceles triangle line AB line AC line CD line MN LUDOLPH VAN CEULEN mean proportional number of sides parallel lines parallelogram perpendicular points of division PROBLEM prove quadrilateral Query 11 radii radius ratio rectilinear figure regular polygon ABCDEF remaining sides Remark right angles right-angular triangle Sect semicircle side AB side AC similar triangles smaller SOLUTION square feet square inches square seconds tangent third line three angles three sides trapezoid trian triangle ABC triangle are equal vertex zoid
Populære avsnitt
Side 195 - Upon a given straight line to describe a segment of a circle, which shall contain an angle equal to a given rectilineal angle.
Side 94 - If two triangles have two sides of the one equal to two sides of the other, each to each, but the...
Side 211 - HDG, the corresponding sides are proportional (page 70, 4thly) ; therefore we have the proportion ED : DG =AD : HD (II.) This last proportion has the first ratio common with the first proportion ; consequently the two remaining ratios are in a geometrical proportion (Theory of Prop., Prin. 3d) ; that is, we have AD : HD = DG: BD; and as, in...
Side 173 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds.
Side 176 - 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.
Side 175 - The side of a regular hexagon inscribed in a circle is equal to the radius of the circle.
Side 129 - Now, since the areas of similar polygons are to each other as the squares of their homologous sides...