New Plane and Solid Geometry |
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Page 31
... altitude as the perpendicular drawn from the opposite vertex to the base , produced if necessary . In general , any side may be taken as the base ; but in an isosceles triangle , unless otherwise specified , the side which is not one of ...
... altitude as the perpendicular drawn from the opposite vertex to the base , produced if necessary . In general , any side may be taken as the base ; but in an isosceles triangle , unless otherwise specified , the side which is not one of ...
Page 32
... altitude , and BAC the vertical angle . 74. If two straight lines , AB and CD , B are cut by a line EF , called a transversal , the angles are named as follows : c , d , e , and fare called interior , angles , and a , b , g , and h ...
... altitude , and BAC the vertical angle . 74. If two straight lines , AB and CD , B are cut by a line EF , called a transversal , the angles are named as follows : c , d , e , and fare called interior , angles , and a , b , g , and h ...
Page 47
... altitude corresponding to these bases is the perpendicular distance between them . A Rhomboid is a parallelogram ... altitudes of the triangle formed by the bisectors . 4 S ces : PROP . XXXI . THEOREM 108 RECTILINEAR FIGURES 47.
... altitude corresponding to these bases is the perpendicular distance between them . A Rhomboid is a parallelogram ... altitudes of the triangle formed by the bisectors . 4 S ces : PROP . XXXI . THEOREM 108 RECTILINEAR FIGURES 47.
Page 52
... altitude of one are equal respectively to the base and altitude of the other . PROP . XXXVII . THEOREM 112. The diagonals of a rectangle are equal . Draw rectangle ABCD ; draw lines AC and BD . We then have : Given AC and BD the ...
... altitude of one are equal respectively to the base and altitude of the other . PROP . XXXVII . THEOREM 112. The diagonals of a rectangle are equal . Draw rectangle ABCD ; draw lines AC and BD . We then have : Given AC and BD the ...
Page 115
... altitudes of triangles OBC and OAD , respec- tively , prove OR OS = BC AD PROP . XVII . THEOREM 245. The homologous altitudes of two similar triangles are in the same ratio as any two homologous sides . Draw A ABC , and line AD L BC ...
... altitudes of triangles OBC and OAD , respec- tively , prove OR OS = BC AD PROP . XVII . THEOREM 245. The homologous altitudes of two similar triangles are in the same ratio as any two homologous sides . Draw A ABC , and line AD L BC ...
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Common terms and phrases
ABC and A'B'C adjacent angles altitude angle formed angles are equal apothem arc BC base and altitude bisector bisects centre chord circle circumference circumscribed coincide construct Converse of Prop diagonals diameter diedral angle distance Draw line equal parts occur equal respectively equally distant equilateral triangle exterior angle faces frustum Given line given point homologous sides hypotenuse intersecting isosceles trapezoid isosceles triangle LAOB lateral area lateral edges line drawn lines be drawn measured by arc middle point number of sides oblique lines opposite parallel parallelogram parallelopiped perimeter perpendicular plane MN polyedron prism Proof proportional Prove pyramid quadrilateral radii radius rectangle regular polygon rhombus right angles right triangle secant segments slant height spherical polygon spherical triangle square straight line surface tangent tetraedron THEOREM trapezoid triedral vertex vertices volume