Elements of Geometry and Trigonometry from the Works of A.M. Legendre: Adapted to the Course of Mathematical Instruction in the United States |
From inside the book
Results 1-5 of 69
Page 59
... RADIUS is a straight line drawn from the centre to any point of the circumference . 3. A DIAMETER is a straight line drawn through the centre and terminating in the circumference . All radii of the same circle are equal . All diameters ...
... RADIUS is a straight line drawn from the centre to any point of the circumference . 3. A DIAMETER is a straight line drawn through the centre and terminating in the circumference . All radii of the same circle are equal . All diameters ...
Page 60
... circumference touches all of the sides of the polygon . о © о POSTULATE . A circumference can be described from any point as a centre , and with any radius . PROPOSITION I. THEOREM . Any diameter divides the circle , 60 GEOMETRY .
... circumference touches all of the sides of the polygon . о © о POSTULATE . A circumference can be described from any point as a centre , and with any radius . PROPOSITION I. THEOREM . Any diameter divides the circle , 60 GEOMETRY .
Page 61
... radius CD . angle ACD , we have In the tri- AD less than ( B. I. , P. the sum of AC and CD VII . ) . But this sum is equal to AB ( D. 3 ) hence , AB is greater than AD ; which was to be proved . B PROPOSITION III . THEOREM . A straight ...
... radius CD . angle ACD , we have In the tri- AD less than ( B. I. , P. the sum of AC and CD VII . ) . But this sum is equal to AB ( D. 3 ) hence , AB is greater than AD ; which was to be proved . B PROPOSITION III . THEOREM . A straight ...
Page 64
... radius which is perpendicular to the chord AB : then will this radius bisect the chord AB , and also the arc AGB . For , draw the radii CA and CB . Then , the right - angled triangles CDA and CDB will have the hypothenuse CA equal to CB ...
... radius which is perpendicular to the chord AB : then will this radius bisect the chord AB , and also the arc AGB . For , draw the radii CA and CB . Then , the right - angled triangles CDA and CDB will have the hypothenuse CA equal to CB ...
Page 65
... radius perpendicular to the chord . But two points determine the position of a straight line ( A. 11 ) : hence , any straight line which passes through two of these points , will pass through the third , and be perpendicular to the ...
... radius perpendicular to the chord . But two points determine the position of a straight line ( A. 11 ) : hence , any straight line which passes through two of these points , will pass through the third , and be perpendicular to the ...
Other editions - View all
Common terms and phrases
AB² AC² altitude angle ACB apothem axis base and altitude base multiplied BC² bisect centre chord circumference coincide cone consequently convex surface corresponding cosec cosine Cotang cylinder denote diameter distance divided draw drawn edges equal bases equal in volume equal to AC equal to half equally distant Formula frustum given angle given line greater hence homologous hypothenuse included angle intersection less Let ABC logarithm lower base mantissa mean proportional measured by half number of sides opposite parallelogram perimeter perpendicular plane MN polyedral angle polyedron prism PROPOSITION XI proved pyramid quadrant radii radius rectangle regular polygons right-angled triangle Scholium segment semi-circumference side BC similar sine slant height sphere spherical angle spherical excess spherical polygon spherical triangle straight line tangent THEOREM triangle ABC triangular prisms triedral angle upper base vertex vertices whence