Elements of Geometry and Trigonometry from the Works of A.M. Legendre: Adapted to the Course of Mathematical Instruction in the United States |
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Page 61
... radii of the same circle are equal . All diameters are also equal , and each is double the radius . 4. An ARC is any part of a circumference . 5. A CHORD is a straight line joining the extremities of an arc . Any chord belongs to two ...
... radii of the same circle are equal . All diameters are also equal , and each is double the radius . 4. An ARC is any part of a circumference . 5. A CHORD is a straight line joining the extremities of an arc . Any chord belongs to two ...
Page 64
... radii to these points , we should have three equal straight lines drawn from the same point to the same straight line ; which is impossible ( B. I. , P. XV . , C. 2 ) : hence , AB can not meet the circumference in more than two points ...
... radii to these points , we should have three equal straight lines drawn from the same point to the same straight line ; which is impossible ( B. I. , P. XV . , C. 2 ) : hence , AB can not meet the circumference in more than two points ...
Page 65
... radii CD and OG . The triangles ACD and EOG have all the sides of the one equal to the corre- sponding sides of the ... radii CA , CD , radii CA , CD , and CH . Now , the sides AC , CH , of the triangle ACH , are equal to the sides AC ...
... radii CD and OG . The triangles ACD and EOG have all the sides of the one equal to the corre- sponding sides of the ... radii CA , CD , radii CA , CD , and CH . Now , the sides AC , CH , of the triangle ACH , are equal to the sides AC ...
Page 66
... radii CA and CB . Then , the right - angled triangles CDA and CDB have the hypothenuse CA equal to CB , and the side CD com- C G D ' mon ; the triangles are , therefore , equal in all respects : hence , AD is equal to DB . Again ...
... radii CA and CB . Then , the right - angled triangles CDA and CDB have the hypothenuse CA equal to CB , and the side CD com- C G D ' mon ; the triangles are , therefore , equal in all respects : hence , AD is equal to DB . Again ...
Page 73
... radii . Let the circumferences , whose centres are C and D , intersect at A : then CD is less than the sum , and greater than the difference of the radii of the two circles . For , draw AC the triangle ACD . than the sum of and AD ...
... radii . Let the circumferences , whose centres are C and D , intersect at A : then CD is less than the sum , and greater than the difference of the radii of the two circles . For , draw AC the triangle ACD . than the sum of and AD ...
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Common terms and phrases
AB² ABCD AC² adjacent angles altitude angles is equal apothem base and altitude bisects centre chord circle circumference circumscribed cone consequently convex surface corresponding Cosine Cotang cylinder denote diagonals diameter distance divided draw drawn edges equally distant equilateral feet find the area formula frustum given angle given line given point greater hence homologous hypothenuse included angle interior angles intersection less Let ABC logarithm lower base mantissa mean proportional measured by half number of sides parallel parallelogram parallelopipedon perimeter perpendicular plane angles plane MN polyedral angle polyedron prism PROPOSITION proved pyramid quadrant radii radius rectangle regular polygon right angles right-angled triangle Scholium secant segment side BC similar sine slant height solution sphere spherical angle spherical polygon spherical triangle square Tang tangent THEOREM triangle ABC triangular prisms triedral angle upper base vertex vertices whence