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 150
... Denote the area of the given inscribed polygon by P , the area of the given circumscribed polygon by P , and the areas of the inscribed and circumscribed polygons having double the number of sides , respectively by p ' and P ' . 1o ...
... Denote the area of the given inscribed polygon by P , the area of the given circumscribed polygon by P , and the areas of the inscribed and circumscribed polygons having double the number of sides , respectively by p ' and P ' . 1o ...
Page 152
... denoting the number of units in the required area by T , we have , approximately , T = 3.141592 ; that is , the area of a circle whose radius is 1 , is 3.141592 . Scholium . For practical computation , the value of T is taken equal to ...
... denoting the number of units in the required area by T , we have , approximately , T = 3.141592 ; that is , the area of a circle whose radius is 1 , is 3.141592 . Scholium . For practical computation , the value of T is taken equal to ...
Page 155
... Denote its area by area CA , its radius by R , and the area of a circle whose by ( P. XII . , S. ) . radius is 1 , by Then , because the areas of circles are to each other as the squares of A their radii ( P. XIII . ) , we have , area ...
... Denote its area by area CA , its radius by R , and the area of a circle whose by ( P. XII . , S. ) . radius is 1 , by Then , because the areas of circles are to each other as the squares of A their radii ( P. XIII . ) , we have , area ...
Page 156
... Denote its circumference by circ . CA , its radius by R , and its diameter by D. From the last Proposition , we have , area CA = TR2 ; and , from Proposition XIV . , we have , area CA = circ . CA x R ; A hence , circ . CA x R = « R2 ...
... Denote its circumference by circ . CA , its radius by R , and its diameter by D. From the last Proposition , we have , area CA = TR2 ; and , from Proposition XIV . , we have , area CA = circ . CA x R ; A hence , circ . CA x R = « R2 ...
Page 194
... denote the parallelopipe- ABCD , and altitude Am ; since the are to each other as two whole num- E AG : P :: AE : Am . F G But , by hypothesis , we have , AG AL : AE AO ; : therefore ( B. II . , P. IV . , C. ) , : AL : P :: AO : Am ...
... denote the parallelopipe- ABCD , and altitude Am ; since the are to each other as two whole num- E AG : P :: AE : Am . F G But , by hypothesis , we have , AG AL : AE AO ; : therefore ( B. II . , P. IV . , C. ) , : AL : P :: AO : Am ...
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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