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 125
... homologous sides ; and the polygons are to each other as the squares of any two homologous sides . 1o . Let ABCDE and FGHIK be similar polygons : then will their perimeters be to each other as any two homologous sides . For , any two ...
... homologous sides ; and the polygons are to each other as the squares of any two homologous sides . 1o . Let ABCDE and FGHIK be similar polygons : then will their perimeters be to each other as any two homologous sides . For , any two ...
Page 126
... homologous sides . For , let the poly- gons be divided into homologous triangles ( P. XXVI . , C. 1 ) ; then , because the homologous triangles ABC and FGH are like parts of the gons will be to • B E polygons to which they belong , the ...
... homologous sides . For , let the poly- gons be divided into homologous triangles ( P. XXVI . , C. 1 ) ; then , because the homologous triangles ABC and FGH are like parts of the gons will be to • B E polygons to which they belong , the ...
Page 128
... angles OBD and OCA equal , because each is measured by half of the arc AD : hence , they are aimilar , and consequently , their homologous sides are proportional ; whence , OB : OC :: OD : OA ; which was to be proved . Cor . From the ...
... angles OBD and OCA equal , because each is measured by half of the arc AD : hence , they are aimilar , and consequently , their homologous sides are proportional ; whence , OB : OC :: OD : OA ; which was to be proved . Cor . From the ...
Page 129
Adapted to the Course of Mathematical Instruction in the United States Charles Davies. homologous sides are proportional : hence , OC : 01 :: OA OD ; : which was to be proved . Cor . From the above proportion , we have , A 02 = OC × OD ...
Adapted to the Course of Mathematical Instruction in the United States Charles Davies. homologous sides are proportional : hence , OC : 01 :: OA OD ; : which was to be proved . Cor . From the above proportion , we have , A 02 = OC × OD ...
Page 135
... homologous sides of the given polygons Find a square equal to the sum or difference of the squares on A and B ; and let X be a side of that square . On X as a side , homologous to A or B , construct a polygon similar to the given ...
... homologous sides of the given polygons Find a square equal to the sum or difference of the squares on A and B ; and let X be a side of that square . On X as a side , homologous to A or B , construct a polygon similar to the given ...
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Common terms and phrases
AB² AC² adjacent angles altitude angle ACB apothem Applying logarithms base and altitude bisect centre chord circle circumference circumscribed coincide cone consequently convex surface corresponding cosec Cosine Cotang cylinder denote diagonals diameter distance divided draw drawn edges equally distant feet find the area Formula frustum given angle given straight line greater hence homologous hypothenuse included angle interior angles intersection less Let ABC log sin lower base mantissa mean proportional measured by half number of sides opposite parallel parallelogram parallelopipedon perimeter perpendicular plane MN polyedral angle polyedron prism PROPOSITION proved pyramid quadrant radii radius rectangle regular polygons right angles right-angled triangle Scholium secant segment semi-circumference side BC similar sine six right slant height sphere spherical polygon spherical triangle square Tang tangent THEOREM triangle ABC triangular prisms triedral angle upper base vertex vertices whence