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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Results 6-10 of 29
Page 90
... ABCD ; join the point A with the points of intersection D and and B : then will both AD and АВ be tangent to the given circle , and there will be two solutions . For , the angles ABC and ADC are right angles ( P. XVIII . , C. 2 ) ...
... ABCD ; join the point A with the points of intersection D and and B : then will both AD and АВ be tangent to the given circle , and there will be two solutions . For , the angles ABC and ADC are right angles ( P. XVIII . , C. 2 ) ...
Page 94
... ABCD and EFGH have equal bases and equal altitudes : then will the parallelograms be equal . For , let them be so placed that their lower bases shall coincide ; then , because they have the same altitude , their upper bases will be in ...
... ABCD and EFGH have equal bases and equal altitudes : then will the parallelograms be equal . For , let them be so placed that their lower bases shall coincide ; then , because they have the same altitude , their upper bases will be in ...
Page 96
... ABCD be divi- ded into 7 , and HEFK into 4 rectangles , all of which will be equal , because they have equal bases and equal altitudes ( P. I. ) : hence , we have , ABCD : HEFK :: 7 4 . But we have , by hypothesis , AB : HE :: 7 : 4 ...
... ABCD be divi- ded into 7 , and HEFK into 4 rectangles , all of which will be equal , because they have equal bases and equal altitudes ( P. I. ) : hence , we have , ABCD : HEFK :: 7 4 . But we have , by hypothesis , AB : HE :: 7 : 4 ...
Page 97
... ABCD , so that it shall take the position AEFD . Then , if the rectangles are not pro- portional to their bases , let us sup- pose that 1 D FK C A ABCD : AEFD :: AB A0 ; : Divide AB into in which AO is greater than AE . equal parts ...
... ABCD , so that it shall take the position AEFD . Then , if the rectangles are not pro- portional to their bases , let us sup- pose that 1 D FK C A ABCD : AEFD :: AB A0 ; : Divide AB into in which AO is greater than AE . equal parts ...
Page 98
... ABCD and AEGF be two rectangles : then will ABCD be to AEGF , as AB X AD is to AEX AF . For , place the rectangles so that the angles DAB and EAF shall be opposite or vertical ; then , produce the sides CD and GE till they meet in H ...
... ABCD and AEGF be two rectangles : then will ABCD be to AEGF , as AB X AD is to AEX AF . For , place the rectangles so that the angles DAB and EAF shall be opposite or vertical ; then , produce the sides CD and GE till they meet in H ...
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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