Elements of Geometry and Trigonometry |
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Page 11
... hypothenuse . Thus , in the triangle ABC , right - angled at A , the side BC is the hypothenuse . B A 17. Among the quadrilaterals , we distinguish : The square , which has its sides equal , and its an- gles right - angles . The ...
... hypothenuse . Thus , in the triangle ABC , right - angled at A , the side BC is the hypothenuse . B A 17. Among the quadrilaterals , we distinguish : The square , which has its sides equal , and its an- gles right - angles . The ...
Page 24
... hypothenuse and a side of the one , equal to the hypothenuse and a side of the other , each to each , the remaining parts will also be equal , each to each , and the triangles themselves will be equal . A In the two right angled ...
... hypothenuse and a side of the one , equal to the hypothenuse and a side of the other , each to each , the remaining parts will also be equal , each to each , and the triangles themselves will be equal . A In the two right angled ...
Page 47
... hypothenuses CA , CD , are equal ; and the side AF , the half of AB , is equal to the side DG , the half of DE : hence the triangles are equal , and CF is equal to CG ( Book I. Prop . XVII . ) ; hence , the two equal chords AB , DE ...
... hypothenuses CA , CD , are equal ; and the side AF , the half of AB , is equal to the side DG , the half of DE : hence the triangles are equal , and CF is equal to CG ( Book I. Prop . XVII . ) ; hence , the two equal chords AB , DE ...
Page 64
... hypothenuse CA common , and the side CB = CD ; hence they are equal ( Book I. Prop . XVII . ) ; hence AD is equal to AB , and also the angle CAD to CAB . And as there can be but one line bisecting the angle BAC , it follows , that the ...
... hypothenuse CA common , and the side CB = CD ; hence they are equal ( Book I. Prop . XVII . ) ; hence AD is equal to AB , and also the angle CAD to CAB . And as there can be but one line bisecting the angle BAC , it follows , that the ...
Page 77
... ( AB + BC ) × ( AB - BC ) = AB2 - BC2 . Scholium . This proposition is equivalent to the algebraical formula , ( a + b ) x ( a — b ) = a2 — b2 . G * PROPOSITION XI . THEOREM The square described on the hypothenuse BOOK IV . 27.
... ( AB + BC ) × ( AB - BC ) = AB2 - BC2 . Scholium . This proposition is equivalent to the algebraical formula , ( a + b ) x ( a — b ) = a2 — b2 . G * PROPOSITION XI . THEOREM The square described on the hypothenuse BOOK IV . 27.
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adjacent altitude angle ACB angle BAC ar.-comp base multiplied bisect Book VII centre chord circ circumference circumscribed common cone convex surface cosine cotangent cylinder diagonal diameter dicular distance divided draw drawn equal angles equally distant equations equivalent feet figure find the area formed four right angles frustum given angle given line gles greater homologous sides hypothenuse inscribed circle inscribed polygon intersection less Let ABC logarithm measured by half number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN polyedron polygon ABCDE PROBLEM proportional PROPOSITION pyramid quadrant quadrilateral quantities radii radius ratio rectangle regular polygon right angled triangle S-ABCDE Scholium secant segment similar sine slant height solid angle solid described sphere spherical polygon spherical triangle square described straight line tang tangent THEOREM triangle ABC triangular prism vertex