Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle, and the Geometry of Solids: to which are Added, Edlemnts of Plane and Spherical TrigonometryW.E. Dean, 1836 - 311 sider |
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Resultat 1-3 av 39
Side 137
... ABCD , AEFG be similar and similarly situated , and have the angle DAB common ; ABCD and AEFG are about the same diameter . K E G D H F C For , if not , let , if possible , the parallelogram BD have its diameter AHC in a different ...
... ABCD , AEFG be similar and similarly situated , and have the angle DAB common ; ABCD and AEFG are about the same diameter . K E G D H F C For , if not , let , if possible , the parallelogram BD have its diameter AHC in a different ...
Side 204
... ABCD is equal to the pyramid EFGH . If they are not equal , let the pyramid EFGH exceed the pyramid ABCD by the solid Z. Then , a series of prisms of the same altitude may be de- scribed about the pyramid ABCD that shall exceed it , by ...
... ABCD is equal to the pyramid EFGH . If they are not equal , let the pyramid EFGH exceed the pyramid ABCD by the solid Z. Then , a series of prisms of the same altitude may be de- scribed about the pyramid ABCD that shall exceed it , by ...
Side 207
... ABCD . In the circle AGB inscribe the polygon AGKBLM that shall differ from the circle by a space less than the ... ABCD , and the cylinder ABCD is equal to the parallel- opiped EQ , by hypothesis ; therefore , ES is less than EQ , and ...
... ABCD . In the circle AGB inscribe the polygon AGKBLM that shall differ from the circle by a space less than the ... ABCD , and the cylinder ABCD is equal to the parallel- opiped EQ , by hypothesis ; therefore , ES is less than EQ , and ...
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Vanlige uttrykk og setninger
ABC is equal ABCD adjacent angles altitude angle ABC angle ACB angle BAC angles equal arc AC base BC bisected centre chord circle ABC circumference cosine cylinder demonstrated diameter divided draw Prob equal and similar equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles given straight line greater Hence hypotenuse inscribed join less Let ABC Let the straight magnitudes meet multiple opposite angle parallel parallelogram parallelopiped perpendicular polygon prism PROP proposition quadrilateral radius ratio rectangle contained rectilineal figure remaining angle right angled triangle SCHOLIUM segment semicircle shewn side BC sine solid angle solid parallelopiped spherical angle spherical triangle square straight line BC THEOR third touches the circle triangle ABC wherefore