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 páginas |
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Página 195
... parallelopiped . Now the solid parallelopiped CM is equal ( 5. 2. Sup . ) to the solid parallelopiped CP , because they are upon the same base , and their insisting straight lines AF , AO , CD , CR ; LM , LP , BH , BQ are terminated in ...
... parallelopiped . Now the solid parallelopiped CM is equal ( 5. 2. Sup . ) to the solid parallelopiped CP , because they are upon the same base , and their insisting straight lines AF , AO , CD , CR ; LM , LP , BH , BQ are terminated in ...
Página 197
... parallelopiped GK upon the base FH , one of whose in- sisting lines is FD , whereby the solids CD , GK must be of the same alti- tude . Therefore the solid AB is equal ( 7. 3. Sup . ) to the solid GK , be- cause they are upon equal ...
... parallelopiped GK upon the base FH , one of whose in- sisting lines is FD , whereby the solids CD , GK must be of the same alti- tude . Therefore the solid AB is equal ( 7. 3. Sup . ) to the solid GK , be- cause they are upon equal ...
Página 206
... parallelopiped having equal bases and altitudes , are equal to one another . Let ABCD be a cylinder , and EF a parallelopiped having equal bases , viz . the circle AGB and the parallelogram EH , and having also equal al- titudes ; the ...
... parallelopiped having equal bases and altitudes , are equal to one another . Let ABCD be a cylinder , and EF a parallelopiped having equal bases , viz . the circle AGB and the parallelogram EH , and having also equal al- titudes ; the ...
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Términos y frases comunes
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