Elements of Geometry and Trigonometry |
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Page 72
This product will give the number of superficial units in the surface ; because , for one unit in height , there are as many superficial units as there are linear units in the base ; for two units in height twice as many ; for three ...
This product will give the number of superficial units in the surface ; because , for one unit in height , there are as many superficial units as there are linear units in the base ; for two units in height twice as many ; for three ...
Page 76
... the difference , AC - AB must be equal to the difference AE - AF , which gives BC = EF ; but IG is equal to BC , and DG to EF , since the lines are parallel ; therefore IGDH is equal to a square described on BC .
... the difference , AC - AB must be equal to the difference AE - AF , which gives BC = EF ; but IG is equal to BC , and DG to EF , since the lines are parallel ; therefore IGDH is equal to a square described on BC .
Page 80
Adding AD2 to each , and observing that the right angled trian- gles ABD , ADC , give AD2 + BD2 = AB2 , and AD2 + CD2 - AC2 , we have AB2 - BC2 + AC2-2BC x CD . Secondly . When the perpendicular AD falls without the triangle ABC ...
Adding AD2 to each , and observing that the right angled trian- gles ABD , ADC , give AD2 + BD2 = AB2 , and AD2 + CD2 - AC2 , we have AB2 - BC2 + AC2-2BC x CD . Secondly . When the perpendicular AD falls without the triangle ABC ...
Page 81
consequently the triangle ABC gives AB2 + BC - 2AE2 + 2BE2 . The triangle ADC gives , in like manner . AD2 + DC2-2AE2 + 2DE2 . Adding the corresponding members together , and observing that BE and DE are equal , we shall have AB2 + AD + ...
consequently the triangle ABC gives AB2 + BC - 2AE2 + 2BE2 . The triangle ADC gives , in like manner . AD2 + DC2-2AE2 + 2DE2 . Adding the corresponding members together , and observing that BE and DE are equal , we shall have AB2 + AD + ...
Page 82
And by reason of the common ratio OE : OF , those two proportions give AE CF B :: EG FH . It may be proved in the E A F H same manner , that EG : FH :: GB : HD , and so on ; hence the lines AB , CD , are cut proportionally by the ...
And by reason of the common ratio OE : OF , those two proportions give AE CF B :: EG FH . It may be proved in the E A F H same manner , that EG : FH :: GB : HD , and so on ; hence the lines AB , CD , are cut proportionally by the ...
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Common terms and phrases
ABCD adjacent altitude base become Book called centre chord circle circumference circumscribed common cone consequently construction contained corresponding cosine Cotang cylinder described diameter difference distance divided draw drawn equal equation equivalent evident expressed extremities fall figure follows formed formulas four frustum give given gles greater half hence homologous included inscribed intersection less likewise logarithm manner means measured meet middle multiplied opposite parallel parallelogram parallelopipedon pass perpendicular plane polygon prism PROBLEM Prop proportional PROPOSITION pyramid quantities radii radius ratio reason rectangle regular remaining right angles Scholium segment shown sides similar sine solid solid angle sphere spherical triangle square straight line Suppose surface taken tang tangent THEOREM third triangle triangle ABC vertex whole
Bibliographic information
| Title | Elements of Geometry and Trigonometry |
| Author | Adrien Marie Legendre |
| Editor | Charles Davies |
| Translated by | David Brewster |
| Edition | 4 |
| Publisher | A.S. Barnes and Company, 1834 |
| Original from | the University of California |
| Digitized | 14 Oct 2010 |
| Length | 359 pages |
| Export Citation | BiBTeX EndNote RefMan |