Elements of Geometry and Trigonometry |
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Page 72
... 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 units in height , three ...
... 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 units in height , three ...
Page 76
... 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 . And those two squares being taken away from the whole square , there re- mains the two ...
... 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 . And those two squares being taken away from the whole square , there re- mains the two ...
Page 80
... give AD + BD2 - AB2 , and AD2 + CD2 = AC2 , we have AB2 = BC2 + AC2-2BC x CD . Secondly . When the perpendicular AD falls without the triangle ABC , we have BD = CD - BC ; and consequently BD2 = CD2 + BC2-2CD × BC ( Prop . IX ...
... give AD + BD2 - AB2 , and AD2 + CD2 = AC2 , we have AB2 = BC2 + AC2-2BC x CD . Secondly . When the perpendicular AD falls without the triangle ABC , we have BD = CD - BC ; and consequently BD2 = CD2 + BC2-2CD × BC ( Prop . IX ...
Page 81
... gives AB2 + BC2-2AE2 + 2BE2 . The triangle ADC gives , in like manner . AD2 + DC2-2AE2 + 2DE2 . E Adding the corresponding members together , and observing that BE and DE are equal , we shall have AB2 + AD2 + DC2 + BC2-4AE2 + 4DE ?. But ...
... gives AB2 + BC2-2AE2 + 2BE2 . The triangle ADC gives , in like manner . AD2 + DC2-2AE2 + 2DE2 . E Adding the corresponding members together , and observing that BE and DE are equal , we shall have AB2 + AD2 + DC2 + BC2-4AE2 + 4DE ?. But ...
Page 82
... give AE : CF B :: EG FH . It may be proved in the G E A C F H same manner , that EG : FH :: GB : HD , and so on ; hence the lines AB , CD , are cut proportionally by the parallels AC , EF , GH , & c . PROPOSITION XVI . THEOREM ...
... give AE : CF B :: EG FH . It may be proved in the G E A C F H same manner , that EG : FH :: GB : HD , and so on ; hence the lines AB , CD , are cut proportionally by the parallels AC , EF , GH , & c . PROPOSITION XVI . THEOREM ...
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Common terms and phrases
adjacent adjacent angles altitude angle ACB angle BAC ar.-comp base multiplied bisect Book VII centre chord circ circumference circumscribed common cone convex surface Cosine cylinder diagonal diameter dicular distance divided draw drawn 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 number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular perpendicular let fall 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 Cotang slant height solid angle solid described sphere spherical polygon spherical triangle square described straight line tang tangent THEOREM triangle ABC triangular prism vertex