Elements of GeometryHilliard, Gray,, 1841 - 235 σελίδες |
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Σελίδα v
... give to it all the exact- ness and precision of which it is susceptible . Perhaps I might have attained this object by calling a straight line that which can have only one position between two given points . For , from this essential ...
... give to it all the exact- ness and precision of which it is susceptible . Perhaps I might have attained this object by calling a straight line that which can have only one position between two given points . For , from this essential ...
Σελίδα xii
... gives also A : B :: C : D A : C :: B : D ; and , since the ratios A : C , B : D , are equal , we obtain B + A : D + C : : A : C , or : : B : D , B - AD - C :: A : C , or : : B : D , :: a result which may be thus enunciated ; In any ...
... gives also A : B :: C : D A : C :: B : D ; and , since the ratios A : C , B : D , are equal , we obtain B + A : D + C : : A : C , or : : B : D , B - AD - C :: A : C , or : : B : D , :: a result which may be thus enunciated ; In any ...
Σελίδα 14
... gives A = A + B ' ; therefore A + B + C = A + B + C ' . Moreover , since , by hypothesis , we have AC < AB , and , consequently , C'B ' < AC ' , it will be seen , that , in the triangle AC'B ' , the angle at A , designated by A ' , is ...
... gives A = A + B ' ; therefore A + B + C = A + B + C ' . Moreover , since , by hypothesis , we have AC < AB , and , consequently , C'B ' < AC ' , it will be seen , that , in the triangle AC'B ' , the angle at A , designated by A ' , is ...
Σελίδα 49
... gives a quadruple square ( fig . 103 ) , a triple Fig . 103 . line a square nine times as great , and so on . THEOREM . 174. The area of any parallelogram is equal to the product of its base by its altitude . Demonstration . The ...
... gives a quadruple square ( fig . 103 ) , a triple Fig . 103 . line a square nine times as great , and so on . THEOREM . 174. The area of any parallelogram is equal to the product of its base by its altitude . Demonstration . The ...
Σελίδα 51
... gives BC = EF ; but , on account of the parallels , IG = BC , and DG = EF , therefore HIGD is equal to the square described upon BC . These two parts being taken from the whole square , there remain the two rectangles BCGI , EFIH ...
... gives BC = EF ; but , on account of the parallels , IG = BC , and DG = EF , therefore HIGD is equal to the square described upon BC . These two parts being taken from the whole square , there remain the two rectangles BCGI , EFIH ...
Άλλες εκδόσεις - Προβολή όλων
Συχνά εμφανιζόμενοι όροι και φράσεις
ABC fig adjacent angles altitude angle ACB angle BAC base ABCD bisect centre chord circ circular sector circumference circumscribed common cone consequently construction convex surface Corollary cube cylinder Demonstration diagonals diameter draw drawn equal angles equiangular equilateral equivalent faces figure formed four right angles frustum GEOM given point gles greater hence homologous sides hypothenuse inclination intersection isosceles triangle join less Let ABC let fall Let us suppose line AC mean proportional measure the half meet multiplied number of sides oblique lines opposite parallelogram parallelopiped perimeter perpendicular plane MN polyedron prism produced proposition radii radius ratio rectangle regular polygon right angles Scholium sector segment semicircle semicircumference side BC similar solid angle sphere spherical polygons spherical triangle square described straight line tangent THEOREM third three angles triangle ABC triangular prism triangular pyramids vertex vertices whence
Δημοφιλή αποσπάσματα
Σελίδα 67 - The square of the hypothenuse is equal to the sum of the squares of the other two sides ; as, 5033 402+302.
Σελίδα 9 - If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are congruent.
Σελίδα 65 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.
Σελίδα 160 - ABC (fig. 224) be any spherical triangle ; produce the sides AB, AC, till they meet again in D. The arcs ABD, ACD, will be...
Σελίδα 168 - In any spherical triangle, the greater side is opposite the greater angle ; and conversely, the greater angle is opposite the greater side.
Σελίδα 157 - CIRCLE is a plane figure bounded by a curved line, all the points of which are equally distant from a point within called the centre; as the figure ADB E.
Σελίδα 8 - Any side of a triangle is less than the sum of the other two sides...
Σελίδα 82 - The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their areas are to each other as the squares of those sides (Prop.
Σελίδα 29 - Two equal chords are equally distant from the centre ; and of two unequal chords, the less is at the greater distance from the centre.
Σελίδα 182 - CD, &c., taken together, make up the perimeter of the prism's base : hence the sum of these rectangles, or the convex surface of the prism, is equal to the perimeter of its base multiplied by its altitude.