Elements of GeometryHilliard, Gray,, 1841 - 235 Seiten |
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Seite 90
... circ . CA and circ . OB ( fig . 165 ) , the circumferences of the circles whose radii are CA and OB we say that circ . CA circ . OB :: CA : OB . For , if this proportion be not true , CA will be to OB as circ . CA is to a fourth term ...
... circ . CA and circ . OB ( fig . 165 ) , the circumferences of the circles whose radii are CA and OB we say that circ . CA circ . OB :: CA : OB . For , if this proportion be not true , CA will be to OB as circ . CA is to a fourth term ...
Seite 91
... circ . CA is to a circumference greater than circ OB ; for , if this were the case , we should have , by inversion , OB : CA :: a circumference greater than circ . OB : circ . CA , or , which is the same thing , :: OB CA circ . OB : a ...
... circ . CA is to a circumference greater than circ OB ; for , if this were the case , we should have , by inversion , OB : CA :: a circumference greater than circ . OB : circ . CA , or , which is the same thing , :: OB CA circ . OB : a ...
Seite 92
... circ . CA. Fig . 167 If CA × circ . CA ( fig . 167 ) be not the area of the circle of which CA is the radius , this quantity will be the measure of a circle either greater or less . Let us suppose , in the first place , that it is the ...
... circ . CA. Fig . 167 If CA × circ . CA ( fig . 167 ) be not the area of the circle of which CA is the radius , this quantity will be the measure of a circle either greater or less . Let us suppose , in the first place , that it is the ...
Seite 93
... circ . CA ; circ . CA2 ′′ × CA. or hence Multiplying each member by CA , we have or CA × circ . CA = « × CÅ , -2 surf . CA = × CA ; therefore , the surface of a circle is equal to the product of the square of the radius by the constant ...
... circ . CA ; circ . CA2 ′′ × CA. or hence Multiplying each member by CA , we have or CA × circ . CA = « × CÅ , -2 surf . CA = × CA ; therefore , the surface of a circle is equal to the product of the square of the radius by the constant ...
Seite 165
... circ.-DE. This property must be reciprocal between the two triangles , since they are described in the same manner ... circ . — BC , circ . - AC , circ.-AB. Indeed , the angle D , for example , has for its measure the arc MI ; but MI + ...
... circ.-DE. This property must be reciprocal between the two triangles , since they are described in the same manner ... circ . — BC , circ . - AC , circ.-AB. Indeed , the angle D , for example , has for its measure the arc MI ; but MI + ...
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Häufige Begriffe und Wortgruppen
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
Beliebte Passagen
Seite 67 - The square of the hypothenuse is equal to the sum of the squares of the other two sides ; as, 5033 402+302.
Seite 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.
Seite 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'.
Seite 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...
Seite 168 - In any spherical triangle, the greater side is opposite the greater angle ; and conversely, the greater angle is opposite the greater side.
Seite 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.
Seite 8 - Any side of a triangle is less than the sum of the other two sides...
Seite 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.
Seite 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.
Seite 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.