Elements of GeometryHilliard, Gray,, 1841 - 235 Seiten |
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Seite 44
... equivalent whose surfaces are equal . Two figures may be equivalent , however dissimilar ; thus a circle may be equivalent to a square , a triangle to a rectangle , & c . The denomination of equal figures will be restricted to those ...
... equivalent whose surfaces are equal . Two figures may be equivalent , however dissimilar ; thus a circle may be equivalent to a square , a triangle to a rectangle , & c . The denomination of equal figures will be restricted to those ...
Seite 45
... equivalent . Demonstration . Let AB ( fig . 96 ) be the common base of the Fig . 96 . two parallelograms ABCD , ABEF ; since they are supposed to have the same altitude , the sides DC , FE , opposite to the bases , will be situated in a ...
... equivalent . Demonstration . Let AB ( fig . 96 ) be the common base of the Fig . 96 . two parallelograms ABCD , ABEF ; since they are supposed to have the same altitude , the sides DC , FE , opposite to the bases , will be situated in a ...
Seite 46
... equivalent . 167. Corollary . Every parallelogram ABCD ( fig . 97 ) is equivalent to a rectangle of the same base and altitude . Fig . 98 . THEOREM . 168. Every triangle ABC ( fig . 98 ) is half of a parallelogramı ABCD of the same base ...
... equivalent . 167. Corollary . Every parallelogram ABCD ( fig . 97 ) is equivalent to a rectangle of the same base and altitude . Fig . 98 . THEOREM . 168. Every triangle ABC ( fig . 98 ) is half of a parallelogramı ABCD of the same base ...
Seite 50
... equivalent to the parallelogram ADKL , and has for its measure EF × AL . But AL = DK ; and , since the triangle IBL is equal to the triangle KCI , the side BL = CK ; therefore AB + CDAL + DK = 2 AL ; thus AL is half the sum of the sides ...
... equivalent to the parallelogram ADKL , and has for its measure EF × AL . But AL = DK ; and , since the triangle IBL is equal to the triangle KCI , the side BL = CK ; therefore AB + CDAL + DK = 2 AL ; thus AL is half the sum of the sides ...
Seite 53
... equivalent to the square AH , double of the triangle HBC . It may be demonstrated , in the same man- ner , that the rectangle CDEG is equivalent to the square AI ; but the two rectangles BDEF , CDEG , taken together , make the square ...
... equivalent to the square AH , double of the triangle HBC . It may be demonstrated , in the same man- ner , that the rectangle CDEG is equivalent to the square AI ; but the two rectangles BDEF , CDEG , taken together , make the square ...
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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.