Elements of GeometryHilliard, Gray,, 1841 - 235 páginas |
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Página 24
... diameter , and every straight line , as AB , which passes through the centre , and is terminated each way by the circumference , is called a diameter . By the definition of a circle the radii are all equal , and all the diameters also ...
... diameter , and every straight line , as AB , which passes through the centre , and is terminated each way by the circumference , is called a diameter . By the definition of a circle the radii are all equal , and all the diameters also ...
Página 25
... diameter AB ( fig . 49 ) bisects the circle and its cir- Fig . 49 , cumference . Demonstration . If the figure AEB be applied to AFB , so that the base AB may be common to both , the curved line AEB must fall exactly upon the curved ...
... diameter AB ( fig . 49 ) bisects the circle and its cir- Fig . 49 , cumference . Demonstration . If the figure AEB be applied to AFB , so that the base AB may be common to both , the curved line AEB must fall exactly upon the curved ...
Página 26
... diameter AB being equal to the diameter EF , the semicircle AMDB may be applied exactly to the semicircle ENGF , and then the curved line AMDB will coincide entirely with the curved line ENGF ; but the portion AMD being sup- posed equal ...
... diameter AB being equal to the diameter EF , the semicircle AMDB may be applied exactly to the semicircle ENGF , and then the curved line AMDB will coincide entirely with the curved line ENGF ; but the portion AMD being sup- posed equal ...
Página 35
... diameter AE , and the radii CB , CD . The angle BCE , being the exterior angle of the triangle ABC , is equal to the sum of the two opposite interior angles , CAB , ABC . But , the triangle BAC being isosceles , the angle CAB = ABC ...
... diameter AE , and the radii CB , CD . The angle BCE , being the exterior angle of the triangle ABC , is equal to the sum of the two opposite interior angles , CAB , ABC . But , the triangle BAC being isosceles , the angle CAB = ABC ...
Página 36
... diameter AD ; the angle BAD is a right angle ( 110 ) , and has for its measure the half of the semicircumference AMD ; the angle DAC has for its measure the half of DC ; therefore BAD + DAC , or BAC , has for its measure the half of AMD ...
... diameter AD ; the angle BAD is a right angle ( 110 ) , and has for its measure the half of the semicircumference AMD ; the angle DAC has for its measure the half of DC ; therefore BAD + DAC , or BAC , has for its measure the half of AMD ...
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Términos y frases comunes
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
Pasajes populares
Página 67 - The square of the hypothenuse is equal to the sum of the squares of the other two sides ; as, 5033 402+302.
Página 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.
Página 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'.
Página 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...
Página 168 - In any spherical triangle, the greater side is opposite the greater angle ; and conversely, the greater angle is opposite the greater side.
Página 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.
Página 8 - Any side of a triangle is less than the sum of the other two sides...
Página 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.
Página 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.
Página 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.