An Elementary Treatise on Geometry: Simplified for Beginners Not Versed in Algebra. Part I, Containing Plane Geometry, with Its Application to the Solution of Problems, Del 1Carter, Hendee, 1834 - 190 sider |
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Side 22
... prove this , for instance , of the C M A -B two lines CA , BM , which have the part MA common ? A. The common part MA belongs to the line MB as well as to the line AC , and therefore MC and AB are , in this case , but the continuation ...
... prove this , for instance , of the C M A -B two lines CA , BM , which have the part MA common ? A. The common part MA belongs to the line MB as well as to the line AC , and therefore MC and AB are , in this case , but the continuation ...
Side 23
... prove this of the two angles ADE , CDE , formed by the line ED , meet- ing the line AC , at the point D ? A M E D A. Because , if at D you erect the perpendicular DM , the two angles , ADE and CDE , occupy exactly the same space , as ...
... prove this of the two angles ADE , CDE , formed by the line ED , meet- ing the line AC , at the point D ? A M E D A. Because , if at D you erect the perpendicular DM , the two angles , ADE and CDE , occupy exactly the same space , as ...
Side 24
... prove it ? A. Because , if you add the same angle a , first to b , and then to e , the sum will , in both A cases , be the same ; namely , equal to two right angles ; which could not be , if the angle b were not equal to the angle e ...
... prove it ? A. Because , if you add the same angle a , first to b , and then to e , the sum will , in both A cases , be the same ; namely , equal to two right angles ; which could not be , if the angle b were not equal to the angle e ...
Side 25
... prove that the triangle a b c is equal to the triangle ABC ? ΔΔ A. By applying the side ab to its equal AB , the side ac will fall upon AC , and be upon BC ; because the angles at a and A , b and B , are respectively equal ; and as the ...
... prove that the triangle a b c is equal to the triangle ABC ? ΔΔ A. By applying the side ab to its equal AB , the side ac will fall upon AC , and be upon BC ; because the angles at a and A , b and B , are respectively equal ; and as the ...
Side 27
... proved that they must be parallel ; and if they are parallel , they cannot meet each other . Q. From a point without a ... prove PE B D -N R it by this diagram ? The line IF is bisected in O , and , from that point O , a perpendicular OP ...
... proved that they must be parallel ; and if they are parallel , they cannot meet each other . Q. From a point without a ... prove PE B D -N R it by this diagram ? The line IF is bisected in O , and , from that point O , a perpendicular OP ...
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An Elementary Treatise on Geometry: Simplified for Beginners Not ..., Del 1 Francis Joseph Grund Uten tilgangsbegrensning - 1834 |
An Elementary Treatise on Geometry: Simplified for Beginners Not ..., Del 1 Francis Joseph Grund Uten tilgangsbegrensning - 1834 |
An Elementary Treatise on Geometry: Simplified for Beginners Not ..., Del 1 Francis Joseph Grund Uten tilgangsbegrensning - 1832 |
Vanlige uttrykk og setninger
adjacent angles angle ABC angle ACB angle x basis bisected called centre chord circle whose radius circum circumference circumscribed circles consequently degrees DEMON diagonal diameter dividing the product draw the lines equal angles equal sides equal triangles exterior angle feet figure ABCDEF found by multiplying fourth term geometrical proportion given angle given circle given straight line given triangle gles height hypothenuse inches isosceles triangle length let fall line AB line AC line CD line MN mean proportional measures half number of sides parallel lines parallelogram ABCD perpendicular points of division PROBLEM prove quadrilateral radii radius rectangle rectilinear figure regular polygon ABCDEF Remark rhombus right angles right-angled triangle second term Sect semicircle side AB side AC similar triangles smaller SOLUTION subtended tangent third line third term three angles three sides trapezoid triangle ABC triangles are equal Truth vertex
Populære avsnitt
Side 2 - District Clerk's Office. BE IT REMEMBERED, that on the tenth day of August, AD 1829, in the fifty-fourth year of the Independence of the United States of America, JP Dabney, of the said district, has deposited in this office the title of a book, the right whereof he claims as author, in the words following, to wit...
Side 78 - If two triangles have two sides of the one equal to two sides of the other, each to each, but the...
Side 2 - CLERK'S OFFIcE. BE it remembered, that on the eleventh day of November, AD 1830, in the fiftyfifth year of the Independence of the United States of America, Gray & Bowen, of the said district, have deposited in this office the title of a book, the right whereof they claim as proprietors, in the words following, to wit...
Side 136 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds.
Side 121 - 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.
Side 137 - The side of a regular hexagon inscribed in a circle is equal to the radius of the circle.
Side 127 - The areas of two regular polygons of the same number of sides are to each other as the squares of their radii, or as the squares of their apothems.
Side 154 - A, with a radius equal to the sum of the radii of the given circles, describe a circle.
Side 90 - ... any two triangles are to each other as the products of their bases by their altitudes.
Side 137 - P is at the center of the circle. II. 18. The sum of the arcs subtending the vertical angles made by any two chords that intersect, is the same, as long as the angle of intersection is the same. 19. From a point without a circle two straight lines are drawn cutting the convex and concave circumferences, and also respectively parallel to two radii of the circle. Prove that the difference of the concave and convex arcs intercepted by the cutting lines, is equal to twice the arc intercepted by the radii.