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 26
... consequently the two straight lines , AB , CD , can neither be converging nor diverging , and therefore they must be parallel . Q. Can two straight lines which meet each other , be perpendicular to the same straight line ? A. No. Q. Why ...
... consequently the two straight lines , AB , CD , can neither be converging nor diverging , and therefore they must be parallel . Q. Can two straight lines which meet each other , be perpendicular to the same straight line ? A. No. Q. Why ...
Side 29
... consequently they are parallel to each other . Q. There is one more case , and that is : If the two straight lines AB , CD ( in our last figure ) , are cut by a third line MN , so as to make the sum of the two interior angles AEF and ...
... consequently they are parallel to each other . Q. There is one more case , and that is : If the two straight lines AB , CD ( in our last figure ) , are cut by a third line MN , so as to make the sum of the two interior angles AEF and ...
Side 30
... to the line EF ; consequently the line CD cannot be parallel to it ; because through the point B only one line can be drawn parallel to the line EF . ( Truth X. ) QUERY X. Can you now tell the relation which the 30 GEOMETRY .
... to the line EF ; consequently the line CD cannot be parallel to it ; because through the point B only one line can be drawn parallel to the line EF . ( Truth X. ) QUERY X. Can you now tell the relation which the 30 GEOMETRY .
Side 32
... consequently these two triangles are equal ; and the side OP , opposite to the angle c , in the triangle MPO , is equal to the side MI , opposite to the equal angle d , in the triangle MPI . In precisely the same manner I can prove that ...
... consequently these two triangles are equal ; and the side OP , opposite to the angle c , in the triangle MPO , is equal to the side MI , opposite to the equal angle d , in the triangle MPI . In precisely the same manner I can prove that ...
Side 33
... consequently parallel to each other . QUERY XIII . What is the sum of all the angles in every triangle equal to ? A. To two right angles . Q. How do you prove this ? A. By drawing , through the vertex of the angle b , a straight B line ...
... consequently parallel to each other . QUERY XIII . What is the sum of all the angles in every triangle equal to ? A. To two right angles . Q. How do you prove this ? A. By drawing , through the vertex of the angle b , a straight B line ...
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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 - 1832 |
An Elementary Treatise on Geometry: Simplified for Beginners Not Versed in ... Francis Joseph Grund Uten tilgangsbegrensning - 1831 |
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 78 - If two triangles have two sides of the one equal to two sides of the other, each to each, but the...
Side 2 - District Clerk's Office. BE IT REMEMBERED, That on the seventh day of May, AD 1828, in the fifty-second year of the Independence of the UNITED STATES OF AMERICA, SG Goodrich, of the said District, has deposited in this office the...
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 154 - Upon a given straight line to describe a segment of a circle, which shall contain an angle equal to a given rectilineal angle.
Side 138 - 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 116 - The side of a regular hexagon inscribed in a circle is equal to the radius of the circle.
Side 101 - Now, since the areas of similar polygons are to each other as the squares of their homologous sides...
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 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.