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 11
... a curved surface ; and by stretching the board , so as to make the wire fall upon it , you have a plane . + Legendre calls all geometrical figures polygons . C gles , and the lines are said to be perpendicular GEOMETRY . 11 DEFINITIONS.
... a curved surface ; and by stretching the board , so as to make the wire fall upon it , you have a plane . + Legendre calls all geometrical figures polygons . C gles , and the lines are said to be perpendicular GEOMETRY . 11 DEFINITIONS.
Side 12
... gles , and the lines are said to be perpendicular to eac other . Any angle smaller than a right angle is called acute , and when greater than a right angle , an obtuse angle . * Two lines which , lying in the same plane , and how- ever ...
... gles , and the lines are said to be perpendicular to eac other . Any angle smaller than a right angle is called acute , and when greater than a right angle , an obtuse angle . * Two lines which , lying in the same plane , and how- ever ...
Side 24
... gles is a right angle . Q. And what is the sum of all the angles , a , b , c , d , e , f , around the same point , equal to ? A. To four right angles . Q. Why ? A. Because if , through that point , you draw a perpendicu- lar to any of ...
... gles is a right angle . Q. And what is the sum of all the angles , a , b , c , d , e , f , around the same point , equal to ? A. To four right angles . Q. Why ? A. Because if , through that point , you draw a perpendicu- lar to any of ...
Side 28
... gles AEF and EFD , or the angles BEF and EFC , equal , what relation would the lines B AB , CD , then bear to each other ? A. They would still be parallel . Q. How can you prove this ? because EFD and A. If the angle AEF is equal to the ...
... gles AEF and EFD , or the angles BEF and EFC , equal , what relation would the lines B AB , CD , then bear to each other ? A. They would still be parallel . Q. How can you prove this ? because EFD and A. If the angle AEF is equal to the ...
Side 46
... gles bear to each other ? A. They coincide with each other in all their parts ; that is , they are equal to each other . B Fig . I. A Q. How can you prove it ? A. Because , if , in a triangle , ABC , for instance , you have the sides AB ...
... gles bear to each other ? A. They coincide with each other in all their parts ; that is , they are equal to each other . B Fig . I. A Q. How can you prove it ? A. Because , if , in a triangle , ABC , for instance , you have the sides AB ...
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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.