Plane and Solid Geometry |
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Αποτελέσματα 1 - 5 από τα 43.
Σελίδα
angles . int . . . . interior . . . triangle . isos . . . . isosceles . . . triangles . rt . . . . right .
. . parallelogram . st . . . . straight . . . parallelograms . sup . . . supplementary . . .
circle . © . . circles . Q . E . D . . . quod erat demonstrandum ( which was to be ...
angles . int . . . . interior . . . triangle . isos . . . . isosceles . . . triangles . rt . . . . right .
. . parallelogram . st . . . . straight . . . parallelograms . sup . . . supplementary . . .
circle . © . . circles . Q . E . D . . . quod erat demonstrandum ( which was to be ...
Σελίδα 49
A parallelogram has its opposite sides parallel . Upper Base Trapezoid
Parallelogram Lower Base Square Rhombus Rectangle A rhombus is an
equilateral parallelogram , whose angles are oblique . A rectangle is a
parallelogram , whose ...
A parallelogram has its opposite sides parallel . Upper Base Trapezoid
Parallelogram Lower Base Square Rhombus Rectangle A rhombus is an
equilateral parallelogram , whose angles are oblique . A rectangle is a
parallelogram , whose ...
Σελίδα 50
The altitude of a parallelogram or trapezoid is the perpendicular distance
between the two bases . PROPOSITION XXXII . THEOREM 134 . The opposite
sides and angles of a parallelogram are equal . Hyp . ABCD is a parallelogram .
To prove ...
The altitude of a parallelogram or trapezoid is the perpendicular distance
between the two bases . PROPOSITION XXXII . THEOREM 134 . The opposite
sides and angles of a parallelogram are equal . Hyp . ABCD is a parallelogram .
To prove ...
Σελίδα 51
Arthur Schultze, Frank Louis Sevenoak. PROPOSITION XXXIII . THEOREM 138 .
If the opposite sides of a quadrilateral are equal , the figure is a parallelogram . m
. Hyp . In quadrilateral ABDC AB = CD ; AD = BC . To prove AD | | BC ; AB | | CD ...
Arthur Schultze, Frank Louis Sevenoak. PROPOSITION XXXIII . THEOREM 138 .
If the opposite sides of a quadrilateral are equal , the figure is a parallelogram . m
. Hyp . In quadrilateral ABDC AB = CD ; AD = BC . To prove AD | | BC ; AB | | CD ...
Σελίδα 52
If two opposite sides of a parallelogram are produced by the same length in
opposite directions and their ends joined to the nearest vertices , another
parallelogram is formed . 140 . REMARK . — Lines may be shown to be parallel
by proving ...
If two opposite sides of a parallelogram are produced by the same length in
opposite directions and their ends joined to the nearest vertices , another
parallelogram is formed . 140 . REMARK . — Lines may be shown to be parallel
by proving ...
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Συχνά εμφανιζόμενοι όροι και φράσεις
ABCD altitude angle angles are equal base bisect bisector called chord circle circumference circumscribed coincide common cone construct contains corresponding cylinder diagonals diameter diedral angles difference distance divide draw drawn equal equidistant equivalent exterior angle faces figure Find formed four frustum geometrical given circle given line given point greater Hence homologous hypotenuse inches included inscribed intersecting isosceles triangle lateral edges length less limit line joining measured median meet midpoints opposite sides parallel parallel lines parallelogram passing perimeter perpendicular plane polyedron polygon prism PROBLEM Proof PROPOSITION prove pyramid quadrilateral radii radius ratio rectangle regular polygon respectively right angles right triangle School segments sides similar sphere spherical triangle square straight line surface tangent THEOREM third transform triangle triangle are equal vertex vertices
Δημοφιλή αποσπάσματα
Σελίδα 45 - If two triangles have two sides of one equal respectively to two sides of the other, but the included angle of the first triangle greater than the included angle of the second, then the third side of the first is greater than the third side of the second.
Σελίδα 119 - In any proportion, the product of the means is equal to the product of the extremes.
Σελίδα 180 - 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'.
Σελίδα 31 - The median to the base of an isosceles triangle is perpendicular to the base.
Σελίδα 305 - A cylinder is a solid bounded by a cylindrical surface and two parallel planes; the bases of a cylinder are the parallel planes; and the lateral surface is the cylindrical surface.
Σελίδα 337 - A spherical polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles. The bounding arcs are the sides of the polygon ; the...
Σελίδα 250 - A straight line perpendicular to one of two parallel planes is perpendicular to the other also.
Σελίδα 297 - A truncated triangular prism is equivalent to the sum of three pyramids whose common base is the base of the prism, and whose vertices are the three vertices of the inclined section.
Σελίδα 105 - I. When the given point, A, is in the circumference. HINT. — What is the angle formed by a radius and a tangent at its extremity ? II. When the given point, A, is without the circle. \ Construction. Join A, and 0 the center of the given circle. On OA as a diameter, construct a circumference, intersecting the given circumference in B and C.
Σελίδα 276 - An oblique prism is equivalent to a right prism whose base is a right section of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism. Hyp. GM is a right section of oblique prism AD', and OM ' a right prism whose altitude is equal to a lateral edge of AD'. To prove AD' =s= GM'. Proof. The lateral edges of GM