Plane and Solid Geometry |
Αναζήτηση στο βιβλίο
Αποτελέσματα 1 - 5 από τα 100.
Σελίδα 11
... is called equiangular if all its angles are equal . Right Obtuse Acute
Equiangular 64 . The base of a triangle is the side on which the figure is
supposed to stand . The base of an isosceles triangle is that side which is equal
to no other ; the ...
... is called equiangular if all its angles are equal . Right Obtuse Acute
Equiangular 64 . The base of a triangle is the side on which the figure is
supposed to stand . The base of an isosceles triangle is that side which is equal
to no other ; the ...
Σελίδα 15
The bisector of the vertical angle of an isosceles triangle bisects the base . Ex .
38 . If , upon the sides of an angle , equal distances be laid off from the vertex ,
and the ends be joined to any point in the bisector of the angle , these lines are ...
The bisector of the vertical angle of an isosceles triangle bisects the base . Ex .
38 . If , upon the sides of an angle , equal distances be laid off from the vertex ,
and the ends be joined to any point in the bisector of the angle , these lines are ...
Σελίδα 27
... and may be omitted at a first reading . PROPOSITION XIV . THEOREM 101 .
The base angles of PARALLEL LINES 27.
... and may be omitted at a first reading . PROPOSITION XIV . THEOREM 101 .
The base angles of PARALLEL LINES 27.
Σελίδα 28
The base angles of an isosceles triangle are equal . Hyp . In A ABC , AB = AC .
To prove ZB = 2C . Hint . — Let AD be the bisector of ZBAC , and prove the
equality of the two triangles . 102 . Cor . An equilateral triangle is equiangular . Ex
. 100 .
The base angles of an isosceles triangle are equal . Hyp . In A ABC , AB = AC .
To prove ZB = 2C . Hint . — Let AD be the bisector of ZBAC , and prove the
equality of the two triangles . 102 . Cor . An equilateral triangle is equiangular . Ex
. 100 .
Σελίδα 29
The bisectors of the base angles of an isosceles triangle are equal .
PROPOSITION XV . THEOREM 103 . If two angles of a triangle are equal , the
sides opposite these angles are equal . [ Converse of XIV . ] BD Hyp . In A ABC ,
ZB = LC .
The bisectors of the base angles of an isosceles triangle are equal .
PROPOSITION XV . THEOREM 103 . If two angles of a triangle are equal , the
sides opposite these angles are equal . [ Converse of XIV . ] BD Hyp . In A ABC ,
ZB = LC .
Τι λένε οι χρήστες - Σύνταξη κριτικής
Δεν εντοπίσαμε κριτικές στις συνήθεις τοποθεσίες.
Άλλες εκδόσεις - Προβολή όλων
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
Πληροφορίες βιβλιογραφίας
| Τίτλος | Plane and Solid Geometry Plane and Solid Geometry, Frank Louis Sevenoak |
| Συγγραφείς | Arthur Schultze, Frank Louis Sevenoak |
| Εκδότης | Macmillan, 1902 |
| Μέγεθος | 370 σελίδες |
| Εξαγωγή αναφοράς | BiBTeX EndNote RefMan |