The Elements of GeometryLeach, Shewell & Sanborn, 1894 - 378 σελίδες |
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Σελίδα 70
... circumference , all points of which are equally distant from a point within , called the centre ; as ABCD . Any portion of the circumference , as AB , is called an arc . A radius is a straight line drawn from the centre to the circumference ...
... circumference , all points of which are equally distant from a point within , called the centre ; as ABCD . Any portion of the circumference , as AB , is called an arc . A radius is a straight line drawn from the centre to the circumference ...
Σελίδα 71
... circumference is understood , unless the contrary is specified . A segment of a circle is the portion included between an arc and its chord ; as AMBN . A semicircle is a segment equal to one - half the circle . A sector of a circle is ...
... circumference is understood , unless the contrary is specified . A segment of a circle is the portion included between an arc and its chord ; as AMBN . A semicircle is a segment equal to one - half the circle . A sector of a circle is ...
Σελίδα 72
... circumference ; as ABCD . In such a case , the circle is said to be circumscribed about the polygon . A polygon is ... circumference . B A C D Let AC be a diameter of the circle ABCD . To prove that AC bisects the circle and its ...
... circumference ; as ABCD . In such a case , the circle is said to be circumscribed about the polygon . A polygon is ... circumference . B A C D Let AC be a diameter of the circle ABCD . To prove that AC bisects the circle and its ...
Σελίδα 73
... circumference . PROPOSITION II . THEOREM . 153. A straight line cannot intersect a circumference in more than two points . M A B Let O be the centre of a circle , and MN any straight line . To prove that MN cannot intersect the ...
... circumference . PROPOSITION II . THEOREM . 153. A straight line cannot intersect a circumference in more than two points . M A B Let O be the centre of a circle , and MN any straight line . To prove that MN cannot intersect the ...
Σελίδα 74
... circumference . A B ' M M ' Let C and C ' be the centres of the equal circles AMB and A'M'B ' ; and let To prove LACBA'C'B ' . arc AB = arc A'B ' . Superpose the sector ABC upon A'B'C ' in such a way that C shall coincide with its equal ...
... circumference . A B ' M M ' Let C and C ' be the centres of the equal circles AMB and A'M'B ' ; and let To prove LACBA'C'B ' . arc AB = arc A'B ' . Superpose the sector ABC upon A'B'C ' in such a way that C shall coincide with its equal ...
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Συχνά εμφανιζόμενοι όροι και φράσεις
ABCD adjacent angles altitude angles are equal approach the limit arc BC area ABC base and altitude BC² bisector bisects centre chord circle circumference circumscribed cone of revolution construct the triangle Converse of Prop cylinder diagonals diameter diedral Draw BC equal respectively equally distant equilateral triangle equivalent exterior angle Find the area frustum given point given straight line Hence homologous hypotenuse intersection isosceles triangle lateral area lateral edges Let ABC measured by arc middle point number of sides O-ABC parallelogram parallelopiped perimeter perpendicular to MN plane MN polyedral polyedron prism produced PROPOSITION prove pyramid quadrilateral radii radius rectangle regular polygon rhombus right angles right triangle secant line segment similar slant height sphere spherical polygon spherical triangle square surface tangent tetraedron THEOREM trapezoid triangle ABC triangles are equal triangular prism triedral vertex volume Whence
Δημοφιλή αποσπάσματα
Σελίδα 165 - Any two rectangles are to each other as the products of their bases by their altitudes.
Σελίδα 39 - If two triangles have two sides of one equal respectively to two sides of the other, but the included angle of the first 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 65 - The straight line joining the middle points of two sides of a triangle is parallel to the third side, and equal to half of it.
Σελίδα 172 - 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. To prove that Proof. A Let the triangles ABC and ADE have the common angle A. A ABC -AB X AC Now and A ADE AD X AE Draw BE.
Σελίδα 122 - In any proportion the terms are in proportion by Composition ; that is, the sum of the first two terms is to the first term as the sum of the last two terms is to the third term.
Σελίδα 355 - Similar cylinders are to each other as the cubes of their altitudes, or as the cubes of the diameters of their bases.
Σελίδα 52 - Two triangles are congruent if two sides and the included angle of one are equal respectively to two sides and the included angle of the other.
Σελίδα 140 - If two polygons are composed of the same number of triangles, similar each to each and similarly placed, the polygons are similar.
Σελίδα 123 - In any proportion the terms are in proportion by composition and division ; that is, the sum of the first two terms is to their difference as the sum of the last two terms to their difference.
Σελίδα 207 - S' denote the areas of two © whose radii are R and R', and diameters D and D', respectively. Then, | = "* § = ££ = £• <§337> That is, the areas of two circles are to each other as the squares of their radii, or as the squares of their diameters.