Plane and Solid GeometryLongmans, Green and Company, 1898 - 210 sider |
Inni boken
Resultat 1-5 av 18
Side 46
... Chord is a straight line which joins any two points on the circumference , as BC . A B D When a chord passes through the centre , it has its greatest length , and is called the Diameter . 129. A Radius is a straight line drawn from the ...
... Chord is a straight line which joins any two points on the circumference , as BC . A B D When a chord passes through the centre , it has its greatest length , and is called the Diameter . 129. A Radius is a straight line drawn from the ...
Side 47
... chord , as AMB , Fig . 1 . 135. A Semicircle is a segment equal to one - half the circle , as ADC , Fig . 1 . 136. A ... chords of that circumference BK . I. § 140. ] 47 THE CIRCLE .
... chord , as AMB , Fig . 1 . 135. A Semicircle is a segment equal to one - half the circle , as ADC , Fig . 1 . 136. A ... chords of that circumference BK . I. § 140. ] 47 THE CIRCLE .
Side 48
James Howard Gore. circumference , and its sides are chords of that circumference , as ABC , Fig . 1 . A polygon is inscribed in a circle when its sides are chords of the circle , as ABC , Fig . 1 . A circle is inscribed in a polygon ...
James Howard Gore. circumference , and its sides are chords of that circumference , as ABC , Fig . 1 . A polygon is inscribed in a circle when its sides are chords of the circle , as ABC , Fig . 1 . A circle is inscribed in a polygon ...
Side 49
... chords . A B A B ' M M ' Let ABM and A'B'M ' be two equal circles , in which 40 = ZC . To prove that the arc AB : arc A'B ' and the chord AB = chord A'B ' . 1. Place the circle ABM upon A'B'M ' so that their centres may coincide , and A ...
... chords . A B A B ' M M ' Let ABM and A'B'M ' be two equal circles , in which 40 = ZC . To prove that the arc AB : arc A'B ' and the chord AB = chord A'B ' . 1. Place the circle ABM upon A'B'M ' so that their centres may coincide , and A ...
Side 50
... chord bisects the chord and its subtended arcs . D C B In the circle ADB , let the diameter CD be perpendicular to the chord AB . To prove that DC bisects AB and its subtended arcs . Let O be the centre of the circle , and join OA and ...
... chord bisects the chord and its subtended arcs . D C B In the circle ADB , let the diameter CD be perpendicular to the chord AB . To prove that DC bisects AB and its subtended arcs . Let O be the centre of the circle , and join OA and ...
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Vanlige uttrykk og setninger
ABCD AC² acute angle AD² adjacent adjacent angles altitude angle formed angles are equal apothem arc BC base and altitude bisect bisector called centre chord circumference circumscribed cone cylinder diagonals diameter diedral angles distance divided draw drawn ECDH equally distant equilateral equivalent EXERCISES faces four right angles frustum given point given straight line hence homologous homologous sides hypotenuse inscribed polygon interior angles intersection isosceles triangle join lateral area lateral edges Let ABC lune mean proportional measured by one-half middle point number of sides parallelogram parallelopiped perimeter perpendicular polyedral angle polyedron PROPOSITION XI prove pyramid Q.E.D. PROPOSITION quadrilateral radii radius ratio rectangle rectangular parallelopiped regular polygon right triangle SCHOLIUM segments semiperimeter sphere spherical angle spherical polygon spherical triangle surface tangent THEOREM triangle ABC triangles are equal triangular triangular prism V-ABC vertex vertical angle
Populære avsnitt
Side 46 - PERIPHERY of a circle is its entire bounding line ; or it is a curved line, all points of which are equally distant from a point within called the centre.
Side 105 - ... any two parallelograms are to each other as the products of their bases by their altitudes. PROPOSITION V. THEOREM. 403. The area of a triangle is equal to half the product of its base by its altitude.
Side 82 - If any number of quantities are proportional, any antecedent is to its consequent as the sum of all the antecedents is to the sum of all the consequents. Let a : b = c : d = e :f Now ab = ab (1) and by Theorem I.
Side 192 - A sphere is a solid bounded by a surface all points of which are equally distant from a point within called the centre.
Side 108 - Two triangles having 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.
Side 146 - A STRAIGHT line is perpendicular to a plane, when it is perpendicular to every straight line which it meets in that plane.
Side 30 - In an isosceles triangle, the angles opposite the equal sides are equal.
Side 80 - 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.
Side 79 - If the product of two quantities is equal to the product of two others, one pair may be made the extremes, and the other pair the means, of a proportion. Let ad = ос.
Side 148 - Equal oblique lines from a point to a plane meet the plane at equal distances from the foot of the perpendicular ; and of two unequal oblique lines the greater meets the plane at the greater distance from the foot of the perpendicular.