A graduated course of problems in practical plane and solid geometry |
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Página 57
... hexagon 6 99 " " 29 A heptagon 7 99 99 99 An octagon 8 99 " " 29 A nonagon 9 " " " " 99 A decagon 10 " " " " An un - decagon 11 99 " " 12 " " 99 A do - decagon Problem 62- ( A . ) To inscribe any regular polygon ( say a pentagon ) in a ...
... hexagon 6 99 " " 29 A heptagon 7 99 99 99 An octagon 8 99 " " 29 A nonagon 9 " " " " 99 A decagon 10 " " " " An un - decagon 11 99 " " 12 " " 99 A do - decagon Problem 62- ( A . ) To inscribe any regular polygon ( say a pentagon ) in a ...
Página 59
... regular penta- gon will be inscribed within a given circle A. Problem 64 . To inscribe a regular hexagon within a given circle A. F B A C E 1. Draw any diameter BC . 2. With B and C as centres , and the PRACTICAL GEOMETRY . 59.
... regular penta- gon will be inscribed within a given circle A. Problem 64 . To inscribe a regular hexagon within a given circle A. F B A C E 1. Draw any diameter BC . 2. With B and C as centres , and the PRACTICAL GEOMETRY . 59.
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... hexagon is inscribed in the given circle A. Problem 65 . To inscribe a regular heptagon within a given circle A. 1. Draw any radius AB , and from B , with BA as radius , describe an arc CAD cutting the circumference in points C and D ...
... hexagon is inscribed in the given circle A. Problem 65 . To inscribe a regular heptagon within a given circle A. 1. Draw any radius AB , and from B , with BA as radius , describe an arc CAD cutting the circumference in points C and D ...
Página 66
... hexagon on a given line AB . 1. With points A and B as centres , and radius AB , describe the arcs intersecting at C. D B 2. From the point C , with CA as radius , describe the circle . 3. From D and E , with the same radius , cut off F ...
... hexagon on a given line AB . 1. With points A and B as centres , and radius AB , describe the arcs intersecting at C. D B 2. From the point C , with CA as radius , describe the circle . 3. From D and E , with the same radius , cut off F ...
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... hexagon . Problem 79 . To construct a regular hexagon , its diameter AB being given . 1. Bisect AB in C ( Pr . 1 ) . F A B H 2. Through the point A draw a line DE perpendicular to AB ( Pr . 2 ) . 3. On CA , as an altitude , construct an ...
... hexagon . Problem 79 . To construct a regular hexagon , its diameter AB being given . 1. Bisect AB in C ( Pr . 1 ) . F A B H 2. Through the point A draw a line DE perpendicular to AB ( Pr . 2 ) . 3. On CA , as an altitude , construct an ...
Términos y frases comunes
altitude angles to xy Atlas axis base BC Pr Bisect the angle bound in cloth circumference cone construct a triangle cube curve cylinder decagon describe a circle describe an arc describe the arc diameter BC distance divide draw a line draw a tangent draw lines edge ellipse equal circles equal in area equilateral triangle given circle given line given point given square ABCD given straight line given triangle ABC heptagon horizontal plane hyperbola inclined inscribe intersection isosceles triangle Join line BC line CD line of bisection lines parallel Maps mean proportional meeting number of equal octahedron parallel to xy parallelogram pentagon perpendicular to xy Philips plan and elevation PLANE GEOMETRY plane of projection point F points of division prism Problem projectors pyramid radii radius rectangle rectilineal figure regular polygon represent required circle rhombus right angles semicircle SOLID GEOMETRY square pyramid trapezium vertical plane
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Página 193 - A cone is a solid figure described by the revolution of a right-angled triangle about one of the sides containing the right angle, which side remains fixed. If the fixed side be equal to the other side containing the right angle, the cone is called a right-angled cone ; if it be less than the other side, an obtuse-angled ; and if greater, an acute-angled cone. XIX. The axis of a cone is the fixed straight line about which the triangle revolves.
Página 123 - A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less.