## Elements of Geometry |

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Page vii

This section is followed by an appendix , in which we have demonstrated that the circle is greater than any rectilineal figure of the same

This section is followed by an appendix , in which we have demonstrated that the circle is greater than any rectilineal figure of the same

**perimeter**. The first section of the second part contains the properties of planes and of solid ... Page 2

6 ) , and the lines taken together make the

6 ) , and the lines taken together make the

**perimeter**of the polygon . 14. The polygon of three sides is the most simple of these figures , and is called a triangle ; that of four sides is called a quadrilateral ; that of five sides ... Page 17

Every convex polygon is such , that a straight line , however drawn , cannot meet the

Every convex polygon is such , that a straight line , however drawn , cannot meet the

**perimeter**in more than two points . THEOREM . 69. If two straight lines AB , CD ( fig . 36 ) , are perpendicular Fig 36 . to a third FG , these two ... Page 66

The

The

**perimeters**of similar polygons are as their homologous sides , and their surfaces are as the squares of these sides ... c . , the**perimeter**of the first figure is to the sum of the consequents FG + GH + HI , & c . , the**perimeter**of ... Page 69

It may be demonstrated , also , that the surface of a triangle is equal to its

It may be demonstrated , also , that the surface of a triangle is equal to its

**perimeter**multiplied by half of the radius of the inscribed circle . For the triangles AOB , BOC , AOC ( fig . 87 ) , which have Fig .### What people are saying - Write a review

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### Common terms and phrases

ABC fig ABCD adjacent altitude applied base called centre chord circ circle circumference circumscribed common cone consequently construction contained convex surface Corollary cylinder Demonstration described diameter difference distance divided draw drawn entire equal equivalent example extremities faces feet figure follows formed four give given greater half hence inclination inscribed intersection isosceles join less let fall manner mean measure meet moreover multiplied namely opposite parallel parallelogram parallelopiped pass perimeter perpendicular plane plane angles polyedron polygon prism PROBLEM produced proportional proposition pyramid radii radius ratio reason rectangle regular polygon respect right angles Scholium sector segment similar solid angle Solution sphere spherical square straight line suppose surface taken tangent THEOREM third triangle ABC vertex vertices whence