## Elements of Geometry |

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Page iii

Moreover , the work is

Moreover , the work is

**divided**into two parts , one treating of plane figures , and the other of solids ; and the subdivisions of each part are denominated sections . As a knowledge of algebraical signs and the theory of proportions is ... Page vi

This work is

This work is

**divided**into eight sections , four of which treat of plane geometry , and four of solid geometry . The first section , entitled first principles , & c.contains the properties of straight lines which meet those of ... Page ix

A indicates the quotient arising from the magnitude represented by A being

A indicates the quotient arising from the magnitude represented by A being

**divided**by that represented by B , or A**divided**by B. A = B signifies that the magnitude represented by A is equal to that represented by B , or A equal to B. A ... Page 16

42 ) , be the proposed polygon ; if , from the vertex of the angle A , we draw to the vertices of the opposite angles the diagonals AC , AD , AE , & c . , it is evident , that the polygon will be

42 ) , be the proposed polygon ; if , from the vertex of the angle A , we draw to the vertices of the opposite angles the diagonals AC , AD , AE , & c . , it is evident , that the polygon will be

**divided**into five triangles , if it have ... Page 33

Let us now imagine the arc AB to be

Let us now imagine the arc AB to be

**divided**into equal parts , of which each shall be less than DO ; there will be at least one point of division between D and 0 ; let I be this point , and join CI ; the arcs AB , AI , will be to each ...### 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