An Elementary Treatise on Plane and Solid Geometry |
From inside the book
Results 1-5 of 13
Page viii
... common measure of two lines ( 156 ) ; or of two arcs ( 157 ) , · 39 40 40 40 41 41 41 42 42 · 43 44 CHAPTER X ... Altitude of parallelogram , of triangle , of trapezoid ( 171 ) , Cases of similar triangles ( 172-180 ) , • • 15 49 49 50 ...
... common measure of two lines ( 156 ) ; or of two arcs ( 157 ) , · 39 40 40 40 41 41 41 42 42 · 43 44 CHAPTER X ... Altitude of parallelogram , of triangle , of trapezoid ( 171 ) , Cases of similar triangles ( 172-180 ) , • • 15 49 49 50 ...
Page 81
... common altitude OM . The sum of the areas of the triangles , or the area of the polygon is , consequently , half the product of the sum of the sides , AB , BC , & c . by the common altitude OM ; that is , half the product of the ...
... common altitude OM . The sum of the areas of the triangles , or the area of the polygon is , consequently , half the product of the sum of the sides , AB , BC , & c . by the common altitude OM ; that is , half the product of the ...
Page 82
... altitudes the radii OA , OM , ON , & c . The sum of the areas of these triangles , or the area of the sector is , then , half the product of the sum of the bases AM , MN , NP , & c . by the common altitude OA ; that is , half the ...
... altitudes the radii OA , OM , ON , & c . The sum of the areas of these triangles , or the area of the sector is , then , half the product of the sum of the bases AM , MN , NP , & c . by the common altitude OA ; that is , half the ...
Page 85
... common to both , and the triangles DBC , DBM are equivalent because they have the same base BD and the same altitude , by § 82 , their vertices C and M being in the line CM parallel to this base . 290. Problem . To find a square ...
... common to both , and the triangles DBC , DBM are equivalent because they have the same base BD and the same altitude , by § 82 , their vertices C and M being in the line CM parallel to this base . 290. Problem . To find a square ...
Page 110
... altitude . Proof . a . The area of each of the parallelograms ABFG , BCGH , & c . , which compose the convex surface of the right prism ( fig . 164 ) is , by § 247 , the product of its base AB , BC & c . , by the common altitude AF ...
... altitude . Proof . a . The area of each of the parallelograms ABFG , BCGH , & c . , which compose the convex surface of the right prism ( fig . 164 ) is , by § 247 , the product of its base AB , BC & c . , by the common altitude AF ...
Contents
86 | |
87 | |
93 | |
95 | |
101 | |
103 | |
104 | |
107 | |
28 | |
34 | |
38 | |
40 | |
45 | |
52 | |
54 | |
63 | |
70 | |
77 | |
78 | |
83 | |
111 | |
116 | |
117 | |
119 | |
120 | |
123 | |
132 | |
134 | |
139 | |
145 | |
Other editions - View all
Common terms and phrases
ABC fig adjacent angles angle BAC arc BC base and altitude bisect centre chord circumference common altitude construct convex surface Corollary DEF fig Definitions denote diameter divided Draw equal arcs equal distances equiangular with respect equilateral equivalent frustum given angle given circle given line given polygon given sides given square gles greater half the product Hence homologous sides hypothenuse infinite number infinitely small Inscribed Angle inscribed circle isosceles Let ABCD line AB fig line BC lines drawn mean proportional number of sides oblique lines parallel lines parallel to BC parallelogram parallelopipeds perimeter perpendicular plane MN polyedron polygon ABCD &c Problem Proof pyramid or cone radii radius rectangles regular polygon right triangle Scholium sector segment side BC similar polygons similar triangles solid angle Solution sphere spherical polygon spherical triangle straight line tangent Theorem triangles ABC triangular prism vertex vertices whence
Popular passages
Page 68 - The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their areas are to each other as the squares of those sides (Prop.
Page 127 - Every section of a sphere, made by a plane, is a circle, Let AMB be a section, made by a plane, in the sphere whose centre is C.
Page 71 - Rectangles of the same altitude are to each other as their bases, and rectangles of the same base are to each other as their altitudes. 245.
Page 20 - The sum of the three angles of any triangle is equal to two right angles.
Page xv - The first term of a ratio is called the antecedent, and the second term the consequent.
Page 83 - ... we suppose the error A to be of any magnitude whatever. 286. Definition. Similar sectors and similar segments are such as correspond to similar arcs. 287. Theorem. Similar sectors are to each other as the squares of their radii. Proof. The similar sectors AOB, A'OB ' (fig. 136) are, by the same reasoning as in t5 97, the same parts of their respective circles, which the angle O= O...
Page 31 - Theorem. In the same circle, or in equal circles, equal arcs are subtended by equal chords.
Page 87 - To construct a parallelogram equivalent to a given square, and having the sum of its base and altitude equal to a given line.
Page 99 - B, from the plane. 320. Theorem. Oblique lines drawn from a point to a plane at equal distances from the perpendicular are equal; and of two oblique lines unequally distant the more remote is the greater.
Page 78 - Similar triangles are to each other as the squares of their homologous sides. Proof. In the similar .triangles ABC, A'B'C