Plane and Solid Geometry |
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Page 49
... Rhombus Rectangle Square A rhombus is an equilateral parallelogram , whose angles are oblique . A rectangle is a parallelogram , whose angles are right angles . A square is an equilateral rectangle . E 132. DEF . An isosceles trapezoid ...
... Rhombus Rectangle Square A rhombus is an equilateral parallelogram , whose angles are oblique . A rectangle is a parallelogram , whose angles are right angles . A square is an equilateral rectangle . E 132. DEF . An isosceles trapezoid ...
Page 52
... rhombus are perpendicular to each other . Ex . 199. If the diagonals of a parallelogram are perpendicular to each other , the figure is a rhombus , or a square . Ex . 200. If each half of the diagonal of a parallelogram is bisected ...
... rhombus are perpendicular to each other . Ex . 199. If the diagonals of a parallelogram are perpendicular to each other , the figure is a rhombus , or a square . Ex . 200. If each half of the diagonal of a parallelogram is bisected ...
Page 53
... rhombus . Ex . 202. The diagonals of a rectangle are equal . Ex . 203. State and prove the converse of the preceding exercise . Ex . 204. If the ends of two diameters of a circle be joined in succes- sion a rectangle is formed . Ex ...
... rhombus . Ex . 202. The diagonals of a rectangle are equal . Ex . 203. State and prove the converse of the preceding exercise . Ex . 204. If the ends of two diameters of a circle be joined in succes- sion a rectangle is formed . Ex ...
Page 70
... rhombus . Ex . 272. The lines joining the midpoints of the sides of a rhombus , taken in order , enclose a rectangle . Ex . 273. The lines joining the midpoints of the sides of any quadri- lateral , taken in order , enclose a ...
... rhombus . Ex . 272. The lines joining the midpoints of the sides of a rhombus , taken in order , enclose a rectangle . Ex . 273. The lines joining the midpoints of the sides of any quadri- lateral , taken in order , enclose a ...
Page 100
... rhombus . * Ex . 385. If from any point in the circumference of a circle , chords be drawn to the vertices of an inscribed equilateral tri- angle , the longest chord equals the sum of the smaller chords . ( 162 ) * Ex . 386. If the ...
... rhombus . * Ex . 385. If from any point in the circumference of a circle , chords be drawn to the vertices of an inscribed equilateral tri- angle , the longest chord equals the sum of the smaller chords . ( 162 ) * Ex . 386. If the ...
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Common terms and phrases
ABCD altitude angles are equal bisect bisector chord circumference circumscribed cone construct a triangle cylinder diagonals diagram for Prop diameter diedral angles divide draw drawn equiangular equiangular polygon equilateral triangle equivalent exterior angle face angles Find the area Find the radius Find the volume frustum given circle given line given point given triangle Hence HINT homologous homologous sides hypotenuse inches inscribed intersecting isosceles triangle joining the midpoints lateral area lateral edges line joining mean proportional median opposite sides parallel lines parallelogram parallelopiped perimeter perpendicular plane MN polyedral angle polyedron prism PROPOSITION prove Proof quadrilateral radii ratio rectangle regular polygon respectively equal rhombus right angles right triangle SCHOLIUM segment similar triangles sphere spherical polygon spherical triangle square straight line surface tangent THEOREM trapezoid triangle ABC triangle are equal triedral vertex
Popular passages
Page 45 - If two triangles have two sides of one equal respectively to two sides of the other, but the included angle of the first triangle greater than the included angle of the second, then the third side of the first is greater than the third side of the second.
Page 119 - In any proportion, the product of the means is equal to the product of the extremes.
Page 180 - The areas of two triangles which have 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. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.
Page 31 - The median to the base of an isosceles triangle is perpendicular to the base.
Page 305 - A cylinder is a solid bounded by a cylindrical surface and two parallel planes; the bases of a cylinder are the parallel planes; and the lateral surface is the cylindrical surface.
Page 337 - A spherical polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles. The bounding arcs are the sides of the polygon ; the...
Page 250 - A straight line perpendicular to one of two parallel planes is perpendicular to the other also.
Page 297 - A truncated triangular prism is equivalent to the sum of three pyramids whose common base is the base of the prism, and whose vertices are the three vertices of the inclined section.
Page 105 - I. When the given point, A, is in the circumference. HINT. — What is the angle formed by a radius and a tangent at its extremity ? II. When the given point, A, is without the circle. \ Construction. Join A, and 0 the center of the given circle. On OA as a diameter, construct a circumference, intersecting the given circumference in B and C.
Page 276 - An oblique prism is equivalent to a right prism whose base is a right section of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism. Hyp. GM is a right section of oblique prism AD', and OM ' a right prism whose altitude is equal to a lateral edge of AD'. To prove AD' =s= GM'. Proof. The lateral edges of GM