The Elements of GeometryLeach, Shewell & Sanborn, 1894 - 378 σελίδες |
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Σελίδα 46
... is a parallelogram whose angles are not right angles , and whose adjacent sides are equal . A Rectangle is a parallelogram whose angles are right angles . A Square is a rectangle whose sides are equal . 46 - BOOK I. PLANE GEOMETRY . -
... is a parallelogram whose angles are not right angles , and whose adjacent sides are equal . A Rectangle is a parallelogram whose angles are right angles . A Square is a rectangle whose sides are equal . 46 - BOOK I. PLANE GEOMETRY . -
Σελίδα 47
Webster Wells. A Square is a rectangle whose sides are equal . Rhomboid . Rhombus . Rectangle . Square . THEOREM . PROPOSITION XXXVII . 104. The opposite sides of a parallelogram are equal . B A Let ABCD be a parallelogram . To prove AB ...
Webster Wells. A Square is a rectangle whose sides are equal . Rhomboid . Rhombus . Rectangle . Square . THEOREM . PROPOSITION XXXVII . 104. The opposite sides of a parallelogram are equal . B A Let ABCD be a parallelogram . To prove AB ...
Σελίδα 53
... square are equal . EXERCISES . 42. If the diagonals of a parallelogram are equal , the figure is a rectangle . 43. If two adjacent sides of a quadrilateral are equal , and the diagonal bisects their included angle , the other two sides ...
... square are equal . EXERCISES . 42. If the diagonals of a parallelogram are equal , the figure is a rectangle . 43. If two adjacent sides of a quadrilateral are equal , and the diagonal bisects their included angle , the other two sides ...
Σελίδα 68
... square . 90. If D is the middle point of the side BC of the triangle ABC , and BE and CF are the perpendiculars from ... square ABCD , such that AE = BF = CG = DH , prove that the figure EFGH is a square . 100. If on the diagonal BD of a ...
... square . 90. If D is the middle point of the side BC of the triangle ABC , and BE and CF are the perpendiculars from ... square ABCD , such that AE = BF = CG = DH , prove that the figure EFGH is a square . 100. If on the diagonal BD of a ...
Σελίδα 69
Webster Wells. 100. If on the diagonal BD of a square ABCD a distance BE is taken equal to AB , and EF is drawn perpendicular to BD , meeting AD at F , prove that AF EF = ED . = 101. Prove the theorem of § 127 by drawing lines from any ...
Webster Wells. 100. If on the diagonal BD of a square ABCD a distance BE is taken equal to AB , and EF is drawn perpendicular to BD , meeting AD at F , prove that AF EF = ED . = 101. Prove the theorem of § 127 by drawing lines from any ...
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Συχνά εμφανιζόμενοι όροι και φράσεις
ABCD adjacent angles altitude angles are equal approach the limit arc BC area ABC base and altitude BC² bisector bisects centre chord circle circumference circumscribed cone of revolution construct the triangle Converse of Prop cylinder diagonals diameter diedral Draw BC equal respectively equally distant equilateral triangle equivalent exterior angle Find the area frustum given point given straight line Hence homologous hypotenuse intersection isosceles triangle lateral area lateral edges Let ABC measured by arc middle point number of sides O-ABC parallelogram parallelopiped perimeter perpendicular to MN plane MN polyedral polyedron prism produced PROPOSITION prove pyramid quadrilateral radii radius rectangle regular polygon rhombus right angles right triangle secant line segment similar slant height sphere spherical polygon spherical triangle square surface tangent tetraedron THEOREM trapezoid triangle ABC triangles are equal triangular prism triedral vertex volume Whence
Δημοφιλή αποσπάσματα
Σελίδα 165 - Any two rectangles are to each other as the products of their bases by their altitudes.
Σελίδα 39 - If two triangles have two sides of one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second.
Σελίδα 65 - The straight line joining the middle points of two sides of a triangle is parallel to the third side, and equal to half of it.
Σελίδα 172 - 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. To prove that Proof. A Let the triangles ABC and ADE have the common angle A. A ABC -AB X AC Now and A ADE AD X AE Draw BE.
Σελίδα 122 - In any proportion the terms are in proportion by Composition ; that is, the sum of the first two terms is to the first term as the sum of the last two terms is to the third term.
Σελίδα 355 - Similar cylinders are to each other as the cubes of their altitudes, or as the cubes of the diameters of their bases.
Σελίδα 52 - Two triangles are congruent if two sides and the included angle of one are equal respectively to two sides and the included angle of the other.
Σελίδα 140 - If two polygons are composed of the same number of triangles, similar each to each and similarly placed, the polygons are similar.
Σελίδα 123 - In any proportion the terms are in proportion by composition and division ; that is, the sum of the first two terms is to their difference as the sum of the last two terms to their difference.
Σελίδα 207 - S' denote the areas of two © whose radii are R and R', and diameters D and D', respectively. Then, | = "* § = ££ = £• <§337> That is, the areas of two circles are to each other as the squares of their radii, or as the squares of their diameters.