Elements of Plane Geometry: For the Use of SchoolsLewis & Sampson, 1844 - 96 sider |
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Side 74
... about the equal angle proportional , are similar ( Prop . 17 ) . : In the same manner it may be shown that the remain- ing triangles are similar . PROP . XXII . THEOREM . The perimeters ( B. 74 [ BOOK V. SIMILAR POLYGONS .
... about the equal angle proportional , are similar ( Prop . 17 ) . : In the same manner it may be shown that the remain- ing triangles are similar . PROP . XXII . THEOREM . The perimeters ( B. 74 [ BOOK V. SIMILAR POLYGONS .
Side 75
... perimeter of the first polygon , is to the sum of all the consequents , the perimeter of the other polygon , as either antecedent is to its consequent ( B. IV . Prop . 6 ) ; that is , the perimeter of ABCDE : the perimeter of FGHIK ...
... perimeter of the first polygon , is to the sum of all the consequents , the perimeter of the other polygon , as either antecedent is to its consequent ( B. IV . Prop . 6 ) ; that is , the perimeter of ABCDE : the perimeter of FGHIK ...
Side 88
... perimeter multiplied by half the line drawn from the centre of the circumscribed circle perpendicular to any side . Let ABCDEF be the regular polygon , OG , OH , lines drawn from the centre of the circumscribed circle perpen- dicular to ...
... perimeter multiplied by half the line drawn from the centre of the circumscribed circle perpendicular to any side . Let ABCDEF be the regular polygon , OG , OH , lines drawn from the centre of the circumscribed circle perpen- dicular to ...
Side 89
... perimeter become equal to the circumference of the circle , and the lines OG , OH , become radii ; hence this rule : The area of a circle is equal to its circumference multiplied by half its radius . * PROP . VII . THEOREM . The perimeters ...
... perimeter become equal to the circumference of the circle , and the lines OG , OH , become radii ; hence this rule : The area of a circle is equal to its circumference multiplied by half its radius . * PROP . VII . THEOREM . The perimeters ...
Side 90
... perimeters are to each other as BC : KL ( B. V. Prop . 22 ) , hence the perimeters are to each other as BO : KS . Again , since BC : KL :: BO : KS , we have BC2 : KL2 : BO2 : KS2 ( B. IV . Prop . 7 ) ; and since the sur- faces of the ...
... perimeters are to each other as BC : KL ( B. V. Prop . 22 ) , hence the perimeters are to each other as BO : KS . Again , since BC : KL :: BO : KS , we have BC2 : KL2 : BO2 : KS2 ( B. IV . Prop . 7 ) ; and since the sur- faces of the ...
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Elements of Plane Geometry: For the Use of Schools Nicholas Tillinghast Uten tilgangsbegrensning - 1844 |
Elements of Plane Geometry: For the Use of Schools Nicholas Tillinghast Uten tilgangsbegrensning - 1844 |
Elements of Plane Geometry: For the Use of Schools - Primary Source Edition Nicholas Tillinghast Ingen forhåndsvisning tilgjengelig - 2013 |
Vanlige uttrykk og setninger
ABCD adjacent angles allel alternate angles altitude angles ABD angles is equal antecedent and consequent B. I. Ax centre circle whose radius circumference circumscribed circumscribed circle common measure Converse of Prop describe an arc diameter divided draw the line equal angles equal B. I. Prop equal chords equal Prop equal respectively equally distant equiangular equivalent feet four numbers given angle given line given point given side half hence the triangles hypotenuse included angle inscribed angle Let ABC linear units longer than AC multiplied number of sides number of square oblique lines opposite parallel parallelogram perimeter perpendicular PROBLEM prove quadrilateral radii rectangle regular polygons respectively equal right angles Prop right-angled triangle Scholium sides AC similar subtended tangent THEOREM three sides triangle ABC triangles are equal vertex
Populære avsnitt
Side 31 - A circle is a plane figure bounded by a curved line, every point of which is equally distant from a point within called the center.
Side 63 - The square described on the hypothenuse of a right-angled triangle is equivalent to the sum of the squares described on the other two sides.
Side 71 - 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'.
Side 53 - In any proportion, the product of the means is equal to the product of the extremes.
Side 89 - 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.
Side 54 - In a series of equal ratios, any antecedent is to its consequent, as the sum of all the antecedents is to the sum of all the consequents. Let a: 6 = c: d = e :/. Then, by Art.
Side 83 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Side 59 - The area of a parallelogram is equal to the product of its base and its height: A = bx h.
Side 16 - Conversely, if two angles of a triangle are equal, the sides opposite them are also equal, and the triangle is isosceles.
Side 61 - From this proposition it is evident, that the square described on the difference of two lines is equivalent to the sum of the squares described on the lines respectively, minus twice the rectangle contained by the lines.