Plane and Solid GeometryGinn & Company, 1900 |
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Resultat 1-5 av 69
Side 20
... Let the diagonals AC , BD of the quadrilateral ABCD meet at O , and let AO be equal to CO , BO to DO . To prove that figure ABCD is a parallelogram . Proof . The ABO and DCO are equal . For and AO = CO , BO = DO , LAOB = 2 COD . = D ...
... Let the diagonals AC , BD of the quadrilateral ABCD meet at O , and let AO be equal to CO , BO to DO . To prove that figure ABCD is a parallelogram . Proof . The ABO and DCO are equal . For and AO = CO , BO = DO , LAOB = 2 COD . = D ...
Side 21
... Let ABCD be a rhombus , and let AC meet BD at O. To prove that AC is 1 to BD , and that Z BAO = 2 DAO , etc. The OAB and OAD are equal . OA is common , AB = AD , Proof . § 150 For § 170 and OB = OD . $ 184 D .. LBAO △ DAO . B = § 128 ...
... Let ABCD be a rhombus , and let AC meet BD at O. To prove that AC is 1 to BD , and that Z BAO = 2 DAO , etc. The OAB and OAD are equal . OA is common , AB = AD , Proof . § 150 For § 170 and OB = OD . $ 184 D .. LBAO △ DAO . B = § 128 ...
Side 22
... Let ABCD be a quadrilateral , E , F , G , H the middle points of the sides , taken in order . To prove that the figure EFGH is a . EF and HG are to AC , EF Proof . and .. figure EFGH is a □ . Draw the diagonals AC , BD . Ꭰ . G $ 189 ...
... Let ABCD be a quadrilateral , E , F , G , H the middle points of the sides , taken in order . To prove that the figure EFGH is a . EF and HG are to AC , EF Proof . and .. figure EFGH is a □ . Draw the diagonals AC , BD . Ꭰ . G $ 189 ...
Side 23
... Let ABCD be an isosceles trapezoid , and let EF , FG , GH , HE join the middle points of its sides , taken in order . To prove that the figure EFGH is a rhombus or a square . SDG CQ Proof . Through the middle points F , H of the legs H ...
... Let ABCD be an isosceles trapezoid , and let EF , FG , GH , HE join the middle points of its sides , taken in order . To prove that the figure EFGH is a rhombus or a square . SDG CQ Proof . Through the middle points F , H of the legs H ...
Side 24
... Let the quadrilateral EFKL be formed by the bisectors of the angles of the rhomboid ABCD . To prove that EFKL is a rectangle . Proof . Since DC and AB are || , 2 CDA + Z BAD = 180 ° . : . LEDA + LEAD = 90 ° . .. ZDEA is a rt . Z. .. Z ...
... Let the quadrilateral EFKL be formed by the bisectors of the angles of the rhomboid ABCD . To prove that EFKL is a rectangle . Proof . Since DC and AB are || , 2 CDA + Z BAD = 180 ° . : . LEDA + LEAD = 90 ° . .. ZDEA is a rt . Z. .. Z ...
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Vanlige uttrykk og setninger
AABC AB² AC² altitude angle apothem arc AC base BD² bisect bisector centre chord circumference circumscribed cone Const describe diagonals diameter dihedral directrix distance Draw AC drawn EFGH ellipse equal equidistant equilateral triangle equivalent Find the area Find the locus frustum given circle given length given plane given point given radius Hence hypotenuse inches intersection isosceles trapezoid isosceles triangle latus rectum Let ABCD median meet middle point parallel pentagon perimeter perpendicular plane MN produced prove that Proof pyramid Q. E. D. Ex Q. E. F. Discussion quadrilateral radii radius rectangle regular hexagon regular polygon respectively rhombus S-ABC segment sides similar sphere square feet straight line surface tangent tetrahedron trihedral vertex vertices Нур
Populære avsnitt
Side 6 - If two triangles have two sides of the 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. Given A ABC and A'B'C...
Side 90 - The sum of the squares of two sides of a triangle is equal to twice the square of half the third side increased by twice the square of the median upon that side.
Side 121 - The square constructed upon the difference of two straight lines is equivalent to the sum of the squares constructed upon these two lines, diminished by twice the rectangle of these lines. Let AB and AC be the two straight lines, and BC their difference.
Side 121 - The square constructed upon the sum of two straight lines is equivalent to the sum of the squares constructed upon these two linec, increased by twice the rectangle of these lines.
Side 17 - The sum of the perpendiculars dropped from any point in the base of an isosceles triangle to the legs, is equal to the altitude upon one of the arms.
Side 17 - The sum of the perpendiculars from any point within an equilateral triangle to the three sides is equal to the altitude of the triangle (Fig.
Side 145 - The sides of a triangle are 10 feet, 17 feet, and 21 feet. Find the areas of the parts into which the triangle is divided by the bisector of the angle formed by the first two sides.
Side 53 - Prove that the locus of the vertex of a triangle, having a given base and a given angle at the vertex, is the arc which forms with the base a segment capable of containing the given angle (§ 318).
Side 187 - I cannot see why it is so very important to know that the lines drawn from the extremities of the base of an isosceles triangle to the middle points of the opposite sides are equal! The knowledge doesn't make life any sweeter or happier, does it?
Side 120 - 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'.