Plane and Solid Geometry: Teacher's Edition |
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9 Ax AABC AB² AC and BD AC² altitude apothem arc BC base BC² bisector bisects chord circumference circumscribed cone of revolution congruent diagonal diameter dihedral distance draw an arc Find the area Find the lateral Find the locus Find the volume frustum given frustum given line given point given sphere Given the line hexagon hypotenuse Iden inscribed lateral area lune whose angle mid-point number of degrees parallel perimeter perpendicular plane MN point equidistant prove that Proof Q. E. F. Proof quadrilateral radius draw radius equal rectangle rectangular parallelepiped regular polygon regular pyramid Required to construct Required to find respectively right circular cylinder segment sides slant height Solution spherical excess spherical polygon spherical surface straight line tangent tetrahedron trihedral V-ABC vertex πα
Popular passages
Page 356 - The line joining the mid-points of two sides of a triangle is parallel to the third side, and equal to half the third side.
Page 58 - 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...
Page 64 - Prove that the middle point of the hypotenuse of a right triangle is equidistant from the three vertices.
Page 69 - On the same base, and on the same side of it, there cannot be two triangles...
Page 192 - The bisector of an angle of a triangle divides the opposite side into segments proportional to the adjacent sides. 144. Theorem. The bisector of an exterior angle of a triangle divides the opposite side produced into segments proportional to the other two sides.
Page 311 - Two triangles are equal if two sides and the included angle of the one are equal respectively to two sides and the included angle of the other (sas = sas). Hyp. In A ABC and A'B'C', AB = A'B', BC = B'C', and Z B = Z B'. To prove A ABC = A A'B' C'. Proof. Apply A ABC to A A'B'C' so that BC shall coincide with B'C'.
Page 51 - An exterior angle of a triangle is equal to the sum of the two opposite interior angles.
Page 326 - The area of a triangle is equal to half the product of its base by its altitude.
Page 236 - The square constructed on the sum of two lines is equivalent to the sum of the two squares constructed on these lines, increased by twice the rectangle of these lines.
Page 237 - 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.