Geometry - solid figures solved problems
      Solid figures, solved problems examples
Example:   An isosceles triangle, of leg b = 2 that is inclined to the base at a = 30°, rotates around its base bisector. Find the volume of obtained solid of revolution.
Solution:  Given  b = 2 and a = 30°.    Vsolid of revolution = ?
Example:   Find the volume of the solid generated by rotating an equilateral triangle ABC of the side a = 2 around the axis which passes through vertex C parallel to base AB
Solution:  Given a = 2.   Vsolid of revolution = ?
Example:   In a regular square pyramid of height 2 and side of the base 1 inscribed is a cube. The side of the cube is? 
Solution:  Given  h = 2 and  a = 1.    x = ?
Example:   Height of a cone is 12 and the volume 324p. Find the central angle of the sector to which is unrolled the side surface of the cone.
Solution:  Given  h = 12 and  V = 324p.    a =  ?
Example:   An isosceles trapezoid with bases lengths, 3n and n, whose base angle is 45°, rotates around greater base. Find the volume of the solid of revolution.
Solution:  Given bases lengths, 3n and n,  and base angle 45°.    Vsolid of revolution = ?
Example:   By rotating the rhomb around its longer diagonal generated is solid whose volume is two times smaller than the volume of the solid generated by rotating the rhomb around its shorter diagonal. The ratio of diagonals d1 : d2, (d1 > d2 ), is? 
Solution:  Given  V1 : V2 = 1 : 2.    d1 : d2 = ?
Example:   In a sphere of radius 2 cm inscribed is a cone of the height 3 cm. The ratio of volumes of the cone and the sphere is?
Solution:  Given  R = 2 cm and h = 3 cm.    Vconic : Vsphere = ?
Example:   The slant height of a cone is 3 cm and is inclined to the base of the cone at the angle 60°. The surface of the sphere inscribed in the cone is?
Solution:  Given s = 3 cm and a = 60°.    S sphere = ?   The cross section through the apex and center of the base of the cone is the equilateral triangle.
Example:   A channel’s cross section has the form of the isosceles trapezoid of bottom side 2 m and upper parallel side 4 m and altitude 1 m. How much water is in the channel if it is fulfilled to the half of its height?
Solution:  Given,  c = 4 m, a = 2 m, h = 1 m,  and  l = 100 m.    V = ?
Geometry and use of trigonometry contents - B
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