Subjects
×
  • ENSB Solutions
  • Basic Mathematics
  • Algebra
  • Trigonometry
  • Analytic Geometry
  • Plane Geometry
  • Solid Geometry
  • Differential Calculus
  • Integral Calculus
  • Differential Equation
  • Solid Geometry Solutions

    Topics || Problems

    A water tank has a cylindrical shape with a diameter of 2 m and a perpendicular height of 3 m. Since it was already old and leaking, it is to be replaced by another tank of the same capacity but in the form of a frustum of a cone. The diameters of the ends of the frustum are designed to be 1 m and 2 m respectively. What must be its height?
    Solution:

    The volume of the shapes must be equal thus, \(V_{cy} = V_{fcone}\)

    Volume of the cylinder: \(V_{cy} = \pi r^2h\)

    \(V_{cy} = \pi (1)^2 (3)\)

    \(V_{cy} = 3\pi \)

    Volume of the frustum of a cone: \(V_{fc} = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1r_2)\)

    \(V_{fc} = \frac{1}{3} \pi h ((0.5)^2 + (1)^2 +(0.5)(1))\)

    \(V_{fc} = \frac{7}{12}\pi h\)

    \(3\pi = \frac{7}{12}\pi h\)

    \(h = 5.14 ~ m\)