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The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of the frustum be 16 cm, find its volume, the slant surface and the total surface.

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If you open math book you will find a lot of mathematical questions but
youshould have solve this question.
It is from class 10 rd sharma chapter 16 mensuration .

In this question we have to find the volume,the slant surface and
total surface from the given informations that the perimeters of
the ends of a frustum of a right circular cone are 44 cm and 33 cm.
If the height of the frustum be 16 cm.

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1 Answer

  1. Given,

    Perimeter of the upper end = 44 cm

    2 π r1 = 44

    2(22/7) r1 = 44

    r1 = 7 cm

    Perimeter of the lower end = 33 cm

    2 π r2 = 33

    2(22/7) r2 = 33

    r2 = 21/4 cm

    Now,

    Let the slant height of the frustum of a right circular cone be L

    L = 16.1 cm

    So, the curved surface area of the frustum cone = π(r1 + r2)l

    = π(7 + 5.25)16.1

    Curved surface area of the frustum cone = 619.65 cm3

    Next,

    The volume of the frustum cone = 1/3 π(r22 + r12 + r1 r2 )h

    = 1/3 π(72 + 5.252 + (7) (5.25) ) x 16

    = 1898.56 cm3

    Thus, volume of the cone = 1898.56 cm3

    Finally, the total surface area of the frustum cone

    = π(r1 + r2) x L + π r12 + π r22

    = π(7 + 5.25) × 16.1 + π72 + π5.252

    = π(7 + 5.25) × 16.1 + π(72 + 5.252) = 860.27 cm2

    Therefore, the total surface area of the frustum cone is 860.27 cm2

     

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