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A frustum of a right circular cone has a diameter of base 20 cm, of top 12 cm and height 3 cm. Find the area of its whole surface and volume.

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Rd sharma class 10 maths exercise-16.3 mensuration.

In this question we have to find the area
of its whole surface and volume given that
a frustum of a right circular cone has a diameter
of base 20 cm, of top 12 cm and height 3 cm.

 

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1 Answer

  1. Given,

    Base diameter of cone (d1) = 20 cm

    So the radius (r1) = 20/2 cm = 10 cm

    Top diameter of Cone (d2) = 12 cm

    So, the radius (r2) = 12/2 cm = 6 cm

    Height of the cone (h) = 3 cm

    Volume of the frustum of a right circular cone = 1/3 π(r22 + r12 + r1 r2 )h

    = π/3(102 + 62 + 10 × 6)3

    = 616 cm3

    Let ‘L’ be the slant height of cone, then we know that

    L = √(r1 – r21)2 + h2

    L = √(10 – 6)2 + 32

    L = √(16 + 9)

    L = 5cm

    So, the slant height of cone (L) = 5 cm

    Thus,

    Total surface area of the frustum = π(r1 + r2) x L + π r12 + π r22

    = π(10 + 6) × 5 + π × 102 + π × 62

    = π(80 + 100 + 36)

    = π(216)

    = 678.85 cm2

     

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