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A frustum of a cone is such that one of its ends is hemispherical with a radius of 21cm and the other top end is circular with a radius of 10.5cm .The perpendicular distance between the centres of the circular parts is 20cm. Find;

  1. The slant length of the original cone. 
  2. The slant length of the frustum. 
  3. The surface area of the frustum. 

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  1. Slant length of original cone
      21  = 20+x
    10.5       x
    21x =10.5(20+x)
    21x =210 + 10.5x
    10.5x = 210 
    10.5      10.5
    x = 20cm
    Triangle of slant length of original cone
    Hy= (40)2 + (21)
    Hy2 = 2041
    Hy = 45.177cm

  2. slant length of frustum
    Hy2 = h2 + b2
    Hy2 = (20)+ (10.5)
    Hy= 510.25
    Hy = 22.588cm
  3. S.A = πRL − πrl
          = (22/7 × 21 × 45.177) − (22/7 × 10.5 × 22.588)
          =2981.682 − 745.404
          =2236.278cm
    πr2 = 22/× 10.5 × 10.5
          = 346.5cm2  
    2πr2 = 2 × 22/7 × 10.5 × 10.5
           =693cm
    T.S.A =3275.778cm2
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