Question

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. 

Answer 1

1. 
   
     21  = 20+x
   10.5       x
   21x =10.5(20+x)
   21x =210 + 10.5x
   10.5x = 210 
   10.5      10.5
   x = 20cm
   
   Hy² = (40)² + (21)² 
   Hy² = 2041
   Hy = 45.177cm
2. 
   
   Hy² = h² + b²
   Hy² = (20)² + (10.5)² 
   Hy² = 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
         =2236.278cm² 
   πr² = ²²/₇ × 10.5 × 10.5
         = 346.5cm²  
   2πr² = 2 × ²²/₇ × 10.5 × 10.5
          =693cm² 
   T.S.A =3275.778cm²