A joker’s cap is in the form of a right circular cone of base radius 3 dm and height 4 dm. Find the area of the sheet required to make 35 such caps.

Asked by Topperlearning User | 17th Oct, 2017, 09:16: AM

Expert Answer:

Given: Radius (r) = 3 dm, height (h) = 4 dm. Let slant height = l     

begin mathsize 12px style l squared space equals space straight r squared space plus space straight h squared
l space equals space square root of space straight r squared space plus space straight h squared end root
l space equals space square root of 3 squared space plus space 4 squared end root
l italic space equals space 5 space dm
Lateral space surface space area space equals space πr l
equals 22 over 7 cross times 3 cross times 5
equals fraction numerator 330 over denominator 7 space dm squared end fraction
Aera space of space sheet space required space to space make space 35 space such space caps space equals space 35 cross times 330 over 7 space equals space 1650 space dm squared
end stylebegin mathsize 12px style l squared space equals space straight r squared space plus space straight h squared
l space equals space square root of space straight r squared space plus space straight h squared end root
l space equals space square root of 3 squared space plus space 4 squared end root
l italic space equals space 5 space dm
Lateral space surface space area space equals space πr l
equals 22 over 7 cross times 3 cross times 5
equals fraction numerator 330 over denominator 7 space dm squared end fraction
Aera space of space sheet space required space to space make space 35 space such space caps space equals space 35 cross times 330 over 7 space equals space 1650 space dm squared
end style

  

 

 

Answered by  | 17th Oct, 2017, 11:16: AM