Theory:

The diameter of the wafer cone is \(10 \ cm\) and, its height is \(12 \ cm\). Calculate the curved surface area of \(20\) such wafer cones.
 
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Solution:
 
Diameter of the cone \((d)\) \(=\) \(10 \ cm\)
 
Radius of the cone \((r)\) \(=\) d2=102=5 \(cm\)
 
Height of the cone \((h)\) \(=\) \(12 \ cm\)
 
Let us first find the slant height of the cone.
 
\(l = \sqrt{r^2 + h^2}\)
 
\(l = \sqrt{5^2 + 12^2}\)
 
\(l = \sqrt{25 + 144}\)
 
\(l = \sqrt{169}\)
 
\(l = 13\) \(cm\)
 
Curved surface area of the cone \(=\) \(\pi r l\) sq. units
 
\(=\) 227×5×13
 
\(=\) 14307
 
\(=\) \(204.28\) \(cm^2\)
 
The curved surface area of a cone is \(204.28 \ cm^2\).
 
Curved surface area of \(20\) cones:
 
\(=\) \(20 \times 204.28\)
 
\(=\) \(4085.6\)
 
Therefore, the curved surface area of \(20\) wafer cones is \(4085.6 \ cm^2\).
 
Important!
The value of \(\pi\) should be taken as \(\frac{22}{7}\) unless its value is shared in the problem.