What length of tape is required for covering the edges of semicircular disc of radius 10 cm

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Solution

Dear Student,

$L e n g t h o f T a p e r e q u i r e d = p e r i m e t e r o f t h e s e m i c i r c l e = πr + 2 r \Rightarrow Length of tape = (π + 2) r = (\frac{22}{7} + 2) 10 (given r = 10 cm) \Rightarrow Length of tape = (\frac{22 + 14}{7}) 10 = (\frac{36}{7}) 10 = \frac{360}{7} cm = 51.4285 cm$

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