Question

# The thickness of a hemispherical bowl is $$0.25cm$$. The inner radius of the bowl is$$5cm$$. Find the outer curved surface area of the bowl. (Take $$\pi =\cfrac { 22 }{ 7 }$$)

Solution

## Let $$r,R$$ and $$w$$ be the inner and outer radii and thickness of the hemispherical bowl respectively.Given that $$r=5cm$$, $$w=0.25cm$$$$\therefore$$ $$R=r+w=5+0.25$$ $$=5.25cm$$Now, outer surface area of the bowl $$=2\pi { R }^{ 2 }\quad =2\times \cfrac { 22 }{ 7 } \times 5.25\times 5.25$$Thus, the outer surface area of the bowl $$=173.25sq.cm$$Mathematics

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