Question

# A cubical block of side $7$ cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

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Solution

## Given,Side of a cubical block which is surmounted by a hemisphere $=7cm$Here, the greatest diameter the hemisphere can have $=\text{Sideofcube}=7cm$.Now, Surface area of the solid so formed $\text{=T.S.Aofcubicalblock+C.S.Aofhemisphere-Areaofbaseofhemisphere}$Step 1: To calculate T.S.A of cubical block T.S.A of cubical block $=6{\text{a}}^{2}\text{(a=side=7cm)}\phantom{\rule{0ex}{0ex}}{\text{=6×(7)}}^{2}\text{}\phantom{\rule{0ex}{0ex}}=6×49\phantom{\rule{0ex}{0ex}}=294{\text{cm}}^{2}\text{}$Step 2 : To calculate C.S.A of hemisphere C.S.A of hemisphere $=2{\mathrm{\pi r}}^{2}\left(\mathrm{here}\mathrm{r}=\frac{7}{2}=3.5\mathrm{cm}\right)\phantom{\rule{0ex}{0ex}}=2×\frac{22}{7}×{\left(3.5\right)}^{2}=2×\frac{22}{7}×12.25\phantom{\rule{0ex}{0ex}}=77{\mathrm{cm}}^{2}$Step 3 : To calculate Area of base of hemisphere Area of base of hemisphere$={\mathrm{\pi r}}^{2}\left(\mathrm{here}\mathrm{r}=\frac{7}{2}=3.5\mathrm{cm}\right)\phantom{\rule{0ex}{0ex}}=\frac{22}{7}×{\left(3.5\right)}^{2}=2×\frac{22}{7}×12.25\phantom{\rule{0ex}{0ex}}=38.5{\mathrm{cm}}^{2}$Step 4: Surface area of the solid so formed $\text{=T.S.Aofcubicalblock+C.S.Aofhemisphere-Areaofbaseofhemisphere}\phantom{\rule{0ex}{0ex}}=294+77-38.5{\text{cm}}^{2}\phantom{\rule{0ex}{0ex}}=332.5{\text{cm}}^{2}$Therefore, the greatest diameter the hemisphere can have is $7\text{cm}$ and the surface area of the solid so formed is $332.5{\text{cm}}^{2}$.

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