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.

Answer 1

Given,

Side of a cubical block which is surmounted by a hemisphere $=7cm$

Here, the greatest diameter the hemisphere can have $=\text{Side of cube}=7cm$.

Now, Surface area of the solid so formed $\text{=T.S.A of cubical block + C.S.A of hemisphere - Area of base of hemisphere}$

Step 1: To calculate T.S.A of cubical block

T.S.A of cubical block

$$\begin{gathered} =6{\text{a}}^2\text{(a=side=7cm)} \\ {\text{=6× (7)}}^2 \\ =6\times 49 \\ =294{\text{cm}}^2 \end{gathered}$$

Step 2 : To calculate C.S.A of hemisphere

C.S.A of hemisphere

$$\begin{gathered} =2{\mathrm{\pi r}}^2(\mathrm{here}\mathrm{r}=\frac{7}{2}=3.5\mathrm{cm}) \\ =2\times \frac{22}{7}\times {(3.5)}^2=2\times \frac{22}{7}\times 12.25 \\ =77{\mathrm{cm}}^2 \end{gathered}$$

Step 3 : To calculate Area of base of hemisphere

Area of base of hemisphere

$$\begin{gathered} ={\mathrm{\pi r}}^2(\mathrm{here}\mathrm{r}=\frac{7}{2}=3.5\mathrm{cm}) \\ =\frac{22}{7}\times {(3.5)}^2=2\times \frac{22}{7}\times 12.25 \\ =38.5{\mathrm{cm}}^2 \end{gathered}$$

Step 4: Surface area of the solid so formed

$$\begin{gathered} \text{=T.S.A of cubical block + C.S.A of hemisphere - Area of base of hemisphere} \\ =294+77-38.5{\text{cm}}^2 \\ =332.5{\text{cm}}^2 \end{gathered}$$

Therefore, the greatest diameter the hemisphere can have is $7\text{cm}$ and the surface area of the solid so formed is $332.5{\text{cm}}^2$.