Question

(i) A hemisphere of maximum possible diameter is placed over a cuboidal block of side 7 cm. Find the surface area of the solid so formed.
[CBSE 2017]

(ii) A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of ₹5 per 100 sq cm. [Use $\mathrm{\pi }$ = 3.14] [CBSE 2015]

Answer 1

(i) Disclaimer: It is written cuboid in the question but it should be cube.

From the figure, it can be observed that the maximum diameter possible for such hemisphere is equal to the cube’s edge, i.e., 7 cm.

Radius (r) of hemispherical part = = 3.5 cm

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part

− Area of base of hemispherical part

= 6 (Edge)² − = 6 (Edge)² +

(ii)

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{the}\mathrm{edge}\mathrm{of}\mathrm{the}\mathrm{cubical}\mathrm{block},a=10\mathrm{cm} \\ \mathrm{The}\mathrm{largest}\mathrm{diameter}\mathrm{of}\mathrm{the}\mathrm{hemisphere}=a=10\mathrm{cm} \\ \mathrm{Also},\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{hemisphere},r=\frac{10}{2}=5\mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{Total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{solid}=\mathrm{TSA}\mathrm{of}\mathrm{cube}+\mathrm{CSA}\mathrm{of}\mathrm{hemisphere}-\mathrm{Area}\mathrm{of}\mathrm{circle} \\ =6a^2+2\mathrm{\pi }{r}^2-\mathrm{\pi }{r}^2 \\ =6a^2+\mathrm{\pi }{r}^2 \\ =6\times 10\times 10+3.14\times 5\times 5 \\ =600+78.5 \\ =678.5{\mathrm{cm}}^2 \\ \mathrm{As},\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{painting}\mathrm{the}\mathrm{solid}=₹5\mathrm{per}100{\mathrm{cm}}^2 \\ \mathrm{So},\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{painting}\mathrm{the}\mathrm{solid}=678.5\times \frac{5}{100}\approx ₹33.92 \end{gathered}$$

hence, the cost of painting the total surface area of the solid is ₹33.92.