Chapter 12 Surface Areas and Volumes Exercise-01
EXERCISE 12.1
Unless stated otherwise, take $\pi=\dfrac{22}{7}$.
1. 2 cubes each of volume $64 \mathrm{~cm}^{3}$ are joined end to end. Find the surface area of the resulting cuboid.
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Solution2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $14 \mathrm{~cm}$ and the total height of the vessel is $13 \mathrm{~cm}$. Find the inner surface area of the vessel.
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Solution
It can be observed that radius $(r)$ of the cylindrical part and the hemispherical part is the same (i.e., $7 cm$ ).
Height of hemispherical part $=$ Radius $=7 cm$
Height of cylindrical part $(h)=13-7=6 cm$
Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part
$=2 \pi r h+2 \pi r^{2}$
Inner surface area of vessel $=2 \times \dfrac{22}{7} \times 7 \times 6+2 \times \dfrac{22}{7} \times 7 \times 7$
$ \begin{aligned} & =44(6+7)=44 \times 13 \\ & =572 cm^{2} \end{aligned} $
3. A toy is in the form of a cone of radius $3.5 \mathrm{~cm}$ mounted on a hemisphere of same radius. The total height of the toy is $15.5 \mathrm{~cm}$. Find the total surface area of the toy.
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Solution
It can be observed that the radius of the conical part and the hemispherical part is same (i.e., $3.5 cm$ ).
Height of hemispherical part $=$ Radius $(r)=3.5=\dfrac{7}{2} cm$
Height of conical part $(h)=15.5-3.5=12 cm$
Slant height $(l)$ of conical part $=\sqrt{r^{2}+h^{2}}$
$ \begin{aligned} & =\sqrt{(\dfrac{7}{2})^{2}+(12)^{2}}=\sqrt{\dfrac{49}{4}+144}=\sqrt{\dfrac{49+576}{4}} \\ & =\sqrt{\dfrac{625}{4}}=\dfrac{25}{2} \end{aligned} $
Total surface area of toy $=$ CSA of conical part + CSA of hemispherical part
$ \begin{aligned} & =\pi r l+2 \pi r^{2} \\ & =\dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{25}{2}+2 \times \dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{2} \\ & =137.5+77=214.5 cm^{2} \end{aligned} $
4. A cubical block of side $7 \mathrm{~cm}$ 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
From the figure, it can be observed that the greatest diameter possible for such hemisphere is equal to the cube’s edge, i.e., $7 cm$.
Radius $(r.$ ) of hemispherical part $=\dfrac{7}{2}=3.5 cm$
Total surface area of solid = Surface area of cubical part + CSA of hemispherical part
- Area of base of hemispherical part
$=6(\text{ Edge })^{2}+2 \pi r^{2}-\pi r^{2}=6(\text{ Edge })^{2}+\pi r^{2}$
Total surface area of solid $=6(7)^{2}+\dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{2}$
$ =294+38.5=332.5 cm^{2} $
5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter $l$ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
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Solution
Diameter of hemisphere $=$ Edge of cube $=I$
Radius of hemisphere $=\dfrac{l}{2}$
Total surface area of solid = Surface area of cubical part + CSA of hemispherical part
- Area of base of hemispherical part
$=6(\text{ Edge })^{2}+2 \pi r^{2}-\pi r^{2}=6(\text{ Edge })^{2}+\pi r^{2}$
Total surface area of solid $=6 l^{2}+\pi \times(\dfrac{l}{2})^{2}$
$ \begin{aligned} & =6 l^{2}+\dfrac{\pi l^{2}}{4} \\ & =\dfrac{1}{4}(24+\pi) l^{2} \text{ unit }^{2} \end{aligned} $
6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10). The length of the entire capsule is $14 \mathrm{~mm}$ and the diameter of the capsule is $5 \mathrm{~mm}$. Find its surface area.
Fig. 12.10
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Solution
It can be observed that
Radius $(r)$ of cylindrical part $=$ Radius $(r)$ of hemispherical
part
$ =\dfrac{\text{ Diameter of the capsule }}{2}=\dfrac{5}{2} $
Length of cylindrical part $(h)=$ Length of the entire capsule $-2 \times r$
$=14-5=9 cm$
Surface area of capsule $=2 \times$ CSA of hemispherical part + CSA of cylindrical part
$ \begin{aligned} & =2 \times 2 \pi r^{2}+2 \pi r h \\ & =4 \pi(\dfrac{5}{2})^{2}+2 \pi(\dfrac{5}{2})(9) \\ & =25 \pi+45 \pi \\ & =70 \pi mm^{2} \\ & =70 \times \dfrac{22}{7} \\ & =220 mm^{2} \end{aligned} $
7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1 \mathrm{~m}$ and $4 \mathrm{~m}$ respectively, and the slant height of the top is $2.8 \mathrm{~m}$, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per $\mathrm{m}^{2}$. (Note that the base of the tent will not be covered with canvas.)
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Solution
Given that,
Height $(h)$ of the cylindrical part $=2.1 m$
Diameter of the cylindrical part $=4 m$
Radius of the cylindrical part $=2 m$
Slant height $(I)$ of conical part $=2.8 m$
Area of canvas used $=$ CSA of conical part + CSA of cylindrical part $=\pi r l+2 \pi r h$
$=\pi \times 2 \times 2.8+2 \pi \times 2 \times 2.1$
$=2 \pi[2.8+2 \times 2.1]=2 \pi[2.8+4.2]=2 \times \dfrac{22}{7} \times 7$
$=44 m^{2}$
Cost of $1 m^{2}$ canvas $=$ Rs 500
Cost of $44 m^{2}$ canvas $=44 \times 500=22000$
Therefore, it will cost Rs 22000 for making such a tent.
8. From a solid cylinder whose height is $2.4 \mathrm{~cm}$ and diameter $1.4 \mathrm{~cm}$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $\mathrm{cm}^{2}$.
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Solution
Given that,
Height $(h)$ of the conical part $=$ Height $(h)$ of the cylindrical part $=2.4 cm$
Diameter of the cylindrical part $=1.4 cm$
Therefore, radius $(r)$ of the cylindrical part $=0.7 cm$
Slant height $(l)$ of conical part $=\sqrt{r^{2}+h^{2}}$
$ \begin{aligned} & =\sqrt{(0.7)^{2}+(2.4)^{2}}=\sqrt{0.49+5.76} \\ & =\sqrt{6.25}=2.5 \end{aligned} $
Total surface area of the remaining solid will be
$=$ CSA of cylindrical part + CSA of conical part + Area of cylindrical base
$=2 \pi r h+\pi r l+\pi r^{2}$
$=2 \times \dfrac{22}{7} \times 0.7 \times 2.4+\dfrac{22}{7} \times 0.7 \times 2.5+\dfrac{22}{7} \times 0.7 \times 0.7$
$ \begin{aligned} & =4.4 \times 2.4+2.2 \times 2.5+2.2 \times 0.7 \\ & =10.56+5.50+1.54=17.60 cm^{2} \end{aligned} $
The total surface area of the remaining solid to the nearest $cm^{2}$ is $18 cm^{2}$.
9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. If the height of the cylinder is $10 \mathrm{~cm}$, and its base is of radius $3.5 \mathrm{~cm}$, find the total surface area of the article.
Fig. 12.11
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Solution
Radius $(r)$ of cylindrical part $=$ Radius $(r)$ of hemispherical part $=3.5 cm$
Height of cylindrical part $(h)=10 cm$
Surface area of article $=$ CSA of cylindrical part $+2 \times$ CSA of hemispherical part
$=2 \pi r h+2 \times 2 \pi r^{2}$
$=2 \pi \times 3.5 \times 10+2 \times 2 \pi \times 3.5 \times 3.5$
$=70 \pi+49 \pi$
$=119 \pi$
$=17 \times 22=374 cm^{2}$
