NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1- Surface Areas and Volumes

1. Two cubes each of volume $64\,\,c{{m}^{3}}$ are joined end to end. Find the surface area of the resulting cuboids.

Ans: Given: Volume of cubes \[=\text{ }64\text{ }c{{m}^{3}}\]

\[{{\left( Edge \right)}^{3}}=\text{ }64\]

Edge \[=\text{ }4\text{ }cm\]

If cubes are joined end to end, the dimensions of the resulting cuboid will be \[4\text{ }cm,4cm,8\text{ }cm\].

Surface area of cuboids:

$\Rightarrow 2\left( lb+bh+lh \right)$

$\Rightarrow 2\left( \left( 4 \right)\left( 4 \right)+\left( 4 \right)\left( 8 \right)+\left( 4 \right)\left( 8 \right) \right)$

$\Rightarrow 2\left( 16+32+32 \right)$

$\Rightarrow 2\left( 16+64 \right)$

$\Rightarrow 2\left( 80 \right)=160\,c{{m}^{2}}$

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $14$cm and the total height of the vessel is $13$cm. Find the inner surface area of the vessel. $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$.

Ans:

It can be observed that the radius \[\left( r \right)\]of the cylindrical part and the hemispherical part is the same (i.e.,\[7\text{ }cm\]).

Height of hemispherical part \[=Radius=7cm\]

Height of cylindrical part \[\left( h \right)=13-7=6\text{ }cm\]

Inner surface area of the vessel : \[\Rightarrow CSA\text{ }of\text{ }cylindrical\text{ }part+CSA\text{ }of\text{ }hemispherical\text{ }part\]

$\Rightarrow 2\pi rh+2\pi {{r}^{2}}$

$\Rightarrow \left[ 2\left( \dfrac{22}{7} \right)\left( 7 \right)\left( 6 \right) \right]+\left[ 2\left( \dfrac{22}{7} \right)\left( 7 \right)\left( 7 \right) \right]$

$\Rightarrow 44\left( 6+7 \right)$

$\Rightarrow 44\times 13=572\,c{{m}^{2}}$

3. A toy is in the form of a cone of radius $3.5cm$ mounted on a hemisphere of same radius. The total height of the toy is $15.5\,cm$. Find the total surface area of the toy. $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$

Ans:

It can be observed that the radius of the conical part and the hemispherical part is the same (i.e., 3.5 cm). 

Height of hemispherical part \[=Radius\left( r \right)=3.5=\dfrac{7}{2}\text{ }cm\]

Height of conical part \[\left( h \right)=5.5-3.5=12\text{ }cm\]

Slant height (l) of conical part:

$\Rightarrow \sqrt{{{r}^{2}}+{{h}^{2}}}$

$\Rightarrow \sqrt{{{\left( \dfrac{7}{2} \right)}^{2}}+{{\left( 12 \right)}^{2}}}=\sqrt{\dfrac{49}{4}+144}$

$\Rightarrow \sqrt{\dfrac{625}{4}}=\dfrac{25}{2}$

Total surface area of toy:

$\Rightarrow CSA\text{ }of\text{ }conical\text{ }part+CSA\text{ }of\text{ }hemispherical\text{ }part$

$\Rightarrow \pi rl+2\pi {{r}^{2}}$

$\Rightarrow \left[ \left( \frac{22}{7} \right)\left( \frac{7}{2} \right)\left( \frac{25}{2} \right) \right]+\left[ 2{{\left( \frac{7}{2} \right)}^{2}}\frac{22}{7} \right]$

$\Rightarrow 137.5+77=214.5\,c{{m}^{2}}$

4. 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. $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$

Ans:

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

Radius (r) of hemispherical part\[=\dfrac{7}{2}=3.5cm\]

Total surface area of solid:

Surface area of cubical part + CSA of hemispherical = Area of base of hemispherical part

$\Rightarrow 6{{\left( edge \right)}^{2}}+2\pi {{r}^{2}}-\pi {{r}^{2}}$

$\Rightarrow 6{{\left( edge \right)}^{2}}+\pi {{r}^{2}}$

\[\Rightarrow 6{{\left( 7 \right)}^{2}}+\left( \dfrac{22}{7} \right){{\left( \dfrac{7}{2} \right)}^{2}}\]

$\Rightarrow 294+38.5=332.5\,c{{m}^{2}}$

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.

Ans:

Diameter of hemisphere \[=Edge\text{ }of\text{ }cube=L\] 

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

$\Rightarrow 6{{\left( edge \right)}^{2}}+2\pi {{r}^{2}}-\pi {{r}^{2}}$

$\Rightarrow 6{{\left( edge \right)}^{2}}+\pi {{r}^{2}}$

$\Rightarrow 6\left( {{L}^{2}} \right)+\frac{\pi {{L}^{2}}}{4}$

$\Rightarrow 6{{L}^{2}}+\dfrac{\pi {{L}^{2}}}{4}$

$=\dfrac{1}{4}\left( 24+\pi  \right){{L}^{2}}$

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its end (see the given figure). The length of the entire capsule is $14$ mm and the diameter of the capsule is 5 mm. Find its surface area [use $\pi =\frac{22}{7}$]

Ans: It can be observed that,

Radius r of cylindrical part = Radius( r) of hemispherical part

$=\dfrac{Diameter\,of\,the\,capsule}{2}$

$=\dfrac{5}{2}$

Length of  cylindrical part ( h )=Length of the entire capsule$-2\times r$ 

$=14-5$

$=9\,mm$

Surface area of capsule:

2CSA of hemispherical part+CSA of cylindrical part

$\Rightarrow 2\left( 2\pi {{r}^{2}} \right)+2\pi rh$

$\Rightarrow 4\pi {{\left( \dfrac{5}{2} \right)}^{2}}+2\pi \left( \dfrac{5}{2} \right)\left( 9 \right)$

$\Rightarrow 25\pi +45\pi $

$\Rightarrow 70\left( \dfrac{22}{7} \right)$

$\Rightarrow 220\,\,m{{m}^{2}}$

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 \[\mathbf{2}.\mathbf{1}\text{ }\mathbf{m}\]and \[\mathbf{4}\text{ }\mathbf{m}\] respectively, and the slant height of the top is\[\mathbf{2}.\mathbf{8}\text{ }\mathbf{m}\], 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 \[\mathbf{Rs}\text{ }\mathbf{500}\text{ }\mathbf{per}\text{ }{{\mathbf{m}}^{2}}\]. (Note that the base of the tent will not be covered with canvas.) $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$

Ans:

Given that,

Height (h) of the cylindrical part \[=2.1m\]

Diameter of the cylindrical part \[=4m\]

Radius of the cylindrical part \[=2m\]

Slant height (l) of conical part \[=2.8m\]

Area of canvas used:

CSA of conical part+CSA of cylindrical part

$\Rightarrow \pi rl+2\pi rh$

$\Rightarrow \pi \left( 2 \right)\left( 2.8 \right)+2\pi \left( 2 \right)\left( 2.1 \right)$

$\Rightarrow 2\pi \left[ 2.8+2\left( 2.1 \right) \right]$ 

$\Rightarrow 2\pi \left[ 2.8+4.2 \right]=2\left( \dfrac{22}{7} \right)\left( 7 \right)$

$\Rightarrow 44{{m}^{2}}$

Cost of \[1\text{ }{{m}^{2}}\]canvas = \[Rs\text{ }500\]

Cost of \[44\text{ }{{m}^{2}}\]canvas \[=44\times 500=22000\]

Therefore, it will cost \[Rs\text{ }22000\]for making such a tent.

8. From a solid cylinder whose height is \[\mathbf{2}.\mathbf{4}\text{ }\mathbf{cm}\] and diameter \[\mathbf{1}.\mathbf{4}\text{ }\mathbf{cm}\], 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 \[\mathbf{c}{{\mathbf{m}}^{2}}\]. [ Use $\pi =\frac{22}{7}$ ]

Ans:

Given:

Height (h) of the conical part=Height (h) of the cylindrical part\[=2.4\text{ }cm\]

Diameter of the cylindrical part = 1.4 cm

Therefore, radius (r) of the cylindrical part \[=0.7cm\]

Slant height (l) of conical part:

$\Rightarrow \sqrt{{{r}^{2}}+{{h}^{2}}}$

$\Rightarrow \sqrt{{{\left( 0.7 \right)}^{2}}+{{\left( 2.4 \right)}^{2}}}$

$\Rightarrow \sqrt{0.49+5.76}$

$\Rightarrow \sqrt{0.49+5.76}$

$\Rightarrow \sqrt{6.25}=2.5$

Total surface area of the remaining solid will be:

CSA of cylindrical part+CSA of conical part +Area of cylindrical base $

$\Rightarrow 2\pi rh+\pi rl+\pi {{r}^{2}}$

$\Rightarrow \left( 2\times \dfrac{22}{7}\times 0.7\times 2.4 \right)+\left( \dfrac{22}{7}\times 0.7\times 2.5 \right)+\left( \dfrac{22}{7}\times 0.7\times 0.7 \right)$

$\Rightarrow \left( 4.4\times 2.4 \right)+\left( 2.2\times 2.5 \right)+\left( 2.2\times 0.7 \right)$

$\Rightarrow 10.56+5.50+1.54=17.60~\text{c}{{\text{m}}^{2}}$

The total surface area of the remaining solid to the nearest $\text{c}{{\text{m}}^{2}}$ is $18~\text{c}{{\text{m}}^{2}}$

9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the given figure. If the height of the cylinder is \[\mathbf{10}\text{ }\mathbf{cm}\], and its base is of radius \[\mathbf{3}.\mathbf{5}\text{ }\mathbf{cm}\], find the total surface area of the article. $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$

Ans:

Given: Radius (r) of cylindrical part =Radius( r) of hemispherical part=3.5 cm

Height of cylindrical part (h) \[=10\text{ }cm\]

Surface area of article:

CSA of cylindrical part+2  CSA of hemispherical part

$\Rightarrow 2\pi rh+2\left( 2\pi {{r}^{2}} \right)$

$\Rightarrow 2\pi \left( 3.5 \right)\left( 10 \right)+2\left( 2\pi  \right){{\left( 3.5 \right)}^{2}}$

$\Rightarrow 70\pi +49\pi $

$\Rightarrow 119\pi $

$\Rightarrow 17\left( 22 \right)=374\,c{{m}^{2}}$

Conclusion

To sum up, it is imperative that you understand NCERT Solutions for Class 10 Maths Ex 12.1 Chapter 12 - Surface Areas and Volumes. Concentrate on comprehending the ideas behind surface area, volume, and cuboids. Get proficient at precisely applying formulas to solve issues quickly. Vedantu's all-inclusive solutions support complete preparation by offering precise explanations and step-by-step instructions. Class 10 Ex 12.1 will build a strong foundation for further mathematical concepts.

Class 10 Maths Chapter 12: Exercises Breakdown

CBSE Class 10 Maths Chapter 12 Other Study Materials

Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.