NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes - Exercise 13.1

Exercise 13.1

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

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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:

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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:

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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:

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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:

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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:

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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:

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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:

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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 $

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

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1

Opting for the NCERT solutions for Ex 13.1 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.1 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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NCERT Solutions for Class 10 Maths Chapter 13 Exercises