NCERT Solutions for Class 10 Maths Chapter 13 - Exercise

Exercise 13.4

1. A drinking glass is in the shape of a frustum of a cone of height\[\mathbf{14}\text{ }\mathbf{cm}\]. The diameters of its two circular ends are \[\mathbf{4}\text{ }\mathbf{cm}\] and \[\mathbf{2}\text{ }\mathbf{cm}\]. Find the capacity of the glass. $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$

Ans:

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Radius (${{r}_{1}}$) of upper base of glass $=\dfrac{4}{2}=2cm$

Radius (${{r}_{2}}$) of lower base of glass$=\dfrac{2}{2}=1cm$ 

Height = $14cm$

Capacity of glass = Volume of frustum of cone

\[=\dfrac{1}{3}\pi\text{h}\left[ \text{r}_{1}^{2}+\text{r}_{2}^{2}+{{\text{r}}_{1}}{{\text{r}}_{2}} \right]\]

\[=\dfrac{1}{3}\pi \text{h}\left[ {{(2)}^{2}}+{{(1)}^{2}}+(2)(1) \right]\]

\[=\dfrac{1}{3}\times \dfrac{22}{7}\times 14(4+1+2)\]

\[=\dfrac{308}{3}\]

\[=102\dfrac{2}{3}~\text{c}{{\text{m}}^{3}}\]

Therefore, the capacity of the glass \[=102\dfrac{2}{3}~\text{c}{{\text{m}}^{3}}\]

2. The slant height of a frustum of a cone is \[\mathbf{4}\text{ }\mathbf{cm}\] and the perimeters (circumference) of its circular ends are \[\mathbf{18}\text{ }\mathbf{cm}\] and \[\mathbf{6}\text{ }\mathbf{cm}\]. Find the curved surface area of the frustum.

Ans:

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Perimeter of upper circular end of frustum $=18 cm$

\[\Rightarrow 2\pi {{r}_{1}}\text{ }=18\]

\[\Rightarrow {{r}_{1}}=\dfrac{9}{\pi }\]

Perimeter of lower circular end of frustum $=6 cm$

\[\Rightarrow 2\pi {{r}_{2}}\text{ }=6\]

\[\Rightarrow {{r}_{2}}=\dfrac{3}{\pi }\]

Slant height (l) of frustum $=4 cm$

CSA of frustum:

$\Rightarrow \pi \left( {{r}_{1}}+{{r}_{2}} \right)l$

$\Rightarrow \pi \left( \dfrac{9}{\pi }+\dfrac{3}{\pi } \right)4$

$\Rightarrow 12\left( 4 \right)=48c{{m}^{2}}$

Therefore, the curved surface area of the frustum$=48c{{m}^{2}}$

3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the figure given below). If its radius on the open side is\[\mathbf{10}\text{ }\mathbf{cm}\], radius at the upper base is \[\mathbf{4}\text{ }\mathbf{cm}\] and its slant height is\[\mathbf{15}\text{ }\mathbf{cm}\], find the area of material used for making it. $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$

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Ans:

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Radius (${{r}_{2}}$) at upper circular end \[=\text{ }4\text{ }cm\]

Radius (${{r}_{1}}$) at lower circular end \[=10\text{ }cm\]

Slant height (l) of frustum \[=15\text{ }cm\]

Area of material used for making the fez:

$\Rightarrow$ CSA of frustum+Area of upper circular end

$\Rightarrow \pi \left( {{r}_{1}}+{{r}_{2}} \right)l+\pi r_{2}^{2}$ 

$\Rightarrow \pi (10+4)15+\pi (4)^2$

$\Rightarrow \pi (14)15+16\pi $

$\Rightarrow 210\pi +16\pi =\dfrac{226\times 22}{7}$

$\Rightarrow 710\dfrac{2}{7}~\text{c}{{\text{m}}^{2}}$

Therefore, the area of material used for making it is $710\dfrac{2}{7}~\text{c}{{\text{m}}^{2}}$.

4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height \[16\text{ }cm\] with radii of its lower and upper ends as \[\mathbf{8}\text{ }\mathbf{cm}\] and \[\mathbf{20}\text{ }\mathbf{cm}\] respectively. Find the cost of the milk which can completely fill the container, at the rate of \[\mathbf{Rs}.\mathbf{20}\]per liter. Also find the cost of metal sheet used to make the container, if it costs\[\mathbf{Rs}.\mathbf{8}\text{ }\mathbf{per}\text{ }\mathbf{100}\text{ }\mathbf{c}{{\mathbf{m}}^{2}}\]. Take \[\mathbf{\pi }\text{ }=\text{ }\mathbf{3}.\mathbf{14}\]

Ans:

Radius (\[{{r}_{1}}\]) of upper end of container \[=20\text{ }cm\]

Radius (\[{{r}_{2}}\]) of lower end of container \[=8\text{ }cm\]

Height (h) of container \[=16\text{ }cm\]

Slant height (l) of frustum $=\sqrt{{{\left( {{r}_{1}}-{{r}_{2}} \right)}^{2}}+{{h}^{2}}}$

$=\sqrt{{{(20-8)}^{2}}+{{(16)}^{2}}}$

$=\sqrt{{{(12)}^{2}}+{{(16)}^{2}}}=\sqrt{144+256}$

$=20~\text{cm}$

Capacity of container $=$ Volume of frustum 

$=\dfrac{1}{3}\pi \text{h}\left[ \text{r}_{1}^{2}+\text{r}_{2}^{2}+{{\text{r}}_{1}}{{\text{r}}_{2}} \right]$

$=\dfrac{1}{3}\times 3.14\times 16\times \left[ {{(20)}^{2}}+{{(8)}^{2}}+(20)(8) \right]$

$=\dfrac{1}{3}\times 3.14\times 16(400+64+160)$

$=\dfrac{1}{3}\times 3.14\times 16\times 624$

$=10449.92~\text{c}{{\text{m}}^{3}}$

$=10.45$ litres

Cost of 1 litre milk$=\operatorname{Rs}20$

Cost of \[10.45\] litre milk $=\operatorname{Rs} 209$ 

Area of metal sheet used to make the container $=\pi \left( {{\text{r}}_{1}}+{{\text{r}}_{2}} \right)\text{l}+\pi {{\text{r}}_{2}}^{2}$

$ =\pi (20+8)20+\pi {{(8)}^{2}} $ 

$=560\pi +64\pi =624\pi \text{c}{{\text{m}}^{2}}$

Cost of $100~\text{c}{{\text{m}}^{2}}$ metal sheet$=Rs\,8$

Cost of $624\pi \text{c}{{\text{m}}^{2}}$ metal sheet $= \dfrac{624\times 3.14\times 8}{100}=\operatorname{Rs}156.75$

Therefore, the cost of the milk which can completely fill the container is \[Rs\text{ }209\]  and the cost of metal sheet used to make the container is \[Rs156.75\]

5. A metallic right circular cone \[\mathbf{20}\text{ }\mathbf{cm}\] high and whose vertical angle is \[\mathbf{60}{}^\circ \] is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a wire of diameter$\dfrac{1}{16}cm$, find the length of the wire. $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$

Ans:

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In $\vartriangle \text{AEG},$

$\dfrac{\text{EG}}{\text{AG}}=\tan {{30}^{{}^\circ }}$

$\text{EG}=\dfrac{10}{\sqrt{3}}~\text{cm}=\dfrac{10\sqrt{3}}{3}$

In \[\vartriangle \text{ABD},\]

\[\dfrac{\text{BD}}{\text{AD}}=\tan {{30}^{{}^\circ }}\]

\[\text{BD}=\dfrac{20}{\sqrt{3}}=\dfrac{20\sqrt{3}}{3}~\text{cm}\]

Radius $\left( {{r}_{1}} \right)$ of upper end of frustum 

$=\dfrac{10\sqrt{3}}{3}~\text{cm}$

Radius $\left( {{r}_{2}} \right)$ of lower end = $\dfrac{20\sqrt{3}}{3}$cm

 Height $(h)$  $=10~\text{cm}$

Volume of frustum:

$\Rightarrow\dfrac{1}{3}\pi\text{h}\left[\text{r}_{1}^{2}+\text{r}_{2}^{2}+{{\text{r}}_{1}}{{\text{r}}_{2}} \right]$

$\Rightarrow\left.\dfrac{1}{3}\times\pi\times10[{{\left(\dfrac{10\sqrt{3}}{3}\right)}^{2}}+{{\left(\dfrac{20\sqrt{3}}{3}\right)}^{2}}+\dfrac{(10\sqrt{3})(20\sqrt{3})}{3\times 3} \right]$

$\Rightarrow \dfrac{10}{3}\pi \left[ \dfrac{100}{3}+\dfrac{400}{3}+\dfrac{200}{3} \right]$

$\Rightarrow \dfrac{10}{3}\times \dfrac{22}{7}\times \dfrac{700}{3}$

$\Rightarrow \dfrac{22000}{9}~\text{c}{{\text{m}}^{3}}$

Radius \[(r)\] of wire$=\dfrac{1}{16}\times \dfrac{1}{2}=\dfrac{1}{32}~\text{cm}$

Let the length of wire be $l$

Volume of wire $=Area\,\,of\,cross\,section\times Length$

$=\left( \pi {{r}^{2}} \right)\times (l)$

$=\pi \times {{\left( \dfrac{1}{32} \right)}^{2}}\times l$

Volume of frustum = Volume of wire

$\Rightarrow \dfrac{22000}{9}=\dfrac{22}{7}{{\left( \dfrac{1}{32} \right)}^{2}}l$

$\Rightarrow \dfrac{7000}{9}\left( 1024 \right)=l$

\[\Rightarrow l=796444.44~\text{cm}\]

\[\Rightarrow l=7964.44~\text{m}\]

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

Exercise 13.4

Opting for the NCERT solutions for Ex 13.4 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 13.4 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 Other Exercises