NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes || CBSE Class 10 Maths Chapter 13 Surface Areas and Volumes solutions Ex 13.4

Explore the NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes (English Medium), with detailed explanations for Exercise 13.4. This resource is designed to simplify complex Surface Areas and Volumes problems, helping Class 10 Maths students grasp the subject effortlessly. For any queries regarding NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4, feel free to leave a comment, and we’ll respond as soon as possible.

NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes || CBSE Class 10 Maths Chapter 13 Surface Areas and Volumes solutions Ex 13.4
Exercise 13.4

NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes || CBSE Class 10 Maths Chapter 13 Surface Areas and Volumes solutions Ex 13.4

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1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass

Answer

Given:
Radius $ \left(\mathrm{r}_{1}\right) $ [Upper base] $ =\frac{4}{2}=2 \mathrm{~cm} $
Radius $ \left(\mathrm{r}_{2}\right) $ [Lower base] $ =\frac{2}{2}=1 \mathrm{~cm} $
Capacity of glass $ = $ Volume of frustum of cone
$=\frac{1}{3} \pi \mathrm{h}\left(\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}+\mathrm{r}_{1} \mathrm{r}_{2}\right)$
$=\frac{1}{3} \pi \mathrm{h}\left[(2)^{2}+(1)^{2}+(2)(1)\right]$
$=\frac{1}{3} * \frac{22}{7} * 14[4+1+2]$
$=\frac{308}{3}$
$=102 \frac{2}{3} \mathrm{~cm}^{3}$
Hence, the capacity of the glass $ =102 \frac{2}{3} \mathrm{~cm}^{3} $

NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes || CBSE Class 10 Maths Chapter 13 Surface Areas and Volumes solutions Ex 13.4

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

Answer

According to the question,
Perimeter of upper circular end of frustum $ =18 \mathrm{~cm} $
$\left[2 \pi r_{1}=18\right]$
$r_{1}=\frac{9}{\pi}$
Perimeter of lower end of frustum $ =6 \mathrm{~cm} $
$2 \pi \mathrm{r}^{2}=6$
$\mathrm{r}_{2}=\frac{3}{\pi}$
Slant height (1) of frustum $ =4 \mathrm{~cm} $
CSA of frustum $ =\pi\left(r_{1}+r_{2}\right) l $
$ =\pi\left(\frac{9}{\pi}+\frac{3}{\pi}\right) * 4$
$=12 * 4$
$=48 \mathrm{~cm}^{2} $

3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig. 13.24). If its radius on the open side is 10 cm, the radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Answer

Given:
Radius ($ \mathrm{r}_{2} $) at upper circular end $ =4 \mathrm{~cm} $
Radius $ \left(\mathrm{r}_{1}\right) $ at lower circular end $ =10 \mathrm{~cm} $
Slant height (1) of frustum $ =15 \mathrm{~cm} $
Area of material used for making the fez $ =\mathrm{CSA} $ of frustum $+$ Area of upper circular end
$=\pi\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right) l+\pi \mathrm{r}_{2}^{2}$
$=\pi(10+4) \times 15+\pi(4)^{2}$
$=\pi(14) \times 15+16 \pi$
$=210 \pi+16 \pi$
$=\frac{226 * 22}{7}$
$=710 \frac{2}{7} \mathrm{~cm}^{2}$
Hence, the area of material used $ =710 \frac{2}{7} \mathrm{~cm}^{2} $.

NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes || CBSE Class 10 Maths Chapter 13 Surface Areas and Volumes solutions Ex 13.4

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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 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per liter.
Also find the cost of metal sheet used to make the container, if it costs Rs 8 per $ 100 \mathrm{~cm}^{2} $ (Take $ \pi=3.14 $)

Answer

Radius ($ \mathrm{r}_{1} $) of upper end of container $ =20 \mathrm{~cm} $
Radius ($ \mathrm{r}_{2} $) of lower end of container $ =8 \mathrm{~cm} $
Height (h) of container $ =16 \mathrm{~cm} $
Slant height (l) of frustum $ =\sqrt{\left(\left(r_{1}-r_{2}\right)_{2}+h_{2}\right)} $
where $ r_{1} $ and $ r_{2}\left(r_{1} > r_{2}\right) $ are radii of frustum and $ h $ is height of frustum
$=\sqrt{12 \times 12+16\times 16}$
$=\sqrt{144+256}$
$=20 \mathrm{~cm}$
The capacity of container = Volume of a frustum
$=\frac{1}{3} \pi \mathrm{h}\left(\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}+\mathrm{r}_{1} \mathrm{r}_{2}\right)$
$=\frac{1}{3} \times 3.14 \times 16 \times\left[(20)^{2}+(8)^{2}+(20)(8)\right]$
$=\frac{1}{3} \times 3.14 \times 16(400+64+160)$
$=\frac{1}{3} \times 3.14 \times 16 \times 624$
$=10449.92 \mathrm{~cm}^{3}$
$=10.45 \text { liters }$
Cost of 1 liter milk $ = $ Rs 20
Cost of 10.45 liters' milk $ =10.45 \times 20 $
$=\text {Rs } 209$
Area of metal sheet used to make the container $ = $ Curved surface area of frustum $+$ area of lower base
$=\pi\left(r_{1}+r_{2}\right) l+\pi r_{2}^{2}$
$=\pi(20+8) 20+\pi(8)^{2}$
$ =560 \pi+64 \pi $
$ =624 \pi \ \mathrm{cm}^{2} $
Cost of $ 100 \mathrm{~cm}^{2} $ metal sheet $ = $ Rs 8
Cost of $ 624 \pi \mathrm{~cm}^{2} $ metal sheet $ =\frac{624 \times 3.14 \times 8}{100} $
$ =156.75 $ Rs.

NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes || CBSE Class 10 Maths Chapter 13 Surface Areas and Volumes solutions Ex 13.4

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5. A metallic right circular cone 20 cm high and whose vertical angle is $ 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 be drawn into a wire of diameter $ \frac{1}{16} \mathrm{~cm} $ find the length of the wire.

Answer

The figure is given below:

In $ \triangle \mathrm{AEG} $,
$ \tan \theta=\frac{\text { perpendicular }}{\text { base }} $
$\frac{E G}{A G}=\tan 30^{\circ}$
$\Rightarrow E G=\tan 30^{\circ} \times \mathrm{AG}$
$\Rightarrow E G=\left(\frac{1}{\sqrt{3}}\right) \times 10 \mathrm{~cm} \quad\left[\tan 30^{\circ}=\frac{ 1 }{ \sqrt{3} }\right]$
$ \mathrm{EG}=\frac{10}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} $
$ \Rightarrow \mathrm{EG}=\frac{10 \sqrt{3}}{3} $
In $ \triangle \mathrm{ABD} $,
$ \frac{B D}{A D}=\tan 30^{\circ} $
$ \Rightarrow \mathrm{BD}=\tan 30^{\circ} \times \mathrm{AD} $
$ \Rightarrow \mathrm{BD}=\left(\frac{1}{\sqrt{3}}\right) \times 20 \mathrm{~cm} $
$ \Rightarrow \mathrm{BD}=\frac{20}{\sqrt{3}} $
$ \Rightarrow \mathrm{BD}=\frac{20 \sqrt{3}}{3} $
Radius $ \left(\mathrm{r}_{1}\right)=\frac{10 \sqrt{3}}{3} \mathrm{~cm} $
Radius $ \left(\mathrm{r}_{2}\right)=\frac{20 \sqrt{3}}{3} \mathrm{~cm} $
$ \operatorname{Height}(\mathrm{h})=10 \mathrm{~cm} $
Volume of frustum $ =(\frac{ 1 }{ 3 }) \pi h\left(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right) $
$ =\frac{1}{3} \times \pi \times 10\left[\left(\frac{10 \sqrt{3}}{3}\right)^{2}+\left(\frac{20 \sqrt{3}}{3}\right)^{2}+\frac{10 \sqrt{3} \times 20 \sqrt{3}}{3 \times 3}\right] $
$ =\frac{10}{3} \pi\left[\frac{100}{3}+\frac{400}{3}+\frac{200}{3}\right] $
$ =\frac{10}{3} \times \frac{22}{7} \times \frac{700}{3} $
$ =\frac{22000}{9} \mathrm{~cm}^{3} $
Radius (r) of wire $ =\frac{1}{16} \times \frac{1}{2}=\frac{1}{32} \mathrm{~cm} $
Let the length of wire be $l$.
Volume of wire $ = $ Area of cross-section $ \times $ Length
$=\left(\pi r^{2}\right)(l)$
$=\pi \times\left(\frac{1}{32}\right)^{2} * 1$
Volume of frustum = Volume of wire
$\frac{22000}{9}=\frac{22}{7} \times\left(\frac{1}{32}\right)^{2} * 1$
$\frac{7000}{9} \times 1024=1$
$1=796444.44 \mathrm{~cm}$
$=7964.44 \text { meters }$

NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes || CBSE Class 10 Maths Chapter 13 Surface Areas and Volumes solutions Ex 13.4

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NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes || CBSE Class 10 Maths Chapter 13 Surface Areas and Volumes solutions Ex 13.4

Download the Math Ninja App Now