Examples - Chapter 12 Surface Areas And Volumes class 10 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Examples (Revised) : Chapter 12 - Surface Areas And Volumes - Ncert Solutions class 10 - Maths

Example 1 :

Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere (see Fig 12.6). The entire top is $5 \mathrm{~cm}$ in height and the diameter of the top is $3.5 \mathrm{~cm}$. Find the area he has to colour. (Take $\pi=\frac{22}{7}$ )

Solution :

This top is exactly like the object we have discussed in Fig. 12.5. So, we can conveniently use the result we have arrived at there. That is :
$
\text { TSA of the toy = CSA of hemisphere }+ \text { CSA of cone }
$
$
\begin{aligned}
\text { Now, the curved surface area of the hemisphere } & =\frac{1}{2}\left(4 \pi r^2\right)=2 \pi r^2 \\
& =\left(2 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}\right) \mathrm{cm}^2
\end{aligned}
$

Also, the height of the cone $=$ height of the top-height (radius) of the hemispherical part
$
=\left(5-\frac{3.5}{2}\right) \mathrm{cm}=3.25 \mathrm{~cm}
$

So, the slant height of the cone $(l)=\sqrt{r^2+h^2}=\sqrt{\left(\frac{3.5}{2}\right)^2+(3.25)^2} \mathrm{~cm}=3.7 \mathrm{~cm}$ (approx)
Therefore, CSA of cone $=\pi r l=\left(\frac{22}{7} \times \frac{3.5}{2} \times 3.7\right) \mathrm{cm}^2$
This gives the surface area of the top as
$
\begin{aligned}
& =\left(2 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}\right) \mathrm{cm}^2+\left(\frac{22}{7} \times \frac{3.5}{2} \times 3.7\right) \mathrm{cm}^2 \\
& =\frac{22}{7} \times \frac{3.5}{2}(3.5+3.7) \mathrm{cm}^2=\frac{11}{2} \times(3.5+3.7) \mathrm{cm}^2=39.6 \mathrm{~cm}^2 \text { (approx.) }
\end{aligned}
$

You may note that 'total surface area of the top' is not the sum of the total surface areas of the cone and hemisphere.

Example 2 :

he decorative block shown in Fig. 12.7 is made of two solids - a cube and a hemisphere. The base of the block is a cube with edge $5 \mathrm{~cm}$, and the hemisphere fixed on the top has a diameter of $4.2 \mathrm{~cm}$. Find the total surface area of the block. (Take $\pi=\frac{22}{7}$ )

Solution : The total surface area of the cube $=6 \times(\text { edge })^2=6 \times 5 \times 5 \mathrm{~cm}^2=150 \mathrm{~cm}^2$. Note that the part of the cube where the hemisphere is attached is not included in the surface area.
So,
$
\begin{aligned}
\text { the surface area of the block }= & \text { TSA of cube }- \text { base area of hemisphere } \\
& + \text { CSA of hemisphere } \\
= & 150-\pi r^2+2 \pi r^2=\left(150+\pi r^2\right) \mathrm{cm}^2 \\
= & 150 \mathrm{~cm}^2+\left(\frac{22}{7} \times \frac{4.2}{2} \times \frac{4.2}{2}\right) \mathrm{cm}^2 \\
= & (150+13.86) \mathrm{cm}^2=163.86 \mathrm{~cm}^2
\end{aligned}
$

Example 3 :

 wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in Fig. 12.8. The height of the entire rocket is $26 \mathrm{~cm}$, while the height of the conical part is $6 \mathrm{~cm}$. The base of the conical portion has a diameter of $5 \mathrm{~cm}$, while the base diameter of the cylindrical portion is $3 \mathrm{~cm}$. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take $\pi=3.14$ )

Solution :

Denote radius of cone by $r$, slant height of cone by $l$, height of cone by $h$, radius of cylinder by $r^{\prime}$ and height of cylinder by $h^{\prime}$. Then $r=2.5 \mathrm{~cm}, h=6 \mathrm{~cm}, r^{\prime}=1.5 \mathrm{~cm}$, $h^{\prime}=26-6=20 \mathrm{~cm}$ and 

$
l=\sqrt{r^2+h^2}=\sqrt{2.5^2+6^2} \mathrm{~cm}=6.5 \mathrm{~cm}
$

Here, the conical portion has its circular base resting on the base of the cylinder, but the base of the cone is larger than the base of the cylinder. So, a part of the base of the cone (a ring) is to be painted.
$
\begin{aligned}
& \text { So, the area to be painted orange }=\text { CSA of the cone }+ \text { base area of the cone } \\
& \text { - base area of the cylinder } \\
& =\pi r l+\pi r^2-\pi\left(r^{\prime}\right)^2 \\
& =\pi\left[(2.5 \times 6.5)+(2.5)^2-(1.5)^2\right] \mathrm{cm}^2 \\
& =\pi[20.25] \mathrm{cm}^2=3.14 \times 20.25 \mathrm{~cm}^2 \\
& =63.585 \mathrm{~cm}^2 \\
&
\end{aligned}
$

Now, the area to be painted yellow $=\mathrm{CSA}$ of the cylinder
+ area of one base of the cylinder
$
\begin{aligned}
& =2 \pi r^{\prime} h^{\prime}+\pi\left(r^{\prime}\right)^2 \\
& =\pi r^{\prime}\left(2 h^{\prime}+r^{\prime}\right) \\
& =(3.14 \times 1.5)(2 \times 20+1.5) \mathrm{cm}^2 \\
& =4.71 \times 41.5 \mathrm{~cm}^2 \\
& =195.465 \mathrm{~cm}^2
\end{aligned}
$

Example 4:

Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end (see Fig. 12.9). The height of the cylinder is $1.45 \mathrm{~m}$ and its radius is $30 \mathrm{~cm}$. Find the total surface area of the bird-bath. (Take $\pi=\frac{22}{7}$ )

Solution :

Let $h$ be height of the cylinder, and $r$ the common radius of the cylinder and hemisphere. Then, 

$\begin{aligned}
\text { the total surface area of the bird-bath } & =\text { CSA of cylinder }+ \text { CSA of hemisphere } \\
& =2 \pi r h+2 \pi r^2=2 \pi r(h+r) \\
& =2 \times \frac{22}{7} \times 30(145+30) \mathrm{cm}^2 \\
& =33000 \mathrm{~cm}^2=3.3 \mathrm{~m}^2
\end{aligned}$

Example 5 :

Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder (see Fig. 12.12). If the base of the shed is of dimension $7 \mathrm{~m} \times 15 \mathrm{~m}$, and the height of the cuboidal portion is $8 \mathrm{~m}$, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of $300 \mathrm{~m}^3$, and there are 20 workers, each of whom occupy about $0.08 \mathrm{~m}^3$ space on an average. Then, how much air is in the shed? (Take $\pi=\frac{22}{7}$ )

Solution :

The volume of air inside the shed (when there are no people or machinery) is given by the volume of air inside the cuboid and inside the half cylinder, taken together.
Now, the length, breadth and height of the cuboid are $15 \mathrm{~m}, 7 \mathrm{~m}$ and $8 \mathrm{~m}$, respectively. Also, the diameter of the half cylinder is $7 \mathrm{~m}$ and its height is $15 \mathrm{~m}$.

So, the required volume $=$ volume of the cuboid $+\frac{1}{2}$ volume of the cylinder
$
=\left[15 \times 7 \times 8+\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 15\right] \mathrm{m}^3=1128.75 \mathrm{~m}^3
$

Next, the total space occupied by the machinery $=300 \mathrm{~m}^3$
And the total space occupied by the workers $=20 \times 0.08 \mathrm{~m}^3=1.6 \mathrm{~m}^3$
Therefore, the volume of the air, when there are machinery and workers
$
=1128.75-(300.00+1.60)=827.15 \mathrm{~m}^3
$

Example 6:

A juice seller was serving his customers using glasses as shown in Fig. 12.13 The inner diameter of the cylindrical glass was $5 \mathrm{~cm}$, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was $10 \mathrm{~cm}$, find the apparent capacity of the glass and its actual capacity. (Use $\pi=3.14$.)

Solution :

Since the inner diameter of the glass $=5 \mathrm{~cm}$ and height $=10 \mathrm{~cm}$,
$
\begin{aligned}
\text { the apparent capacity of the glass } & =\pi r^2 h \\
& =3.14 \times 2.5 \times 2.5 \times 10 \mathrm{~cm}^3=196.25 \mathrm{~cm}^3
\end{aligned}
$

But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass.
$
\text { i.e., } \quad \text { it is less by } \frac{2}{3} \pi r^3=\frac{2}{3} \times 3.14 \times 2.5 \times 2.5 \times 2.5 \mathrm{~cm}^3=32.71 \mathrm{~cm}^3
$

So, the actual capacity of the glass $=$ apparent capacity of glass - volume of the hemisphere
$
\begin{aligned}
& =(196.25-32.71) \mathrm{cm}^3 \\
& =163.54 \mathrm{~cm}^3
\end{aligned}
$

Example 7:

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is $2 \mathrm{~cm}$ and the diameter of the base is $4 \mathrm{~cm}$. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take $\pi=3.14$ )

Solution :

Let BPC be the hemisphere and $A B C$ be the cone standing on the base of the hemisphere (see Fig. 12.14). The radius BO of the hemisphere (as well as of the cone $)=\frac{1}{2} \times 4 \mathrm{~cm}=2 \mathrm{~cm}$.
So,
$
\begin{aligned}
\text { volume of the toy } & =\frac{2}{3} \pi r^3+\frac{1}{3} \pi r^2 h \\
& =\left[\frac{2}{3} \times 3.14 \times(2)^3+\frac{1}{3} \times 3.14 \times(2)^2 \times 2\right] \mathrm{cm}^3=25.12 \mathrm{~cm}^3
\end{aligned}
$

Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of the base of the right circular cylinder $=\mathrm{HP}=\mathrm{BO}=2 \mathrm{~cm}$, and its height is
$
\mathrm{EH}=\mathrm{AO}+\mathrm{OP}=(2+2) \mathrm{cm}=4 \mathrm{~cm}
$

So, the volume required $=$ volume of the right circular cylinder - volume of the toy
$
\begin{aligned}
& =\left(3.14 \times 2^2 \times 4-25.12\right) \mathrm{cm}^3 \\
& =25.12 \mathrm{~cm}^3
\end{aligned}
$

Hence, the required difference of the two volumes $=25.12 \mathrm{~cm}^3$.