Examples (Revised) - Chapter 13 - Surface Areas & Volumes - Ncert Solutions class 9 - Maths

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Chapter 11: Surface Areas & Volumes - NCERT Solutions for Class 9 Maths

Example 1 :

Find the curved surface area of a right circular cone whose slant height is $10 \mathrm{~cm}$ and base radius is $7 \mathrm{~cm}$.
Solution : Curved surface area $=\pi r l$
$
\begin{aligned}
& =\frac{22}{7} \times 7 \times 10 \mathrm{~cm}^2 \\
& =220 \mathrm{~cm}^2
\end{aligned}
$

Example 2 :

The height of a cone is $16 \mathrm{~cm}$ and its base radius is $12 \mathrm{~cm}$. Find the curved surface area and the total surface area of the cone (Use $\pi=3.14$ ).
Solution :

Here, $h=16 \mathrm{~cm}$ and $r=12 \mathrm{~cm}$.
So, from $l^2=h^2+r^2$, we have
$
l=\sqrt{16^2+12^2} \mathrm{~cm}=20 \mathrm{~cm}
$

Example 3 :

A corn cob (see Fig. 11.5), shaped somewhat like a cone, has the radius of its broadest end as $2.1 \mathrm{~cm}$ and length (height) as $20 \mathrm{~cm}$. If each $1 \mathrm{~cm}^2$ of the surface of the cob carries an average of four grains, find how many grains

Fig. 11.5 you would find on the entire cob.
Solution :

Since the grains of corn are found only on the curved surface of the corn cob, we would need to know the curved surface area of the corn cob to find the total number of grains on it. In this question, we are given the height of the cone, so we need to find its slant height.
Here, $\quad l=\sqrt{r^2+h^2}=\sqrt{(2.1)^2+20^2} \mathrm{~cm}$
$
=\sqrt{404.41} \mathrm{~cm}=20.11 \mathrm{~cm}
$

Therefore, the curved surface area of the corn cob $=\pi r l$
$
=\frac{22}{7} \times 2.1 \times 20.11 \mathrm{~cm}^2=132.726 \mathrm{~cm}^2=132.73 \mathrm{~cm}^2 \text { (approx.) }
$

Number of grains of corn on $1 \mathrm{~cm}^2$ of the surface of the corn cob $=4$
Therefore, number of grains on the entire curved surface of the cob
$
=132.73 \times 4=530.92=531 \text { (approx.) }
$

So, there would be approximately 531 grains of corn on the cob.

Example 4 :

Find the surface area of a sphere of radius $7 \mathrm{~cm}$.
Solution :

The surface area of a sphere of radius $7 \mathrm{~cm}$ would be
$
4 \pi r^2=4 \times \frac{22}{7} \times 7 \times 7 \mathrm{~cm}^2=616 \mathrm{~cm}^2
$

Example 5 :

Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius $21 \mathrm{~cm}$.
Solution :

The curved surface area of a hemisphere of radius $21 \mathrm{~cm}$ would be
$
=2 \pi r^2=2 \times \frac{22}{7} \times 21 \times 21 \mathrm{~cm}^2=2772 \mathrm{~cm}^2
$

(ii) the total surface area of the hemisphere would be
$
3 \pi r^2=3 \times \frac{22}{7} \times 21 \times 21 \mathrm{~cm}^2=4158 \mathrm{~cm}^2
$

Example 6 :

The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of $7 \mathrm{~m}$. Find the area available to the motorcyclist for riding.
Solution :

Diameter of the sphere $=7 \mathrm{~m}$. Therefore, radius is $3.5 \mathrm{~m}$. So, the riding space available for the motorcyclist is the surface area of the 'sphere' which is given by
$
\begin{aligned}
4 \pi r^2 & =4 \times \frac{22}{7} \times 3.5 \times 3.5 \mathrm{~m}^2 \\
& =154 \mathrm{~m}^2
\end{aligned}
$

Example 7 :

A hemispherical dome of a building needs to be painted (see Fig. 11.9). If the circumference of the base of the dome is $17.6 \mathrm{~m}$, find the cost of painting it, given the cost of painting is ₹ 5 per $100 \mathrm{~cm}^2$.
Solution :

Since only the rounded surface of the dome is to be painted, we would need to find the curved surface area of the hemisphere to know the extent of painting that needs to be done. Now, circumference of the dome $=17.6 \mathrm{~m}$. Therefore, $17.6=2 \pi r$.
So, the radius of the dome $=17.6 \times \frac{7}{2 \times 22} \mathrm{~m}=2.8 \mathrm{~m}$
The curved surface area of the dome $=2 \pi r^2$

$
\begin{aligned}
& =2 \times \frac{22}{7} \times 2.8 \times 2.8 \mathrm{~m}^2 \\
& =49.28 \mathrm{~m}^2
\end{aligned}
$

Now, cost of painting $100 \mathrm{~cm}^2$ is $₹ 5$.
So, cost of painting $1 \mathrm{~m}^2=₹ 500$
Therefore, cost of painting the whole dome
$
\begin{aligned}
& =₹ 500 \times 49.28 \\
& =₹ 24640
\end{aligned}
$

Example 8:

The height and the slant height of a cone are $21 \mathrm{~cm}$ and $28 \mathrm{~cm}$ respectively. Find the volume of the cone.
Solution :

From $l^2=r^2+h^2$, we have
$
r=\sqrt{l^2-h^2}=\sqrt{28^2-21^2} \mathrm{~cm}=7 \sqrt{7} \mathrm{~cm}
$

So, volume of the cone $=\frac{1}{3} \pi r^2 h=\frac{1}{3} \times \frac{22}{7} \times 7 \sqrt{7} \times 7 \sqrt{7} \times 21 \mathrm{~cm}^3$
$
=7546 \mathrm{~cm}^3
$

Example 9 :

Monica has a piece of canvas whose area is $551 \mathrm{~m}^2$. She uses it to have a conical tent made, with a base radius of $7 \mathrm{~m}$. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately $1 \mathrm{~m}^2$, find the volume of the tent that can be made with it.
Solution :

Since the area of the canvas $=551 \mathrm{~m}^2$ and area of the canvas lost in wastage is $1 \mathrm{~m}^2$, therefore the area of canvas available for making the tent is $(551-1) \mathrm{m}^2=550 \mathrm{~m}^2$.
Now, the surface area of the tent $=550 \mathrm{~m}^2$ and the required base radius of the conical tent $=7 \mathrm{~m}$
Note that a tent has only a curved surface (the floor of a tent is not covered by canvas!!).

Therefore, curved surface area of tent $=550 \mathrm{~m}^2$.
That is,
$
\pi r l=550
$
or,
$
\frac{22}{7} \times 7 \times l=550
$
or,
$
l=3 \frac{550}{22} \mathrm{~m}=25 \mathrm{~m}
$

Now,
$
l^2=r^2+h^2
$

Therefore,
$
\begin{aligned}
h=\sqrt{l^2-r^2} & =\sqrt{25^2-7^2} \mathrm{~m}=\sqrt{625-49} \mathrm{~m}=\sqrt{576} \mathrm{~m} \\
& =24 \mathrm{~m}
\end{aligned}
$

So, the volume of the conical tent $=\frac{1}{3} \pi r^2 h=\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 \mathrm{~m}^3=1232 \mathrm{~m}^3$.

Example 10 :

Find the volume of a sphere of radius $11.2 \mathrm{~cm}$.
Solution :

Required volume $=\frac{4}{3} \pi r^3$
$
=\frac{4}{3} \times \frac{22}{7} \times 11.2 \times 11.2 \times 11.2 \mathrm{~cm}^3=5887.32 \mathrm{~cm}^3
$

Example 11:

A shot-putt is a metallic sphere of radius $4.9 \mathrm{~cm}$. If the density of the metal is $7.8 \mathrm{~g} \mathrm{per} \mathrm{cm}^3$, find the mass of the shot-putt.
Solution :

Since the shot-putt is a solid sphere made of metal and its mass is equal to the product of its volume and density, we need to find the volume of the sphere.
Now, volume of the sphere $=\frac{4}{3} \pi r^3$
$
\begin{aligned}
& =\frac{4}{3} \times \frac{22}{7} \times 4.9 \times 4.9 \times 4.9 \mathrm{~cm}^3 \\
& =493 \mathrm{~cm}^3 \text { (nearly) }
\end{aligned}
$

Further, mass of $1 \mathrm{~cm}^3$ of metal is $7.8 \mathrm{~g}$.
Therefore, mass of the shot-putt $=7.8 \times 493 \mathrm{~g}$
$
=3845.44 \mathrm{~g}=3.85 \mathrm{~kg} \text { (nearly) }
$

Example 12 :

A hemispherical bowl has a radius of $3.5 \mathrm{~cm}$. What would be the volume of water it would contain?
Solution :

The volume of water the bowl can contain
$
\begin{aligned}
& =\frac{2}{3} \pi r^3 \\
& =\frac{2}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5 \mathrm{~cm}^3=89.8 \mathrm{~cm}^3
\end{aligned}
$