Exercise 11.1 (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

Assume $\quad \pi=\frac{22}{7}$ unless ^stated otherwise.
Ex 11.1 Question 1.

Diameter of the base of a cone is $10.5 \mathrm{~cm}$ and its slant height is $\mathbf{1 0} \mathbf{~ c m}$. Find its curved surface area and its total surface area.

Answer.

Diameter $=10.5 \mathrm{~cm}$

$\Rightarrow$ Radius $(r)=\frac{10.5}{2}=\frac{21}{4} \mathrm{~cm}$

Slant height of cone ${ }^{(l)}=10 \mathrm{~cm}$

Curved surface area of cone $=\pi r l$
$=\frac{22}{7} \times \frac{21}{4} \times 10=165 \mathrm{~cm}^2$

Total surface area of cone $=\pi r(l+r)$
$=\frac{22}{7} \times \frac{21}{4}\left(10+\frac{21}{4}\right)$
$=\frac{22}{7} \times \frac{21}{4} \times \frac{61}{4}=251.625 \mathrm{~cm}^2$
Ex 11.1 Question 2.

Find the total surface area of a cone, if its slant height is $\mathbf{2 1} \mathrm{cm}$ and diameter of the base is $24 \mathrm{~cm}$.

Answer.

Slant height of cone ${ }^{(l)}=21 \mathrm{~m}$

Diameter of cone $=24 \mathrm{~m}$

$\Rightarrow$ Radius of cone $(r)=\frac{24}{2}=12 \mathrm{~m}$

Total surface area of cone $=\pi r(l+r)$
$
\begin{aligned}
& =\frac{22}{7} \times 12(21+12) \\
& =\frac{264}{7} \times 33=1244.57 \mathrm{~m}^2
\end{aligned}
$
Ex 11.1 Question 3.

Curved surface area of a cone is $308 \mathrm{~cm}^2$ and its slant height is $14 \mathrm{~cm}$. Find (i) radius of the base and (ii) total surface area of the cone.

Answer.

(i) Slant height of cone ${ }^{(l)}=14 \mathrm{~cm}$

Curved surface area of cone $=308 \mathrm{~cm}^2$
$
\begin{aligned}
& \Rightarrow \pi r l=308 \\
& \Rightarrow \frac{22}{7} \times r \times 14=308
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow r=\frac{308 \times 7}{14 \times 22} \\
& \Rightarrow r=7 \mathrm{~cm}
\end{aligned}
$
(ii) Total surface area of the cone
$=$ Curved surface area + Area of circular base
$
\begin{aligned}
& =308+\pi r^2 \\
& =308+\frac{22}{7} \times(7)^2 \\
& =462 \mathrm{~cm}^2
\end{aligned}
$

Ex 11.1 Question 4.

A conical tent is $10 \mathrm{~m}$ high and the radius of its base is $\mathbf{2 4} \mathrm{m}$. Find:
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of a $\mathbf{m}^{\mathbf{2}}$ canvas is Rs. 70 .

Answer.

Height of the conical tent ${ }^{(h)}=10 \mathrm{~m}$

Radius of the conical tent $(r)=24 \mathrm{~m}$
(i) Slant height of the tent $(l)=\sqrt{r^2+h^2}$
$
\begin{aligned}
& =\sqrt{(24)^2+(10)^2} \\
& =\sqrt{576+100}=\sqrt{676}=26 \mathrm{~m}
\end{aligned}
$
(ii) Canvas required to make the tent
$=$ Curved surface area of the tent $=\pi r l$
$
=\frac{22}{7} \times 24 \times 26=\frac{13728}{7} \mathrm{~m}^2
$
$\because$ Cost of $1 \mathrm{~m}^2$ canvas $=$ Rs. 70

$\begin{aligned}
& \therefore \text { Cost of } \frac{13728}{7} \mathrm{~m}^2 \text { canvas } \\
& =70 \times \frac{13728}{7}=\text { Rs. } 137280
\end{aligned}$

Ex 11.1 Question 5.

What length of tarpaulin $3 \mathrm{~m}$ wide will be required to make conical tent of height $8 \mathrm{~m}$ and base radius $6 \mathrm{~m}$ ? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately $\mathbf{2 0} \mathbf{c m}$. (Use $\pi=3.14$ )

Answer.

Height of the conical tent ${ }^{(h)}=8 \mathrm{~m}$ and Radius of the conical tent ${ }^{(r)}=6 \mathrm{~m}$

Slant height of the tent $(l)=\sqrt{r^2+h^2}$
$
=\sqrt{(6)^2+(8)^2}=\sqrt{36+64}=\sqrt{100}=10 \mathrm{~m}
$

Area of tarpaulin $=$ Curved surface area of tent $=\pi r l=3.14 \times 6 \times 10=188.4 \mathrm{~m}^2$

Width of tarpaulin $=3 \mathrm{~m}$

Let Length of tarpaulin $=\mathrm{L}$
$\therefore$ Area of tarpaulin $=$ Length $\times$ Breadth
$
=L \times 3=3 \mathrm{~L}
$

Now According to question,
$
\begin{aligned}
& 3 L=188.4 \\
& \Rightarrow L=\frac{1884.4}{3}=62.8 \mathrm{~m}
\end{aligned}
$

The extra length of the material required for stitching margins and cutting is $20 \mathrm{~cm}=0.2 \mathrm{~m}$.

So the total length of tarpaulin bought is $(62.8+0.2) \mathrm{m}=63 \mathrm{~m}$

Ex 11.1 Question 6.

The slant height and base diameter of a conical tomb are $\mathbf{2 5} \mathrm{m}$ and $14 \mathrm{~m}$ respectively. Find the cost of whitewashing its curved surface at the rate of Rs. 210 per $100 \mathrm{~m}^2$.

Answer.

Slant height of conical tomb $(l)$
$=25 \mathrm{~m}$, Diameter of tomb $=14 \mathrm{~m}$
$\therefore$ Radius of the tomb $(r)=\frac{14}{2}=7 \mathrm{~m}$

Curved surface are of tomb $=\pi r l$
$
=\frac{22}{7} \times 7 \times 25=550 \mathrm{~m}^2
$
$\because$ Cost of white washing $100 \mathrm{~m}^2$
$=$ Rs. 210
$\therefore$ Cost of white washing $1 \mathrm{~m}^2=\frac{210}{100}$
$\therefore$ Cost of white washing $550 \mathrm{~m}^2$
$
=\frac{210}{100} \times 550=\text { Rs. } 1155
$

Slant height of the cone $(l)=\sqrt{r^2+h^2}$
$
\begin{aligned}
& =\sqrt{(7)^2+(24)^2} \\
& =\sqrt{49+576}=\sqrt{625}=25 \mathrm{~cm}
\end{aligned}
$

Area of sheet required to make a cap
$
\begin{aligned}
& =\text { CSA of cone }=\pi r l \\
& =\frac{22}{7} \times 7 \times 25=550 \mathrm{~cm}^2
\end{aligned}
$
$\therefore$ Area of sheet required to make 10 caps $=10 \times 550=5500 \mathrm{~cm}^2$

Ex 11.1 Question 8.

A bus stop is barricaded from the remaining part of the road, by using $\mathbf{5 0}$ hollow cones made of recycled cardboard. Each cone has a base diameter of $40 \mathrm{~cm}$ and height $1 \mathrm{~m}$. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per $\mathrm{m}^2$, what will be the cost of painting all these cones? (Use $\pi=3.14$ and take $\sqrt{1.04}=1.02$ )

Answer.

Diameter of cone $=40 \mathrm{~cm}$
$\Rightarrow$ Radius of cone $(r)=\frac{40}{2}=20 \mathrm{~cm}$
$
=\frac{20}{100} \mathrm{~m}=0.2 \mathrm{~m}
$

Height of cone ${ }^{(h)}=1 \mathrm{~m}$

Slant height of cone $(l)=\sqrt{r^2+h^2}$
$
=\sqrt{(0.2)^2+(1)^2}=\sqrt{1.04} \mathrm{~m}
$

Curved surface area of cone $=\pi r l$
$
\begin{aligned}
& =3.14 \times 0.2 \times \sqrt{1.04} \\
& =0.64056 \mathrm{~m}^2
\end{aligned}
$
$\because$ Cost of painting $1 \mathrm{~m}^2$ of a cone
$
=\text { Rs. } 12
$
$\therefore$ Cost of painting $0.64056 \mathrm{~m}^2$ of a cone
$
=12 \times 0.64056=\text { Rs. } 7.68672
$
$\therefore$ Cost of painting of 50 such cones
$
=50 \times 7.68672=\text { Rs. } 384.34 \text { (approx.) }
$