EXERCISE 11.1
$$\text { Assume } \pi=\frac{22}{7} \text {, unless stated otherwise. }$$
1. Diameter of the base of a cone is $10.5 \mathrm{~cm}$ and its slant height is $10 \mathrm{~cm}$. Find its curved surface area.
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Solution
$(\frac{10.5}{2}) cm$
Radius $(r)$ of the base of cone $=\quad=5.25 cm$
Slant height (I) of cone $=10 cm$
CSA of cone $=\pi r l$
$=(\frac{22}{7} \times 5.25 \times 10) cm^{2}=(22 \times 0.75 \times 10) cm^{2}=165 cm^{2}$
Therefore, the curved surface area of the cone is $165 cm^{2}$.
2. Find the total surface area of a cone, if its slant height is $21 \mathrm{~m}$ and diameter of its base is $24 \mathrm{~m}$.
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Solution
Radius $(r)$ ofSlant height $[.$ Assume $.\pi=\frac{22}{7}]$
Total
$=\pi r(r+I)$ the base of cone $=$
(I) of cone $=21 m$
surface $\quad=12 m$ area of cone
$=[\frac{22}{7} \times 12 \times(12+21)] m^{2}$
$=(\frac{22}{7} \times 12 \times 33) m^{2}$
$=1244.57 m^{2}$
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.
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Solution
(i) Slant height (I) of cone $=14 cm$
Let the radius of the circular end of the cone be $r$. We
know, CSA of cone $=$ nrl
(308) $cm^{2}=(\frac{22}{7} \times r \times 14) cm$
$\Rightarrow r=(\frac{308}{44}) cm=7 cm$
Therefore, the radius of the circular end of the cone is $7 cm$.
(ii) Total surface area of cone $=$ CSA of cone + Area of base
$\begin{aligned}& =\pi r l+\pi r^{2} \\& =[308+\frac{22}{7} \times(7)^{2}] cm^{2} \\& =(308+154) cm^{2} \\& =462 cm^{2}\end{aligned}$
Therefore, the total surface area of the cone is $462 cm^{2}$.
4. A conical tent is $10 \mathrm{~m}$ high and the radius of its base is $24 \mathrm{~m}$. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of $1 \mathrm{~m}^{2}$ canvas is ₹ 70 .
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Solution
(i) Let $A B C$ be a conical tent.
Height ( $h$ ) of conical tent $=10 m$
Radius ( $r$ ) of conical tent $=24 m$
Let the slant height of the tent be I.
In $\triangle A B O, A B^{2}=A O^{2}+$
$BO^{2}$
$\begin{aligned}& I^{2}=h^{2}+r^{2} \\& =(10 m)^{2}+(24 m)^{2} \\& =676 m^{2} \\& \therefore I=26 m\end{aligned}$
Therefore, the slant height of the tent is $26 m$.
(ii) CSA of tent $=\pi rl$
$\begin{aligned}& =(\frac{22}{7} \times 24 \times 26) m^{2} \\& =\frac{13728}{7} m^{2}\end{aligned}$
Cost of $1 m^{2}$ canvas $=$ Rs 70
Cost of 7 canvas $=$
$\frac{13728}{7} m^{2} \quad Rs(\frac{13728}{7} \times 70)$
$=$ Rs 137280
Therefore, the cost of the canvas required to make such a tent is Rs 137280 .
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 $20 \mathrm{~cm}$ (Use $\pi=3.14$ ).
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Solution
Height $(h)$ of conical tent $=8 m$
Radius ( $r$ ) of base of tent $=6 m$
Slant height (I) of tent $=\sqrt{r^{2}+h^{2}}$
$=(\sqrt{6^{2}+8^{2}}) m=(\sqrt{100}) m=10 m$
CSA of conical tent $=\pi r \mid$
$=(3.14 \times 6 \times 10) m^{2}$
$=188.4 m^{2}$
Let the length of tarpaulin sheet required be $I$.
As $20 cm$ will be wasted, therefore, the effective length will be $(I-0.2 m)$.
Breadth of tarpaulin $=3 m$
Area of sheet $=$ CSA of tent $[(I$
$-0.2 m) \times 3] m=188.4 m^{2} I-$
$0.2 m=62.8 m I=63 m$
Therefore, the length of the required tarpaulin sheet will be $63 m$.
6. The slant height and base diameter of a conical tomb are $25 \mathrm{~m}$ and $14 \mathrm{~m}$ respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per $100 \mathrm{~m}^{2}$.
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Solution
Slant height (I) of conical tomb $=25 m$
Base radius $(r)$ of tomb $=\frac{14}{2}=7 m$
CSA of conical tomb $=\pi r l$
$=(\frac{22}{7} \times 7 \times 25) m^{2}$
$=550 m^{2}$
Cost of white-washing $100 m^{2}$ area $=$ Rs 210
Cost of white-washing $550 m^{2}$ area $=Rs(\frac{210 \times 550}{100})$
$=Rs 1155$
Therefore, it will cost Rs 1155 while white-washing such a conical tomb.
7. A joker’s cap is in the form of a right circular cone of base radius $7 \mathrm{~cm}$ and height $24 \mathrm{~cm}$. Find the area of the sheet required to make 10 such caps.
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Solution
Radius ( $r$ ) of conical cap $=7 cm$
Height (h) of conical cap $=24 cm$
Slant height (I) of conical cap $=\sqrt{r^{2}+h^{2}}$.
$=[\sqrt{(7)^{2}+(24)^{2}}] cm=(\sqrt{625}) cm=25 cm$
CSA of 1 conical cap $=\pi r l$
$=(\frac{22}{7} \times 7 \times 25) cm^{2}=550 cm^{2}$
CSA of 10 such conical caps $=(10 \times 550) cm^{2}=5500 cm^{2}$ Therefore, $5500 cm^{2}$ sheet will be required.
8. A bus stop is barricaded from the remaining part of the road, by using 50 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 ₹ 12 per $\mathrm{m}^{2}$, what will be the cost of painting all these cones? (Use $\pi=3.14$ and take $\sqrt{1.04}=1.02)$
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Solution
Radius ( $r$ ) of cone $={ }^{\frac{40}{2}=20 cm}=0.2 m$
Height ( $h$ ) of cone $=1 m$
Slant height (I) of cone $=\sqrt{h^{2}+r^{2}}$
$=[\sqrt{(1)^{2}+(0.2)^{2}}] m=(\sqrt{1.04}) m=1.02 m$
CSA of each cone $=\pi r l$
$=(3.14 \times 0.2 \times 1.02) m^{2}=0.64056 m^{2} CSA$
of 50 such cones $=(50 \times 0.64056) m^{2}$
$=32.028 m^{2}$
Cost of painting $1 m^{2}$ area $=$ Rs 12
Cost of painting $32.028 m^{2}$ area $=Rs(32.028 \times 12)$
$=$ Rs 384.336
$=$ Rs 384.34 (approximately)
Therefore, it will cost Rs 384.34 in painting 50 such hollow cones.
