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 $165 cm^{2}$.

Solution

Radius $(r)$ of Slant 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

$=[\frac{22}{7} \times 12 \times(12+21)] m^{2}$

$=(\frac{22}{7} \times 12 \times 33) m^{2}$

$=1244.57 m^{2}$

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 $462 cm^{2}$.

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 .

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 $63 m$.

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.

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.

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.