Question

answer the above pic
class 10 ​

Answer 1

Answer:

• $44 \: {m}^{2}$
• ₹22000

Step-by-step explanation:

Given :-

• Height of cylindrical part = 2.1 m
• Diameter of cylindrical part = 4m
• Slant height of conical part = 2.8 m

To find :-

• Canvas used for making tent.
• Cost of canvas of tent at the rate of ₹500

Formula used :-

• $CSA \: of \: cylinder \: = \: 2\pi \: \: rh$
• $CSA \: of \: cone \: = \pi \: rl$

Solution :-

$=> \: radius \: = \frac{diameter}{2} \\ = > \: \frac{4}{2} = 2m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Required surface area =

CSA of cone + CSA of cylinder

$= > \: \pi \: rl \: + 2\pi \: rh$

$= > \: \pi \: \: r(l + 2h)$

$= > \: \frac{22}{7} \times 2 \times (2.8 + 2 \times 2.1)$

$= > \: \frac{44}{7} (2.8 + 4.2)$

$= > \frac{44}{7}\times7 \: \\ = > 44 \: {m}^{2}$

Cost of canvas of tent at the rate of ₹500 per metre square is =

$Therefore, \: 44 \times 500 = ₹22000$

Answer 2

$\LARGE\textbf{\underline{\underline{Given :-}}}$

$\large\texttt{↦ Height of the cylindrical shape = 2.1 m . }\\\\\large\textsf{↦ Diameter of the cylindrical shape = 4 m .}\\\\\large\textsf{↦ Slant height of the cone shape = 2.8 m .}\\\\\large\textsf{↦ Rate of the canvas of the tent = ₹ 500 / m . }$

$\LARGE\textbf{\underline{\underline{To find :-}}}$

$\large\textsf{↦ Total cost of the canvas of the tent = ?}$

$\LARGE\textbf{\underline{\underline{Formula :-}}}$

$\large: \: \Longrightarrow\underline{\boxed{\textsf\textcolor{green}{ C.S.A _ { ( \: Cone \: )} = \pi r l }}}\\\\\large: \: \Longrightarrow\underline{\boxed{\textsf\textcolor{green}{ C.S.A _ { ( \: Cylinder \: )}= 2 \pi r h }}}$

$\LARGE\textbf{\underline{\underline{Method :-}}}$

• First we have to find the C.S.A of the cone and C.S.A of the cylinder .

• Then we have to add them together .

• And then multiply the sum by the cost of per square meter .

• * C.S.A. = Curved Surface Area .

$\LARGE\textbf{\underline{\underline{Solution :-}}}$

$\large\bigstar\textsf\textcolor{orange}{ \: \: \: C.S.A _ { ( \: Cone \: )} = \cfrac{22}{7} × 2 × 2.8 }\\\\\\\large: \: \Longrightarrow\textsf{= \cfrac{22}{7} × 5.6 }\\\\\\\large: \: \Longrightarrow\textsf{= \cancel\cfrac{22}{7} × \cancel{5.6} }\\\\\\\large: \: \Longrightarrow\textsf{= 22 × 0.8}\\\\\\\large: \: \Longrightarrow\underline{\boxed{\textsf\textcolor{purple}{ C.S.A _ { ( \: Cone \: )}= 17.6 m^{2} }}}$

$\large\bigstar\textsf{\textcolor{orange}{ \: \: \: C.S.A _ { ( \: Cylinder\: )} = 2 × \cfrac{22}{7} × 2 × 2.1 }}\\\\\\\large: \: \Longrightarrow\textsf{= \cfrac{22}{7}× 8.4 }\\\\\\\large: \: \Longrightarrow\textsf{= \cancel\cfrac{22}{7}×\cancel{8.4} }\\\\\\\large: \: \Longrightarrow\textsf{= 22 × 1.2 }\\\\\\\large: \: \Longrightarrow\underline{\boxed{\textsf\textcolor{purple}{ C.S.A _ { ( \: Cylinder \: ) }= 26.4 m^{2} }}}\\\\\\\bigstar\large\textsf\textcolor{orange}{ \: \: \: C.S.A _ { ( \: Cone + Cylinder \: )} = 17.6m^{2} + 26.4^{2} }\\\\\\\large: \: \Longrightarrow\underline{\boxed{\textsf\textcolor{purple}{= 44 m²}}}\\\\\\\bigstar\large\textsf\textcolor{orange}{\: \: \: Cost of the canvas = 44 × 500}\\\\\\\large: \: \Longrightarrow\underline{\boxed{\textsf\textcolor{purple}{= ₹ 22,000}}}$