Question

6. A circus tent is cylindrical to a height of 4m and conical above it . If its diameter is 105 m and its slant height is 40m, the total area of the canvas required in m2 is
a) 1760 b) 2640 c) 3960 d) 7920​

Answer 1

AnsWer :

⠀

$</p><p>\large{\purple{\textsf{TSA of the canvas required is \textbf{7920}} \: \sf m^2}.}$

Step-by-step explanation:

$\bullet\:\:\textsf{Radius of the cylindrical part of tent = \textbf{105/2 m}}$

$\bullet\:\:\textsf{Slant height (l) = \textbf{40 m}}$

________________________

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$\bigstar\:\tt TSA \: of \: tent = Lateral \: suface \: of \: cone + Lateral \: suface \: cylinder\:\bigstar$

$\tiny:\implies\sf TSA \: of \: tent = 2 \pi r h + \pi r l$

$\tiny:\implies\sf TSA \: of \: tent = \bigg \lgroup \sf{ 2 \times \dfrac{22}{7} \times \dfrac{105}{2} \times 4} \bigg\rgroup + \bigg \lgroup \dfrac{22}{7} \times \dfrac{105}{2} \times 40\bigg \rgroup$

$\tiny:\implies\sf TSA \: of \: tent = 1320 \: + \: 6600$

$\tiny:\implies \underline {\boxed{\sf TSA \: of \: tent = 7920 \: m^{2}}}$

⠀

$</p><p>\therefore\:\underline{\textsf{TSA of the canvas required is \textbf{7920}} \: \sf m^2}.$

Answer 2

Answer:

The total area of the canvas required is d) 7920.

Step-by-step explanation:

Given :-

• A circus tent is cylindrical to a height of 4 m and conical above it.
• Its diameter is 105 m and its slant height is 40 m.

To find :-

• The total area of the canvas required.

Solution :-

Formula used :

★${\boxed{\sf{CSA\:of\: cylinder=2\pi\:rh}}}$

★ ${\boxed{\sf{CSA\:of\:cone=\pi\:rl}}}$

In case of cylinder ,

• Diameter = 105 m
• Height (h)= 4 m

Then,

• Radius(r) of the cylinder = 105/2 = 52.5 m.

CSA of the cylinder ,

= 2πrh

= [2× (22/7)× 52.5×4] m²

= 9240/7 m²

= 1320 m²

In case of conical part,

Radius of cylinder = Radius of cone

• Radius (r)= 52.5 m
• Slant height (l)= 40 m

CSA of conical part,

= πrl

= [(22/7) × 52.5 × 40 ] m²

= 46200/7 m²

= 6600 m²

Therefore,

Total area of the canvas required,

= CSA of cylinder + CSA of cone

= (1320 + 6600) m²

= 7920 m²

Therefore, the total area of the canvas required is 7920 m².