Question

Yhe lower part of a circus tent is cylinderical aand the upper part is conical.the diameter of the base of tent is 56 m and the height of the cylindrical part is 15 m.the totaal height of the tent is 60m .how much canvas is required for the tent?

Answer 1

Canvas required for the tent = Csa of cylinder + Csa of cone.
CSA of cylinder = $\pi$×r²×h
Now, height of cylinder = 15m
Radius= 28m
Applying the above formula, we get 1320m²

CSA of cone = $\pi$×r×l
⇒22/7×28×53= 4664m² ⇒1320+4664=5984m²


⇒

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Radius :-

$\tt{\rightarrow\dfrac{Diameter}{2}}$

$\tt{\rightarrow\dfrac{56}{2}}$

= 28 m

Height :-

= 15 m

${\boxed{\sf\:{CSA\;of\;Cylinder=2\pi rh}}}$

$\tt{\rightarrow 2\times\dfrac{22}{7}\times 28\times 15}$

= 2640 m²

${\boxed{\sf\:{CSA\;of\;Cone=\pi rl}}}$

$\textbf{\underline{Slant\;height\;of\;Cone}}$

= √r² + h²

=√28² + 45²

= √784 + 2025

= √2809

$\textbf{\underline{Slant\;height\;of\;Cone=53\;m}}$

${\boxed{\sf\:{CSA\;of\;Cone=\pi rl}}}$

$\tt{\rightarrow\dfrac{22}{7}\times 28\times 53}$

= 4664 m²

Total area of the cylindrical and conical :-

= 2640 + 4664

= 7304 m²