Question

a conical tent is 10m high and the radius of its base is 24m .find the curved surface area of the tent ​

Answer 1

Answer:

Given, h = 10 m , r = 24 m

We knows that

Curved surface area of the cone

$= \pi \: r \: l$

$l \: = \sqrt{r {}^{2} \: + h {}^{2} }$

$l \: = \sqrt{10 {}^{2} + 24 {}^{2} }$

$l = \sqrt{100 + 576}$

$l = \sqrt{676}$

$l = 26cm$

Now,

Curved surface area of the tent

$=\pi \: r \: l$

$= 22 \div 7 \times 24 \times 26$

$= 13728 \: \div \: 7$

$= 1961.15 \: \: approx$

Answer 2

Given:

➨A conical tent is 10 m high, Height (h) = 10m. And, the radius of its base is 24 m, radius (r) = 24m.

Need to find:

➨The CSA(Curved surface area) of the conical tent.

Solution:

➨ Finding slant Height of the conical tent, let slant height be l.

⠀

Therefore,

⠀

$\begin{gathered}:\implies\sf (l)^2 = (h)^2 + (r)^2 \\\\\\:\implies\sf l^2 = 10^2 + 24^2 \\\\\\:\implies\sf l^2 = 100 + 576\\\\\\:\implies\sf l^2 = 676\\\\\\:\implies\sf l = \sqrt{676}\\\\\\:\implies{\underline{\boxed{\frak{\pink{l = 26\;m}}}}}\;\bigstar\end{gathered}$

$\therefore{\underline{\sf{Hence, \; slant\; height\; of \; conical\; tent \; is \; \bf{26\;m }.}}}∴$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀

⠀

$\begin{gathered}\dag\;{\underline{\frak{As\;we\;know\;that,}}}\\ \\\end{gathered}$

$\star\;\boxed{\sf{\purple{CSA_{\:(cone)} = \pi rl}}}⋆$

where,

➥r is the radius of the cone and l is the slant Height of the cone.

⠀

Therefore,

$\begin{gathered}:\implies\sf CSA_{\:(tent)} = \bigg(\dfrac{22}{7} \times 24 \times 26\bigg)\\\\\\:\implies\sf CSA_{\:(tent)} = \bigg(\dfrac{22}{7} \times 624 \bigg)\\\\\\:\implies{\underline{\boxed{\frak{\purple{ CSA_{\:(tent)} = 1961\;m^2}}}}}\;\bigstar\end{gathered}$

$\therefore{\underline{\sf{Hence, \;CSA\; of \; conical\; tent \; is \; \bf{1961\;m^2 }.}}}∴$

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$\begin{gathered}\qquad\qquad\boxed{\bf{\mid{\overline{\underline{\blue{\bigstar\: Formulae\:related\:to\:cone :}}}}}\mid}\\\\\end{gathered}$

⠀

$\sf Area\:of\:base = \bf{\pi r^2}$

$\sf Curved\:surface\:area\:of\:cone = \bf{\pi rl}$

$\sf Total\:surface\:area\:of\:cone = Area\:of\:base + CSA = \pi r^2 + \pi rl = \bf{\pi r(r + l)}$

$\sf Volume\:of\:cone = \bf{\dfrac{1}{3} \pi r^2 h}$