The volume of a conical tent is 1232 , m^3 an | vedclass

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(D) Given: Volume of cone $V = 1232 \, m^3$,Base area $A = 154 \, m^2$,Width of canvas $w = 2 \, m$.Step $1$: Find the radius $r$.$A = \pi r^2 = 154$$

Vedclass · Accepted Answer

$275$

Vedclass · Answer

(D) Given: Volume of cone $V = 1232 \, m^3$,Base area $A = 154 \, m^2$,Width of canvas $w = 2 \, m$.Step $1$: Find the radius $r$.$A = \pi r^2 = 154$$\frac{22}{7} 	imes r^2 = 154 \Rightarrow r^2 = \frac{154 	imes 7}{22} = 49 \Rightarrow r = 7 \, m$.Step $2$: Find the height $h$.$V = \frac{1}{3} \pi r^2 h = 1232$$\frac{1}{3} 	imes 154 	imes h = 1232 \Rightarrow h = \frac{1232 	imes 3}{154} = 8 	imes 3 = 24 \, m$.Step $3$: Find the slant height $l$.$l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \, m$.Step $4$: Find the area of the canvas (Curved Surface Area).$CSA = \pi r l = \frac{22}{7} 	imes 7 	imes 25 = 550 \, m^2$.Step $5$: Find the length of the canvas.$Length = \frac{Area}{Width} = \frac{550}{2} = 275 \, m$.

The volume of a conical tent is $1232 \, m^3$ and the area of its base is $154 \, m^2$. Find the length of the canvas required to build the tent,if the canvas is $2 \, m$ in width. (in $m$) (Take $\pi = \frac{22}{7}$)

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