Monica has a piece of canvas whose area is 55 | vedclass

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(C) The total area of the canvas is $551 \, m^2$. Since $1 \, m^2$ is lost in wastage,the area available for the tent is $551 \, m^2 - 1 \, m^2 = 550

Vedclass · Accepted Answer

$1232$

Vedclass · Answer

(C) The total area of the canvas is $551 \, m^2$. Since $1 \, m^2$ is lost in wastage,the area available for the tent is $551 \, m^2 - 1 \, m^2 = 550 \, m^2$.The curved surface area of a conical tent is given by $\pi rl$. Since the floor is not covered,we have $\pi rl = 550 \, m^2$.Given the radius $r = 7 \, m$,we substitute the values: $\frac{22}{7} 	imes 7 	imes l = 550$.Solving for $l$,we get $22l = 550$,so $l = \frac{550}{22} = 25 \, m$.Using the relation $l^2 = r^2 + h^2$,we find the height $h$: $h = \sqrt{l^2 - r^2} = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24 \, m$.The volume of the conical tent is $V = \frac{1}{3} \pi r^2 h$.Substituting the values: $V = \frac{1}{3} 	imes \frac{22}{7} 	imes 7^2 	imes 24 = \frac{1}{3} 	imes 22 	imes 7 	imes 24 = 22 	imes 7 	imes 8 = 1232 \, m^3$.

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