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A conical tent with a capacity of 600 m^{3} stands on a circular base of area 160 m^{2} Find in m^{2} the area of the canvas.

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#### Solution

Area of the circular base = 160m^{2}

`pir^2 = 160`

⇒ `r = sqrt((160 xx 7)/22)`

⇒ `r = sqrt(50.909)` = 7.134 m

Therefore , radius = 7.134 m

Capacity or Volume of the tent = 600m^{3}

`1/3pir^2h = 600`

⇒ `1/3 xx 22/7 xx 7.13 xx 7.13 xx h = 600`

⇒ `h = (600 xx 3 xx 7)/(7.13 xx 7.13 xx 22)`

⇒ h = 11.265m

Therefore , vertical height = 11.265 m

We know slant height (l) =

l = `sqrt(r^2 + h^2)`

⇒ `l = sqrt(7.134^2 + 11.265^2)`

⇒ `l = sqrt(177.624) = 13.327`

Therefore , slant height = 13.327 m

The curved surface area = `pirl = 22/7 xx 7.134 xx 13.327` = 298.9m^{2}

Hence , the area of the canvas = 298.9 m^{2}

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