Q. 144.3( 11 Votes )

# A tent is made in

Answer :

Given: Diameter of base of frustum = 20 m

Diameter of top of frustum = 6 m

∴ Radius of base = R = 20/2 = 10 m

∴ Radius of top = r = 6/2 = 3 m

Height of frustum = h_{f} = 24 m

Height of tent = h_{t} = 28 m

∴ height of cone = h_{c} = h_{t} -h_{f} = 28-24 = 4 m

Formula: Total surface area of frustum = πr^{2} + πR^{2} + π(R + r)l_{f} m^{2}

Total surface area of cone = πrl_{c}

Where l_{f} = slant height of frustum & l_{c} = slant height of cone

∴ l_{f} = 25 m

For slant height of cone consider right angled ΔABC from figure

AB = h_{c} = 4 m ; BC = r = 3 m ; AC = l_{c}

By pythagoras theorm

AB^{2} + BC^{2} = AC^{2}

∴ 4^{2} + 3^{2} = l_{c}^{2}

∴ l_{c} = 5

Since length cannot be negative l_{c} = 5 m

Since for tent we don’t require the upper circle of frustum and the lower circle of frustum hence we should subtract their area as we don’t require canvas for that.

Surface area of upper circle = πr^{2}

Surface area of lower circle = πR^{2}

∴ Surface area of frustum for which canvas is required = πr^{2} + πR^{2} + π(R + r)l_{f} - πr^{2}-πR^{2}cm^{2}

= π(R + r)l_{f}m^{2}

= (22/7) × (10 + 3) × 25 m^{2}

= (22/7) × 325 m^{2}

= 1021.4285 m^{2}

Surface area of cone = πrl_{c} m^{2}

= (22/7) × 3 × 5 m^{2}

= (22/7) × 15 m^{2}

= 47.1428 m^{2}

∴ Quantity of canvas required = surface area of frustum + surface area of cone

= 1021.4285 + 47.1428 m^{2}

= 1068.5713 m^{2}

∴ Quantity of canvas required = 1068.5713 m^{2}

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