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 = hf = 24 m


Height of tent = ht = 28 m


height of cone = hc = ht -hf = 28-24 = 4 m


Formula: Total surface area of frustum = πr2 + πR2 + π(R + r)lf m2


Total surface area of cone = πrlc


Where lf = slant height of frustum & lc = slant height of cone




lf = 25 m


For slant height of cone consider right angled ΔABC from figure


AB = hc = 4 m ; BC = r = 3 m ; AC = lc


By pythagoras theorm


AB2 + BC2 = AC2


42 + 32 = lc2


lc = 5


Since length cannot be negative lc = 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 = πr2


Surface area of lower circle = πR2


Surface area of frustum for which canvas is required = πr2 + πR2 + π(R + r)lf - πr2-πR2cm2


= π(R + r)lfm2


= (22/7) × (10 + 3) × 25 m2


= (22/7) × 325 m2


= 1021.4285 m2


Surface area of cone = πrlc m2


= (22/7) × 3 × 5 m2


= (22/7) × 15 m2


= 47.1428 m2


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


= 1021.4285 + 47.1428 m2


= 1068.5713 m2


Quantity of canvas required = 1068.5713 m2


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