Q. 15

# A tent consists o

Answer : Given: Diameter of base of frustum = 26 m

Diameter of top of frustum = 14 m

Radius of base = R = 26/2 = 13 m

Radius of top = r = 14/2 = 7 m

Height of frustum = hf = 8 m

height of cone = hc = 12 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 = 10 m

Consider right angled ΔABC from figure

AB = hc = 12 m ; BC = r = 7 m ; AC = lc

By pythagoras theorm

AB2 + BC2 = AC2

122 + 72 = lc2

lc = 13.892

Since length cannot be negative lc = 13.892 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

= 3.14 × (13 + 7) × 10 m2

= 3.14 × 200 m2

= 628 m2

Surface area of cone = πrlc m2

= 3.14 × 7 × 13.892 m2

= 3.14 × 97.244 m2

= 305.346 m2

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

= 628 + 305.346 m2

= 933.346 m2

Quantity of canvas required = 933.346 m2

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