# A tent consists of a frustum of a cone, surmounted by a cone. If the diameter of the upper and lower circular ends of the frustum be 14m and 26m respectively. The height of the frustum be 8m and the slant height of the surmounted conical portion be 12m, find the area of canvas required to make the tent.(Assume that the radii of the upper circular end of the frustum and the base of surmounted conical portion are equal)

1
by hardiknegi99

2015-12-27T14:06:01+05:30
Diameter of upper end = 14 m...radius , r₂ = 14/2 = 7m
diameter of lower end = 26m...radius , r₁ = 26/2 = 13m
height , h = 8m
∴slant height , l = \sqrt{(r2 ^{2} -r1 ^{2} } +h ^{2}
= \sqrt{(13 ^{2} - 7 ^{2} } +8 ^{2}
= \sqrt{(169 - 49} } +64 }
= \sqrt{120 } +64 }
= \sqrt(184)
=13.56m
∴curved surface area= π(r₁+r₂)l
=π×(13+7)×13.56

=π×20×13.56

=852.3m²

now,diameter, d, of cone = 14m...radius,r=7m
slant height, l = 12m
∴curved surface area=πrl
=π×7×12
=π×84
=264m²

∴area of canvas required = curved surface area of frustum + curved surface                                              area of cone
=[852.3+264]m²
=1116.3m²
ANS:1116.3 m² of canvas is required

mark as brainiliest plz if it helped
sry as i haven't given in symbols..for sqrt().....in the above equations..coz they weren't coming when i saves the answer.nd the value of pi,π=22/7

triche