A tent is made in the form of a frustum of a cone surmounted by another cone. The diameters of the base and the top of the frustum are 20 m and 6 m, respectively, and the height is 24 m. If the height of the tent is 28 m and the radius of the conical part is equal to the radius of the top of the frustum, find the quantity of canvas required.
For the lower portion of the tent:
Diameter of the base= 20 m
Radius, R, of the base = 10 m
Diameter of the top end of the frustum = 6 m
Radius of the top end of the frustum = r = 3 m
Height of the frustum = h = 24 m
Slant height = l
$=\sqrt{h^{2}+(R-r)^{2}}$
$=\sqrt{24^{2}+(10-3)^{2}}$
$=\sqrt{576+49}$
$=\sqrt{625}=25 \mathrm{~m}$
For the conical part:
Radius of the cone's base = r = 3 m
Height of the cone = Total height - Height of the frustum = 28-24 = 4 m
Slant height, $L$, of the cone $=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5 \mathrm{~m}$
Total quantity of canvas = Curved surface area of the frustum + curved surface area of the conical top
$=(\pi l(R+r))+\pi L r$
$=\pi(l(R+r)+L r)$
$=\frac{22}{7}(25 \times 13+5 \times 3)$
$=\frac{22}{7}(325+15)=1068.57 \mathrm{~m}^{2}$
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