A tent consists of a frustum of a cone capped by a cone.

Solution:

The height of the frustum cone is h = 8 m. The radii of the end circles of the frustum are r1 = 13m and r2 =7m.

The slant height of the frustum cone is

$l=\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}$

$=\sqrt{(13-7)^{2}+8^{2}}$

$=\sqrt{100}$

 

$=10$ meter

The curved surface area of the frustum is

$S_{1}=\pi\left(r_{1}+r_{2}\right) \times l$

$=\pi \times(13+7) \times 10$

$=\pi \times 20 \times 10$

 

$=200 \pi \mathrm{m}^{2}$

The base-radius of the upper cap cone is 7m and the slant height is 12m. Therefore, the curved surface area of the upper cap cone is

$S_{2}=\pi \times 7 \times 12$

$=\frac{22}{7} \times 7 \times 12$

$=22 \times 12$

 

$=264 \mathrm{~m}^{2}$

Hence, the total canvas required for the tent is

$S_{1}+S_{2}=200 \pi+264$

$=892.57 \mathrm{~m}^{2}$

Hence total canvas is $892.57 \mathrm{~m}^{2}$