Question

What length of trapaulin 3m wide is required to make a conical tent of
height 8m and base radius 6m? Assume that the extra length of material
that will be required for stitching margins and wastage in cutting is
approximately 20 cm.
(use n = 3.14)​

Answer 1

Trapaulin will be csa .

so , finding csa ;

CSA =

$\pi \: r \: l(where \: l = \sqrt{ {h}^{2} + {r}^{2} }$

Radius = 6m

height = 8m

slanting edge = l

we know,

$l = \sqrt{ {h}^{2} + {r}^{2} } \\ l = \sqrt{ {8}^{2} + {6}^{2} } \\ l = \sqrt{64 + 36} \\ l = \sqrt{100} \\ l = 10$

l=10m

$csa = \pi \: r \: l\\ csa = (3.14 \times 6 \times 10) \\ csa = 188. {40m}^{2}$

CSA=188.4m^2

Now,

Area of trapezium = area of tent

length×breadth = area of tent

lenth × 3= Area of tent .

$length = \frac{1}{3} (area \: of \: tent) \\ length = \frac{188.4}{3} \\ l = 62.8$

Length = 62.8

Now given margin is 20cm

Total length = length calculated + margin

TL = 62.8+20cm

$tl = 62.8m + 20 \frac{1}{100}m \\ tl = 62.8 + 0.2 \\ tl = 63m$

Answer 2

Answer:

• Height = 8m
• Radius = 6m

$\\ \sf \: Slant \: \: height \: = \sqrt{ {r}^{2} + {h}^{2} } \\ \\ \\ \implies \sf \: \sqrt{ {6}^{2} + {8}^{2} } \\ \\ \\ \implies \sf \: \sqrt{100} \\ \\ \\ \implies \sf \blue{10 \: m}$

____________________________

$\\ \large{ \boxed{ \sf{ \green{curved \: \: surface \: \: area = \pi \: \:rl}}}} \\ \\ \\ \implies \sf \: \frac{22}{7} \times 6 \times 10 \\ \\ \\ \implies \sf \red{188.4 \: {m}^{2} }$

• Effective length ( 1 - 0.2 ) m
• Breadth = 3m

$\\ \sf \: Area \: \: of \: \: sheet = Csa \: \: of \: \: tent \\ \\ \\ \sf ( \: 1 - 0.2 \: m) \times 3 = 188.4 \\ \\ \\ \implies \sf \: 1 - 0.2 = 62.8 \\ \\ \\ \sf\implies \pink{l = 63m}$