Question

What length of canvas $3m$ wide will be required to make a conical tent of height $8m$ and radius of base $6m$. (use $\pi =3.14$)

Answer 1

3 m wide canvas, height 8 m & radius 6 m

$\to$ curved surface Area of tent = $\pi .rl.$

$r=6m,h=8m$

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

curved surface Area $=\pi .rl=3.14\times 6\times 10=188.4m^2$

Now,

Area of canvas = Area of tend

$L\times breadth=188.4m^2$

$L\times 3m=188.4m^2$

$L=\frac{188.4}{3}$

$L=62.8meter$