Question

Plzzzz solve this ,help

Answer 1

Answer:

990.3 $m^{3}$

Step-by-step explanation:

area of canvass used = TSA of cone + margins

551 = TSA + 1

TSA = 1

TSA of cone = $\pi r^{2} + \pi rl$

$\pi r^{2} + \pi rl$ = 550

$\frac{22}{7}  *7^{2}  + \frac{22}{7} * 7 *l = 550\\ \\l = 18 m\\\\ h = \sqrt{7^{2} + 18^{2}  }$

h = 19.3 m

volume = $\frac{1}{3} \pi r^{2}$

= 990.3 $m^{3}$