Question

the diameter of a roller is 84cm and its length is 120cm .it 500 revolution to complete and move over playground .find the area of playground​

Answer 1

Answer:

$surface\: area\: of \: roller \: is \: = \: 2\pi rh + 2\pi {r}^{2}$

$area \: is \: 2 \times \frac{22}{7} \times 84 \times 120 + 2 \times \frac{22}{7} \times \: 84 \times 84$

$2 \times 22 \times 12 + 2 \times 22 \times 12 \times 84$

$44880 {cm}^{2}$

$surface \: area \: \times number \: of \: revolutions = \: area \: of \: playground$

$44880 \times 500 = 22440000 \: {cm}^{2}$

$1 {cm}^{2} = \frac{1}{10000}$

$22440000 {cm}^{2} = \frac{22440000}{10000}$

$cancellation$

$2244 {m}^{2}$

Answer 2

hope it's help u dear..!!!