Question

Find the area of the road if the road roller takes 750 complete revolutions to move once over to level a road in which the the diameter of a road roller is 84 cm and length is 1 m.

Answer 1

Given:-

⇒D=84 cm,l=1m

⇒r=42cm

⇒Area of road covered in 1 revolution =

$2\pi r l = 2 \times \frac{22}{7} \times 42 \times 1 = \frac{264}{100}sq \: m$

⇒Area of road covered in 750 revolution =

$750 \times ( \frac{264}{100}) = 1980 \: sq \: m$

Answer 2

Step-by-step explanation:

Ncert solutions

Grade 8

Mathematics

Science

Chapters in NCERT Solutions - Mathematics , Class 8

Exercises in Mensuration

Question 9

Q9) A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.

Solution 9:

A road roller is cyclindrical in shape, hence one revolution is equal to the curved surface area of cylinder.

1 revolution = 2\pi rh2πrh

r\ =\ \frac{84}{2\ }cm\r =

2

84

cm

= 42cm

h = 1m = 100cm

CSA = 2\times\frac{22}{7}\times42\times1002×

7

22

×42×100

= 26400 cm\ ^2cm

2

750 revolutions of road roller completely level a road.

Area covered by road roller = 750\times26,400\ cm^2750×26,400 cm

2

= 19800000cm\ ^2cm

2

= 1980\ \ m^21980 m

2