Question

10. A road roller takes 900 complete revolutions to move once to level a road fully. Find the area of the road, if the diameter of the cylindrical wheel of the road roller is 98 cm and length is 1.25 m. ​

Answer 1

Answer:

To Find:-

• Area of the Road

Given:-

• Diameter of the roller= 98cm
• Therefore Radius= 98÷2=49cm
• Length/Height=1.25m
• Total 900 revolution

Solution:-

We know when a road roller is rolling

it top amd bottom doesn't touches ground,only Sides do it

Therefore we've to find C.S.A or Curved Surface Area

We know

C.S.A of am Figure= Base Perimeter×Height

that is 2πrh

therefore one C.S.A= 1 revolution

First of all,find C.S.A ,later multiply it with 900

Before, convert it into metre

Radius= 49cm= 0.49m

Substituting the values

Take π= 22÷7

$2 \times \frac{22}{7} \times 0.49 \times 1.25 \\ \frac{44}{7} \times \frac{49}{100} \times \frac{125}{100} \\ {44} \times \frac{7}{100} \times \frac{125}{100} = \frac{38500}{10000} \\ = 3.85$

One Revolution= 3.85

900 revolution= 3.85×900

Therefore total Area of the road = 3465m²

Hope it helps

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Answer 2

Answer:

breadth of the field is 33