How i solve the question of class 9th ncert math of Surface areas and Volumes chapter of exercise 13.2 of question no 4 . I think it is very important question of class 9th give me the tricky way for solving this question The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? (Assume π = 22/7)
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The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? (Assume π = 22/7) Q.4
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A roller is shaped like a cylinder.
Let h be the height of the roller and r be the radius.
h = Length of roller = 120 cm
Radius of the circular end of roller = r = (84/2) cm = 42 cm
Now, CSA of roller = 2πrh
= 2×(22/7)×42×120
= 31680 cm2
Area of field = 500×CSA of roller
= (500×31680) cm2
= 15840000 cm2
= 1584 m2.
Therefore, area of playground is 1584 m2. Answer!