Question

The diameter of a garden roller is 1.4 m and it is 2 m long. If it needs 100 revolutions to
level the garden once . Then What will be the area of the garden ?

A. 176 m ^2
B. 88 m ^ 2
C. 880 m ^ 2​

Answer 1

Answer :

Area of the garden is 880m²

Explanation :

The roller is in the form of cylinder

And the area covered by the roller will the curved surface area

Area of the garden :

→ Curved surface area of the roller × no. of revolutions

Or,

→ 2πrh × 100

Here,

• ‘r' (radius) = 1.4 ÷ 2 [since, r = ½ of d ]
• ‘h’ (height) = 2m
• Taking ‘π’ as 22/7

So,

→ (2 • 22/7 • 0.7 • 2) × 100

→ 8.8 × 100

→ 880m²

Area of the garden is 880m²

____________________

Fomula used :

• Curved surface area of a cylinder : 2πrh

Answer 2

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Given :

• Diameter of roller = 1.4 m
• Length of roller = 2 m
• Revolution need to level the ground = 100

To find :

• Area of garden

Solution :

✿ Distance covered in one revolution will be equal to the CSA of cylinder and the distance covered in 100 revolutions will be equal to the area of garden.

We know that ,

${\boxed{\sf \ Curved \ surface \ area \ of \ cylinder \ = \ 2\pi rh}}$

here,

• r is radius
• h is height

${\sf{➣ \ Radius \ = \ \frac{Diameter}{2} }}$

${\sf{➣ \ Radius \ = \ \frac{1.4}{2} }}$

${\sf{➣ \ Radius \ = \ .7 m }}$

Now ,

${\sf{↣ \ CSA \ = \ 2 \times \frac{22}{\cancel7} \times \cancel{.7} \times 2}}$

${\sf{↣ \ CSA \ = \ 2 \times 22 \times .1 \times 2}}$

${\sf{↣ \ CSA \ = \ 8.8 {m}^{2} }}$

Distance covered in one revolution = 8.8 m²

Similarly,

↪ Distance covered in 100 revolutions = 100×8.8

↪ Distance covered in 100 revolutions = 880 m²

Thus the are of garden is 880m².

Hence option (c) is correct.

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