Question

8. The lateral surface area of a hollow cylinder is 4224 cm.
It is cut along its height and formed a rectangular sheet
of width 33 cm. Find the perimeter of rectangular sheet?

9. 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.​

Answer 1

Answer:

8. Perimeter = 322 cm.

9. Area = 1980 m²

Step-by-step explanation:

8.

Lateral surface area of hollow cylinder = 4224 cm²

Height of a hollow cylinder = 33 cm

C.S.A of hollow cylinder = 2πrh

=> 4224 = 2 * 22/7 * r * 33

=> r = (64 * 7)/22 cm

Length of rectangular sheet = 2πr

=> l = 2 * (22/7) * (64 * 7)/22

=> l = 128 cm

Perimeter of rectangular sheet = 2(l + b)

=> 2(128 + 33)

=> 322 cm

Therefore, Perimeter = 322 cm.

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9.

It is given in the question that,

Diameter = 84 cm

Hence,

Radius = (84/2) = 42 cm.

Given,

Length (h) = 1 m = 100 cm

Now,

Lateral surface area = 2πrh

=> 2 * 22/7 * 42 * 100

=> 26400 cm²

Hence, the area of the road will be

= lateral surface area * number of rotations

= 26400 * 750

= 19800000 cm²

= 1980 m²

Therefore, Required Area of the road = 1980 m².

Hope it helps!

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

8. 322 cm

9. 1980 m^2

$\huge\star{\green{\underline{\mathfrak{Explanation :-}}}}$

8.

CSA or LSA of hollow cylinder = 4224 cm^2.

$2\pi \: r \: h = 4224 \: {cm}^{2}$

As the width of the rectangular sheet is 33 cm, then the height of the cylinder is 33 cm.

$2 \times \dfrac{22}{7} \times r \times 33 = 4224$

$r = 1221\:\:\cancel {4224}\times \dfrac{7}{22} \times \dfrac{1}{\cancel{2}} \times \dfrac{1}{33}$

$r = 2112 \times \dfrac{7}{22} \times \dfrac{1}{33}$

$r = 20.36 \: cm$

Now, the circumference of the hollow cylinder' s base = length of the rectangular sheet.

$2\pi \: r = l$

$2 \times \dfrac{22}{7} \times 20.36 = l$

$l = 128 \: cm$

Now, perimeter of the rectangular sheet :-

$= 2(l + b)$

$= 2(128 + 33)$

$= 322 \: cm$

9.

The length of the roller = 1 m

The diameter of the roller = 84 cm

So, the radius of the roller = 42 cm =$\dfrac{42}{100} \: m$

It takes 750 complete revolutions to level a road.

So, the area of the road :-

$= 750 \times LSA of the roller$

$= 750 \times (2\pi \: r \: h)$

$= 750 \times (2 \times \dfrac{22}{\cancel {7}} \times \dfrac{\cancel {42}\:\:6}{100} \times 1)$

$= 750 \times 2 \times 22 \times \dfrac{6}{100} \times 1$

$= \dfrac{1500 \times 132}{100}$

$= 15 \times 132$

$= 1980 \: {m}^{2}$

So, the area of the road = 1980 m^2

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