Question

17. Play grounds offer kids fresh air friends fun and exercise. But it's important to make sure that

faulty equipment improper surfaces and unsafe behaviour don't ruin the fun. In this order, Jaipur

Nagar Nigam announced a online tender for levelling the roads of Dawarka Das Park A

contractor Vicky Sain get this work from Jaipur Nagar Nigam. He has a roller of diameter of 84

cm and its length is 120 cm. It takes 500 complete revolutions to move once to level a

playground

Then, give the answer of following questions

(i) Find the area (in m2

) covered in 1 revolution.
(a) 2.168 (c) 4.168

(b) 3.168 (d) 5.168

(ii) Write the formula to find the curved surface area of cylinder

(a) π r

2h (c)2 πr(r + h)

(b) 2 π rh (d) None of these

(iii) Find the area( in m2

) of playground

(a) 1484 (c) 1684

(b) 1584 (d) 1784

(iv) If area of one revolution is 5.168 m2

.Then find the area (in m?

) of playground

(a) 1584 (c) 3584

(b) 2584 (d) 4584

(v) If area of playground is 3500 m2

, then find the area of one revolution

(a) 7 m2

(c) 6 m2

(b) 5 m2

(d) 4 m2
​

Answer 1

Answer:

r=42cm

h=120cm

a) area covered in one revolution=2πrh

=2×(22/7) ×42× 120 =44*6*120=31680cm^2 =3.168m^2

b) formula for csa= 2πrh

c) area of playground= total revolution× c. s. a = 500×3.168 = 1584m^2

d) area of playground= 500×5.168=2584m^2

e) area of one revolution= area÷total revolution

=3500÷500 = 7m^2

Answer 2

r=42cm

h=120cm

a) area covered in one revolution=2+rh

=2×(22/7) ×42× 120 =44*6*120=31680cm^2

=3.168m^2

b) formula for csa= 2+rh

c) area of playground= total revolutionx c. s. a = 500×3.168 = 1584m^2

d) area of playground= 500×5.168-2584m^2

e) area of one revolution area total revolution

=3500÷500 = 7m^2