Question

If perpendicular height of road roller is 35 m and base radius is 7 m then find the area of the ground covered by road roller in 10 revolutions? *​

Answer 1

Answer:

Radius of the roller = 35 cm

Length of the roller = 3 m = 300 cm

Curved Surface Area of the Roller (Cylindrical shape) = begin mathsize 16px style 2 πrh space space equals space 2 space cross times 22 over 7 cross times 35 space cross times space 300 space equals space 66000 space cm squared end style

Therefore the area covered by the roller in one revolution = 66000 cm2

Therefore the area covered by the roller in 30 revolutions = 66000 × 30 = 1980000 cm2

Therefore the area of the play ground = 1980000 cm2 = 198 m2

Cost of levelling the ground = 5 x 198 = Rs. 990

Length of the roller = 120 cm

Curved Surface Area of the Roller (Cylindrical shape) = begin mathsize 16px style 2 πrh space space equals space 2 space cross times 22 over 7 cross times 42 space cross times space 120 space equals space 31680 space cm squared end style

Therefore the area covered by the roller in one revolution = 31680 cm2

Therefore the area covered by the roller in 1000 revolutions = 31680 × 1000 = 31680000 cm2

Therefore the area of the play ground = 31680000 cm2 = 3168 m2

→ l × b = 3168

→ l × 32 = 3168

→ l = 99 m

Step-by-step explanation:

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Answer 2

Answer:

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Step-by-step explanation:

Height of the cone is not given

Assuming height of the cone =8m

Let radius of base of cone =r

Curved surface area of cone =188.4cm

2

πrl=188.4cm

2

l=

r

2

+h

2

=

r

2

+8

2

=3.14×r×

r

2

+64

=188.4

=r

r

2

+64

=60

Squaring both sides, we get

r

2

(r

2

+64)=3600

r

4

+64r

2

−3600=0

r

4

+100r

2

−36r

2

−3600=0

(r

2

+100)(r

2

−36)=0

r

2

−36=0

r

2

=36

r=6 m