Question

8. The inner circumference of a circular track is 220 m.
The track is 7 m wide everywhere. Calculate the cost
of putting fence along the outer circle at the rate of 2
per metre.
[Use A = 22/7]

,,,,,,,,,


9. The radius of a wheel of a vehicle is 42 cm. How many
revolutions will it complete in a journey of 19.8 km?​

Answer 1

Answer:

8) Rs. 528

9) 7500

Step-by-step explanation:

8. Circumference of inner side = 220 m

                                       2πr       = 220

                                              r    = 220 × 7 ÷ 44

                                                  r = 35 m

Now, width of track  = 7 m

∴ Outer radius = 35 + 7 ⇒ 42 m

Therefore outer circumference  = 2πR

                                                     = 2 × 22 ÷ 7  × 42

                                                     = 264 m

∴ Cost of fencing = Rs. (264×2) = Rs. 528

             

9. Radius of the wheel = r = 42 cm (given)

Circumference of wheel = Circumference of circle = 2πr

=2×  22 ÷ 7  × 42 = 264

Circumference of wheel is 264 cm.

Again, distance travelled by wheel = 19.8 km =1980000 cm  (Convert km into cm)

Now, number of revolutions taken by a wheel = 1980000 ÷ 264

                                                                             = 7500

Answer 2

inner circumference=2πr=220m

r=220/2π=110/π

r=(110/22)×7=35m

outer radius=inner radius+7=35+7=42m

cost of fence along the outer circle will be=circumference of outer circle×2

=(2×22/7×42)×2

=528Rs

9)No of revolution =Total journey/2πr

=(19.8×1000)/(2×22/7×0.42m)=7500