Question

a circular road runs around a circular ground if the difference between the circumference of the outer circle and the inner circle is 66 m the width of road.​

Answer 1

Required Answer:-

Let the radius of the outer circle be R and radius of the inner circle (circular ground) be r. We know,

${\boxed{ \rm{Circumference \: of \: \: circle = 2\pi r}}}$

Then, according to question

➛2πR - 2πr = 66 m

➛ 2π(R - r) = 66 m

➛ 2 × 22/7 (R - r) = 66 m

➛ R - r = 66 × 7/22 × 2 m

➛ R - r = 3 × 7/2 m

➛ R - r = 10.5 m

Here, the width of the road is the difference between the radius of the larger circle and the radius of the smaller circle. And hence, the answer is R - r = 10.5 m

Answer 2

Answer:

33 / π m

Step-by-step explanation:

Let radius of inner edge of the road be r1.

Let radius of outer edge of the road be r2.

Outer circumference = 2 π r1

Inner circumference = 2 π R2

Given,

difference of circumference = 66m

2 π r2 - 2 π r1 = 66

2 π ( r2 - r1 ) = 66

r2 - r1 = 66 / 2 π

width of road = r2 - r1 = 33 / π m