Question

a race track is in the form of a ring whose inner circumference is 220m and outer circumference is 264m. find the width of the track.​

Answer 1

Step-by-step explanation:

let's inner radius = r

&outer radius = r'

width d = r' - r

according to question,

iner circumference = 2πr

220 = 2πr

110 = (22/7)×r

5×7 = r

r = 35m

& outer circumference = 2πr'

264 = 2πr'

132 = (22/7)×r'

6×7 = r'

r' = 42m

so, width of the track = r'-r

= 42-35 - 7m answer

Answer 2

Step-by-step explanation:

let the radius of outer side= R

and the let radius of inner side=r

inner circumfrence=220

2πr = 220

πr = 110 -(1)

r= 110×7/22

r=35m

putting equation (1)

πr=110 ( both of side multiple by r)

r(πr)=110×r

πr^2= 110×35= 3850m^2 - (a)

outer circumference= 264m

2πR= 264

πR=132 -(2)

R=132×7/22

R=42m

similarly

putting equation (2)

R(πR)= 132

πR^2= 132×R

πR^2= 132×42

πR^2=5544m^2 -(A)

area of track = (A)-(a)= 5544 -3850=1694 m^2