Question

hca racetrack is in the form of a ring whose inner circumference is 528 m and the outer circumference is 616 m. find the width of the track​​

Answer 1

$\huge\underbrace\red{Answer}$

Given :-

★Inner circumference of ring = 528cm

★Outer circumference of ring = 616cm

To Find :-

★Radius of inner and outer circumference of ring

★Width of the track

Solution :-

★Let the inner and outer radii of the track be r and R metres

We know that,

Inner circumference of ring = 528cm

Formula of Circumference = 2πr

∴ 2πr = 528cm

By substituting values,

$2 \times ( \frac{22}{7} ) \times r = 528cm$

$\frac{44}{7} \times r = 528cm$

$r = 528cm \times \frac{7}{44}$

$r = 84cm$

∴ The Radius of inner circumference of the ring is 84cm

Outer circumference of ring = 616cm

2πR = 616cm

By substituting values,

$2 \times \frac{22}{7} \times R = 616cm$

$\frac{44}{7} \times R = 616cm$

$R = 616cm \times \frac{7}{44}$

$R = 98cm$

∴ The Radius of outer circumference of the ring is 98cm

Now,

Width of the track = Radius of outer circumference of ring - Radius of inner circumference of ring

= (R-r)

= 98cm - 84cm

= 14cm

Hence, The width of the track is 14cm

Answer 2

Refer The Attachment ⬆️

$\huge{\pink{\underline{\underline{\bf{\bigstar\;SoluTion:}}}}}$

Let the inner and outer radii of the track be r meters and R meters respectively.

Then,

= 2πr = 528 m

= 2 × (22/7) × r = 528

= r = (528/ (2 × (22/7)))

= r = (528/ 2) × (7/22)

= r = (264/2) × (7/11)

= r = 132 × 0.6364

= r = 84 m

And,

= 2πR = 616 m

= 2 × (22/7) × R = 616

= R = (616/ (2 × (22/7)))

= R = (616/ 2) × (7/22)

= R = (308/2) × (7/11)

= R = 154 × 0.6364

= R = 98 m

Now,

(R – r) = (98 – 84)

☞ 14 m

Hence, The width of the track is $\fbox\red{\sf{14cm}}$

$\green{\sf{\star\; MrMonarque}}$

Hope It Helps You ✌️