Question

a race track is in the form of a ring whose inner circumference is 220 m and the circumference is 440 m find the width of the track​

Answer 1

Solution:

Given:

⇒ Inner circumference = 220 m

⇒ Outer circumference = 440 m

To find:

⇒ Width of track

Formula used:

Circumference = 2πr

Let inner circumference be r₁ and outer circumference be r₂.

So, inner circumference = 2πr₁

$\sf{\implies 220=2\times \dfrac{22}{7}\times r_{1}}$

$\sf{\implies r_{1}=\dfrac{220\times 7}{2\times 22}}$

$\sf{\implies r_{1}=\dfrac{1540}{44}}$

$\sf{\implies r_{1}=35\;m}$

⇒ Outer circumference = 2πr₂

$\sf{\implies 440=2\times \dfrac{22}{7}\times r_{2}}$

$\sf{\implies r_{2}=\dfrac{440\times 7}{2\times 22}}$

$\sf{\implies r_{2}=\dfrac{3080}{44}}$

$\sf{\implies r_{2}=70\;m}$

⇒ Width of road = Outer circumference - Inner circumference

⇒ Width of road = 70 - 35

⇒ Width of road = 35 m

Answer 2

$\large{\mathfrak{\red{\underline{Question:}}}}$

A race track is in the form of a ring whose inner circumference is 220 m and the circumference is 440 m. Find the width of the track​?

$\large{\mathfrak{\red{\underline{Answer:}}}}$

$\star{\bf{\red{\underline{Given:}}}}$

• Inner circumference = 220 m
• Outer circumference = 440 m
• Width of track = ?

$\sf{So,\;Let\;inner\;circumference\;be\;r.}$

$\sf{Outer\;circumference\;be\;R.}$

$\sf{\implies Inner\;circumference=2\pi r}$

$\sf{\implies 220 = 2\times \dfrac{22}{7} \times r}$

$\sf{\implies 1540=44\times r}$

$\sf{\implies r=\dfrac{1540}{44}}$

${\boxed{\boxed{\red{\bf{\implies r=35\;m}}}}}$

$\sf{\implies Outer\;circumference=2\pi R}$

$\sf{\implies 440 = 2\times \dfrac{22}{7} \times R}$

$\sf{\implies 3080=44\times R}$

$\sf{\implies R=\dfrac{3080}{44}}$

${\boxed{\boxed{\red{\bf{\implies R=70\;m}}}}}$

$\sf{Now,\;as\;we\;know,}$

$\sf{\implies Width\;of\;the\;track=Outer\;circumference-Inner\;circumference}$

$\sf{\implies Width\;of\;the\;track=70-35}$

$\large{\boxed{\boxed{\red{\bf{\implies Width\;of\;the\;track=35\;m}}}}}$