Question

A race track is in the form of a ring whose inner circumference is of 396 m and
outer circumference is of 440 m. Find the width and area of track.​

Answer 1

Answer:

The width of the track is difference in the 2 radii and is equal to 14m.

Answer 2

❏ Question:-

@ A race track is in the form of a ring whose inner circumference is of 396 m and outer circumference is of 440 m. Find the width and area of track.

❏ Solution:-

✦Given:-

• Inner circumference($C_{inn}$) = 396 cm.

• Outer Circumference($C_{out}$) = 440 cm.

✦To Find:-

• The width(d) of that track = ?.

• Area(A) of that track. = ?.

✦ Explanation:-

Let, the Inner radius and the outer radius of the circular track is r and R .

Now, The inner circumference of the track is ,

$\sf\longrightarrow C_{inn}=2\pi r$

The Outer circumference of the track is ,

$\sf\longrightarrow C_{out}=2\pi R$

Now,

$\sf\longrightarrow C_{out}- C_{inn}=2\pi R-2\pi r$

$\sf\longrightarrow 440-396=2\times \dfrac{22}{7} (R- r)$

$\sf\longrightarrow \dfrac{\cancel{44}\times7}{\cancel{22}\times\cancel2}= (R- r)$

$\sf\longrightarrow \boxed{(R- r) = 7\:cm}...........(i)$

➝ Now, The width of the track is,

$\sf\longrightarrow d=R-r$

$\sf\longrightarrow\boxed{\red{ d= 7 \:cm}}$

∴ Width of the track is = 7 cm.

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➔ 2nd Part of the problem

$\huge \star$

$\sf\longrightarrow C_{out}+C_{inn}=2\pi R-2\pi r$

$\sf\longrightarrow 440+396=2\times \dfrac{22}{7} (R+r)$

$\sf\longrightarrow \dfrac{\cancel{836}\times7}{\cancel{22}\times\cancel2}= (R+ r)$

$\sf\longrightarrow 19\times7 (R+r)$

$\sf\longrightarrow\boxed{ (R+r)=133\:cm }..................(ii)$

Now, adding $\bf eq^n$ (i) & (ii), we get.

$\sf\longrightarrow (R-r)+(R+r)=7+133$

$\sf\longrightarrow R-\cancel{r}+R+\cancel{r}=140$

$\sf\longrightarrow 2R=140$

$\sf\longrightarrow R=\frac{\cancel{140}}{\cancel2}$

$\sf\longrightarrow\boxed{\large{ \red{R=70\:cm}}}$

Putting the value ,R = 70 cm, in $\bf eq^n$ (i) , we get.

$\sf\longrightarrow R-r=7$

$\sf\longrightarrow 70-r=7$

$\sf\longrightarrow r=70-7$

$\sf\longrightarrow\boxed{\large{ \red{r=63\:cm}}}$

Now,

➝ Area of the path,

$\sf\longrightarrow A_{path}=A_{out}-A_{in}$

$\sf\longrightarrow A_{path}=\pi R^2-\pi r^2$

$\sf\longrightarrow A_{path}=\pi (R^2- r^2)$

$\sf\longrightarrow A_{path}=\frac{22}{7}\times (R+r)\times(R-r)$

$\sf\longrightarrow A_{path}=\frac{22}{7}\times 133\times7$

$\sf\longrightarrow A_{path}=22\times 133$

$\sf\longrightarrow \boxed{\large{\blue{A_{path}=2926\:cm^2}}}$

∴ Area of the track is = 2926 cm².

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❏ Formula used:-

✦CIRCLE✦

❚ For a Circle of radius r ,

(1)Area is given by,

$\sf\longrightarrow \boxed{Area=\pi r{}^{2}}$

(2) Circumference is given by,

$\sf\longrightarrow\boxed{ Circumference=2\pi r=\pi d}$

where, d=2r= diameter of the circle.

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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