Question

a race track is inthe form of ring whose inner circumference is 528 m and outer circumference is 616m find the width of the track​

Answer 1

AnswEr :

• Inner Circumference = 528 m
• Outer Circumference = 616 m
• Find the Width of the Track?

Witdth of the Track will be (R - r) i.e. (Outer Radius – Inner Radius)

• Calculation of Outer Radius :

$\longrightarrow \tt Outer \:Circumference = 2\pi R \\ \\\longrightarrow \tt 616 = 2 \times \dfrac{22}{7} \times R \\ \\\longrightarrow \tt \cancel{616} \times \dfrac{7}{ \cancel{2 \times 22}} = R \\ \\\longrightarrow \tt 14 \times 7= R \\ \\\longrightarrow \blue{\tt R = 98\:m}$

• Calculation of Inner Radius :

$\longrightarrow \tt Inner\:Circumference = 2\pi r \\ \\\longrightarrow \tt 528 = 2 \times \dfrac{22}{7} \times r \\ \\\longrightarrow \tt \cancel{528} \times \dfrac{7}{ \cancel{2 \times 22}} = r \\ \\\longrightarrow \tt 12 \times 7= r \\\\\longrightarrow \blue{\tt r = 84\:m}$

$\rule{300}{1}$

• Width of the Racing Track :

↠ Width = Outer Radius – Inner Radius

↠ Width = 98m – 84m

↠ Width = 14m

⠀

∴ Width of the Racing Track will be 14m.

$\rule{300}{2}$

• S H O R T C U T ⠀T R I C K :

$\begin{tabular}{|c |c | c|}\cline{1-3}Radius & Circumference & Area \\\cline{1-3}3.5 & 22 & 38.5 \\7 & 44 &154\\10.5 & 66&346.5 \\14 & 88&616\cline{1-2}\cline{1-3}\end{tabular}$

This is Table, which we need to Remember:

• Radius Increase by 3.5, then Circumference will Increase by 22
• Radius Increase by 3.5, the Area will Increase by Square of Increase : 2² = 4 times, 3² = 9 times.

_______________________________

• Let's Head to the Question :

⇒ Track Circumference = Outer Circumference - Inner Circumference

⇒ Track Circumference = 616m - 528m

⇒ Track Circumference = 88m

See From the Above Table, The One whose Circumference is 88, Radius is 14.

∴ Width of Racing Track will be 14m.

#answerwithquality #BAL

Answer 2

$\bf{\Huge{\underline{\boxed{\bf{\blue{ANSWER\::}}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}}$

A race track is in the form of ring whose inner circumference is 528m and outer circumference is 616m.

$\bf{\Large{\underline{\bf{To\:find\::}}}}}$

The width of the track.

$\bf{\Large{\underline{\boxed{\bf{\orange{Explanation\::}}}}}}$

$\bf{We\:have\begin{cases}\sf{The\:inner\:circumference=528m}\\ \sf{The\:outer\:circumference=616m}\end{cases}}$

$\bf{\large{\underline{\boxed{\bf{\red{Inner\:Circumference\:of\:ring\::}}}}}}$

We know that formula of the circumference of circle: 2πr

$\longmapsto\bf{2\pi r=528m}$

$\longmapsto\bf{2*\frac{22}{7} *r=528}$

$\longmapsto\bf{\frac{44}{7} *r=528}$

$\longmapsto\bf{r=\frac{\cancel{528}*7}{\cancel{44}} }$

$\longmapsto\bf{r=12*7}$

$\longmapsto\bf{r=84m}$

$\bf{\large{\underline{\boxed{\bf{\red{Outer\:Circumference\:of\:ring\::}}}}}}$

$\longmapsto\bf{2\pi R=616m}$

$\longmapsto\bf{2*\frac{22}{7} *R=616}$

$\longmapsto\bf{\frac{44}{7} *R=616}$

$\longmapsto\bf{R=\frac{\cancel{616}*7}{\cancel{44}} }$

$\longmapsto\bf{R=14*7}$

$\longmapsto\bf{R=98m}$

Now,

• $\bf{\huge{\underline{\sf{\pink{The\:width\:of\:the\:track\::}}}}}$

→Width=Outer circumference of Radius-Inner circumference of radius

→ Width = 98m - 84m

→ Width = 14m

Thus,

The width of the track is 14m.