# Problems relating with mensuration.

• Jul 21st 2008, 03:31 AM
prashantvrm
Problems relating with mensuration.
Plz solve the whole question as soon as possible.

1)The circumference of a circular field is 440 m.A 10 m wide road runs around it.Find the cos of leveling the road at the rate of 14 paisa per square meter.
1 paisa = Rs100
Its the indian currency.
Or leave it where we have to find the cost.

2)A race track is in the form of a ring whose inner circumference is 440 m and the outer circumference is 506 m.Find the wedth of the track and also the area
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• Jul 21st 2008, 04:27 AM
kalagota
Quote:

Originally Posted by prashantvrm
Plz solve the whole question as soon as possible.

1)The circumference of a circular field is 440 m.A 10 m wide road runs around it.Find the cos of leveling the road at the rate of 14 paisa per square meter.
1 paisa = Rs100
Its the indian currency.
Or leave it where we have to find the cost.

since the circumference is 440m, you can get the radius of the circular field by $\displaystyle C = 2\pi r \leftrightarrow r = \frac{C}{2\pi}$..
since the road is 10m wide, then the outer radius (including the road) is $\displaystyle R= r+10$

so the area of the circular road is $\displaystyle A_{road} = \pi R^2 - \pi r^2 = \pi (R^2 - r^2)$
i believe you can continue from here,, :)

Quote:

Originally Posted by prashantvrm
2)A race track is in the form of a ring whose inner circumference is 440 m and the outer circumference is 506 m.Find the wedth of the track and also the area
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inner: $\displaystyle r = \frac{C_{in}}{2\pi}$
outer: $\displaystyle R = \frac{C_{out}}{2\pi}$

width:$\displaystyle R-r$
area: $\displaystyle \pi R^2 - \pi r^2 = \pi (R^2 - r^2)$