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Math Help - Problems relating with mensuration.

  1. #1
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    Angry 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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  2. #2
    MHF Contributor kalagota's Avatar
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    Taguig City, Philippines
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    Quote Originally Posted by prashantvrm View Post
    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 C = 2\pi r \leftrightarrow r = \frac{C}{2\pi}..
    since the road is 10m wide, then the outer radius (including the road) is R= r+10

    so the area of the circular road is 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 View Post
    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
    ================================================== =========
    inner: r = \frac{C_{in}}{2\pi}
    outer: R = \frac{C_{out}}{2\pi}

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