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Math Help - Easy but Frustrating Problem

  1. #1
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    Easy but Frustrating Problem

    A circular dartboard is divided into 20 equal sectors, one of which is shown in the diagram. O is the centre of the circle. The areas for scoring double and treble are marked A and B respectively.

    Find the ratio area A:area B in the form n:1, giving n correct to 1 d.p

    OM:99mm
    MN:8mm
    OP:162mm
    PQ:8mm

    Thanks guys
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  2. #2
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    the circle with radius OM has area \pi 99^2
    the circle with radius ON has area \pi (99+8)^2
    the circle with radius OP has area \pi 162^2
    the circle with radius OQ has area \pi (162+8)^2

    the sector OM has area \frac{\pi 99^2}{20}
    the sector ON has area \frac{\pi (99+8)^2}{20}
    the sector OP has area \frac{\pi 162^2}{20}
    the sector OQ has area \frac{\pi (162+8)^2}{20}
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  3. #3
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    Hello, alibond07!

    You need to know the area of a circle and some common sense.
    Exactly where is your difficulty?


    A circular dartboard is divided into 20 equal sectors,
    one of which is shown in the diagram. O is the centre of the circle.
    The areas for scoring double and treble are marked A and B respectively.

    Find the ratio (area A) : (area B) in the form n : 1, giving n correct to 1 d.p

    OM: 99mm . MN: 8mm . OP: 162mm . PQ: 8mm
    This is a diagram of the radii:
    Code:
          : - - - - 107 - - - :
          : - - -99 - - :
                          (A)             (B)
          O ----------- M --- N ------- P --- Q
    
          : - - - - - - - 162 - - - - - :
          : - - - - - - - - 170 - - - - - - - :

    The circle with radius  OQ = 170 has area: . \pi(170^2)

    The circle with radius  OP = 162 has area: . \pi(162^2)

    . . The area of ring A is: . 28,\!900\pi - 26,\!244\pi \:=\:2656\pi


    The circle with radius ON = 107 has area: . \pi(107^2)

    The circle with radius OM = 99 has area: . \pi(99^2)

    . . The area of ring B is: . 11,\!449\pi - 9801\pi \:=\:1648\pi


    Hence: . \frac{\text{ring A}}{\text{ring B}} \:=\:\frac{2656\pi}{1648\pi} \;=\;1.611650485


    Therefore: . \text{(area A)} : \text{(area B)} \;\approx\;1.6:1

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