Question 1101920: An archery target is constructed of four concentric circles such that the area of the inner circle is equal to the area of each of the three rings (blank spaces between the edges of the circles). If the radius of the outer circle is 6 m, find the width of band a, the second band from the outer circle, in m.
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! The target described looks like this
 .
The ratio of the areas of similar 2-dimensional shapes
is the square of the ratio of corresponding length measures.
If a circle has  times the radius of another circle,
it has  times its area.
Obviously, it goes both ways: the ratio of the radi of two circles, is the square root of the ratio of their areas.
If a circle has  times the area of a smaller circle,
the radius of the larger circle is  times the radius of a smaller circle.
That is what happens with the inner and outer circles in the target.
If the area of the inner circle is  ,
the area of the next circle is  ,
the area of the next circle is  , and
the area of the outer circle is  .
So, the radius of the inner circle is  .
The areas of the circles, inner to outer, are in the ratio
   ,
and that means that the areas of the radii are in the ratio
     .
The radii of the circles in the target, in m, are
,  ,  , and  .
So, the width of band a, the second band from the outer circle, in m, is
    .
The width is about  .
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