Question 1101920
The target described looks like this
{{{drawing(300,300,-2.1,2.1,-2.1,2.1,
circle(0,0,1),circle(0,0,1.414),
circle(0,0,1.732),circle(0,0,2),
arrow(1.414,0,1.732,0),arrow(1.732,0,1.414,0),
locate(1.5,0,"?")
)}}} .
The ratio of the areas of similar 2-dimensional shapes
is the square of the ratio of corresponding length measures.
If a circle has {{{2}}} times the radius of another circle,
it has {{{2^2=4}}} 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 {{{4}}} times the area of a smaller circle,
the radius of the larger circle is {{{sqrt(4)=2}}} 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 {{{A}}} ,
the area of the next circle is {{{A+A=2A}}} ,
the area of the next circle is {{{A+2A=3A}}} , and
the area of the outer circle is {{{A+3A=4A}}} .
So, the radius of the inner circle is {{{6m/2=3m}}} .
The areas of the circles, inner to outer, are in the ratio
{{{1}}}{{{":"}}}{{{2}}}{{{":"}}}{{{3}}}{{{":"}}}{{{4}}} ,
and that means that the areas of the radii are in the ratio
{{{sqrt(1)}}}{{{"="}}}{{{1}}}{{{":"}}}{{{sqrt(2)}}}{{{":"}}}{{{sqrt(3)}}}{{{":"}}}{{{2}}}{{{"="}}}{{{sqrt(4)}}} .
The radii of the circles in the target, in m, are
{{{3}}} , {{{3sqrt(2)}}} , {{{3sqrt(3)}}} , and {{{6}}} .
So, the width of band a, the second band from the outer circle, in m, is
{{{3sqrt(3)-3sqrt(2)}}}{{{"="}}}{{{3(sqrt(3)-sqrt(2))}}}{{{"="}}}{{{approximately}}}{{{0.954}}} .
The width is about {{{highlight(0.954m)}}} .