Question 1190325
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Let the equilateral triangle be ABC with base AC and peak B. Let BD be the altitude of the triangle to side AC.  Let EF be the side of the square parallel to base AC but not on AC; EF intersects BD at G.<br>
Let r = OG be the radius of the inner circle.<br>
Triangle BGF is a 30-60-90 right triangle; and the length of GF is the radius of the circle, r; so {{{BG=r*sqrt(3)}}}.<br>
BO is the radius of the outer circle; and {{{BO=OG+BG=r+r*sqrt(3) = r(1+sqrt(3))}}}<br>
The radii of the inner and outer circle are then {{{r}}} and {{{r(1+sqrt(3))}}}, so the ratio of the areas of the inner and outer circles is<br>
ANSWER: {{{(1/(1+sqrt(3)))^2}}}<br>
Re-format as needed/desired....<br>