Question 824332
The formula for area of a circle is:
{{{A = pi*r^2}}}
So the area for the larger circle will be:
{{{A[L] = pi*(2+sqrt(3))^2}}}
Using the {{{(a+b)^2 = a^2+2ab+b^2}}} pattern to square the radius:
{{{A[L] = pi*((2)^2+2(2)(sqrt(3))+(sqrt(3))^2)}}}
Simplifying...
{{{A[L] = pi*(4+4sqrt(3)+3)}}}
{{{A[L] = pi*(7+4sqrt(3))}}}<br>
Repeating this for the smaller circle:
{{{A[S] = pi*(2-sqrt(3))^2}}}
Using the {{{(a-b)^2 = a^2-2ab+b^2}}} pattern to square the radius:
{{{A[S] = pi*((2)^2-2(2)(sqrt(3))+(sqrt(3))^2)}}}
Simplifying...
{{{A[S] = pi*(4-4sqrt(3)+3)}}}
{{{A[S] = pi*(7-4sqrt(3))}}}<br>
Now we find the ratio of these areas:
{{{A[L]/A[S] = (pi*(7+4sqrt(3)))/(pi*(7-4sqrt(3)))}}}
Simplifying... The {{{pi}}}'s cancel:
{{{A[L]/A[S] = (7+4sqrt(3))/(7-4sqrt(3))}}}
This may be an acceptable answer.<br>
But it does have a square root in the denominator. Usually denominators should be rationalized:
{{{A[L]/A[S] = ((7+4sqrt(3))/(7-4sqrt(3)))((7+4sqrt(3))/(7+4sqrt(3)))}}}
Using the {{{(a+b)^2}}} pattern on top and the {{{(a+b)(a-b)=a^2-b^2}}} pattern on the bottom:
{{{A[L]/A[S] = ((7)^2+2(7)(4sqrt(3))+(4sqrt(3))^2)/((7)^2-(4sqrt(3))^2)}}}
Simplifying...
{{{A[L]/A[S] = (49+56sqrt(3)+48)/(49-48)}}}
{{{A[L]/A[S] = (97+56sqrt(3))/1}}}
{{{A[L]/A[S] = 97+56sqrt(3)}}}