<PRE><B>Question 1120521</B>
&lt;br&gt;
Let P be the midpoint of chord AB -- i.e., it is the center of the semicircle with AB as a diameter.&lt;br&gt;
Let r be the radius of that semicircle.  That means AB = 2r; so what we are looking for is the value of (2r)^2 = 4r^2.&lt;br&gt;
Segment OP is perpendicular to chord AB, so triangle OPB is a right triangle.  PB=r and OB=6, so&lt;br&gt;
(1) {{{(OP)^2+r^2=6^2}}}&lt;br&gt;
We can use Stewart&#39;s Theorem to find (OP)^2 in terms of r:&lt;br&gt;
{{{3r^2+4r^2 = 7(OP)^2+84}}}
{{{7(OP)^2 = 7r^2-84}}}
(2) {{{(OP)^2 = r^2-12}}}&lt;br&gt;
Substituting (2) in (1)...&lt;br&gt;
{{{(r^2-12)+r^2 =36}}}
{{{2r^2=48}}}&lt;br&gt;
And, remembering that the number we are looking for is 4r^2, the answer is&lt;br&gt;
AB^2 = 96</PRE>